问:
Determining the kernel of an isogeny
Let $$g:(x,y)\rightarrow \Big(\frac{x^2+x+3}{x+1},\frac{x^2y+2xy+15y}{x^2+2x+1}\Big)$$
be a map $E_1\rightarrow E_2$ where
$$E_1/\mathbb{F}_{17}:y^2=x^3+1$$
$$E_2/\mathbb{F}_{17}:y^2=x^3-10$$

I want to determine the kernel of $g$. Apparently the answer is $(-1,0)$ and the point at infinity. I really have no idea why that is. I thought to compute the kernel you had to compute
$\frac{x^2+x+3}{x+1}=0$ and $\frac{x^2y+2xy+15y}{x^2+2x+1}=0$, but than you get different solutions. I'd appreciate it if someone could explain how I could compute this, thank you in advance!

答:
This answer is here only to give the correction of the statement in the OP.
It is useful to use computer algebra systems like sage to figure out the structure.

First of all, i will restate the question as follows, let us compute the isogeny from $E_1$ defined as in the OP over the field $F=\Bbb F_{17}=$GF(17) which has the kernel given by the two points, the "infinity point" $O_1$, and the affine point $(-1,0)$. Here, $O_1$ is the neutral point for the operation, the infinity point, tacitly taken so in such situations. It is the point not showing up in the given affine form of the elliptic curve, but rather in its homogenized version $Y^2Z=X^3+Z^3$. The point "at infinity" is $O_1=[X:Y:Z]=[0:1:0]$. This is the only rational point on the curve with $Z=0$. Each other rational point has $Z\ne 0$, we may and do norm this component as $Z=1$, and then $x:=X/Z$, $y:=Y/Z$ are satisfying the given affine equation
$$
E_1\ :\qquad y^2 = x^3 +1\ .
$$
The point $P_1=(-1,0)$ is on $E_1$, and it is a $2$-torsion point.
(Since $-1$ is a root of $x^3+1$.)
It turns out that $|E_1(F)|=18$. In fact sage can compute isogenies between two elliptic curves, so i wanted just to check the situation, insert it in my collection of quick code snippets, and let the already existing comments be the answer. Well, there are some problems with this plan...

```<code>sage: E1 = EllipticCurve(GF(17), [0, 1])
sage: E2 = EllipticCurve(GF(17), [0, -10])
sage: P1 = E1( (-1, 0) )
sage: phi = E1.isogeny(P1)
sage: phi.codomain()
Elliptic Curve defined by y^2 = x^3 + 2*x + 5 over Finite Field of size 17
sage: phi.codomain().is_isomorphic(E2)
False
sage: phi.rational_maps()
((x^2 + x + 3)/(x + 1), (x^2*y + 2*x*y - 2*y)/(x^2 + 2*x + 1))
</code>```

In words: We initialize the curves E1, E2 to correspond to the Weierstraß equations of $E_1$, $E_2$ from the OP, take $P_1$ to be the point that should a posteriori generate the kernel, ask for an isogeny from E1 to some other curve that has the kernel $\{O_1,P_1\}$, look at its codomain, and check that it is isomorphic to E2. Well, this codomain, the elliptic curve
$$
E_3\ :\qquad y^2 = x^3 + 2x + 5\ ,
$$
is not isomorphic to E2. But the delivered rational map coincides with the rational map $g$ displayed in the OP. (Note that modulo $17$ we have $x^2y+2xy-2y=x^2y+2xy+15y$.)

So $g$ is in fact a map $E_1\to E_3$, and we can easily check this: Let $\xi=\xi(x,y)$ and $\eta=\eta(x,y)$ be the components of $g$...
$$
\begin{aligned}
\xi&:=\frac{x^2+x+3}{x+1}\ ,\\
\eta&:=y\cdot \frac{x^2+2x-2}{(x+1)^2}\ .
\qquad\text{Then for $(x,y)\in E(F)$:}\\[3mm]
\eta^2&=y^2\cdot\frac{(x^2+x-2)^2}{(x+1)^4}
=(x^3+1)\cdot\frac{(x^2+x-2)^2}{(x+1)^4}
=(x^2-x+1)\cdot\frac{(x^2+x-2)^2}{(x+1)^3}\\
&=\xi^3+2\xi+5\ .
\end{aligned}
$$

Now we have above a correction of the statement at some place that "is not important for the solution".

The kernel of $\phi$ is computed as follows. We write $\phi$ in the homogenized form, as in the comment = full answer of Mindlack, 4 times upvoted,
$$
[X:Y:Z]\overset g\longrightarrow
[\ (X^2+XZ+3Z^2)(X+Z)\ :\ Y(X^2+2XZ-2Z^2)\ :\ (X+Z)^2Z\ ]\ ,
$$
and ask for the points $[X:Y:Z]$ of $E_1$ mapped to the infinity point = neutral point of $E_2$, $O_2=[0:1:0]$, so we need $(X^2+XZ+3Z^2)(X+Z)=(X+Z)^2Z=0$. The two possibilities in $(X+Z)^2Z=0$ lead to $O_1$ and $P_1$ as solutions.

It turns out that $E_1$, $E_2$ are also isogenous, but the kernel of the isogeny $h:E_1\to E_2$ has three elements, $O_1$, $Q_1(0,1)$, $Q_2(0,-1)$. In sage:

```<code>sage: h = E1.isogeny(E1( (0, -1) ), codomain=E2)
sage: h.rational_maps()
((x^3 + 4)/x^2, (-x^3*y + 8*y)/x^3)
</code>```

So an isogeny between $E_1$ and $E_2$ is given by the above map:
$$
h(x,y)=\left(\ \frac 1{x^2}(x^3+4)\ ,\ -\frac y{x^3}(x^3-8)\ \right)\ .
$$

Title: Determining the kernel of an isogeny

Body:
This answer is here only to give the correction of the statement in the OP.
It is useful to use computer algebra systems like sage to figure out the structure.

First of all, i will restate the question as follows, let us compute the isogeny from $E_1$ defined as in the OP over the field $F=\Bbb F_{17}=$GF(17) which has the kernel given by the two points, the "infinity point" $O_1$, and the affine point $(-1,0)$. Here, $O_1$ is the neutral point for the operation, the infinity point, tacitly taken so in such situations. It is the point not showing up in the given affine form of the elliptic curve, but rather in its homogenized version $Y^2Z=X^3+Z^3$. The point "at infinity" is $O_1=[X:Y:Z]=[0:1:0]$. This is the only rational point on the curve with $Z=0$. Each other rational point has $Z\ne 0$, we may and do norm this component as $Z=1$, and then $x:=X/Z$, $y:=Y/Z$ are satisfying the given affine equation
$$
E_1\ :\qquad y^2 = x^3 +1\ .
$$
The point $P_1=(-1,0)$ is on $E_1$, and it is a $2$-torsion point.
(Since $-1$ is a root of $x^3+1$.)
It turns out that $|E_1(F)|=18$. In fact sage can compute isogenies between two elliptic curves, so i wanted just to check the situation, insert it in my collection of quick code snippets, and let the already existing comments be the answer. Well, there are some problems with this plan...

sage: E1 = EllipticCurve(GF(17), [0, 1])
sage: E2 = EllipticCurve(GF(17), [0, -10])
sage: P1 = E1( (-1, 0) )
sage: phi = E1.isogeny(P1)
sage: phi.codomain()
Elliptic Curve defined by y^2 = x^3 + 2*x + 5 over Finite Field of size 17
sage: phi.codomain().is_isomorphic(E2)
False
sage: phi.rational_maps()
((x^2 + x + 3)/(x + 1), (x^2*y + 2*x*y - 2*y)/(x^2 + 2*x + 1))

In words: We initialize the curves E1, E2 to correspond to the Weierstraß equations of $E_1$, $E_2$ from the OP, take $P_1$ to be the point that should a posteriori generate the kernel, ask for an isogeny from E1 to some other curve that has the kernel $\{O_1,P_1\}$, look at its codomain, and check that it is isomorphic to E2. Well, this codomain, the elliptic curve
$$
E_3\ :\qquad y^2 = x^3 + 2x + 5\ ,
$$
is not isomorphic to E2. But the delivered rational map coincides with the rational map $g$ displayed in the OP. (Note that modulo $17$ we have $x^2y+2xy-2y=x^2y+2xy+15y$.)

So $g$ is in fact a map $E_1\to E_3$, and we can easily check this: Let $\xi=\xi(x,y)$ and $\eta=\eta(x,y)$ be the components of $g$...
$$
\begin{aligned}
\xi&:=\frac{x^2+x+3}{x+1}\ ,\\
\eta&:=y\cdot \frac{x^2+2x-2}{(x+1)^2}\ .
\qquad\text{Then for $(x,y)\in E(F)$:}\\[3mm]
\eta^2&=y^2\cdot\frac{(x^2+x-2)^2}{(x+1)^4}
=(x^3+1)\cdot\frac{(x^2+x-2)^2}{(x+1)^4}
=(x^2-x+1)\cdot\frac{(x^2+x-2)^2}{(x+1)^3}\\
&=\xi^3+2\xi+5\ .
\end{aligned}
$$

Now we have above a correction of the statement at some place that "is not important for the solution".

The kernel of $\phi$ is computed as follows. We write $\phi$ in the homogenized form, as in the comment = full answer of Mindlack, 4 times upvoted,
$$
[X:Y:Z]\overset g\longrightarrow
[\ (X^2+XZ+3Z^2)(X+Z)\ :\ Y(X^2+2XZ-2Z^2)\ :\ (X+Z)^2Z\ ]\ ,
$$
and ask for the points $[X:Y:Z]$ of $E_1$ mapped to the infinity point = neutral point of $E_2$, $O_2=[0:1:0]$, so we need $(X^2+XZ+3Z^2)(X+Z)=(X+Z)^2Z=0$. The two possibilities in $(X+Z)^2Z=0$ lead to $O_1$ and $P_1$ as solutions.

It turns out that $E_1$, $E_2$ are also isogenous, but the kernel of the isogeny $h:E_1\to E_2$ has three elements, $O_1$, $Q_1(0,1)$, $Q_2(0,-1)$. In sage:

sage: h = E1.isogeny(E1( (0, -1) ), codomain=E2)
sage: h.rational_maps()
((x^3 + 4)/x^2, (-x^3*y + 8*y)/x^3)

So an isogeny between $E_1$ and $E_2$ is given by the above map:
$$
h(x,y)=\left(\ \frac 1{x^2}(x^3+4)\ ,\ -\frac y{x^3}(x^3-8)\ \right)\ .
$$


问:
What is the relationship between the 2d projection and a 3d point when using 1 point perspective?
I understand the concept of $1$ point projection:

As a $3d$ point gets closer to the vanishing point (further on the $z$ axis), the x and y values of the corresponding $2d$ screen point get closer to the $x,y$ of the vanishing point.

I just don't know what this relationship is.

I assume it will include a $y = m/x$ relationship as the z position will never reach the z value of the vanishing point (infinity).
How I think it will work

In the image I have a diagonal line from the projected point to the vanishing point labled d, for distance, and I think $d = 1/z$ (or some other constant in place of $1$). However, I don't know how I would calculate the values for $ScreenX$, and $ScreenY$.

Please don't give answers only in matrix form as I intend to use them in some code I am writing, and I'm not yet confident enough to turn matrices into single equations.

Edit: I think my question is asking the same as: One-point perspective formula and https://stackoverflow.com/questions/56559793/one-point-perspective-and-point-with-negative-depth

答:
I was talking about this with some people on Discord and one gave me this answer: (I got permission to post it here)

the correct function is: f(x,y,z) = (x/z, y/z)

then z is the depth in the scene and you can move back and forward
with f(x,y,z,t) = (x/(z-t), y/(z-t))
where t is the camera z position

generally, if you have a camera transformation, you apply that before the projection transformation

f(x-a,y-b,z-c) = ((x-a)/(z-c),(y-b)/(z-c)) to translate the camera so that (a,b,c) is the origin

later you might want to apply scaling or rotation to the camera.
but notice that you subtract rather than add to translate the camera.
the same rule applies to scaling and rotation.
generally you transform by the inverse of the camera matrix before you project

ContentLicense: CC BY-SA 4.0


问:
Global diffeomorphism between two surfaces
I tried to show the function $ f: S_a \to \mathbb{S}^2$, given by

$$ f(p) = \dfrac{p}{|p|}$$

is a global diffeomorphism between $S_a$ and $\mathbb{S}^2$ where

$$S_a = \{(x,y,z) : e^{x^2}+ e^{y^2} + e^{z^2} = a\}$$

and $a>3$ is a real number fixed. I already proved that $f$ is a local diffeomorphism, but I'm not able to show that $f$ is injective to conclude $f$ is a global diffeomorphism.
Someone can help me?

答: HINT: Fix $(x,y,z)\in S^2$. Consider the function $g(t)=e^{(tx)^2}+e^{(ty)^2}+e^{(tz)^2}$. Can you show that $g$ is an increasing function of $t\in (0,\infty)$?


问:
Optimization Of Sum of Coefficients
What is the maximum sum of the coefficients of a monic quartic polynomial whose roots are all greater than 1 and multiply to 2145?

I've tried writing out the polynomial, which we know is of the form x^4+ax^3+bx^2+cx+2145. I know that you have to try and use Vieta, so I have written out all the equations in terms of the roots, but nothings coming of it. I keep running about in circles.

答:
Let $f(X)$ be a monic polynomial of degree $n$ such that all roots of $f$ are real and $>1$, and $f(0)=(-1)^n\cdot 2145$. We want to maximize $f(1)$.
In particular, for odd $n$, we find that $f(1)<0$. Hence we need only consider the case of even $n$.

