问:
Profinite completion of locally profinite topological group
I was trying to prove the following lemma with two parts. Let $G$ be a locally profinite topological group. Denote the profinite completion of $G$ by $\hat{G} = \underset{N}{\varprojlim} G/N$, where $N$ runs over the normal open subgroups with finite index. Then,

(i) The canonical map $G \to \hat{G}$ is injective.

(ii) Let $H$ be a profinite group, and $f: G \to H$ injective continuous homomorphism with dense image. Then $f$ induces a topological isomorphism $\hat{G} \xrightarrow{\sim} H$.

Following is what I thought about (i) and my rough idea about (ii).

(i) $G \to \hat{G}$ is induced by the maps $G \to G/N$ where $N$ is normal, open, and of finite index. Hence, proving injectivity is equivalent to proving that the intersection of all normal, open subgroups of finite index is trivial. If $G$ is compact it is easy since being locally profinite, there is an open compact neighborhood basis of $0$, and by compactness each such basis element is of finite index and can be refined to be normal, which implies the result. But $G$ is only known to be locally compact. I am not sure how to proceed.

(ii) By the universal property of profinite completion we have the existence of such a map $\hat{G} \to H$. To prove surjectivity, let $h \in H$. Since $f$ is dense there exist a sequence $g_n \in G$ such that $g_n$ "converges to $h$" meaning $g_n$ is contained in a "decreasing" sequence of neighborhoods of $h$. Then since $\hat{G}$ is complete and hausdorff, there is a unique limit $g$ of $g_n$ in $\hat{G}$. Then $g$ should map to $h$, hence surjectivity. For injectivity, suppose there is a non-zero element $a$ in the kernel. $G \to \hat{G}$ is dense, so there is a sequence $a_n \in G$ such that its image "converge to $a$". Since $a \neq 0$ and $G \to \hat{G}$ is injective by part (i) above, $a_n$ can be chosen "bounded away from $0 \in G$". Now consider $f(a_n)$, by injectivity they are "bounded away from $0 \in H$". Since $H$ is complete and hausdorff $f(a_n)$ should have a unique limit in $H$ not equal to zero, which should be the image of $a$ under $\hat{G} \to H$, but it contradicts the fact that $a$ is in the kernel. So $a = 0$ hence injectivity.

In part (ii) I use terms such as convergence, boundedness without appealing to any metric. But I am sure that these arguments could be better expressed in the realm of topological groups - something that I am new to. It would be very useful if someone could help me refine my ideas.

答: Both of these statements are false.  For (i), consider for instance $G=\mathbb{Q}$ with the discrete topology.  This has no proper subgroups of finite index, so $\hat{G}$ is trivial.  For (ii), consider $G=\mathbb{Z}$ (with the discrete topology).  For any prime $p$, the inclusion map $\mathbb{Z}\to\mathbb{Z}_p$ is an injective continuous homomorphism with dense image, but the induced map $\hat{\mathbb{Z}}\to\mathbb{Z}_p$ is not injective (in fact $\hat{\mathbb{Z}}$ is the product $\prod_{q\text{ prime}}\mathbb{Z}_q$ and the induced map is the projection).  Your proposed argument for injectivity is wrong because $(a_n)$ being bounded away from $0$ does not imply $(f(a_n))$ is bounded away from $0$.  Indeed, for the inclusion $f:\mathbb{Z}\to\mathbb{Z}_p$, the sequence $a_n=p^n$ stays away from $0$ in $\mathbb{Z}$ but $f(a_n)$ converges to $0$ in $\mathbb{Z}_p$.


问:
How many ways are there to color the walls of a pentagonal room using five different colors, so that no two non-adjacent walls have the same color?

How many ways are there to color the walls of a pentagonal room using five different colors, so that no two non-adjacent walls have the same color?

I tried casework, but it got very messy before long. I can't think of any other way to approach the problem. Can anyone help?

Thanks!

答:
Many of these answers are overcomplicated - this comes from a middle school math test (2014 MATHCOUNTS National Team #9).

Here is my solution:

Let the colors be purple, green, red, blue, and black.

Case 1: Upper left color is the same as the bottom color.
There are 5 ways to choose the lower-left color, and 4 ways to then choose the upper left and bottom color. Notice that for the other two sides, we can choose any two different colors that are not the color that we colored the bottom. There are $\binom{4}{2}=6$ ways to do this, but we have to multiply by $2$ to get $12$ because they can be switched. Therefore, the total number of ways for this case is $5*4*12=240$.

Case 2: Upper left color and bottom color are different. There are 5 ways to choose the lower left color, and again $\binom{4}{2}2=12$ ways to choose the upper left and bottom colors. For the other two colors, there are $4*4=16$ theoretical ways, but remember that they can't be the same color, so we have to subtract off $3$ ways for the three possible colors that they can be. The total number of ways for this case is $5*12*(16-3)=780$.

The answer is then $780+240=\boxed{1020}$.

No recursion or graph theory needed.

ContentLicense: CC BY-SA 4.0


问:
How to prove $\int_0^\infty J_\nu(x)^3dx\stackrel?=\frac{\Gamma(1/6)\ \Gamma(1/6+\nu/2)}{2^{5/3}\ 3^{1/2}\ \pi^{3/2}\ \Gamma(5/6+\nu/2)}$?
I am interested in finding a general formula for the following integral:
$$\int_0^\infty J_\nu(x)^3dx,\tag1$$
where $J_\nu(x)$ is the Bessel function of the first kind:
$$J_\nu(x)=\sum _{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+\nu+1)}\left(\frac x2\right)^{2n+\nu}.\tag2$$
Mathematica gives the following result:
$$\int_0^\infty J_\nu(x)^3dx=\frac1{\pi\,\nu}{_3F_2}\left(\frac12,\frac12-\nu,\frac12+\nu;1-\frac\nu2,1+\frac\nu2;\frac14\right)+\frac{\Gamma\left(-\frac\nu2\right)\Gamma\left(\frac{1+3\nu}2\right)\cos\frac{\pi\,\nu}2}{2^{\nu+1}\ \pi^{3/2}\ \Gamma(\nu+1)}{_3F_2}\left(\frac{1-\nu}2,\frac{1+\nu}2,\frac{1+3\nu}2;1+\frac\nu2,1+\nu;\frac14\right),\tag3$$
which can be significantly simplified for odd and half-integer values of $\nu$. The results at those points allow to conjecture another, simpler general formula:
$$\int_0^\infty J_\nu(x)^3dx\stackrel?=\frac{\Gamma\left(\frac16\right)\ \Gamma\left(\frac16+\frac\nu2\right)}{2^{5/3}\ 3^{1/2}\ \pi^{3/2}\ \Gamma\left(\frac56+\frac\nu2\right)},\tag4$$
which agrees with $(3)$ to a very high precision for many different values of $\nu$. It also has an advantage that it is defined for all $\nu>-\frac13$ whereas $(3)$ is undefined at every even $\nu$ and requires calculating a limit at those points.