Let the roots of $f$ be $\alpha_1,\ldots,\alpha_n\in[1,2145]$. Then $f(1)$ is a continuous function $$\phi(\alpha_1,\ldots,\alpha_n)=\prod_k(\alpha_k-1) $$of the $\alpha_k$, and as the set
$$\{\,(\alpha_1,\ldots,\alpha_n)\in[1,2145]^n\mid \prod\alpha_k=2145\,\} $$
is compact, the (clearly positive) maximum of $\phi$ is actually achieved in some point.
Let $(\alpha_1,\ldots,\alpha_n)$ be a maximizer. Then clearly $\alpha_k>1$ for all $k$.
Pick two indieces $1\le i<j\le n$ and
let
$$\beta_k=\begin{cases}\sqrt{\alpha_i,\alpha_j}&k\in\{i,j\}\\\alpha_k&\text{otherwise}\end{cases} $$
Then
$$1\ge \frac{\phi(\beta_1,\ldots,\beta_n)}{\phi(\alpha_1,\ldots,\alpha_n)}=\frac{(\beta_i-1)(\beta_j-1)}{(\alpha_i-1)(\alpha_j-1)}=\frac{\beta_i\beta_j-(\beta_i+\beta_j)+1}{(\alpha_i-1)(\alpha_j-1)}=\frac{\alpha_i\alpha_j-(\beta_i+\beta_j)+1}{(\alpha_i-1)(\alpha_j-1)}.$$
By the AM-GM inequality,
$$\alpha_i+\alpha_j\ge 2\sqrt{\alpha_i\alpha_j}=\beta_i+\beta_j $$
with equality iff $\alpha_i=\alpha_j=\sqrt{\alpha_i\alpha_j}$.
Hence,
$$ \frac{\alpha_i\alpha_j-(\beta_i+\beta_j)+1}{(\alpha_i-1)(\alpha_j-1)}\ge \frac{\alpha_i\alpha_j-(\alpha_i+\alpha_j)+1}{(\alpha_i-1)(\alpha_j-1)}=1$$
with equality iff $\alpha_i=\alpha_j=\sqrt{\alpha_i\alpha_j}$.
Both inequalities together tell us that indeed $$\alpha_i=\alpha_j.$$
As this golds for all $i,j$, we conclude that $\alpha_k=\sqrt[n]{2145}$ for all $k$. This makes
$$f(1)=(\sqrt[n]{2145}-1)^n.$$
It is straightforward to verify that the right hand side is decreasing as a function of $n$, hence the maximum is achieved for $n=2$ and
$$ f(X)=(X-\sqrt{2145})^2$$
and the maximum is
$$ f(1)=(\sqrt{2145}-1)^2\approx 2053.37$$

Title: Optimization Of Sum of Coefficients


问:
How do you work out $\sqrt{2} \pmod{31}$?
If $a^{n}\equiv 1 \pmod p$ with $n$ odd then $a^{n}=a^{1 + \frac{n-1}{2}\cdot 2}\equiv 1\pmod p$ so we have $a \equiv (a^{\frac{1-n}{2}})^{2} \pmod{p}$ and then we have $\sqrt{a}=a^{\frac{1-n}{2}} \pmod{p}$.

From this we have that since $2^{5}\equiv 1 \pmod {31}$ then $2 \equiv (2^{-2})^{2},$ so that $\sqrt{2}\equiv 2^{-2} \pmod {31}$.

I know that $2^{-2}\equiv 8 \pmod {31}$, (and this checks out since $8^2=64 \equiv2 \pmod{31}$) but I don't know how this was obtained and would like to know how do you see that $2^{-2} \equiv 8 \pmod{31}$ without guessing.

答:
Basically you are asking if there is a smart way of extending modulo function to rational powers, or beyond.

The simplest solution is to use offset, but I guess from your description that is not what you have in mind. You want it as an operator in the sense of

$$\mathbb{E}^2a \equiv a^2 \mod m$$

Ok, but then look at this

$$\mathbb{E}^{\frac1{2}}a \equiv a^{\frac1{2}} \mod m$$

Start from the basic property that must be satisfied

$$\mathbb{E} \equiv \mathbb{E}^{\frac1{2}}\mathbb{E}^{\frac1{2}} \equiv \left ( \mathbb{E}^{\frac1{2}} \right )^2$$

This means that you are asking for quadratic residue (if it exists) over, in your case, $31$. You are asking what numbers have quadratic residue $2$ over $31$.

$$x \equiv 2 \mod 31$$
$$\sqrt{x}^2 \equiv 2 \mod 31$$

Notice that $\sqrt{2}$ is defined only by the property from the modulo function perspective:

$$(\sqrt{2})^2 \equiv 2 \mod 31$$

This does mean that $\sqrt{}$ is an abuse on notation, it should be $\mathbb{E}^{\frac1{2}}$.

The result is the series in form of $31k+8$ or $31k+23$ or put it together:

$$8, 23, 39,... 31k \pm 8, k \geq 2$$

Reducing to only $8$ and $23$ is technically wrong because even $\sqrt{x}=39$ satisfies

$$\sqrt{x}^2 \equiv 2 \mod 31$$

and any integer with that property is a square root of $2$ modulo $31$, and you have never specified which set we should choose the result from, so it is implicitly assumed that it is among integers, right?

All roots are defined through the opposite function $(\sqrt[n]{a})^n=a$ but you have to take into account its domain and what $=$ means within that domain.


问:
The Lie algebra of the generators of the projective transformation is isomorphic to the Lie algebra of traceless matrices.
The general projective transformation of the $x$-$y$ plane is given by
$$\tilde{x}=\frac{a_1x+a_2 y+a_3}{a_7x+a_8y+a_9},\quad\tilde{y}=\frac{a_4x+a_5y+a_6}{a_7x+a_8y+a_9}$$
for some constants $a_i\in\mathbb{R}$. From this, we find the infinitesimal generators
$$\begin{array}{lll}
X_1=x\partial_x,& X_2=y\partial_x,& X_3=\partial_x, \\
X_4=x\partial_y,& X_5=y\partial_y,& X_6=\partial_y, \\
X_7=-x^2\partial_x-xy\partial_y,& X_8=-xy\partial_x-y^2\partial_y,& X_9=-x\partial_x-y\partial_y
\end{array}
$$
corresponding respectively to each parameter $a_i$. With the commutator
$$[X_i,X_j]=X_iX_j-X_jX_i,$$
the vector space generated by $\{X_1,\ldots,X_9\}$ forms a Lie algebra. Let us denote this vector space by $\mathfrak{g}$. A basis for that space could be $\{X_1,\ldots,X_8\}$ since $X_9=-X_1-X_5$.

I want to show that it is isomorphic to the Lie algebra
$$\mathfrak{sl}(3,\mathbb{R}):=\{A\in\mathbf{M}_{3\times 3}(\mathbb{R}):\text{trace}(A)=0\}$$
with commutator $[A,B]=AB-BA$, by explicitly giving an isomorphism
$$f:(\mathfrak{sl}(3,\mathbb{R}),[\cdot,\cdot])\longrightarrow(\mathfrak{g},[\cdot,\cdot])$$

I made a few attempts that all failed. Since $X_1+X_5+X_9=0$, we can see a sort of correspondence with traceless matrices, but I don't see how to use that.

EDIT: To get the generators $X_i$, we differentiate with respect to each $a_i$ both $\tilde{x}$ and $\tilde{y}$ and evaluate at the identity transformation, that is at $a_1=1,a_5=1,a_9=1$ and $a_2=a_3=a_4=a_6=a_7=a_8=0$. For example, to get $X_2$, we have
$$\frac{d\tilde{x}}{da_2}=\frac{y}{a_7x+a_8y+a_9},\quad\frac{d\tilde{y}}{da_2}=0$$
which evaluates to $y$ and $0$ respectively so that $X_2=y\partial_x+0\partial_y=y\partial_x$.

答:
I think that some slightly better notation solves your problem.  Let
$$
\begin{array}{lll}
Y_{1,1} = x\partial_x, & Y_{1,2} = x\partial_y, & Y_{1,3} = -x^2\partial_x - xy\partial_y \\
Y_{2,1} = y\partial_x, & Y_{2,2} = y\partial_y, & Y_{2,3} = -xy\partial_x - y^2\partial_y\\
Y_{3,1} = \partial_x, & Y_{3,2} = \partial_y, & Y_{3,3} = -x\partial_x - y\partial_y.
\end{array}
$$
This is essentially the transpose of your setup, but indexed in as a matrix with a pair of coordinates.  In other words,
$$
Y_{i,j} = \partial_{a_{i,j}} = \frac{\partial x}{\partial a_{i,j}} \partial_x + \frac{\partial y}{\partial a_{i,j}} \partial_y \qquad \textrm{for each } 1 \le i,j \le 3.
$$
Now, the generators $Y_{i,j}$ of $\mathfrak{g}$ behave like matrix units under the commutator bracket.  In other words,
$$
[Y_{i,j}, Y_{k,l}] = \delta_{j,k}Y_{i,l} - \delta_{l,i}Y_{k,j}.
$$
Your isomorphism is just the identity map.</p>

ContentLicense: CC BY-SA 3.0


问:
K theory of the wedge of circles
I am interested in finding the $K_0, K_1$ groups of $C(S^1 \vee S^1)$.

We know that $K_0 (C(S^1)) = K_1(C(S^1)) = \mathbb Z$, but is this directly helpful?

答:
If $X$ is a compact space written as $X=A\cup B$, where $A$ and $B$ are closed subsets, with $I=A\cap B$, then the Mayer-Vietoris sequence is an exact sequence of topological $K$-groups
$\require{AMScd}$
\begin{CD}
  K_0(C(X)) @>({\rho_A}_*,{\rho_B}_*)>> K_0(C(A))\oplus K_0(C(B)) @>{\pi_A}_*-{\pi_B}_*>> K_0(C(I))\\
  @A \partial  AA  @.
 @VV\text{ind}V\\
K_1(C(I))
@<<{\pi_A}_*-{\pi_B}_*<
K_1(C(A))\oplus K_1(C(B))
@<<({\rho_A}_*,{\rho_B}_*)<
K_1(C(X))
  \end{CD}

where

$\rho_A:C(X)\to C(A)$,

$\rho_B:C(X)\to C(B)$,

$\pi_A:C(A)\to C(I)$,

$\pi_B:C(B)\to C(I)$,

are the restriction maps.  Regarding the case in point, where $A=B=S^1$, and $I=\{*\}$, we have
\begin{CD}
 K_0(C(X))@>({\rho_A}_*,{\rho_B}_*)>> \mathbb Z\oplus \mathbb Z @>{\pi_A}_*-{\pi_B}_*>> \mathbb Z\\
  @A \partial  AA  @.
 @VV\text{ind}V\\
0
@<<{\pi_A}_*-{\pi_B}_*<
\mathbb Z\oplus \mathbb Z
@<<({\rho_A}_*,{\rho_B}_*)<
K_1(C(X))
  \end{CD}
so we must have that $K_0(C(X))=\mathbb Z$, and $K_1(C(X))=\mathbb Z\oplus \mathbb Z$.

Title: K theory of the wedge of circles


问:
How many "distinct" polynomials are there mod $n$?
(This sounds like a fairly natural question, but I couldn't find it asked on this site before…)

Fix an integer $n$ as the modulus, and consider polynomials (with integer coefficients) modulo $n$.

For any polynomial $f$ and any fixed $k$, we have $f(x) + nx^k \equiv f(x) \pmod n$ for all integers $x$; thus any two polynomials with some coefficient differing by a multiple of $n$ are equivalent in the sense that they give the same value mod $n$ for any integer $x$ as input. So we can assume that each coefficient of $f$ is one of $\{0, 1, 2, \dots, n-1\}$.

When $n$ is prime, we know (from Fermat's little theorem) that $x^n \equiv x \pmod n$ for all integers $x$, which means any polynomial of degree $n$ or greater is equivalent (in the same sense as above) to a polynomial of lower degree. Even when $n$ is not prime, some such relationship probably exists; for example when $n=4$, we can see that $x^4 \equiv x^2 \pmod 4$. (I guess, though this may be wrong and I haven't checked it, that if the prime factorization of $n$ is $n = p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r}$, then $x^n \equiv x^{n/(p_1p_2\cdots p_r)} \pmod n$.)

In any case, whenever $x \equiv y \pmod n$, we have $f(x) \equiv f(y) \pmod n$, so it is enough to consider the polynomial's values at $n$ consecutive integers say $\{0, 1, 2, \dots, n-1\}$. As there are at most $n$ distinct values it can have mod $n$ for each input $n$, there are at most $n^n$ distinct polynomials in this sense.

The question: How many distinct polynomials with integer coefficients are there mod $n$? Here, the meaning of "distinct" is that polynomials $f$ and $g$ are "distinct" if there exists at least one integer $x$ at which $f(x) \not\equiv g(x) \pmod n$. Or, using the terminology from what I found later (see below), we want the number of polynomial functions from $\mathbb{Z}$ to $\mathbb{Z}/n\mathbb{Z}$.

That is the question, but of course before asking it here I tried a few things to find the answer myself, including writing the following program.

Define the "signature" (just a name I made up for writing this program: is there already a standard term for this?) of a polynomial $f$ as the tuple $(f(0), f(1), f(2), …, f(n-1))$ of its values at all $n$ points. We can count the number of distinct polynomials mod $n$ by keeping track of all the encountered signatures, as we start with the zero polynomial and successively add $ax^k$ for each $a$ from $0$ to $n-1$, for increasing $k$ until $x^k$ has the same signature as some already seen polynomial.

```<code>#!/usr/bin/env python3

# Adding the signatures of two polynomials mod n
def add(t1, t2):
    n = len(t1)
    assert n == len(t2)
    return tuple((t1[i] + t2[i]) % n for i in range(n))

def count(n):
    szero = tuple([0]*n)  # The signature of the zero polynomial
    seen = {szero}  # Only the zero polynomial has been seen
    k = 0
    sxk = tuple([1]*n)  # The signature of x^k
    while True:
        news = set()  # New signatures, by adding ax^k for some a
        saxk = szero
        for a in range(1, n):
            saxk = add(saxk, sxk)  # The signature of a*(x^k)
            for s in seen:
                newp = add(saxk, s)  # a*(x^k) + (smaller polynomial)
                if newp not in seen: news.add(newp)
        seen.update(news)
        # Now done with all polynomials of degree k
        k += 1
        sxk = tuple((sxk[i]*i) % n for i in range(n))
        if sxk in seen: break
    return len(seen)

print('n', '\t', 'count(n)')
for n in range(1, 11):  # Gets very slow for n=11
    print(n, '\t', count(n))
</code>```

This prints:

```<code>n    count(n)
1    1
2    4
3    27
4    64
5    3125
6    108
7    823543
8    1024
9    19683
10   12500
</code>```

The good thing about computing these small values is that there is the OEIS, using which you can, to quote Donald Knuth, "compute your way into the literature". And accordingly, searching the OEIS for this sequence $1$, $4$, $27$, $64$, $3125$, $108$, $823543$, $1024$, $19683$ gives A058067 described as "Number of polynomial functions from Z to Z/nZ" which is exactly our question. We also learn from that page that there's a formula

$$
\prod_{k=0}^{n-1} \frac{n}{\gcd(n, k!)}
$$

that fits the numbers we computed and (per that page) was first proved by Aubrey J. Kempner in 1921 in a two-part paper called "Polynomials and their residue systems" (1, 2), and there's what appears to be a nice paper in 2000 by Manjul Bhargava called "The Factorial Function and Generalizations" PDF that generalizes it (and maybe gives a shorter proof).

But I haven't read either of these papers as I ran out of my time budget for this (well, I exceeded it by writing up this question), and I hope someone can give a clear proof/derivation of the formula either using those papers or independently.

Edit: I read the latter paper and understood it enough to answer this question (left an answer below), but I have a follow-up question: Smallest number of residue classes for a polynomial function mod $n$ that separates $0$.

答:
The simplest case is when $n$ is prime: Then the first $p$ monomials are independent and give us $n^n$ polynomial functions.

The next case of interest is when $n$ is a prime power, whci is left as an exercise to the reader. Once we have solved that, the full result follows using the Chinese Remainder Theorem.


问:
Borel Measures on $\mathbb{R}$
My question is the following: do there exist Borel measures $\mu$ and $\nu$ such that, given a differentiable function $f$, we have
$$
f'(0) = \int_\mathbb{R} f d\mu - \int_\mathbb{R} f d\nu?
$$

I've been thinking about the one-to-one correspondence between Borel measures and (left) continuous increasing functions $F$ given by
$$
\mu((a,b]) := F(b) - F(a).
$$
I figure if I can choose an appropriate $F$, then the result would follow. I was also thinking that maybe the Riesz representation theorem is another approach, since differentiation is a linear map. I'd appreciate any help.

答:
This is impossible. The idea is that a function which is compactly supported and uniformly close to 0 must have small integral, but its derivative at 0 can still be big.

Let $\phi$ be some nonnegative smooth cutoff function which has support inside $[-1,1]$ and equals 1 on some neighborhood of 0.  If your equation is to hold for $\phi$, then in particular $\int \phi\,d\mu$ and $\int \phi\,d\nu$ must both be finite.  Now consider the function $f_n(t) = \frac{1}{n} \phi(t) \sin(n^2 t)$.  Note that $|f_n| \le \phi$ and $f_n \to 0$ pointwise (even uniformly), so  $\int f_n\,d\mu \to 0$ and $\int f_n\,d\mu \to 0$ by dominated convergence.   Thus the right side of the proposed equation goes to $0$.  But $f_n'(0) = n \to \infty$.

Title: Borel Measures on $\mathbb{R}$


问:
Find the marginal PDF and name its distribution
My question:

If $X_{1}, X_{2}$ have the joint pdf

$$f(x_{1}, x_{2}) = \begin{cases} 6x_{1}x_{2}^{2} &: 0 <x_{1} < 1, 0 < x_{2} < 1\\ 0 &: \text{else} \end{cases}$$

and $Y_{1} = X_{1}X_{2}, Y_{2} = X_{1}\sqrt{X_{2}}$. what is the marginal pdf of $Y_{1}$, what kind of distribution does it have, and what are the parameters?

My work so far:

First I found the joint pdf of $Y_{1}, Y_{2}$ using

$$f_{{y_{1}}{y_{2}}}\left(y_{1}, y_{2}\right)=f_{x_{1}x_{2}}\left(x_{1}, x_{2}\right)\cdot |J|$$

where J is the Jacobian, and having found $x_{1}=\frac{y_{2}^{2}}{y_{1}}$, $x_{2}=\frac{y_{1}^{2}}{y_{2}^{2}}$ and $|J|=\frac{2}{y_{2}}$,  we have

$$=6\left(\frac{y_{2}^{2}}{y_{1}}\right) \cdot \left(\frac{y_{1}^{2}}{y_{2}^{2}}\right)^{2} \cdot \frac{2}{y_{2}} $$

$$ = \frac{12y_{1}^{3}}{y_{2}^{3}}$$

So,

$$f_{{y_{1}}{y_{2}}}= \begin{cases} \frac{12y_{1}^{3}}{y_{2}^{3}} &: 0 < \frac{y_{2}^{2}}{y_{1}} < 1, 0 < \frac{y_{1}^{2}}{y_{2}^{2}} < 1\\ 0 &: \text{else} \end{cases}$$

Now here is where I am having some problems. I found the marginal pdf of $Y_{1}$ as

$$f_{y_{1}}\left(y_{1}\right)= \int^{\sqrt{y_{1}}}_{y_{1}}\frac{12y_{1}^{3}}{y_{2}^{3}} dy_{2}$$

$$=6y_{1}-6y_{1}^{2}$$.

This is not any specific distribution that I can identify. Any hints would be very helpful. Thanks.

答:
All your efforts are correct.  You should also note that the support is on $(0..1)$, that $6y(1-y)$ is a parabola and symmetric about $y=1/2$.

Less obvious is that $6=4!/2!^2$.

$$\begin{align}f_{\small Y_1}(y) &= 6y\,(1-y)\,\mathbf 1_{\small 0\lt y\lt 1}\\[1ex] &=\dfrac{(2+2)!}{2!~2!}\,y^{2-1}(1-y)^{2-1}\,\mathbf 1_{\small 0\lt y\lt 1} \end{align}$$

This is not any specific distribution that I can identify. Any hints would be very helpful. Thanks.

You beta look for a family of supports defined over $(0..1)$ that are symmetrical (for some parameters, at least).


问:
$5 \mid n^2 - m^2$ is an equivalence relation
How can I show this is an equivalence relation:

$$
n \operatorname{R} m \Longleftrightarrow n^2 - m^2 \textrm{ is divisible by } 5
$$

I know equivalence relations are symmetric, reflexive and transitive.  I'm just not sure how to use this knowledge to prove it.

答:
Recall that the binary relation $ \operatorname{R} $ on $\mathbb{N}^2$ is an equivalence relation iff it is symmetric, reflexive and transitive.

Fix any $n,m\in\mathbb{N}$.

Reflexive ($n \operatorname{R} n$): $n^2-n^2=0$ and $5|0$

Symmetric (if $n \operatorname{R} m$ then $m \operatorname{R} n$): if $5|n^2-m^2=0$ then $5|-(n^2-m^2)=m^2-n^2$

Transitive (if $(n \operatorname{R} m)$ and $(m \operatorname{R} \ell)$ then $n \operatorname{R} \ell$): if $5|n^2-m^2=0$ and $5|m^2-\ell^2=0$ then $5|(n^2-m^2+m^2-\ell^2)=n^2-\ell^2$


问:
Not quite understanding a proof for a function of a random variable.
Theorem

Proof

Why is is that in the decreasing case in the proof, G(y) evaluates to P(X>=s(y) as opposed to P(X<=s(y)? What swaps the sense of the inequality?

答:
As this says: $G(y)=\Pr(r(X) \le y)$

But if $r(x)$ is a decreasing functions, then a smaller value of $x$ corresponds to a larger value of $r(x)$ while a larger value of $x$ corresponds to a smaller value of $r(x)$.

If the inverse of $r(x)$ is $s(y)$ then this means a smaller value of $s(y)$ corresponds to a larger value of $r(s(y))=y$ while a larger value of $s(y)$ corresponds to a smaller value of $r(s(y))=y$. So $s(y)$ is also a decreasing function.

Rather like multiplying by $-1$, this means that the inequality reverses as $r(X) \le Y \iff s(r(X)) \ge s(Y) \iff X \ge s(Y)$, leading to $\Pr(r(X) \le y) = \Pr(X \ge s(y))$. That is $1-\Pr(X \lt s(y))$ but since you have a continuous distribution, it is also $1-\Pr(X \le s(y))=1-F(s(y))$ so you can say in this case $$G(y)=1-F(s(y))$$


问:
Expression for Rank 2 Tensor in Vector Notation
How does one write the following expression

$D_{jk} (r_k \delta_{ij} - r_{i}\delta_{jk} - r_j \delta_{ik})$

in matrix notation?  Is this just

$\textbf{D} (\textbf{r} \times \textbf{I})$?

答:
The given expression in matrix notation should be:

$$
(\textbf{D}-\textbf{D'}- \ \text{trace}(\textbf{D}))\textbf{r}
$$

By applying the Kronecker Delta we have:
$$
D_{jk} (r_k \delta_{ij} - r_{i}\delta_{jk} - r_j \delta_{ik}) \\
=D_{jk} r_k \delta_{ij} - D_{jk}r_{i}\delta_{jk} - D_{jk}r_j \delta_{ik} \\
=D_{ik} r_k - D_{kk}r_{i} - D_{ji}r_j \\
$$

I was not able to find a cross product (Levi-Civita expression) from the given Kronecker Detla Expression.


问:
Example of an abstract elementary class that is not even a pseudo-elementary class
I have read the definition of an abstract elementary class, and I am told that all elementary class and even all pseudo-elementary classes are abstract elementary classes. That raises the question, is there an abstract elementary class which is not even a pseudo-elementary class? And if so, can someone exhibit an example?</p>

答:
First, note that there's a bit of a mismatch of terminology here. A pseudoelementary class is just a class of structures (satisfying some properties). An AEC, by contrast, is a class of structures together with an embedding relation (satisfying some properties). So strictly speaking, it is not the case that each pseudoelementary class is an AEC; rather, if $P$ is a pseudoelementary class then there is an AEC whose underlying class of structures is exactly $P$.

To make your question more precise, let me rephrase it as follows:

What are examples of AECs whose underlying classes of structures are not pseudoelementary?

For a silly example, take the class of all structures of cardinality $\le\aleph_{17}$ (say) and the identity embedding relation. This trivially satisfies the AEC axioms; in particular, its Lowenheim-Skolem number is $\aleph_{17}$.

(An earlier version of this answer made a very silly mistake here: if we take as the embedding relation the usual elementary embedding relation in the above, we do not get an AEC since the chain axiom isn't satisfied - consider a proper elementary chain of length $\omega_{17}$ consisting of structures of cardinality $\aleph_{16}$.)

Of course there are far more interesting ones. Even ignoring the silliness re: the embedding relation, this example is "eventually elementary" in the sense that there is a elementary class which coincides with it once we restrict attention to large enough structures (namely, $Mod(\perp)$). For a better example, pick some $\mathcal{L}_{\omega_1,\omega}$-sentence $\varphi$ and look at the models of $\varphi$ ordered by $\mathcal{L}_{\omega_1,\omega}$-elementary embeddings. This is an AEC, and if you pick $\varphi$ correctly the class of structures in it will not be "eventually pseudoelementary."

Now as a cautionary tale, note that in general regular logics in the sense of non-AEC-based abstract model theory do not generally translate nicely into AECs. Suppose $\mathcal{L}$ is a regular logic and $T$ is a $\mathcal{L}$-theory. The pair $(Mod_\mathcal{L}(T),\preccurlyeq_\mathcal{L})$ is a natural thing to consider and trivially satisfies the AEC axioms except Tarski-Vaught and Lowenheim-Skolem. However, this still leaveslots of room for nastiness. For example, both $\mathcal{L}_{\omega_1,\omega_1}$ and $\mathsf{SOL}$ can detect when an ordering has countable cofinality, and this leads to the Tarski-Vaught axiom failing in their respective "candidate AECs."

This should actually be a promising sign, though: it indicates that the AEC axioms are strict enough to rule out a lot of the nastier logics that abstract model theory runs into, and raises the possibility of an actually nontrivial structure theory (which of course is what emerged).


问:
Let $G$ a nilpotent group such that $10$ divides $|G|$. It is true that $G$ has element of order $10$?
Let $G$ a nilpotent group such that $10$ divides $|G|$. It is true that $G$ has element of order $10$?

We know that $G$ can be expressed like  direct sum of Sylow's subgroups and that $G$ has a normal subgroup of order $10$ but how can we answer to this question?

答:
There are elements of order  $5$ and $2$, by Cauchy.

If I can show those elements commute, then their product will have order $10$.

But nilpotency and finiteness imply that elements of relatively prime order commute.


问:
Formal Proof of "No Largest Number"
I need to create a formal proof of there not being the largest number, with only the definition of "less than" being given with Peano arithmetic. $s(x)$ means successor of $x$ or $x + 1$.

$$
\text{Premise}\\
\forall x\forall y (x \lt y \iff \exists z(x + s(z) = y)\\
\text{Prove}\\
\forall x \exists y (x < y)
$$

I've tried proof by contradiction, but the resources online don't seem to translate well into a formal proof. In particular, the strategy of "assume there is a largest number, add one, done" doesn't really translate with formal proof steps.

答:
So you want to prove that for all $x$, there exists a $y$ such that $x$ is less than $y$. To do that, imagine you are given an $x$ (maybe $x=99$) and try to come up with a $y$ that is greater than $x$. How about $x + 1$ (i.e., $100$ if $x = 99$)? So there's your proof: as $y = x + 1 > x$ for any $x$, then for all $x$, there exists a $y$ such that $x < y$.

Proof by contradiction is best regarded as a last resort: to be used only if working forward from the positive information that you have isn't getting you anywhere.

Title: Formal Proof of "No Largest Number"


问:
Is it really easier to decrease your overall grade in a class than increase it?
I was thinking about how grades work. Obviously most people’s grades are weighted, each type of assignment holds a different value and exams usually count for the most. People always complain about how when they get good grades their overall grade barely increases and when they get bad grades, their grades tank. That got me thinking and I came up with the scenario below.

If I have 5 assignments and the average is 80 and I get a 6th assignment one of 3 things can happen:

I get below an 80, my average drops

I get an 80, my grade stays the same

I get above an 80, my grade increases

Say I get below an 80. If I get a 70, the new average is 78 and 1/3. So, the average decreased by 1 and 2/3. If I get a 90, the new average is 81 and 2/3. So the average increased by 1 and 2/3 (the same exact amount).