Is it possible to prove the formula $(4)$?

Mathematica is able to evaluate $(1)$ for even values of $\nu$ in terms of the Meijer G-function. Plugging those expressions into $(4)$ we get another form of the conjecture:
$$G_{3,3}^{2,1}\left(\frac14\left|\begin{array}{c}2a,1,2-2a\\\frac12,1-a,a\\\end{array}
\right.\right)\stackrel?=\frac{\Gamma\left(\frac16\right)\ \Gamma\left(\frac23-a\right)}{2^{5/3}\ 3^{1/2}\ \pi\ \Gamma\left(\frac43-a\right)}.\tag5$$
Incidentally, the case $a=\frac12$ would positively resolve my another question.

答:
Thank you for posting this question, I enjoyed trying to answer it.

Start with the expression that Mathematica gave you and replace each
argument $\frac14$ of a hypergeometric function with $\frac z4$,
because we will be taking limits. I will call the two hypergeometric
functions $Q_1(z)$ and $Q_2(z)$. Each term can be brought to a closed
form by using identity 16.6.2 from the DLMF.

Setting $a=\frac12$, $b=1-\frac\nu2$, we get
$$ Q_1(z) = (1-z)^{-\frac12} F\left(\frac16,
\frac36,\frac 56; 1-\frac\nu2, 1+\frac\nu2; \frac{-27 z}{4(1-z)^3}
\right), $$
and setting $a=\frac{1+3\nu}{2}$, $b=1+\frac\nu2$, we get
$$ Q_2(z) = (1-z)^{-\frac{1+3\nu}{2}} F\left(\frac{1+3\nu}6,
\frac{3+3\nu}{6}, \frac{5+3\nu}{6}; 1+\frac\nu2, 1+\nu;
\frac{-27z}{4(1-z)^3} \right). $$
(Note that there are 6 possible identities to try per function, one for each possible choice of
$a$ and $b$ from the parameters, so it helps to do this on a
computer.)

The reason this works is that now the point $z=1$ is a singular point
of the hypergeometric functions on the right hand side, and Mathematica
will succeed in finding the limits as $z\to1$. The expression for the
whole integral that you have is
$$ Q =
\frac{2^{\frac43}\pi^{\frac12}}{\Gamma(-\frac16)\Gamma(\frac56+\frac\nu2)\Gamma(\frac56-\frac\nu2)\sin\frac{\nu\pi}2}
\left( -1 + 3^{-\frac{3\nu}2}\cos\left(\frac{\nu\pi}{2}\right)
\frac{\Gamma(\frac{1+3\nu}{2})
\Gamma(\frac56-\frac\nu2)}{\Gamma(\frac{1+\nu}2)\Gamma(\frac56+\frac\nu2)}
\right). $$
Call the large expression in brackets $A$, and then write
$$ A = -1 + B 3^{-\frac{3\nu}{2}}\cos\frac{\pi\nu}{2}, \qquad B = \frac{\Gamma(\frac{1+3\nu}{2})
\Gamma(\frac56-\frac\nu2)}{\Gamma(\frac{1+\nu}2)\Gamma(\frac56+\frac\nu2)}. $$

Now, Mathematica will not simplify $A$ or $B$ on its own, so it needs
help. Set $x=\frac16+\frac\nu2$, and use the multiplication formula to
get
$$
\frac{\Gamma(\frac{1+3\nu}2)}{\Gamma(\frac{1+\nu}{2})\Gamma(\frac56+\frac\nu2)}
= \frac{\Gamma(3x)}{\Gamma(x+\frac13)\Gamma(x+\frac23)} =
\frac{\Gamma(x)}{2\pi} 3^{3x-1/2}. $$
After this, $A$ simplifies to
$$ A = -1 +
\frac{\Gamma(\frac16+\frac\nu2)\Gamma(\frac56-\frac\nu2)}{2\pi}\cos\frac{\pi\nu}{2}
= -1 + \frac{\cos\frac{\pi\nu}{2}}{2\sin(\frac\pi6+\frac{\pi\nu}{2})},
$$
where I've also used the reflection formula for $\Gamma(z)\Gamma(1-z)$ to get rid of the
gamma functions. Some further amount of manual trigonometry yields
$$ A = -\frac{\sqrt{3}}{2}\frac{\sin\frac{\pi\nu}{2}}{\sin(\frac\pi6 +
\frac{\nu\pi}{2})}. $$

Finally, write
$$ \frac{1}{\sin(\frac\pi6+\frac{\pi\nu}{2})} =
\frac{\Gamma(\frac16+\frac\nu2)\Gamma(\frac56-\frac\nu2)}{\pi}, $$
and substitute back. Lots of things cancel, and the answer is
$$ Q = -\frac{3^{1/2}2^{1/3}}{\pi^{1/2}}
\frac{\Gamma(\frac16+\frac\nu2)}{\Gamma(-\frac16)\Gamma(\frac56+\frac\nu2)}. $$

This closed form is equivalent to the one you gave through the
use of $\Gamma(\frac16)\Gamma(-\frac16)=-12\pi$.

P.S.
I would also like to note that the integral
$$ I(\nu,c) = \int_0^\infty J_\nu(x)^2 J_\nu(c x)\,dx $$
and its general form
$$ \int_0^\infty x^{\rho-1}J_\nu(a x) J_\mu(b x) J_\lambda(c x)\,dx $$
appear in Gradshteyn and Ryzhik, and you can find a paper "Some
infinite integrals involving bessel functions, I and II" by
W. N. Bailey, which evaluates this integral in terms of Appell
functions, but only in the case $c>2$ ($|c|>|a|+|b|$), which is where the $F_4$ Appell
function converges. DLMF 16.16.6 actually gives a way to write this
integral as
$$
\frac{\Gamma(\frac{1+3\nu}{2})c^{-1-2\nu}}{\Gamma(1+\nu)^2\Gamma(\frac{1-\nu}{2})}
\,\,\,{}_2F_1\left( \frac{1+\nu}{2}, \frac{1+3\nu}{2}; 1+\nu; x
\right)^2, \qquad x = \frac{1-\sqrt{1-4/c^2}}{2}, $$
but the issue is that this is only correct for $c>2$, and the rhs is
complex for $c<2$. Appell function would only be defined by analytic
continuation in this case anyway, and I didn't find anything useful about
non-principal branches of Appell or hypergeometric functions.

For $c>2$, Mathematica also gives the following:
$$
I(\nu,c) = \frac{\Gamma(\frac{1+3\nu}{2})c^{-1-2\nu}}{\Gamma(\frac{1-\nu}{2})\Gamma(1+\nu)^2}
\,\,\,{}_3F_2\left( \frac{1+\nu}{2}, \frac{1}{2}+\nu,
\frac{1+3\nu}{2}; 1+\nu, 1+2\nu; \frac{4}{c^2} \right), $$
but this is incorrect when $c<2$.