So I immediately thought that idea is bullshit. But I’m wondering if maybe I’m wrong. Here are some factors I thought could prove me wrong:

The consistency of the grades (an average of 50 can mean you have all 50s or you got an equal amount of 100s and 0s, so maybe if you get a 90 in one scenario, it changes your grade differently than if you got a 90 in the other scenario)

The amount of assignments (if you have less assignments, that’s great if your grades are good but if you do badly even on one, your overall grade can suffer)

How much the assignment was weighted (if you did badly on an assignment that’s 5% versus 10% will affect how much your grade drops)

答:
As you rightly noticed, there's no mathematical bias keeping an average low.

However, the difficulty of obtaining the grades necessary to alter your average may be the cause of the perception that an average is easy to lower but hard to raise.

If you already have an average of 80, you need a high mark with a significant relative weight to your existing average to significantly change that average. In most cases, you'd need to get 90+ on the concerned assignment, which means cutting your failure rate by half, which is no small feat. One might even consider it equivalent to being two times better.

On the other hand, lowering your average by an equivalent amount is easy: you just need to get 70-, which is easily achieved by having a bad night of sleep and misreading an exam question.


问:
What is the formal definition of repeated limit using quantifiers?
The basic question is what has been asked in the title. I looked for the definition here, here and here but no definition uses quantifiers.

I tried to formulate the definition but succeeded only partially. For this I considered the following special case,

Let $f:D(\subseteq\mathbb{R}^2)\to\mathbb{R}$ and $(a,b)\in D$. Then we say $\displaystyle\lim_{x\to a}f(x,y)=L(y)$ if $$(\forall y)(\exists L_y\in\mathbb{R})\left[(\forall \varepsilon>0)(\exists \delta>0)\right[|x-a|<\delta\rightarrow |f(x,y)-L_y|<\varepsilon]]$$

答:
The definition for infinite limit is (almost) the following:

$$\exists b\forall \epsilon\exists k_0 \forall n:(n>k_0 \rightarrow|c_{n}-b|<\epsilon) \land \forall n \exists k_2 :(n>k_2) \rightarrow(\exists c_n  \forall \epsilon \exists k_1 \forall m: (m>k_1 \rightarrow|a_{n,m}-c_n|<\epsilon))$$

Here $b$ defines the outer limit and $c_n$ the inner limit which is required to exist after some point $k_2$. If the above is false, then the limit does not exist. We still need to define the domains, which should be obvious. The finite limit is defined replacing the conditions and switching to real or rational functions.

However, this may not be the best way to think about repeated limit as it requires proving what $c_n$ (the first limit) is independently. We can prove that the definition is equivalent to the intuitive definition of what a repeated limit should mean.
$$\exists b \forall \epsilon \exists k_0 \forall n \exists k_1(\epsilon,n)\forall m   : (n>k_0 \land m>k_1) \rightarrow|a_{n,m}-b|<\epsilon$$

Proof:

$\Rightarrow$

Suppose the first definition holds. Let now $\epsilon \in \mathbb R_+$ and $n,m \in \mathbb N$. Now by triangle inequality there are $k_0 (\epsilon),k_1 (n,\epsilon)\in \mathbb N$ such that
$$|a_{n,m}-b|=|a_{n,m}-c_n+c_n-b|\leq|a_{n,m}-c_n|+|c_n-b|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon,$$

when $k_0<n,k_1<m$. This is since, by the first condition we have a $k_0$ such that $|c_n-b|<\frac{\epsilon}{2}, k_0<n$, and by the second condition there is a $k_1(\epsilon,n)$ such that $|a_{n,m}-c_n|<\frac{\epsilon}{2}$, when $k_1<m$. This concludes the proof, we only switch the quantifier order to reflect the dependencies.

$\Leftarrow$

We have two statements to prove.

Proof of the second statement:
We can argue by contradiction, select $\epsilon=1$ and we get a contradiction as $c_n=b$ should satisfy the condition. Constructively, the condition should possibly be added to the second definition.

Proof of the first statement: Let $\epsilon \in \mathbb R_+$, we again use the triangle inequality. By the second statement and the assumption there is a $k_0,k_1$, such that
$$|c_n-b|=|a_{n,m}-c_n|+|a_{n,m}-b|< \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon,$$
when $k_0<n,k_1<m$. Hence the statement is proved.


问:
Obtain non linear solution using neural network
The function f(x)=theta·x where theta is a row vector and x is a column vector, is a linear function. How can I obtain a non-linear function g(x) using a multi layer network, that also takes in x as input and theta as the parameter of the first layer?

If I have a 2 layer network, with a first dense layer of parameter theta and second dense layer with a non-linear activation function, would that suffice to say that I now have a non-linear function?

答:
Mathematically speaking, your argument is insufficient. After all, if the second layer have weights of 0, then $g(x)=0$, which is linear.

That being said, it is highly unlikely that a neural network that was not designed for that purpose ends up being linear, so depending on your needs, that may be sufficient.


问:
Sum of Three Series
I am having a bit of trouble with this problem and was wondering if I can get some advice. Here is the problem:

Suppose $\sum_{n = 1}^{\infty} a_n = A$, $\sum_{n = 1}^{\infty} b_n = B$ and $\sum_{n = 1}^{\infty} c_n = C$ are convergent series , where $A, B, C$ are constants. Evaluate $\sum_{n = 1}^{\infty} (a_n + b_n + c_{n + 1})$ if it exists.

So my attempt to this problem:

I know that the sum can be distributed inside the brackets, such that
\begin{align*}
\sum_{n = 1}^{\infty} (a_n + b_n + c_{n + 1}) = \sum_{n = 1}^{\infty} a_n + \sum_{n = 1}^{\infty} b_n + \sum_{n = 1}^{\infty} c_{n + 1}
\end{align*}

We know that $\sum_{n = 1}^{\infty} a_n = A$ and $\sum_{n = 1}^{\infty} b_n = B$, so
\begin{align*}
\sum_{n = 1}^{\infty} a_n + \sum_{n = 1}^{\infty} b_n + \sum_{n = 1}^{\infty} c_{n + 1} = A + B + \sum_{n = 1}^{\infty} c_{n + 1}
\end{align*}

I am not exactly sure how I am supposed to find $\sum_{n = 1}^{\infty} c_{n + 1}$. Would appreciate some tips and advice. Thanks!

答:
The answer is simply:
$$
A+B+C-c_1
$$
The trick is just to realize than the C sum lacks the first term, which is $c_1$.

The sum is allowed to be performed because the series are convergent. If two of them diverges, the sum do not necessarily will diverge, which by the way, is not the case in here.

Title: Sum of Three Series


问:
Proof for how .9999999...=1
I was just curious for the proof/theorem for a very close decimal (so when you keep adding a decimal) to equal the next integer such as .999999999999....=1. Thank you.</p>

答:
The given number is a geometric series in disguise:

\begin{align*}
0.999\ldots = 9\sum_{n=1}^{\infty}10^{-n} = \frac{9}{10}\left(\frac{1}{1 - \frac{1}{10}}\right) = 1
\end{align*}
which converges because its ratio is $0.1 < 1$.

Title: Proof for how .9999999...=1


问:
Ito differential equation
Define

$$X_t := \left( \begin{matrix} \cos W_t \\ \sin W_t \end{matrix} \right).$$

where  $W = \left( W_t,\mathcal F_t \right) _{t\ge0}$ is a standard Wiener process. Find the Ito differential of X and show that

$$\mathbb EX_t = \left( \begin{matrix} \exp \left( -\frac t2 \right) \\ 0 \end{matrix} \right).$$

答:
By applying Itô's Formula
$$f(W_t)-f(W_0) = \int_0^t f'(W_s) \, dW_s + \frac{1}{2} \int_0^t f''(W_s) \, ds$$

to $f(x) := \cos x$ we obtain

$$\cos(W_t)-1 = -\int_0^t \sin(W_s) \, dW_s -  \frac{1}{2} \int_0^t \cos(W_s) \, ds$$

The mapping $$(t,w) \mapsto M(t,w) := \int_0^t \sin(W_s) \, dW_s$$ is a martingale since it's a stochastic integral with respect to a Brownian motion. This implies $\mathbb{E}M_t=0$. Thus,

$$\mathbb{E}(\cos(W_t))-1 = - \frac{1}{2} \int_0^t \mathbb{E}(\cos(W_s)) \, ds$$

Define $\varphi(s) := \mathbb{E}(\cos(W_s))$. The last equation is equivalent to

$$\varphi(t)-1 = - \frac{1}{2} \int_0^t \varphi(s) \, ds$$

Solve this differential equation and you are done with the first component of $X_t$. Similar argumentation works for $\sin(W_t)$.

Title: Ito differential equation


问:
Fundamental vector field on a vector space
We know that in general if M is a manifold provided with an action of a lie group G, then for X in the Lie algebra of G , we define the vector field $X_M$  by

$X_M(m) = \frac{d}{dt}_{|t=0}e^{-tX}.m $.

I've come across this statement wich I don't know how to prove it:

Let V be a euclidean vector space of dimension n with basis $e_i$, $1 \leq i \leq n$. We denote by $x_i$ the coordinate functions on V, and $\partial_i $ the direvation in the direction of the vector $e_i$.
Suppose that V is provided with an action of a Lie group G.

Why this is true:

If $X \in \mathfrak{g}$,  then the vector field $X_V$ is given by :

$X_V= - \sum_{i < j} (Xe_i,e_j)(x_i \partial_j - x_j \partial_i)$ ?

Thanks!

答:
It isn't true. For example, take $V=\mathbb{R}^2$, $G=\operatorname{GL}(2,\mathbb{R})$, and $X = \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\in\mathfrak{g}$. Then
$$
X_V(x_1,x_2) = \frac{d}{dt}\Big\vert_{t=0}\begin{pmatrix} e^{-t} & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix} = \frac{d}{dt}\Big\vert_{t=0}\begin{pmatrix}e^{-t}x_1 \\ x_2\end{pmatrix} = -x_1\partial_1,
$$
but according to your formula it would be
$$
X_V(x_1,x_2) = -(Xe_1,e_1)(x_1\partial_2-x_2\partial_1) = -(x_1\partial_2-x_2\partial_1).
$$
Perhaps the statement you read has some additional conditions on the group $G$?


问:
Equality of Line Integrals for One Forms
I saw a post about how if the equality between integrals of two functions f and g holds over any domain V, then it implies an equality of the integrands too, so f=g. Can we extend this idea to C1 one forms on Rn? If we have two C1 1-forms that have the same line integral along every piecewise smooth curve Ψ from [a,b] to Rn, then does this imply the one forms are equal too?</p>

答:
This is an excellent question. Yes. Suppose $\omega,\eta$ are $1$-forms on $\Bbb R^n$ and they are not equal at some point $p$. Then (by continuity) there is a neighborhood of $p$ on which they are not equal. By subtracting, we reduce to the case of a single $1$-form $\phi$ which is nonzero at $p$, and therefore nonzero on some neighborhood. Look at the hyperplane $\Pi = \{v\in\Bbb R^n: \phi(p)(v) = 0\}$. $\Pi$ divides $\Bbb R^n$ into two halves, if you like; choose a $w\in\Bbb R^n$ on the positive side of $\Pi$. Let $\epsilon>0$ be suitably small. If you let $\Psi$ be the line segment from $p$ to $p+\epsilon w$, then I claim that $$\int_\Psi\phi > 0,$$
which settles your question (by taking the contrapositive). Can you finish the argument?


问:
Manipulate Equation to get Complicated Expression
Suppose $$a/x+b/y+c/z=1/143$$ for real numbers $a, b,$ and $c$. What is:
$$abc(x/a+y/b+z/c-143)/((11a-x/13)(11b-y/13)(11c-x/13))\:?$$

答:
HINT:

With

$\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=\frac{1}{143}$  (I)

$\frac{1}{xbc}+\frac{1}{yac}+\frac{1}{zab}=\frac{1}{143abc}$ (II)

Expand the expression and do some algebra operation

$\frac{abc(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}−143)}{(11a−\frac{x}{13})(11b−\frac{y}{13})(11c−\frac{z}{13})}$

=$\frac{(xbc+yac+zab-143abc)13^3}{(143a-x)(143b-y)(143c-z)}$

=$\frac{(xbc+yac+zab-143abc)13^3}{143^3abc-143^2(xbc+yac+zab)+143(ayz+bxz+cxy)-xyz}$

=$\frac{(xbc+yac+zab-143abc)13^3}{143^3abc-143^2(xbc+yac+zab)+143xyz(\frac{a}{x}+\frac{b}{y}+\frac{c}{z})-xyz}$  (III)

Compare  (III) with (I), (II), you will get the answer.

 -$\frac{13}{11}$


问:
Expected value and variance from random variable
Given the following

$Y_1 \sim \mathcal{N}(μ, σ^2 )$

and

$Y_2=α+βY_1+U \;where \; Y_1 \;and \;U\;is\;independent\;and\;U∼\mathcal{N}(0,v^2)$

Let $μ=350$ and $σ^2 =12365$

How would i calculate the expected value and variance from $Y_2$ ?
And how can i find the distribution of it?

答:
You still haven't given values for $v, a, b,$
so I have supplied some.

The distribution of $Y_2$
is normal.

In my simulation in R below, I have used a million
iterations, so the sample mean and SD of $Y_2$
should agree with the respective population values
to several significant digits--I guess, good enough to
find mistakes in applying the formulas.

```<code>set.seed(2021)
y1 = rnorm(10^6, 350, 111.198)
u =  rnorm(10^6, 0, 50)
a = 10;  b = 50
y2 = a + b*y1 + u
summary(y2);  sd(y2);  var(y2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  -8248   13765   17511   17513   21263   43899
[1] 5566.788  # aprx SD(Y2)
[1] 30989134  # aprx Var(Y2)
</code>```

Here are histograms of the simulated distributions
along with the corresponding normal density curves.

R code for figure.

```<code>par(mfrow=c(1,3))
hist(y1, prob=T, col="wheat")
 curve(dnorm(x,350,111.198), add=T, lwd=2)
hist(u, prob=T, col="wheat")
 curve(dnorm(x,0,50), add=T, lwd=2)
hist(y2, prob=T, col="skyblue2")
 curve(dnorm(x, 17813, 5566.788), add=T, lwd=2)
par(mfrow=c(1,1))
</code>```

You say in a comment that you don't follow how to
calculate some of the results for $Y_2.$
That is
not a comment that prompts specific explanations.

I have had a quick look at answers by @herbsteinberg and @JLF, which seem fine:
(+1) for each.

Can you say specifically
what you don't understand? If there are several
difficulties, you might start by saying what the first
couple of them are.

Body:
You still haven't given values for $v, a, b,$
so I have supplied some.

The distribution of $Y_2$
is normal.

In my simulation in R below, I have used a million
iterations, so the sample mean and SD of $Y_2$
should agree with the respective population values
to several significant digits--I guess, good enough to
find mistakes in applying the formulas.

set.seed(2021)
y1 = rnorm(10^6, 350, 111.198)
u =  rnorm(10^6, 0, 50)
a = 10;  b = 50
y2 = a + b*y1 + u
summary(y2);  sd(y2);  var(y2)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  -8248   13765   17511   17513   21263   43899
[1] 5566.788  # aprx SD(Y2)
[1] 30989134  # aprx Var(Y2)

Here are histograms of the simulated distributions
along with the corresponding normal density curves.

R code for figure.

par(mfrow=c(1,3))
hist(y1, prob=T, col="wheat")
 curve(dnorm(x,350,111.198), add=T, lwd=2)
hist(u, prob=T, col="wheat")
 curve(dnorm(x,0,50), add=T, lwd=2)
hist(y2, prob=T, col="skyblue2")
 curve(dnorm(x, 17813, 5566.788), add=T, lwd=2)
par(mfrow=c(1,1))

You say in a comment that you don't follow how to
calculate some of the results for $Y_2.$
That is
not a comment that prompts specific explanations.

I have had a quick look at answers by @herbsteinberg and @JLF, which seem fine:
(+1) for each.

Can you say specifically
what you don't understand? If there are several
difficulties, you might start by saying what the first
couple of them are.


问:
help to understand the partial derivative of $f(x,y) = {1 + h(x) \over 1 + (g(y))^2}$
Can you please help me to understand how given this function:

$f(x,y) = {1 + h(x) \over 1 + (g(y))^2}$

Its partial derivative is:

$\frac{\partial f}{\partial x} = {1 \over 1 + (g(y))^2}(1 + h'(x))$

My initial attempt was:
$\frac{\partial f}{\partial x} = {1 \over 1 + (g(y))^2}(h'(x))$

答:
The approach I take (and also show to students) is to replace literally, either mentally or on paper, any occurrence of a variable or function of a variable which is not the one which is involved in the partial differentiation with a constant, say, $ \ C \ . $  This can make it easier to see which portions of the expression will require differentiation and to avoid confusion over how to proceed.  To pull an problem from Stewart (6th Ed., Section 14.3, #36):

$$f(x,y,t,z) \ = \ \frac{xy^2}{t + 2z} \ ; $$

$$\frac{\partial f}{\partial y} \ = \ \frac{\partial}{\partial y} [ \ \frac{C \cdot y^2}{D} \ ] \ = \ \frac{C}{D} \cdot 2y \ = \ \frac{2xy}{t + 2z} \ ;   $$

$$\frac{\partial f}{\partial z} \ = \ \frac{\partial}{\partial z} [ \ \frac{C}{D + 2z} \ ] \ = \ \frac{\partial}{\partial z} [ \ C \cdot (D + 2z)^{-1} \ ] $$

$$= \ \ C \cdot (-1) \cdot (D + 2z)^{-2} \cdot 2 \ = \ -\frac{2xy^2}{(t + 2z)^2} \ .   $$

This may seem like a "training-wheels" approach to some, but I find it particularly helpful in getting students to be clear about what partial differentiation does and where to use it in conjunction with the other rules of differentiation, such as the Product, Quotient, and Chain Rules.

For your function, we have

$$f(x,y) \ = \ \frac{1 + h(x)}{1 + [g(y)]^2} $$

$$ \Rightarrow \ \frac{\partial f}{\partial x} \ = \ \frac{\partial}{\partial x} [ \ \frac{1 + h(x)}{D} \ ] \ = \ \frac{1}{D} \cdot [  \frac{  \partial}{\partial x}h(x) \ ] \   = \  \frac{h'(x)}{1 + [g(y)]^2} \ , $$

$$ \frac{\partial f}{\partial y} \ = \ \frac{\partial}{\partial y} [ \ \frac{C}{1 + [g(y)]^2} \ ] \ = \ C \cdot   \frac{  \partial}{\partial y} [ \ (1 + [g(y)]^2)^{-1} \ ] $$

$$= \ C \cdot (-1) \cdot (1 + [g(y)]^2)^{-2} \cdot [ \  \frac{  \partial}{\partial y} (1 + [g(y)]^2) \ ] $$

$$= \ C \cdot (-1) \cdot (1 + [g(y)]^2)^{-2} \cdot [ \ 2g \cdot \frac{\partial}{\partial y} g(y) \ ] = \  -\frac{2 \ [ \ 1 + h(x) \ ] \cdot g(y) \cdot g'(y) }{( \ 1 + [g(y)]^2 \ )^2}  \ , $$

with the primes in each expression representing differentiation by the appropriate variable.


问:
Question about the determinant of the Jordan algebra
I'm recently reading the book Analysis on Symmetric Cones and meet with a question. In Chapter 2, the author asserts that for given $x\in V$, where $V$ is a Jordan algebra, the function $q=\det (e,x,x^2 ,x^3 ,\dotsb ,x^{r-1} ,e_{r+1} ,\dotsb ,e_n )$, where $n$ is the dimension of $V$ and $r$ is the rank of $V$, is a polynomial. I guess at first that it is a homogeneous function and so it is called a "polynomial", but when I read further, I find that it must be actually a polynomial. I just don't know why it could be a polynomial since it is a function from $V$ to the scalar field $\mathbb{R} $.

答:
A polynomial on a finite-dimensional vector space $V$ is defined to be a polynomial function of any choice of linear coordinates $x_1, \dots, x_n$ on that vector space.

In more detail:

Given any basis $e_1, \dots, e_n$ of a finite-dimensional vector space $V$, we can write any vector $x \in V$ as

$$ x = \sum_{i=1}^n x_i e_i $$

for some unique choice of numbers $x_i \in \mathbb{R}$.  These numbers $x_i$ are thus functions of $x$, called (linear) coordinate functions.

Any function

$$ f \colon V \to \mathbb{R} $$

that is a polynomial in the linear coordinate functions is called a polynomial on $V$.

In particular, the determinant function on a Jordan algebra is a polynomial on that Jordan algebra, in this sense.

The usual determinant of an $n \times n$ matrix is likewise a polynomial function on the space of $n \times n$ matrices.


问:
Combining derivatives of a composition of function
I was studying the proof of a proposition, which states the closure of tangent spaces under addition and scalar multiplication. The proof is straightforward for the most part, however, I can't seem to understand this application of differentiation across the composed functions. What follows is the complete proof.

Proof. Let $(U, x)$ be a chart on $M,$ with $U$ a neighbourhood of $p$.
i) Define the curve
$$
\sigma(t):=x^{-1}((x \circ \gamma)(t)+(x \circ \delta)(t)-x(p))
$$
Note that $\sigma$ is smooth since it is constructed via addition and composition of smooth maps and, moreover:
$$
\begin{aligned}
\sigma(0) &=x^{-1}(x(\gamma(0))+x(\delta(0))-x(p)) \\
&\left.=x^{-1}(x(p))+x(p)-x(p)\right) \\
=x^{-1}(x(p)) \\
&=p
\end{aligned}
$$
Thus $\sigma$ is a smooth curve through $p$. Let $f \in \mathcal{C}^{\infty}(U)$ be arbitrary. Then we have
$$
\begin{aligned}
X_{\sigma, p}(f) &:=(f \circ \sigma)^{\prime}(0) \\
&=\left[f \circ x^{-1} \circ((x \circ \gamma)+(x \circ \delta)-x(p))\right]^{\prime}(0)
\end{aligned}
$$
where $\left(f \circ x^{-1}\right): \mathbb{R}^{d} \rightarrow \mathbb{R}$ and $((x \circ \gamma)+(x \circ \delta)-x(p)): \mathbb{R} \rightarrow \mathbb{R}^{d},$ so by the multivariable
chain rule
$$
=\left[\partial_{a}\left(f \circ x^{-1}\right)(x(p))\right]\left(\left(x^{a} \circ \gamma\right)+\left(x^{a} \circ \delta\right)-x^{a}(p)\right)^{\prime}(0)
$$
$=\left[\partial_{a}\left(f \circ x^{-1}\right)(x(p))\right]\left(\left(x^{a} \circ \gamma\right)^{\prime}(0)+\left(x^{a} \circ \delta\right)^{\prime}(0)\right)$
$=\left(f \circ x^{-1} \circ x \circ \gamma\right)^{\prime}(0)+\left(f \circ x^{-1} \circ x \circ \delta\right)^{\prime}(0)$
$=(f \circ \gamma)^{\prime}(0)+(f \circ \delta)^{\prime}(0)$
$=:\left(X_{\gamma, p} \oplus X_{\delta, p}\right)(f)$

Here lies my doubt: how does the third step follow from the second?

$$
=\left[\partial_{a}\left(f \circ x^{-1}\right)(x(p))\right]\left(\left(x^{a} \circ \gamma\right)+\left(x^{a} \circ \delta\right)-x^{a}(p)\right)^{\prime}(0)
$$
$=\left[\partial_{a}\left(f \circ x^{-1}\right)(x(p))\right]\left(\left(x^{a} \circ \gamma\right)^{\prime}(0)+\left(x^{a} \circ \delta\right)^{\prime}(0)\right)$
$=\left(f \circ x^{-1} \circ x \circ \gamma\right)^{\prime}(0)+\left(f \circ x^{-1} \circ x \circ \delta\right)^{\prime}(0)$
$=(f \circ \gamma)^{\prime}(0)+(f \circ \delta)^{\prime}(0)$
$=:\left(X_{\gamma, p} \oplus X_{\delta, p}\right)(f)$

Here's a link to the original notes.
It comes in Chapter 9, Proposition 9.2

答:
It's not clear exactly what point you mean by "third step", but: the multivariable chain rule says that if $h:\mathbb{R}^d\to\mathbb{R}$ and $k:\mathbb{R}\to\mathbb{R^d}$, then the derivative of $h\circ k:\mathbb{R}\to\mathbb{R}$ at $0\in\mathbb{R}$ is given by
$$
(h\circ k)'(0) = Dh(k(0))\cdot Dk(0) = \partial_ah(k(0))\, (k^a)'(0).
$$
Now apply this with $h = f\circ x^{-1}$ and $k = x\circ \gamma + x\circ\delta - x(p)$ or $k=x\circ \gamma + x\circ\delta$ (depending on which step you mean).


问:
Is there a quick proof as to why the vector space of $\mathbb{R}$ over $\mathbb{Q}$ is infinite-dimensional?
It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?</p>

答: As $\pi$ is trascendent over $\mathbb{Q}$. Then the set $\{1, \pi, \pi^{2},\cdots\}$ is linearly independent.

ContentLicense: CC BY-SA 2.5


问:
Prove that exists only one scalar product $\langle \ , \ \rangle$ in $\mathbb{R}^3$ that has $u, v,w$ as orthonormal basis
Let  $ u = (1, −1, 0), v = (0, 1, 1), w = (1, 1, 0) $ be vectors of $\mathbb{R}^
3$.

Prove that exists only one scalar product $\langle \ , \ \rangle$ in  $\mathbb{R}^3$ that has $u, v,w$ as
orthonormal basis.

Any idea to begin with?

I've think about the idea that the matrix $M(\langle \ , \ \rangle, \mathscr{S} ) = I_3 $, with $\mathscr{S}$ a Sylvester basis. Because of the fact that an euclidian geometry is an orthogonal geometry with Sylvester invvariant $r_0=0, r_+=n, r_-=0$. But I'm not sure at all...

Any help? Thanks!!

答:
Suppose $\langle \cdot, \cdot \rangle'$ inner product for which $u,v,w$ is an orthonormal basis. For each $x \in \mathbb{R}^3$ $$ x= \langle x, u\rangle' u + \langle x, v \rangle' v + \langle x, w \rangle' w .$$ Taking the inner product of this with respect to the original inner product $\langle \cdot, \cdot \rangle$ gives $\langle x ,u \rangle = \langle x ,u \rangle'$, $\langle x ,v \rangle = \langle x ,v \rangle'$ and $\langle x ,w \rangle = \langle x ,w \rangle'$. This implies $\langle \cdot ,\cdot \rangle = \langle \cdot ,\cdot \rangle'$.

A note: there was nothing special here about the fact we were working in $\mathbb{R}^3$ or the specific values of $u,v,w$.


问:
The Lie algebra of the generators of the projective transformation is isomorphic to the Lie algebra of traceless matrices.
The general projective transformation of the $x$-$y$ plane is given by
$$\tilde{x}=\frac{a_1x+a_2 y+a_3}{a_7x+a_8y+a_9},\quad\tilde{y}=\frac{a_4x+a_5y+a_6}{a_7x+a_8y+a_9}$$
for some constants $a_i\in\mathbb{R}$. From this, we find the infinitesimal generators
$$\begin{array}{lll}
X_1=x\partial_x,& X_2=y\partial_x,& X_3=\partial_x, \\
X_4=x\partial_y,& X_5=y\partial_y,& X_6=\partial_y, \\
X_7=-x^2\partial_x-xy\partial_y,& X_8=-xy\partial_x-y^2\partial_y,& X_9=-x\partial_x-y\partial_y
\end{array}
$$
corresponding respectively to each parameter $a_i$. With the commutator
$$[X_i,X_j]=X_iX_j-X_jX_i,$$
the vector space generated by $\{X_1,\ldots,X_9\}$ forms a Lie algebra. Let us denote this vector space by $\mathfrak{g}$. A basis for that space could be $\{X_1,\ldots,X_8\}$ since $X_9=-X_1-X_5$.

I want to show that it is isomorphic to the Lie algebra
$$\mathfrak{sl}(3,\mathbb{R}):=\{A\in\mathbf{M}_{3\times 3}(\mathbb{R}):\text{trace}(A)=0\}$$
with commutator $[A,B]=AB-BA$, by explicitly giving an isomorphism
$$f:(\mathfrak{sl}(3,\mathbb{R}),[\cdot,\cdot])\longrightarrow(\mathfrak{g},[\cdot,\cdot])$$

I made a few attempts that all failed. Since $X_1+X_5+X_9=0$, we can see a sort of correspondence with traceless matrices, but I don't see how to use that.