ContentLicense: CC BY-SA 3.0


问:
Surjective linear transformation of finitely generated free modules and their determinants
I kind of got stuck on Aluffi Algebra Exercise 6.4 of chapter 6 which is:

Let $F$ be a finitely generated free $R$-module and let $\alpha$ be a linear transformation of $F$. Give an example of an injective $\alpha$ that is not surjective; in fact, prove that $\alpha$ not surjective precisely when $\det(\alpha$) is not a unit.

Here is the proof for $\alpha$ not being surjective iff $det(\alpha$) is not a unit (in $R$):

Since $F$ is a finitely generated free module, $F \cong R^n$ thus $\alpha$ corresponds to a $n\times n$ matrix $A$ thus if $\alpha$ is not surjective, there is no right-inverse for $\alpha$, thus for all linear transformation $\beta \in End_R(F)$, we have $\alpha \circ \beta \neq id$ meaning there is no $n\times n$ matrix $B$ such that $AB = I_n$ which implies $\det$($\alpha \circ \beta)$ = $\det(\alpha)\det(\beta) = \det(A)\det(B) \neq 1$ proves that $\det(\alpha)$ is not a unit. Same line of reasoning gives the converse.

Is my proof correct? And also can anyone give an example of a linear transformation of $F$ that injects but not surjects?

Since I am self-studying, any help would be appreciated.

答:
Suppose that $\alpha$ is surjective, since $\alpha$ is injective, it is bijective and has an inverse $\beta$, $det(\alpha)det(\beta)=1$.

Suppose that $det(\alpha)$ is invertible in the basis $(e_1,...,e_n)$, consider the matrix of cofactors of $\alpha$ $C_{ij})$, the matrix of the inverse of $\alpha$ is ${1\over{det(\alpha)}}(C_{ij}^T)$.

https://en.wikipedia.org/wiki/Minor_(linear_algebra)#Inverse_of_a_matrix


问:
Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$
Given a triangle with sides $a, b, c$ prove that: $a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$

I attempted to solve this question as follows:

$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c)\le 3abc$

$a^2(b+c)+b^2(c+a)+c^2(a+b)-(a^3+b^3+c^3)\le 3abc$

$(a+b)(b+c)(c+a)-2abc-3abc\le 3abc$

$(a+b)(b+c)(c+a)-5abc\le 3abc$

$(a+b)(b+c)(c+a)\le 8abc$

And I got stuck here. Could you please explain to me how to finish it off?

答: It is just Schur’s $^{[1]}$ Inequality of degree 3. You can search the web and read about it. It is pretty well known and someone named Aritra12 posted a handout on it recently on Aops.


问:
Describe all ideals of $\mathbb{Z}[x]$ containing $\langle{x-1}\rangle$.
So I know that I have to use the Correspondence Theorem to solve this question, but I have trouble interpreting the results. Let $R$ be a commutative ring with unity and $I$ be an ideal in $R$. As I understand it, there is a bijection
$$\left\{\text{ ideals of } \frac{R}{I}\right\} \leftrightarrow\{\text{ ideals of } R \text{ containing } I\}$$
that preserves inclusion and some other things.

Now, I take $R = \mathbb{Z}[x]$ and $I = \langle{x-1}\rangle$. If we consider the quotient group $\frac{\mathbb{Z}[x]}{\langle{x-1}\rangle}$, then we can think of it as $\mathbb{Z}[1]$ (if we think of the quotient group as adjoining $[x]$ and then requiring the condition that $x-1=0$), which is just $\mathbb{Z}$.

But there are a lot of ideals in $\mathbb{Z}$, so what is the conclusion I am supposed to come to about the ideals in $R$ containing $I$?

答:
Consider $e:\mathbb{Z}[X]\rightarrow \mathbb{Z}$ defined by $e(P)=P(1)$, $e$ is a morphism of ring, and $\ker(e)=\langle x-1\rangle$. We deduce that the ideals of $\mathbb{Z}[X]$ which contain $X-1$ are the inverse image of the ideal of $\mathbb{Z}$ by $e$.

Such ideal is defined by an integer $n$, and is $\{P\in\mathbb{Z}[X]:P(1)$ is a multiple of $n\}$.


答:
Let $I$ be an ideal in $\mathbb Z[X]$ containing $\langle X-1 \rangle$.

Define
$$
J= I \cap \mathbb Z
$$
Show that $J$ is an ideal in $\mathbb Z$ and hence there exists some $n \in \mathbb Z$ such that $J=n\mathbb Z$.

Now note that every $P(X) \in J$ can be written in the form
$$
P(X)=(X-1)Q(X)+a
$$
for some $a \in \mathbb Z$. This gives immediately that $a \in J=n \mathbb Z$.

From here, you can immediately deduce that
$$
I=\langle X-1, n\rangle
$$

Title: Describe all ideals of $\mathbb{Z}[x]$ containing $\langle{x-1}\rangle$.

Body:
So I know that I have to use the Correspondence Theorem to solve this question, but I have trouble interpreting the results. Let $R$ be a commutative ring with unity and $I$ be an ideal in $R$. As I understand it, there is a bijection
$$\left\{\text{ ideals of } \frac{R}{I}\right\} \leftrightarrow\{\text{ ideals of } R \text{ containing } I\}$$
that preserves inclusion and some other things.

Now, I take $R = \mathbb{Z}[x]$ and $I = \langle{x-1}\rangle$. If we consider the quotient group $\frac{\mathbb{Z}[x]}{\langle{x-1}\rangle}$, then we can think of it as $\mathbb{Z}[1]$ (if we think of the quotient group as adjoining $[x]$ and then requiring the condition that $x-1=0$), which is just $\mathbb{Z}$.

But there are a lot of ideals in $\mathbb{Z}$, so what is the conclusion I am supposed to come to about the ideals in $R$ containing $I$?


问:
How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$?
How can the following inequation be proven?

$$a^2 + b^2 + c^2 \ge ab + bc + ca$$

答:
$$q := a^2 + b^2 + c^2 - ab - bc - ca = \frac12 \begin{bmatrix} a\\ b \\ c\end{bmatrix}^\top \underbrace{\begin{bmatrix} 2 & -1 & -1\\ -1 & 2 & -1 \\ -1 & -1 & 2\end{bmatrix}}_{=: {\rm L}} \begin{bmatrix} a\\ b \\ c\end{bmatrix}$$

where matrix $\rm L$ is the Laplacian of the cycle graph with $3$ vertices, whose (signed) incidence matrix is

$${\rm C} = \begin{bmatrix} -1 & \color{blue}{1} & 0\\ 0 & -1 & \color{blue}{1} \\ \color{blue}{1} & 0 & -1\end{bmatrix}$$

Since $\rm L = C^\top C$, we obtain the following sum of squares (SOS) decomposition

$$2q = (\color{blue}{b} - a)^2 + (\color{blue}{c} - b)^2 + (\color{blue}{a} - c)^2 \geq 0$$

which is the SOS decomposition proposed by Mark Bennet. Since matrix $\rm L$ is rank-$2$, a terser SOS decomposition with only $2$ terms can easily be found — say, via the Cholesky decomposition.

Using Macaulay2,

```<code>Macaulay2, version 1.16
with packages: ConwayPolynomials, Elimination, IntegralClosure, InverseSystems, LLLBases, MinimalPrimes, PrimaryDecomposition, ReesAlgebra, TangentCone, Truncations

i1 : needsPackage( "SumsOfSquares" );
--loading configuration for package "NumericalAlgebraicGeometry" from file /Users/rodrigo/Library/Application Support/Macaulay2/init-NumericalAlgebraicGeometry.m2
--loading configuration for package "Bertini" from file /Users/rodrigo/Library/Application Support/Macaulay2/init-Bertini.m2
--warning: symbol "Verbosity" in MinimalPrimes.Dictionary is shadowed by a symbol in SemidefiniteProgramming.Dictionary
--  use the synonym MinimalPrimes$Verbosity

i2 : R = QQ[a,b,c];

i3 : q = a^2 + b^2 + c^2 - a*b - a*c - b*c

      2          2                2
o3 = a  - a*b + b  - a*c - b*c + c

o3 : R

i4 : sosPoly solveSOS q

             1    1  2    3        2
o4 = (1)(a - -b - -c)  + (-)(b - c)
             2    2       4

o4 : SOSPoly

i5 : tex o4

o5 = <span class="math-container">$\texttt{SOSPoly}\left\{\texttt{coefficients}\,\Rightarrow\,\left\{1,\,\frac{3}{4}\right\},\,\texttt{generators}\,\Rightarrow\,\left\{a-\frac{1}{2}\,b-\frac{1}{2}\,c,\,b-c\right
     \},\,\texttt{ring}\,\Rightarrow\,R\right\}$</span>
</code>```

In $\TeX$,

$$\texttt{SOSPoly}\left\{\texttt{coefficients}\,\Rightarrow\,\left\{1,\,\frac{3}{4}\right\},\,\texttt{generators}\,\Rightarrow\,\left\{a-\frac{1}{2}\,b-\frac{1}{2}\,c,\,b-c\right\},\,\texttt{ring}\,\Rightarrow\,R\right\}$$

polynomials sum-of-squares-method macaulay2 graph-laplacian

Body:
$$q := a^2 + b^2 + c^2 - ab - bc - ca = \frac12 \begin{bmatrix} a\\ b \\ c\end{bmatrix}^\top \underbrace{\begin{bmatrix} 2 & -1 & -1\\ -1 & 2 & -1 \\ -1 & -1 & 2\end{bmatrix}}_{=: {\rm L}} \begin{bmatrix} a\\ b \\ c\end{bmatrix}$$

where matrix $\rm L$ is the Laplacian of the cycle graph with $3$ vertices, whose (signed) incidence matrix is

$${\rm C} = \begin{bmatrix} -1 & \color{blue}{1} & 0\\ 0 & -1 & \color{blue}{1} \\ \color{blue}{1} & 0 & -1\end{bmatrix}$$

Since $\rm L = C^\top C$, we obtain the following sum of squares (SOS) decomposition

$$2q = (\color{blue}{b} - a)^2 + (\color{blue}{c} - b)^2 + (\color{blue}{a} - c)^2 \geq 0$$

which is the SOS decomposition proposed by Mark Bennet. Since matrix $\rm L$ is rank-$2$, a terser SOS decomposition with only $2$ terms can easily be found — say, via the Cholesky decomposition.

Using Macaulay2,

Macaulay2, version 1.16
with packages: ConwayPolynomials, Elimination, IntegralClosure, InverseSystems, LLLBases, MinimalPrimes, PrimaryDecomposition, ReesAlgebra, TangentCone, Truncations

i1 : needsPackage( "SumsOfSquares" );
--loading configuration for package "NumericalAlgebraicGeometry" from file /Users/rodrigo/Library/Application Support/Macaulay2/init-NumericalAlgebraicGeometry.m2
--loading configuration for package "Bertini" from file /Users/rodrigo/Library/Application Support/Macaulay2/init-Bertini.m2
--warning: symbol "Verbosity" in MinimalPrimes.Dictionary is shadowed by a symbol in SemidefiniteProgramming.Dictionary
--  use the synonym MinimalPrimes$Verbosity

i2 : R = QQ[a,b,c];

i3 : q = a^2 + b^2 + c^2 - a*b - a*c - b*c

      2          2                2
o3 = a  - a*b + b  - a*c - b*c + c

o3 : R

i4 : sosPoly solveSOS q

             1    1  2    3        2
o4 = (1)(a - -b - -c)  + (-)(b - c)
             2    2       4

o4 : SOSPoly

i5 : tex o4

o5 = $\texttt{SOSPoly}\left\{\texttt{coefficients}\,\Rightarrow\,\left\{1,\,\frac{3}{4}\right\},\,\texttt{generators}\,\Rightarrow\,\left\{a-\frac{1}{2}\,b-\frac{1}{2}\,c,\,b-c\right
     \},\,\texttt{ring}\,\Rightarrow\,R\right\}$

In $\TeX$,

$$\texttt{SOSPoly}\left\{\texttt{coefficients}\,\Rightarrow\,\left\{1,\,\frac{3}{4}\right\},\,\texttt{generators}\,\Rightarrow\,\left\{a-\frac{1}{2}\,b-\frac{1}{2}\,c,\,b-c\right\},\,\texttt{ring}\,\Rightarrow\,R\right\}$$

polynomials sum-of-squares-method macaulay2 graph-laplacian


问:
Is a prime ideal in the polynomial ring over an algebraically closed field prime also in polynomial rings over extension fields?

Let $k \subsetneq K$ a field extension, $k$ algebraically closed, $I \subset k[x_1,\ldots, x_n]$ a prime ideal generated by the polynomials $f_1, \ldots, f_r$. Let $J \subset K[x_1, \ldots, x_n]$ be the ideal generated by $f_1, \ldots, f_r$. Is $J$ then also prime?

Can you provide a proof or a counterexample (or a reference to a book where the proof could be found)?

答: $K[X_1,\dotsc,x_n]/J = K \otimes_k k[x_1,\dotsc,x_n]/I$ is the tensor product of two integral domains over an algebraically closed field, hence also an integral domain (see e.g. here). So we may replace $K$ by any $k$-algebra which is an integral domain.</p>


问:
Strong Induction proof : $x_k = 1/2(x_{k+1}+x_{k-1})-1$ holds for all integers $k∈Z_{≥0}$
I have just started to learn how to do strong induction. I am struggling to fully understand how to properly do it though. I am trying to work on this exercise where I must prove $x_k = 1/2(x_{k+1}+x_{k-1})-1$ holds for all integers $k∈Z_{≥0}$. The terms are: $0,1,4,9,16...$ The term $x_1 = 1$ was given in the question but all the others were solved for.

I have so far defined a base case of $k=0$ and $k=1$ that holds true. I picked $0$ because it's the minimum value of $k$ and I picked $1$ because I thought I needed too, but I am not sure if it was necessary or why it would be necessary.

I then did the strong induction hypothesis where I state that $P(k)$ is true for all integers such that $0≤k≤n$, where $n∈Z$ such that $n≥1$. I picked $1$ because it's the largest base case value, but I am not sure if it is correct since I am unsure about the base case itself.

After this point, I am not sure how to really apply induction. I tried it many times but my proof doesn't look right.

I really need help on solving this, I am more comfortable with weak induction. I plan to use this solution as a reference for when I solve other problems, so I will be using it as an example to refer back too when I get stuck on tougher problems.