EDIT: To get the generators $X_i$, we differentiate with respect to each $a_i$ both $\tilde{x}$ and $\tilde{y}$ and evaluate at the identity transformation, that is at $a_1=1,a_5=1,a_9=1$ and $a_2=a_3=a_4=a_6=a_7=a_8=0$. For example, to get $X_2$, we have
$$\frac{d\tilde{x}}{da_2}=\frac{y}{a_7x+a_8y+a_9},\quad\frac{d\tilde{y}}{da_2}=0$$
which evaluates to $y$ and $0$ respectively so that $X_2=y\partial_x+0\partial_y=y\partial_x$.

答:
Over any field $k$, the group $G=GL_{n+1}(k)$ acts on $Pk^n$ via linear maps on $k^{n+1}$, and the quotient map $q:k^{n+1}-\{0\} \rightarrow Pk^n$, whose fibers are lines (with $0$ removed). In matrices, on the affine subset identifiable with homogeneous coords $\{x_i\}$ with $x_{n+1}=1$, this is
$$
\{g_{ij}\}\cdot \pmatrix{x_1 \cr x_2 \cr \vdots \cr x_n \cr 1}
\;=\; \pmatrix{
g_{11}x_1+g_{12}x_2+\ldots+g_{1,n}x_n+g_{1,n+1}\cdot 1
\cr
g_{21}x_1+g_{22}x_2+\ldots+g_{2,n}x_n+g_{2,n+1}\cdot 1
\cr
\vdots
\cr
g_{n+1,1}x_1+g_{n+1,2}x_2+\ldots+g_{n+1,n}x_n+g_{n+1,n+1}\cdot 1
}
$$
Renormalizing to get the representative with last coordinate $1$, this is
$$
g\cdot x \;=\;
\pmatrix{
{g_{11}x_1+g_{12}x_2+\ldots+g_{1,n}x_n+g_{1,n+1}\cdot 1
\over
g_{n+1,1}x_1+g_{n+1,2}x_2+\ldots+g_{n+1,n}x_n+g_{n+1,n+1}\cdot 1
}
\cr
\vdots
\cr
{g_{n,1}x_1+g_{n,2}x_2+\ldots+g_{n,n}x_n+g_{n,n+1}\cdot 1
\over
g_{n+1,1}x_1+g_{n+1,2}x_2+\ldots+g_{n+1,n}x_n+g_{n+1,n+1}\cdot 1
}
\cr
1
}
$$
Visibly the center acts trivially here, and $SL_{n+1}(k)$ maps finite-to-one to $PGL_{n+1}(k)$, so the Lie algebras are the same. That is, the projective-affine transformation group $PGL_{n+1}(k)$ has the same Lie algebra as $SL_{n+1}(k)$, described however one likes in further detail.</p>


答:
We can be more general than that.  Let $V$ be a vector space and let $u_1,\dots, u_n$ be any basis on $V$.  Let $\langle\cdot, \cdot \rangle_0$ and $\langle\cdot, \cdot \rangle_1$ be two inner products on $V$ such that
$$
\langle u_i,u_j\rangle_0 = \langle u_i,u_j\rangle_1
$$
for all $i,j$.  We want to show that
$$
\langle \cdot, \cdot\rangle_0=\langle \cdot, \cdot\rangle_1
$$
Let $v,w\in V$.  Then we can express it as a linear combination of $u_1,\dots, u_n$:
$$
v=\sum_i v^i u_i,\hspace{0.2in} w=\sum_i w^iu_i
$$
Then
$$
\begin{align}
\langle v,w\rangle_0&=\sum_{i,j}v^iw^j\langle u_i,u_j\rangle_0\\
&=\sum_{i,j}v^iw^j\langle u_i,u_j\rangle_1\\
&=\sum_{i,j}\langle v^iu_i,w^ju_j\rangle_1\\
&=\langle v,w\rangle_1,
\end{align}
$$
which proves the claim.

Body:
Let  $ u = (1, −1, 0), v = (0, 1, 1), w = (1, 1, 0) $ be vectors of $\mathbb{R}^
3$.

Prove that exists only one scalar product $\langle \ , \ \rangle$ in  $\mathbb{R}^3$ that has $u, v,w$ as
orthonormal basis.

Any idea to begin with?

I've think about the idea that the matrix $M(\langle \ , \ \rangle, \mathscr{S} ) = I_3 $, with $\mathscr{S}$ a Sylvester basis. Because of the fact that an euclidian geometry is an orthogonal geometry with Sylvester invvariant $r_0=0, r_+=n, r_-=0$. But I'm not sure at all...

Any help? Thanks!!


问:
Show that every ring with three elements is commutative

Show that every ring with three elements is commutative

I'm unsure how can I proceed with the problem? I found some answers online on $x^3=x$ being commutative, but not sure if this is equivalent to the property I want to show?

答:
There's already an answer, but it assumes you are talking about rings with identity.  In fact the statement holds even for rings without identity.

The additive group is cyclic of order $3$.  Let it be generated by $a$, so that the elements of the ring are $0$, $a$, and $a+a$.  Since $0$ commutes with everything, the only possibility for non-commutativity is for $a$ and $a+a$ not to commute.  But, using the distributive laws, $a(a+a)=a\cdot a + a\cdot a=(a+a)a$, so $a$ and $a+a$ commute, so the ring must be commutative.

In fact, it turns out there are only two distinct rings with three elements: The field with $3$ elements, and the ring with three elements all of whose products are $0$.


问:
$5 \mid n^2 - m^2$ is an equivalence relation
How can I show this is an equivalence relation:

$$
n \operatorname{R} m \Longleftrightarrow n^2 - m^2 \textrm{ is divisible by } 5
$$

I know equivalence relations are symmetric, reflexive and transitive.  I'm just not sure how to use this knowledge to prove it.

答:
I assume your relation is over the set of integers.

Consider arbitrary elements $x, y, z$ in $\mathbb{Z}$. Show that all three equivalence relation properties hold for any such elements.

You could also argue that R is equivalent to (no pun) say $n^2 \equiv m^2 \mod 5$ and since modular congruence is known to be an equivalence relation, R is an equivalence relation.


问:
Finding an equivalent norm on $\ell^1$
Consider the space $\ell^1(\mathbb{N})$ of complex, summable sequences. I'm searching for a norm on $\ell^1(\mathbb{N})$ which is equivalent to the standard $\ell^1$-norm. Is it possible to write down such a norm such that the unit ball in $\ell^1(\mathbb{N})$ w.r.t. the new norm becomes strictly convex?

答:
$\newcommand{\tn}[1]{|\!|\!| #1|\!|\!|}$Recall that $\ell ^1\subseteq \ell ^2$, and that $\Vert x\Vert _2\leq \Vert x\Vert _1$,  for every $x$ in $\ell ^1$.  We may thus define a new  norm  on
$\ell ^1$ by
$$
  \tn x =   \Vert x\Vert _1 +  \Vert x\Vert _2,
  $$
and since
$$
  \Vert x\Vert _1 \leq  \tn x \leq  2\Vert x\Vert _1,
  $$
we see that $\tn \cdot $ is equivalent to $\Vert \cdot \Vert _1$.

Finally, we claim that   $\tn \cdot $ is strictly convex.  To see this, suppose that $x$ and $y$ are linearly independent
vectors in $\ell ^1$ such that $\tn x = \tn y = 1$.  Then
$$
  \tn{x+y} = \Vert x+y\Vert _1 +  \Vert x+y\Vert _2 \leq  $$$$ \leq
  \Vert x\Vert _1 +  \Vert y\Vert _1 +  \Vert x+y\Vert _2 < $$$$ <
  \Vert x\Vert _1 +  \Vert y\Vert _1 +  \Vert x\Vert _2 +  \Vert y\Vert _2 \leq \tn x + \tn y,
  $$
where the strict inequality is due to the strict convexity of $\Vert \cdot \Vert _2$.  This proves that $\tn \cdot $ is strictly convex.

Title: Finding an equivalent norm on $\ell^1$


问:
A Multivariable Distribution ($k$-dimension) with a given sum $m$
I am wondering if there exists a distribution of $k$ positive random variables and their sum is a given deterministic scalar, say $m$. Mathematically, I am finding the following distribution:
$$
\mathbb{P}(X_1,\cdots,X_k),\quad s.t. \quad X_1+\cdots+X_k=m.
$$
I am hoping that this distribution has a parameter which is like the variation parameter $\sigma^2$ in Gaussian distribution, that can characterize the variation of the $k$ random variables.

My thoughts

Entropy might be a good metric to describe the variation of such $k$ random variables.
However, I do not know how to generate those $k$ random variables that satisfy their summation being $m$, and their entropy being, say $h$.

This is just my thoughts. If there are any similar distributions and have the desired parameter, please let me know. Thanks ahead!

答:
Something like a symmetric Dirichlet distribution, i.e. with equal concentration parameter, may be the kind of thing you are looking for.

You can generate them as $k$ Gamma random variables and then divide by their sum.  That would then add up to $1$, so you may want to multiply by $m$.


问:
How many "distinct" polynomials are there mod $n$?
(This sounds like a fairly natural question, but I couldn't find it asked on this site before…)

Fix an integer $n$ as the modulus, and consider polynomials (with integer coefficients) modulo $n$.

For any polynomial $f$ and any fixed $k$, we have $f(x) + nx^k \equiv f(x) \pmod n$ for all integers $x$; thus any two polynomials with some coefficient differing by a multiple of $n$ are equivalent in the sense that they give the same value mod $n$ for any integer $x$ as input. So we can assume that each coefficient of $f$ is one of $\{0, 1, 2, \dots, n-1\}$.

When $n$ is prime, we know (from Fermat's little theorem) that $x^n \equiv x \pmod n$ for all integers $x$, which means any polynomial of degree $n$ or greater is equivalent (in the same sense as above) to a polynomial of lower degree. Even when $n$ is not prime, some such relationship probably exists; for example when $n=4$, we can see that $x^4 \equiv x^2 \pmod 4$. (I guess, though this may be wrong and I haven't checked it, that if the prime factorization of $n$ is $n = p_1^{e_1}p_2^{e_2}\cdots p_r^{e_r}$, then $x^n \equiv x^{n/(p_1p_2\cdots p_r)} \pmod n$.)

In any case, whenever $x \equiv y \pmod n$, we have $f(x) \equiv f(y) \pmod n$, so it is enough to consider the polynomial's values at $n$ consecutive integers say $\{0, 1, 2, \dots, n-1\}$. As there are at most $n$ distinct values it can have mod $n$ for each input $n$, there are at most $n^n$ distinct polynomials in this sense.

The question: How many distinct polynomials with integer coefficients are there mod $n$? Here, the meaning of "distinct" is that polynomials $f$ and $g$ are "distinct" if there exists at least one integer $x$ at which $f(x) \not\equiv g(x) \pmod n$. Or, using the terminology from what I found later (see below), we want the number of polynomial functions from $\mathbb{Z}$ to $\mathbb{Z}/n\mathbb{Z}$.

That is the question, but of course before asking it here I tried a few things to find the answer myself, including writing the following program.

Define the "signature" (just a name I made up for writing this program: is there already a standard term for this?) of a polynomial $f$ as the tuple $(f(0), f(1), f(2), …, f(n-1))$ of its values at all $n$ points. We can count the number of distinct polynomials mod $n$ by keeping track of all the encountered signatures, as we start with the zero polynomial and successively add $ax^k$ for each $a$ from $0$ to $n-1$, for increasing $k$ until $x^k$ has the same signature as some already seen polynomial.

```<code>#!/usr/bin/env python3

# Adding the signatures of two polynomials mod n
def add(t1, t2):
    n = len(t1)
    assert n == len(t2)
    return tuple((t1[i] + t2[i]) % n for i in range(n))

def count(n):
    szero = tuple([0]*n)  # The signature of the zero polynomial
    seen = {szero}  # Only the zero polynomial has been seen
    k = 0
    sxk = tuple([1]*n)  # The signature of x^k
    while True:
        news = set()  # New signatures, by adding ax^k for some a
        saxk = szero
        for a in range(1, n):
            saxk = add(saxk, sxk)  # The signature of a*(x^k)
            for s in seen:
                newp = add(saxk, s)  # a*(x^k) + (smaller polynomial)
                if newp not in seen: news.add(newp)
        seen.update(news)
        # Now done with all polynomials of degree k
        k += 1
        sxk = tuple((sxk[i]*i) % n for i in range(n))
        if sxk in seen: break
    return len(seen)

print('n', '\t', 'count(n)')
for n in range(1, 11):  # Gets very slow for n=11
    print(n, '\t', count(n))
</code>```

This prints:

```<code>n    count(n)
1    1
2    4
3    27
4    64
5    3125
6    108
7    823543
8    1024
9    19683
10   12500
</code>```

The good thing about computing these small values is that there is the OEIS, using which you can, to quote Donald Knuth, "compute your way into the literature". And accordingly, searching the OEIS for this sequence $1$, $4$, $27$, $64$, $3125$, $108$, $823543$, $1024$, $19683$ gives A058067 described as "Number of polynomial functions from Z to Z/nZ" which is exactly our question. We also learn from that page that there's a formula

$$
\prod_{k=0}^{n-1} \frac{n}{\gcd(n, k!)}
$$

that fits the numbers we computed and (per that page) was first proved by Aubrey J. Kempner in 1921 in a two-part paper called "Polynomials and their residue systems" (1, 2), and there's what appears to be a nice paper in 2000 by Manjul Bhargava called "The Factorial Function and Generalizations" PDF that generalizes it (and maybe gives a shorter proof).

But I haven't read either of these papers as I ran out of my time budget for this (well, I exceeded it by writing up this question), and I hope someone can give a clear proof/derivation of the formula either using those papers or independently.

Edit: I read the latter paper and understood it enough to answer this question (left an answer below), but I have a follow-up question: Smallest number of residue classes for a polynomial function mod $n$ that separates $0$.