Thank you for any solutions or help provided,  I really appreciate it.

答:
Due to the use of $2$ previous index values (i.e., $x_k$ and $x_{k-1}$) in the sequence definition compared to the one with the highest index (i.e., $x_{k+1}$), when using strong induction, you correctly stated that you will need to define $2$ base cases (in general, the number of previous values in the definition determines the minimum number of base cases which are required). As your question statement implied (and ironX's question comment also stated), the hypothesis is that, with

$$x_k = \frac{x_{k+1} + x_{k-1}}{2} - 1 \tag{1}\label{eq1A}$$

and initial values of $x_0 = 0$ and $x_1 = 1$, we then have

$$x_k = k^2 \tag{2}\label{eq2A}$$

for all integers $k \ge 0$.

The $2$ initial, base values of $x_0$ and $x_1$ both satisfy \eqref{eq2A}. Next, the induction hypothesis is to assume \eqref{eq2A} holds for all non-negative integers $k \le n$ for some integer $n \ge 1$, as you already stated in your question text. To use this, first rearrange \eqref{eq1A} as

$$\begin{equation}\begin{aligned}
2x_k & = x_{k+1} + x_{k-1} - 2 \\
x_{k+1} & = 2x_k - x_{k-1} + 2
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$

The induction hypothesis has $x_n = n^2$ and $x_{n-1} = (n-1)^2$, so using this and $k = n$ in \eqref{eq3A} gives

$$\begin{equation}\begin{aligned}
x_{n+1} & = 2x_{n} - x_{n-1} + 2 \\
& = 2(n^2) - (n-1)^2 + 2 \\
& = 2n^2 - (n^2 - 2n + 1) + 2 \\
& = 2n^2 - n^2 + 2n - 1 + 2 \\
& = n^2 + 2n + 1 \\
& = (n + 1)^2
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$

This shows \eqref{eq2A} then also holds for $k = n + 1$, completing the induction step. Thus, this proves by strong induction that \eqref{eq2A} is true for all $k \ge 0$.


问:
Show non-standard interpretation on language of arithmetic s.t. for any formula, if $\mathcal{N} \vDash F(n)$, then $\mathcal{M} \vDash F[m]$

Let $\mathcal{L}$ be the language of arithmetic. Let $\Gamma$ be the set of sentences true in the standard interpretation $\mathcal{N}$ [natural numbers]. Let $F(x)$ be any formula. Show that if $\mathcal{N} \vDash F(n)$, for every numeral $n$, then there is a non-standard model $\mathcal{M}$ of $\Gamma$ and a non-standard number $m$ such that $\mathcal{M} \vDash F[m]$

(Hint: use compactness)

The compactness theorem says that if every finite subset of a set of sentences has a model, then the whole set has a model. I'm not sure how that helps.

One common nonstandard interpretation is one with rational numbers $\mathcal{Q}$ instead of natural numbers. However, that is not a model for $\Gamma$. Consider the formula $\forall x \exists y (x < y \, \& \, \lnot \exists z (x < z \, \& \, z < y))$. That is true in $\mathcal{N}$ and therefore is in the set $\Gamma$ but it's not true in $\mathcal{Q}$.

What other kind of non-standard models for $\Gamma$ exist?

Can I get a hint on this?

EDIT: Revised with help from @Noah Schweber.

答:
Consider the set $\Gamma$ as defined in OP's post. Add to this a constant symbol $m$ and, for each numeral $i$, the axiom $m \neq i$. Also, add the axiom $F(m)$.

Show that any finite subset of these axioms is satisfiable (aka has a model). Then, apply the compactness theorem.


问:
Schur Triangularization for real inner product space
I am trying to compare Schur's triangularization theorem for the real and complex case. The theorem for the real case is:

"Let A:X→X be an operator acting in a real inner product space.  Suppose that all eigenvalues of A are real (meaning that A has exactly n= dim X real  eigenvalues,  counting  multiplicities).  Then  there  exists  an orthonormal basis u1,u2,...,un in X such that the matrix of A in this basis is upper triangular.

In other words, any real n×n matrix A with all real eigenvalues can be represented as $T=UTU^{*}=UTU^{T}$,  where U is an orthogonal,  and T is areal upper triangular matrices."

The theorem for complex case is identical except we do not assume existence of eigenvalues at all.

When the theorem is saying that we must have n real eigenvalues counting multiplicity, this is referring to algebraic multiplicity and not geometric multiplicity correct? If that was geometric multiplicity, then we can just diagonalize the matrix. But why do we need to assume we have n eigenvalues (counting multiplicity)? The proof for the complex case is just an induction proof where we only need one eigenvalue.

Also, compared to the complex case, is the theorem saying that the eigenvalues must already exist for us to triangularize the matrix? In the complex case, we can simply pull out a eigenvalue since at least one must exist due to the Fundamental theorem of algebra, but in the real case, there are matrix that has no real eigenvalues such as the rotation matrix. So we can triangularize any matrix in a complex space, the number of matric we can triangularize in the real case is quite restricted right?

答:
You just need the eigenvalues to be real, you don't need the matrix to be diagonalizable. In other words, it's counting algebraic multiplicities. I wouldn't get too hung up on the Theorem as stated, some versions just say "assume $A$ is a real matrix with real eigenvalues".

As you noted, there are real matrices without any real eigenvalues. But once you have a real eigenvalue, you have geometric multiplicity at least one, so you can always produce at least one eigenvector, which suffices to make the induction work.


问:
How to prove if two matrices are similar without explicitly finding the basis matrix?
My understanding is that two matrices A and B are similar if and only if there exists some other matrix M such that A = MBM-1.

Is there a way to prove that two matrices are similar without knowing exactly what the basis matrix is? Because obviously if we did, then the "proof" would be a trivial one.

Specifically I'd like to know how the guy in this YouTube video  comes up with similar matrices off the top of his head without explicitly finding the basis matrix.

答:
al eigenvalues are roots of the characteristic polynomial, let's call it $f(x).$

It is possible to test for multiple roots by simply finding the polynomial gcd of $f(x)$ and the derivative $f'(x)$

If $f(x)$ has no multiple roots, and $A,B$ are two matrices that have the same characteristic $f(x),$  then they are similar. In particular, both diagonalize, and the diagonal entries agree.

Now, suppose that there are repeated roots, same for the two matrices. If the minimal polynomials do not agree, the matrices are not similar. The degree of the $(x- \lambda )$ term in the minimal polynomial is the size of the largest Jordan block for that eigenvalue $\lambda$

As the size of the matrices gets larger, there are more and more possibilities of distinct Jordan forms. Which is to say, if the characteristic polynomials and minimal polynomials agree, there may still be room for the matrices to fail to be similar.


问:
How to judge the isomorphism of two drawings of same graph?
We say two drawings of a graph are isomorphic if there is a homeomorphism of the surface that maps one drawing to the other. But in actual operation, I find it quite difficult to determine whether two drawings are isomorphic. I don’t know how to describe homeomorphism.
For example, the following two drawings both show one graph $K_6$ on the plane:

Are these two drawing isomorphic? why? Is there a more operational way to judge?

The definition of drawing isomorphism comes from the following book, which is also the first monograph of Crossing number.

Crossing numbers of graphs

答:
The big problem with testing if two graphs are isomorphic is that there's so many potential ways to match them up. With graph drawings, there is no combinatorial explosion - at least not when the graph drawing is $2$-connected. (If not, then we can still find isomorphisms between blocks efficiently, but it might be hard to figure out which blocks correspond.)

Treat a graph drawing as a plane embedding of a graph, in which the intersections are merely another type of vertex. If the graph drawing is $2$-connected, then each face has a face walk which visits all the vertices in order around the face, which is unique up to changing the starting point and reversing the order.

Let's make two initial guesses about the drawing isomorphism:

Pick a length-$k$ face of one drawing, and guess which length-$k$ face of the other drawing it corresponds to.

Pick face walks $v_1, \dots, v_k, v_1$ and $w_1, \dots, w_k, w_1$ of the two faces, and guess that $v_i$ corresponds to $w_i$ for $i=1, 2, \dots, k$.

Then if our guesses were right, the rest of the isomorphism is determined. The other face touching edge $v_i v_{i+1}$ must correspond to the other face touching edge $w_i w_{i+1}$, and knowing two corresponding vertices tells us how the rest of their face walks correspond. We can repeat this to grow the region where we know the isomorphism, one face at a time. Either we reach a contradiction (two faces that must match have different lengths, or two vertices that must match have different types), or we find an isomorphism.

We can try every possible way to make these guesses, and see if they work out. In the plane, we have a shortcut: the unbounded faces must correspond.


问:
If $\forall y\in B, y=h(x)$ where $x\in A$, is it true that $\exists y$ s.t. $y=f(x), x\in A$ for every $y\in B$?
I know it is not great grammar, but is it logically incorrect to say something like

If $\forall y\in B, y=h(x)$ where $x\in A$, it is true that $\exists y$ s.t. $y=f(x), x\in A$ for every $y\in B$

答: If you're saying $[\forall y \in B (f(y) = x)] \implies [\forall y \in B (\exists y \in B (f(y) = x))]$, this is indeed true. If you're saying $[\forall y \in B (f(y) = x)] \implies [\exists y \in B (f(y) = x)]$, this is not necessarily true (for consider $B = \emptyset$).


答:
See Tsemo Aristide's answer on the cofactor matrix.

For an example of an injective non-surjective module homomorphism, consider $f(x) = 2x$, $f : \mathbb{Z} \to \mathbb{Z}$, which is an injective $\mathbb{Z}$-module homomorphism but is not surjective.

Title: Surjective linear transformation of finitely generated free modules and their determinants

Body:
I kind of got stuck on Aluffi Algebra Exercise 6.4 of chapter 6 which is:

Let $F$ be a finitely generated free $R$-module and let $\alpha$ be a linear transformation of $F$. Give an example of an injective $\alpha$ that is not surjective; in fact, prove that $\alpha$ not surjective precisely when $\det(\alpha$) is not a unit.

Here is the proof for $\alpha$ not being surjective iff $det(\alpha$) is not a unit (in $R$):

Since $F$ is a finitely generated free module, $F \cong R^n$ thus $\alpha$ corresponds to a $n\times n$ matrix $A$ thus if $\alpha$ is not surjective, there is no right-inverse for $\alpha$, thus for all linear transformation $\beta \in End_R(F)$, we have $\alpha \circ \beta \neq id$ meaning there is no $n\times n$ matrix $B$ such that $AB = I_n$ which implies $\det$($\alpha \circ \beta)$ = $\det(\alpha)\det(\beta) = \det(A)\det(B) \neq 1$ proves that $\det(\alpha)$ is not a unit. Same line of reasoning gives the converse.

Is my proof correct? And also can anyone give an example of a linear transformation of $F$ that injects but not surjects?

Since I am self-studying, any help would be appreciated.


问:
Estimate limit without using L'Hopital
It is possible to estimate the following limit without using L'Hopital Rule?

$$\lim_{x\to \frac{\pi }{3}}\left(\frac{\sin\left(x-\frac{\pi}{3}\right)}{1-2\cos\left(x\right)}\right)$$

I will be happy to get any tip!

答:
This is a variant on Clement C.'s answer, showing a way to further simplify the calculation of the limit.

After recentering the limit around $x\to0$ and showing that $1-2\cos(x-\pi/3)=1-\cos x+\sqrt3\sin x$, multiply numerator and denominator by $1+\cos x$, so that we have

$$\begin{align}
{\sin x\over1-2\cos(x-{\pi\over3})}
&={\sin x\over1-\cos x+\sqrt3\sin x}\cdot{1+\cos x\over1+\cos x}\\
&={\sin x(1+\cos x)\over1-\cos^2x+\sqrt3\sin x(1+\cos x)}\\