答:
Well bearing in must $m^k$ will eventually repeat.  Perhaps not with $m^{\phi(n)}\equiv m^0\equiv 1\pmod n$ but with some priliminary $k$ so that $m^{\phi(n) + k}\equiv m^k$.

So that gives us an upper bound $ax^k \equiv (a-n)x^{k-\phi(n)}$ so $(\phi(n) + 1)(n-1)$ is an upper bound.


问:
Coin Distribution Problem- Combinations
In how many ways can we distribute a single quarter, a single dime, a single nickel and 25 separate cents between $5$ children if

a) we have no restrictions?
b) such that the oldest kid gets either $20$ cents or $25$ cents.

a) $$\;\binom{1+5-1}{1}\times \binom{1+5-1}{1}\times \binom{1+5-1}{1}\times  \binom{25+5-1}{5} = 14,844,375$$

b) My textbook is a spanish translation of the original and the exercise is badly worded and I cannot find it in the english version, so I am confused.

Please help me out.

答:
In how many ways can we distribute a single quarter, a single dime, a single nickel, and $25$ pennies between five children if we have no restrictions?

There are five ways to distribute the quarter, five ways to distribute the dime, and five ways to distribute the nickel.  Since there are no restrictions on the distribution of the pennies, the number of ways they can be distributed is the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 25 \tag{1}$$
in the nonnegative integers, where $x_i$ represents the number of pennies received by the $i$th oldest child.

A particular solution of equation 1 corresponds to the placement of $5 - 1 = 4$ addition signs in a row of $25$ ones.  For instance,
$$1 1 1 1 1 1 1 + 1 1 1 1 1 1 + 1 1 1 1 1 + 1 1 1 1 + 1 1 1$$
corresponds to the solution $x_1 = 7$, $x_2 = 6$, $x_3 = 5$, $x_4 = 4$, $x_5 = 3$.  The number of such solution corresponds the number of ways we can select which $4$ of the $29$ positions required for $25$ ones and four addition signs will be filled with addition signs, which is
$$\binom{25 + 5 - 1}{5 - 1} = \binom{29}{4}$$
Hence, you should have obtained the answer
$$5^3\binom{29}{4}$$

When choices are made independently, you should be multiplying rather than adding.  Addition is used when choices are mutually exclusive.

In how many ways can we distribute a single quarter, a single dime, a single nickel, and $25$ pennies between five children if the oldest kids gets either $20$ cents or $25$ cents?

Consider cases, depending on how many cents the oldest child receives and how that amount is distributed.  Once you have chosen which coins the oldest child has received, distribute the remaining coins to the remaining children.

For instance, if the oldest child receives $25$ cents as by receiving the dime, the nickel, and ten pennies, distribute the quarter and the remaining $15$ pennies to the other four children.

Title: Coin Distribution Problem- Combinations


问:
The "set" of equivalence classes of things.
This question might be silly.

I was wondering wether some form of choice is needed to be on firm footing when talking about the set of equivalence classes of certain things. For instance, when one talks of the set of equivalence classes of extensions of one $R$-module $N$ by another, $M$,
$$\lbrace E\text{ such that there exists a short exact sequence }0\to N\to E\to M\to 0\rbrace/\sim$$
where two short exact sequences involving $E$ and $E'$ are equivalent if there is a commutative diagram
$$\begin{array}{c}
0\to &N&\to E\to &M&\to 0\\
&\Vert&~~~\downarrow\phi&\Vert&\\
0\to &N&\to E'\to &M&\to 0.
\end{array}$$
In order to justify that this is indeed a set, I can only think of the following approach: take a section (a map of sets) $\sigma:M\to E$ of $p:E\to M$ and use this to obtain a bijection of sets $$E\simeq N\times M,~e\mapsto(n(e),p(e))$$
where $n(e)\in N$ is defined by $n(e)+\sigma(p(e))=e$, and argue that there is only a set of $R$-module structures on the set $N\times M$, so that in the above we may choose $E$ to always be $N\times M$ as a set, endowed with different $R$-module structures. If I'm not mistaken, the existence of sections to surjective maps depends on the axiom of choice.

A similar problem arises when one talks about the set of isomorphism classes of all $n$-dimensional vector bundles over a given space $X$, or that of principal $G$-bundles, although in this particular case I think one can directly argue over the set of all open covers of $X$.

I guess the question boils down to the following: is choice needed to show that given a map $X\to Y$ with fibers of constant cardinality, the cardinality of $X$ depends only on that of $Y$ and of the fiber?

答:
First of all, note that since the equivalence classes themselves are not sets, the collection of all equivalence classes is not a set either, it's not even a class. The correct phrasing is to say that there exists a set of representatives, rather than a class of representatives (which might not exist either, by the way). Of course if we want to use Scott's trick then the equivalence classes are cut down to be sets, and then it's a legit thing to say that there is only a set of equivalence classes.

Secondly, if we require that the exact sequences exist, and that the equivalence relation is defined by the existence of isomorphism between $E$ and $E'$. The equivalence classes themselves are well-defined without any appeal to the axiom of choice.

I believe, however, without sitting to verify the details I think it should be consistent to have a proper class of sets (of distinct cardinality) which are the countable union of pairs. The existence of an exact short sequence means that a particular pair is mapped onto $0$. So a proper class of sets can exist which have this sort of sequence, without the existence of an isomorphism, and therefore even using the set theoretical tricks to turns equivalence classes into sets, this is still going to be a proper class.

But that been said, if we want concrete sets $M$ and $N$, then in order to assure that the sections exist we need only the axiom of choice for families of size $|Y|$ whose members are the "constant" cardinality (we can just as well take $|X|$ to be sure, if we want to be sure of this).

If we assume such choice, then all these $E$ (as in the sequences) are isomorphic to subsets of $\mathcal P(N\times M)$, so there can only be a set of those.

For example if $Y$ is countable, then the axiom of countable choice is sufficient, if we also have that the fibers are all finite - or even pairs - then the axiom of choice for countable families of pairs should be sufficient. Usually saying that there is a choice function for finite sets, or for families of pairs, would be a nicer way of saying that (although stronger than required).


答:
Assuming the children are not identical, the distribution of the 3 coins worth more than a penny can happen $5^3$ ways.  Then 25 pennies distributed among 5 children, by the "stars and bars" argument, would be ${25+5-1 \choose 4}$ so for the answer to (a) I get $125 \times \frac{29\cdot 28\cdot 27 \cdot 26}{24} = 2968875$

For part (b), first consider the case where the oldest kid gets $20$ cents.  That means he got a dime and $10$ pennies, or a dime and a nickel and $5$ pennies, or a nickel and $15$ pennies, or $20$ pennies.  I don't see any more clever way of doing this than by calculating the ways for each branching possibility.

If the oldest got a dime and ten pennies:  There are four children left. So the quarter and the nickel have $4^2$ ways to go.  Then there are 15 pennies and 4 children, making ${18 \choose 3}$ ways.  So that gives $\frac{16 \cdot 18 \cdot 17 \cdot 16}{3!} = 13056$ possibilities.

The rest are calculated similarly.

Title: Coin Distribution Problem- Combinations

Body:
In how many ways can we distribute a single quarter, a single dime, a single nickel and 25 separate cents between $5$ children if

a) we have no restrictions?
b) such that the oldest kid gets either $20$ cents or $25$ cents.

a) $$\;\binom{1+5-1}{1}\times \binom{1+5-1}{1}\times \binom{1+5-1}{1}\times  \binom{25+5-1}{5} = 14,844,375$$

b) My textbook is a spanish translation of the original and the exercise is badly worded and I cannot find it in the english version, so I am confused.

Please help me out.


问:
Proving two equivalent propositions
Prove that the propositional formulas:

```<code>    P OR Q OR R
</code>```

and

```<code>(P AND NOT(Q)) OR (Q AND NOT(R)) OR (R AND NOT(P)) OR (P AND Q AND R)
</code>```

are equivalent.

I have tried De Morgan's Laws and Distributive Laws on the right hand side, but I got some formulas that are even more complicated.

答:
Because I was taught this as someone who did electronics, I have a differing notation that has stuck with me. Here $+$ denotes OR, $\cdot$ denotes AND and $\overline{P}$ represents not $P$ etc. Apologies for the inconsistency it just works better in my head. One can show this using distributivity laws only.

Note that

$$ P\cdot \overline{Q} + P\cdot Q \cdot R = P\cdot(\overline{Q}+Q\cdot R)=P\cdot (\overline{Q} + R) = P\cdot \overline{Q}+ P\cdot R$$

Similarly,

$$ Q\cdot \overline{R} + P\cdot Q \cdot R =Q\cdot \overline{R}+ Q\cdot P$$

and

$$ R\cdot \overline{P} + P\cdot Q \cdot R =R\cdot \overline{P}+ R\cdot Q.$$

Thus,

$$\begin{split}P\cdot \overline{Q}+ Q\cdot \overline{R}+ R\cdot \overline{P} + P\cdot Q \cdot R & = P\cdot \overline{Q}+ P\cdot Q \cdot R+ Q\cdot \overline{R}+ P\cdot Q \cdot R+ R\cdot \overline{P} + P\cdot Q \cdot R \\ &= P\cdot \overline{Q}+ P\cdot R + Q\cdot \overline{R}+ Q\cdot P+R\cdot \overline{P}+ R\cdot Q \\ & = P+Q+R\end{split}$$

as desired.

You could also just use a truth table but that is not so elegant.

Title: Proving two equivalent propositions


问:
Drunk squirrel probability riddle
I stumbled upon this question from an old exam, and I was quite curious how to approach it since there are dependent variables at play. It goes like this:

assume that a drunk squirrel is dropped at location i of an 1-dimensional branch of an oak tree,
where i is drawn from a Gaussian $i \sim \mathcal{N}(\mu_i = 0; \sigma_s² = 4)$. After the drop, the squirrel makes a step to the right
by distance d, where $ d \sim \mathcal{N}(0; 1)$ (If d is negative the squirrel moves to the left).

If we write f as the final location of the squirrel, we conclude that$ f \sim \mathcal{N}(i; 1)$. d is also independent of i.

The squirrel ends up at location $f=2$, what is the most likely location $i$ that the squirrel landed on.

My approach was to define the final location as $z = i+d$ which should be normally distributed with the sum of the mean and variances, but then $\mu_z \neq i$ (should be $0$ according to my logic).

Any ideas on how to approach this?

答:
https://stats.stackexchange.com/questions/9071/intuitive-explanation-of-contribution-to-sum-of-two-normally-distributed-random is related

Hint:

Write down the likelihood, proportional to the product of the densities of starting at $i$ and then moving by $2-i$

Find the value of $i$ which maximises this, possibly by taking the derivative of the logarithm of the likelihood, or perhaps by completing the square

This is going to be the same $i$ which minimises $\dfrac{i^2}{4}+\dfrac{(2-i)^2}{1}$

Title: Drunk squirrel probability riddle


问:
sample variance divided by variance is Chi Square
To estimate the sample variance, the following relation is often used:

$$\frac{(n-1)s^2}{\sigma^2} \sim \chi^2(n-1) $$

With $(n-1)$ being the degrees of freedom.

Could someone provide me a formal proof and some intuition for this relation?

答:
Simulation in R: Population mean estimated by sample mean.

```<code>set.seed(2021)
n = 5;  mu = 100;  sg = 15
q4 = replicate(n^5, 4*var(rnorm(n,mu,sg))/sg^2)
mean(q4)
[1] 4.000032  # aprx mean of CHISQ(4) = 4.
</code>```

In the figure below the density function of $\mathsf{Chisq}(\nu=4)$ fits the simulated values, while the
density function of $\mathsf{Chisq}(\nu=5)$ does not.

```<code>hdr = "Mean estimated: CHISQ(4)"
hist(q4, prob=T, br=20, ylim=c(0,.25), col="skyblue2", main=hdr)
 curve(dchisq(x,4), add=T, lwd=2)
 curve(dchisq(x,5), add=T, lwd=2, col="red", lty="dotted")
</code>```

Population mean known:

```<code>set.seed(320)
n = 5;  mu = 100;  sg = 15
q5 = replicate(n^5, sum((rnorm(n,mu,sg)-mu)^2)/sg^2)
mean(q5)
[1] 5.109602  # aprx mean of CHISQ(5) = 5
</code>```

```<code>hdr = "Mean known: CHISQ(5)"
hist(q5, prob=T, br=20, ylim=c(0,.20), col="skyblue2", main=hdr)
 curve(dchisq(x,5), add=T, lwd=2, col="red", lty="dotted")
</code>```

Body:
Simulation in R: Population mean estimated by sample mean.

set.seed(2021)
n = 5;  mu = 100;  sg = 15
q4 = replicate(n^5, 4*var(rnorm(n,mu,sg))/sg^2)
mean(q4)
[1] 4.000032  # aprx mean of CHISQ(4) = 4.

In the figure below the density function of $\mathsf{Chisq}(\nu=4)$ fits the simulated values, while the
density function of $\mathsf{Chisq}(\nu=5)$ does not.

hdr = "Mean estimated: CHISQ(4)"
hist(q4, prob=T, br=20, ylim=c(0,.25), col="skyblue2", main=hdr)
 curve(dchisq(x,4), add=T, lwd=2)
 curve(dchisq(x,5), add=T, lwd=2, col="red", lty="dotted")

Population mean known:

set.seed(320)
n = 5;  mu = 100;  sg = 15
q5 = replicate(n^5, sum((rnorm(n,mu,sg)-mu)^2)/sg^2)
mean(q5)
[1] 5.109602  # aprx mean of CHISQ(5) = 5

hdr = "Mean known: CHISQ(5)"
hist(q5, prob=T, br=20, ylim=c(0,.20), col="skyblue2", main=hdr)
 curve(dchisq(x,5), add=T, lwd=2, col="red", lty="dotted")


问:
I cannot understand how this matrix works or how it is defined
I'm currently reading Ranking a Stream of News and have trouble on page 100 (don't be afraid, the math starts at 99). I cannot understand a matrix they define.

In this article, the authors are providing a mathematical representation -as a graph- of a stream of news consisting of articles and sources. This paragraph should, if I'm understanding correctly, be about how to determine how important edges are. An edge should be more important, for example, when it links an article to a authoritative source or when it links two articles with high similarity to each other.