&={\sin x(1+\cos x)\over\sin^2x+\sqrt3\sin x(1+\cos x)}\\
&={1+\cos x\over\sin x+\sqrt3(1+\cos x)}\\
&\to{1+1\over0+\sqrt3(1+1)}={1\over\sqrt3}\\
\end{align}$$

Title: Estimate limit without using L'Hopital


问:
Mumford Red Book proof of surjective morphism of schemes is stable under base change
I am reading the following proof in the Red Book of Varieties and Schemes by Mumford saying that a surjective morphism of schemes is stable under base change and I came across some things that I didn't understand.

"The maps $r^*$ and $s^*$ define inclusions of fields"

The morphisms of schemes $r$ induces a map on stalks $\mathscr{O}_{S,r(x)}\to \mathscr{O}_{X,x}$, which is a local homomorphism, so we get a map $k(r(x))\to k(x)$. Why should this be an inclusion ? This might be obvious, but my intuition about this $k(x)$ is still very vague.

Edit: This is clear now, morphisms of fields are always injective.

"defining $\alpha^*$ and $\beta^*$ to be the compositions"

Here, he is defining what the map on stalks should be. Does this then immediately define the morphism of sheaves?

Any insight would be very much appreciated.

答:
Your first question has already been handled by Daniel Hast and Severin Schraven in the comments: all morphisms of fields are injective because the kernel must be a proper ideal.

To answer your second question, the map $\mathcal{O}_X\to\alpha_*\mathcal{O}_{\operatorname{Spec} \Omega}$ is specified by a compatible collection of maps $\mathcal{O}_X(U)\to (\alpha_*\mathcal{O}_{\operatorname{Spec} \Omega})(U)$ for $U\subset X$ open. But $(\alpha_*\mathcal{O}_{\operatorname{Spec} \Omega})(U)$ is equal to $0$ when $x\notin U$ and $\Omega$ when $x\in U$, so we can define $\mathcal{O}_X(U)\to (\alpha_*\mathcal{O}_{\operatorname{Spec} \Omega})(U)$ as the zero map when $x\notin U$ and the map $\mathcal{O}_X(U)\to\mathcal{O}_{X,x}\to k(x) \to \Omega$ when $x\in U$. These maps are compatible because the map $\mathcal{O}_X(U)\to\mathcal{O}_{X,x}$ factors as $\mathcal{O}_X(U)\to\mathcal{O}_X(V)\to\mathcal{O}_{X,x}$ for $x\in V\subset U$. So this really is enough to define the morphism you want.


问:
Find the area of the space enclosed by points $(x, y)$ such that fullfill the following condition: $9x^3+9x^2y-45x^2=4y^3+4xy^2-20y^2$
Find the area of the space enclosed by points $(x, y)$ such that fullfill the following condition: $9x^3+9x^2y-45x^2=4y^3+4xy^2-20y^2$

My solution goes as follows:

Factorizing we have:

$(9x^2-4y^2)(x+y-5)=0$

$(3x+2y)(3x-2y)(x+y-5)=0$

Hence we get that the vertices of the triangle are points $(-10, 15), (0, 0)$ and $(2, 3)$ and from this we conclude that the area is $30$.

My question is how was I supposed to think of the particular factorization? Can you propose an intuitive method?

答:
Moving everything to one side, $9x^3+9x^2y-45x^2-4xy^2+20y^2-4y^3=0$. Thinking about factorizations, we see the appearance of multiple $9x$'s and $4y$'s. Take out the $9x^2$ and $4y^2$ first to get $9x^2(x+y-5)+4y^2(-x-y+5)$. This is the woah moment, and the rest follows easily.

Alternatively, you can do something similar to Simon's Favorite Factoring Trick (SFFT), which you can learn more about here: https://artofproblemsolving.com/wiki/index.php/Simon%27s_Favorite_Factoring_Trick.


问:
Equality of two elliptic integrals
by wolfram 1   and wolfram 2

It's true that $\displaystyle \int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{2}{\sqrt{1-\sin^2\theta\sin^2\phi}} d\phi d\theta =\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin\theta\sin\phi}} d\phi d\theta$ ?

If yes,  I need  a way to prove the equality of the following two integrals

I tried everything  but I am unable to convert  into a standard form so How do I solve this problem.

Addition 1: For the second integral
$\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin\theta\sin\phi}} d\phi d\theta =\bigg(\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin\theta}}d\theta\bigg)^2 $ and we can use that $2 \int^{\frac{\pi}{2}}_{0}\frac{1}{\sqrt{\sin x}} \mathrm dx = \frac{\Gamma(1/2)\Gamma(1/4)}{\Gamma(3/4)} = \frac{\Gamma \left( \frac{1}{4}\right)^2}{\sqrt{2\pi}}$

Addition 2: For the first integral

Let $\displaystyle K(k)=\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1-k^2\sin^2 t}}dt$  ( Complete Elliptic Integral of the First Kind). we know that $ \displaystyle K(k)=\frac{\pi}{2}\sum_{n=0}^\infty \left(\frac{(2n)!}{2^{2n}(n!)^2}\right)^2k^{2n}$. Then it's not difficult de show that $\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^2\theta\sin^2\phi}} d\phi d\theta=(\pi /2 )^2 \sum _{n=0}^{\infty }(\frac{(2n)!}{4^n(n!)^2})^3$.

The egality of the two integrals hold if we can calculte $$\sum _{n=0}^{\infty }(\frac{(2n)!}{4^n(n!)^2})^3$$
Wolfram gives

Addition 3:

This link gives

答:
I wish i have found out the following solution (thanks go to Etanche and Jandri).

\begin{align}J&=\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1-\sin^2(\theta)\sin^2 \varphi}}d\varphi d\theta\\
&\overset{z\left(\varphi\right)=\arcsin\left(\sin(\theta)\sin \varphi\right)}=\int_0^{\frac{\pi}{2}} \left(\int_0^ \theta\frac{1}{\sqrt{\sin(\theta-z)\sin(\theta+ z)}}dz\right)d\theta\tag1\\
&=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\pi-u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du \tag2\\
&=\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du+\underbrace{\frac{1}{2}\int_0^{\frac{\pi}{2}} \left(\int_{\frac{\pi}{2}}^{\pi-u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du}_{w=\pi-v}\\
&=\int_0^{\frac{\pi}{2}} \left(\int_{u}^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du\\
&=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv-\int_0^{\frac{\pi}{2}} \left(\int_{0}^{u}\frac{1}{\sqrt{\sin u\sin v}}dv\right)du\\
&=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv-\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{\sin u}}\left(\int_0^{u}\frac{1}{\sqrt{\sin v}}dv\right)du\\
&=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv-\frac{1}{2}\left(\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin v}}dv\right)^2\tag3\\