In the paragraph before, they introduced $G_\omega$ as a stream of news, generally a graph consisting of a set $N$ with articles $n_1, ...$ and a set $S$ with sources $s_1, ...$ (both nodes) and edges sets $E_1$ (source - article) and $E_2$ (article - article).

Let A be the (weighted) adjacency matrix associated with $G_\omega$. We can attribute an identifier to the nodes in $G_\omega$ so that any source precedes the pieces of news. We define the matrix $$A = \begin{bmatrix}O & B\\B^T & \Sigma\end{bmatrix},$$ where $B$ refers to edges from sources to news articles, and $b_{ij} = 1$ if the source $s_i$ emitted article $n_j$ and  is the similarity matrix. Assuming one can learn similarity of sources, the matrix $A$ can be modified in the upper-left corner incorporating
  a submatrix taking into account a source-source information.

I'm not able to understand this paragraph. This may partially be a lack of English skills which makes it hard to read for me, but I got problems with the math as well. Here are my questions:

What is $A$? What is its concrete meaning?

What is $O$? With "$A$ can be modified in the upper-left corner", do they mean they will change $O$?

What "data type" is $B$? I'd think this is a set, "edges from sources to news articles", thus $E_1$. However, they're calculating $B^T$ - what does that mean, if $B$ is a set of edges?

What is $T$? I don't see it anywhere else. Is it some constant like $i$ that I don't know of?

What is $\Sigma$? I know the sum function, but don't see how it's relevant here. I only see that in the paragraph before this one they define $\sigma_{ij}$ as the continous similarity between article $n_i$ and article $n_j$. Would $\Sigma$ be the set of all $\sigma$, just like $N$ is the set of all $n$ (in this article)?

Nota Bene: I am not very familiar with matrices or set theory. Actually, I'm on the level of a just-finished high school student. I'd like some layman explanation, please.

答:
I discussed this on the Electrical Engineering chat (permalink) and ThePhoton explained it to me.

$O$ is the zero matrix. This matrix contains only zeros.

$B$ is the matrix where in one direction (horizontal) the sources $s \in S$ are listed and in the other direction (vertical) the articles $n \in N$. This matrix shows which source produced an article.

$B^T$ is the transpose of $B$ (wrapped over the diagonal).

$\Sigma$ is the similarity matrix. It tells you how much the articles are like each other.

Now the meaning of

Assuming one can learn similarity of sources, the matrix $A$ can be modified in the upper-left corner incorporating a submatrix taking into account a source-source information.

We saw that $\Sigma$ was the similarity matrix. It shows similarity between articles. We could also calculate similarity of sources. 'By default', this is disabled: in the upper left corner of $A$, a zero matrix is put. Now assuming you can calculate similarity of sources, you could replace $O$ with $\Sigma_s$, which would be a matrix determining similarity of sources.

Now we can see that $A$ is a matrix containing both the relations between sources and articles (which articles were produced by which sources), and the relations between different articles (how similar they are), and, if $O$ is replaced, the relations between different sources (how similar they are).


问:
The sums of $3$ and $8$ must give a number greater than $13$
Prove every natural number greater than $13$ can be express as sum of $3$ and $8$ (regardless of order)

答:
If you apply a well-known Euclidean algorithm,  you will get

$$3m+8n=k$$

$$m = 8 z + 3 k, ~n = -3 z - k, ~  z\in\mathbb Z$$

Then, we need

$$\begin{cases} 8z+3k ≥0 \\-3z-k≥0\\ k>13\end{cases}$$

Finally, we have

$$k>13,  ~ \frac{3k}{8}≥z≥\frac k3$$

For $k>13$, there always exist an integer $z$, such that  $\frac{3k}{8}≥z≥\frac k3$ holds.


问:
Let A and B be a Banach algebra with $1\in A\subset B$ (1 is the identity). Let $a\in A $ then show that $Spectrum B\subset SpectrumA$
Let A and B be a Banach algebra with $1\in A\subset B$ (1 is the identity). Let $a\in A $ then show that
$$1).Spectrum_a\ B\subset Spectrum_aA$$
$$2). Boundary(Spectrum_aA)\subset Boundary(Spectrum_aB)$$

First, let's take $x\in Spectrum_a\ B$, I'm going to show that is in $Spectrum_aA$.
since $x\in Spectrum_a\ B$

$a-x*1$ is not invertible. Since $a\in A$ , $_\ $ contains x.
This implies $Spectrum_a\ B\subset Spectrum_aA$.

Is this correct?
And how we show the second part?

答:
Lemma.  Assume that $A$ is a unital Banach algebra and that $\{a_n\}_n\subseteq A$ is a sequence of invertible elements,  converging to
a non-invertible element $a$.  Then $a$ is a topological zero-divisor, meaning that there exists a sequence $\{x_n\}_n$ in
$A$, with $\|x_n\|=1$, and $\displaystyle\lim_{n\to \infty }ax_n = 0$.

Proof.  We begin by claiming that there is no positive constant $c$, such that
$$
  \|ax\|\geq  c\|x\|,\quad\forall x\in  A.
  $$
In order to prove the claim we assume otherwise and hence
$$
  c\|a_n^{-1}\|\leq
  \|aa_n^{-1}\| =
  \|(a-a_n+a_n)a_n^{-1}\| \leq  $$$$ \leq
  \|a-a_n\|\|a_n^{-1}\| +  1,
  $$
from where it follows that
$$
  \|a_n^{-1}\| \leq  \frac 1{c-\|a-a_n\|},
  $$
as long as the denominator is positive, which it sure is for large enough $n$.   This  also shows that   $\|a_n^{-1}\|$ is bounded.  Given any $n$
and $m$ we then have that
$$
  \Vert a_n^{-1}-a_m^{-1}\Vert  =
  \Vert a_n^{-1}(a_m-a_n)a_m^{-1}\Vert  \leq  $$$$ \leq
  \Vert a_n^{-1}\|\|a_m-a_n\|\|a_m^{-1}\Vert  \leq
  \|a_m-a_n\|\big(\sup_k\|a_k^{-1}\Vert \big)^2.
  $$
This says that $\{a_n^{-1}\}_n$ is a Cauchy sequence.   Setting $\displaystyle b= \lim_{n\to \infty }a_n^{-1}$, we then have that
$$
  ab = \big (\lim_{n\to \infty }a_n\big )\big (\lim_{n\to \infty }a_n^{-1}\big ) = \lim_{n\to \infty }a_na_n^{-1} = 1,
  $$
and likewise $ba=1$, so $a$ is invertible, a contradiction,  hence proving the claim.

Consequently, for every $n>0$, the inequality  "$\|ax\|\geq  (1/n)\|x\|$" does not always hold, so there exists some $x_n$
such that
$$
  \|ax_n\|< (1/n)\|x_n\|.
  $$
The conclusion then  follows by replacing the $x_n$ with  $x_n/\|x_n\|$. QED

Back to the question,   suppose that $\lambda $ is in the boundary of $\sigma _A(a)$.  Then $a-\lambda $ is not invertible, but it is
clearly a limit of invertible elements of the form $a-\rho _n$, where $\rho _n$ lies in complement of $\sigma (A)$.  By the Lemma,  we
deduce that $a-\lambda $ is a topological zero-divisor, and  this obviously prevents $a-\lambda $ from being invertible in $B$, so we see that
$\lambda \in \sigma _B(a)$.

This shows that
$$
  \partial \sigma _A(a)\subseteq \sigma _B(a),
  $$
and from this one easily shows that in fact
$$
  \partial \sigma _A(a)\subseteq \partial \sigma _B(a).
  $$


问:
Prove polynomial $n^6-n^5+n^4+n^2-n+1$ is positive for each natural number $n$
How can I prove that the polynomial
$$p(x)=n^6-n^5+n^4+n^2-n+1$$
is positive for every natural number $n$.
I have tried factoring it and got $$(n^4+1)(n^2-n)=(n^4+1)n(n-1),$$ but couldn't get anything else out of it.

Thank you in advance

Body:
\begin{align}
n^6-n^5+n^4+n^2-n+1 &= n^5(n-1) + n^4 + n(n-1) + 1\\
&= n(n^4+1)(n-1)+n^4+1\\
&\ge n^4+1\\&>0
\end{align}


问:
Finding the marginal distributions
I got these 2 models for hourly wage for 2 periods:

The hourly wage for period 1 is normally distributed with mean $µ$ and variance $σ^2$ so $Y_1 \sim N(\mu,\sigma^2)$.

And the hourly wage for period 2 is given by:
$$Y_2=\alpha+\beta Y_1+U$$ where $Y_1$ and $U$ are independent and $U \sim N(0,v^2)$.
Then we assume that $\beta \neq 0$ and let $\mu = 350$ and $\sigma^2=12365$ and $\alpha=350\cdot(1-\beta)$ and $v^2=12365 \cdot (1-\beta^2)$.

Now I have to find the marginal distributions of $Y_1$ and $Y_2$. I have found on Wikipedia that the marginal probability is  $_()=_[_{|}(|)]$ and I have in a previous task found that $(_2|_1)=+⋅_1$. Is that the same value (I'm not totally sure on the notation will be the same) and how can I use it to find the marginal distributions of $Y_1$ and $Y_2$?

答:
in your problem the issue is simplier. You just have a stated linear relation that links $Y_1$ and $Y_2$. On the internet you can also find that "linear combination of Gaussians are still Gaussian" and your $Y_2$ is a linear function of $Y_1$ and $U$ thus it is also gaussian. All you have to do is to substitute values and use the properties of Expectation and Variance

$\mathbb{E}[a+bX]=a+b\mathbb{E}[X]$ (Expectation is a linear operator)

$\mathbb{V}[a+bX+Y]=b^2\mathbb{V}[X]+\mathbb{V}[Y]$ (if X and Y are incorrelated)

Thus

$$\mathbb{E}[Y_2]=350-350\beta+350\beta+0=350$$

$$\mathbb{V}[Y_2]=\beta^2 12365+12365(1-\beta^2)=12365$$

concluding, marginal distribution are

$$Y_1\sim N(350;12365)$$

$$Y_2\sim N(350;12365)$$

...they are identically distributed

If you do a step forward you can see that they are identically distributed but correlated with correlation coefficient $ \rho=\beta$

Title: Finding the marginal distributions


Title: Prove polynomial $n^6-n^5+n^4+n^2-n+1$ is positive for each natural number $n$

Body:
How can I prove that the polynomial
$$p(x)=n^6-n^5+n^4+n^2-n+1$$
is positive for every natural number $n$.
I have tried factoring it and got $$(n^4+1)(n^2-n)=(n^4+1)n(n-1),$$ but couldn't get anything else out of it.

Thank you in advance

Body: $n^6-n^5-n+1+n^4+n^2 = (n^5-1)(n-1)= (n-1)^2(n^4+n^3+n^2+n+1)+n^4+n^2 \ge n^4+n^2\ge 2$.


答:
correct is
$$  (n^4+1)(n^2 - n + 1)  = (n^2 - n \sqrt 2 + 1)(n^2 + n \sqrt 2 + 1)(n^2 - n + 1)$$

all three quadratic factors are positive, check discriminants

the factors of the quartic come from Sophie Germain's identity
$$ a^4 + 4 b^4 = (a^2 +2ab + 2 b^2)  (a^2 -2ab + 2 b^2) $$
with, I guess, $a=n, b = \frac{1}{\sqrt 2}$

Not sure why this is  associated with Germain


答: I guess you’d be solving for n in the inequality $n^6 + n^4 + n^2 + 1 > n^5 + n$.  For $n > 1, n^6 > n^5$ and for all $n, n^4 + n^2 + 1 > n$, so the inequality must be true for all $n$. I don’t think factorising will help you prove it.


问:
Putnam 1967 problem, integration
I have some problems understanding this solution of this problem:

Given $f,g : \mathbb{R} \rightarrow \mathbb{R}$ - continuous functions with period $1$, prove that

$$\lim _{n \rightarrow \infty} \int_0^1 f(x)g(nx) \, dx = \int_0^1 f(x) \, dx \cdot \int_0^1g(x) \, dx$$

It says there that if we split this integral into $n$ parts to get

$$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} \int_{\frac{r}{n}}^{\frac{r+1}{n}} f(x)g(nx) \, dx$$

we will have

$$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} f\left(\frac{r}{n}\right) \int_{\frac{r}{n}}^{\frac{r+1}{n}}g(nx) \, dx$$

because for large $n$, $f$ is roughly constant over the range.

Could you tell me why $f$ is roughly constant over the range for $n$ large enough?

I would really appreciate all your insight.

答:
We have
$$\int_0^1 f(x)g(nx)dx = \sum_{r=0}^{n-1} \int_{\frac{r}{n}}^{\frac{r+1}{n}}f(x)g(nx)dx$$
so we change the variable: $t=nx$ we find
$$\int_0^1 f(x)g(nx)dx = \frac{1}{n}\sum_{r=0}^{n-1} \int_{{r}}^{{r+1}}f(\frac{t}{n})g(t)dt$$

Now for large $n$ and by continuity of $f$ we can approximate $f(\frac{t}{n})$ in the interval $[r,r+1]$ by $f(\frac{r}{n})$ since
$$\lim_{n\to\infty} f(\frac{t}{n})-f(\frac{r}{n})=0\quad \forall t\in[r,r+1]$$and since $g$ is with period $1$ we have
$$\int_0^1 f(x)g(nx)dx \approx \int_{{0}}^{{1}}g(t)dt \frac{1}{n}\sum_{r=0}^{n-1}f(\frac{r}{n}) $$
finally we conclude by the Riemann sum.

Title: Putnam 1967 problem, integration


答:
First prove this for $14,15,16$ because they are the first numbers greater than $13$.

$14=3+3+8$, $15=3+3+3+3+3$, $16 = 8+8$.

Now, if you add $3$ to each of these, you get

$17 = 3+3+3+8$, $18=3+3+3+3+3+3$, $19=8+8+3$

keep going, adding $3$, and you get

$20 = 3+3+3+3+8$, $21=3+3+3+3+3+3+3$, $22=8+8+3+3$

You can clearly see from these examples that all numbers greater than $13$ can be expressed as the sum of $3$s and $8$ because we can just keep adding $3$ to our original trio to get all the numbers.

Title: The sums of $3$ and $8$ must give a number greater than $13$

Body: Prove every natural number greater than $13$ can be express as sum of $3$ and $8$ (regardless of order)