&=\boxed{\frac{1}{2}\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin u\sin v}}dudv}
\end{align}

$(1)$: $\displaystyle dz=\dfrac{\sqrt{\sin^2\theta-\sin^2 z}}{\sqrt{1-\sin^2 z}}d\varphi$, $\sin^2 a-\sin^2 b=\sin(a-b)\sin(a+b)$

$(2)$: Perform the change of variable $u=\theta-z,v=\theta+z$

$(3)$: $\displaystyle\int_0^{\frac{\pi}{2}} f(x)f^\prime(x)dx=\frac{1}{2}\left(f^2\left(\frac{\pi}{2}\right)-f^2(0)\right)$ and $\displaystyle f(x)=\int_0^{x}\frac{1}{\sqrt{\sin u}}du$

(Edit: problem fixed again)

Title: Equality of two elliptic integrals


Title: Strong Induction proof : $x_k = 1/2(x_{k+1}+x_{k-1})-1$ holds for all integers $k∈Z_{≥0}$

Body:
I have just started to learn how to do strong induction. I am struggling to fully understand how to properly do it though. I am trying to work on this exercise where I must prove $x_k = 1/2(x_{k+1}+x_{k-1})-1$ holds for all integers $k∈Z_{≥0}$. The terms are: $0,1,4,9,16...$ The term $x_1 = 1$ was given in the question but all the others were solved for.

I have so far defined a base case of $k=0$ and $k=1$ that holds true. I picked $0$ because it's the minimum value of $k$ and I picked $1$ because I thought I needed too, but I am not sure if it was necessary or why it would be necessary.

I then did the strong induction hypothesis where I state that $P(k)$ is true for all integers such that $0≤k≤n$, where $n∈Z$ such that $n≥1$. I picked $1$ because it's the largest base case value, but I am not sure if it is correct since I am unsure about the base case itself.

After this point, I am not sure how to really apply induction. I tried it many times but my proof doesn't look right.

I really need help on solving this, I am more comfortable with weak induction. I plan to use this solution as a reference for when I solve other problems, so I will be using it as an example to refer back too when I get stuck on tougher problems.

Thank you for any solutions or help provided,  I really appreciate it.

Body:
The problem is to prove that if $x_k=k^2$ and $x_k=\frac{x_{k+1}+x_{k-1}}{2}-1$ then $x_{k+1}=(k+1)^2$, given $x_0=0, x_1=1$

$P(k): x_k=k^2\wedge x_k=\frac{x_{k+1}+x_{k-1}}{2}-1\Rightarrow x_{k+1}=(k+1)^2$

$P(k+1): x_{k+1}=(k+1)^2\wedge x_{k+1}=\frac{x_{k+2}+x_{k}}{2}-1\Rightarrow x_{k+2}=(k+2)^2$

You need to solve for $x_{k+2}$:

$(k+1)^2=\frac{x_{k+2}+k^2}{2}-1\Rightarrow ...x_{k+2}=(k+2)^2$

Therefore, $P(k)\Rightarrow P(k+1)$


答:
Step by step, we have:

$$9(x^3+x^2y-5x^2)=4(y^3+xy^2-5y^2)$$
$$\Rightarrow 9x^2(x + y - 5) = 4y^2 (y + x - 5)$$

and moving everything to one side gives $(9x^2-4y^2)(x + y - 5) = 0$.

Body:
Find the area of the space enclosed by points $(x, y)$ such that fullfill the following condition: $9x^3+9x^2y-45x^2=4y^3+4xy^2-20y^2$

My solution goes as follows:

Factorizing we have:

$(9x^2-4y^2)(x+y-5)=0$

$(3x+2y)(3x-2y)(x+y-5)=0$

Hence we get that the vertices of the triangle are points $(-10, 15), (0, 0)$ and $(2, 3)$ and from this we conclude that the area is $30$.

My question is how was I supposed to think of the particular factorization? Can you propose an intuitive method?


问:
Polynomial with real coefficient which has root $\sqrt{5}+\sqrt{7}$
Is it possible to find a polynomial with real coefficients with a root of $\sqrt{5}+\sqrt{7}$ ?

答:
I'm assuming you mean "rational" or "integer" instead of real.

Let $x=\sqrt{5}+\sqrt{7}$. Then $x^2=5+7+2\sqrt{35}=12+2\sqrt{35}$. Generally, $a+\sqrt{b}$ and $a-\sqrt{b}$ form the roots of a quadratic when $\sqrt{b}\notin\mathbb{Q}$ and $a,b\in\mathbb{Q}$. Applying that here, the sum of the roots of the polynomial with roots $12+2\sqrt{35}$ and $12-2\sqrt{35}$ is $24$. The product of the roots is $12^2-4*35=4$. Therefore $12+2\sqrt{35}$ is one of the roots of $z^2-24z+4=0$ (by Vieta's). This implies that $x^2$ is a root of $z^2-24z+4=0$, and consequently $\boxed{x^4-24x^2+4=0}$.

Check for yourself that this works. In fact, this is the minimal polynomial of $\sqrt{5}+\sqrt{7}$.


问:
Averaging conditional density over an event
I expect this should be a basic property of regular conditional densities/stochastic kernels, but somehow I am having trouble verifying this.

Suppose we have random variables $X$ and $Y$ with (smooth) joint probability density $p(x,y)$ and (smooth) conditional density for X given Y, namely $p(x|y)$. Then, is it true that: $$p(x|y \in A) = \frac{1}{\mathbb{P}(Y \in A)} \int_A p(x|y) p(y) dy $$

答:
I would write something like

$$p(x \mid y \in A)=\dfrac{\int\limits_{y \in A} p(x,y) \,dy}{\int\limits_{z}\int\limits_{y \in A} p(z,y) \,dy\, dz}$$ providing that the denominator is positive, where $z$ ranges over all possible values of $x$

The numerator is pretty much your $\int\limits_A p(x\mid y) \,p(y) \,dy $ and the denominator your $\mathbb{P}(Y \in A)$, so your expression works too.


问:
How to prove $1-\frac12+\frac13-\frac14+\ldots+\frac1{199}-\frac1{200}=\frac1{101}+\frac1{102}+\ldots+\frac1{200}$?
I want to prove $$1-\frac12+\frac13-\frac14+\ldots+\frac1{199}-\frac1{200}=\frac1{101}+\frac1{102}+\ldots+\frac1{200}$$
First I added $\frac12+\frac14+\ldots+\frac1{200}$ to both sides of the equation but that wasn't helpful.
I'm not sure how to prove it I can write it as partial sums:$$\sum_{n=1}^{200}\frac{(-1)^{n+1}}n=\sum_{n=1}^{100}\frac{1}{n+100}$$But I don't see a way to proceed from here.

答:
Hint:

$$\frac12+\frac14+\ldots+\frac1{200} =\frac12\left(1+\frac12+\ldots+\frac1{100}\right)$$

and you may want to add it twice to both sides


问:
Intermediate value existence: $f''(\xi)-f(\xi)(1+2\cot^2\xi)=0$.

Let $f$ be twice continuouly differentiable on $[0,\pi]$. $f(0)f(\pi)<0$. Show that there exists $\xi\in(0,\pi)$, such that $f''(\xi)-f(\xi)(1+2\cot^2\xi)=0$

I do not whether the problem is right, which is recalled by my classmate. Clearly, some zero of $f$ exists. $\cot\xi$ where to stand? Of course, is should not be $\pi/2$.

答: Here is a similar question from IMC2013,It can help you solve your question.


问:
Why does $\sum_{n=1}^{∞}e^{-\frac{an}{x}}=\frac{1}{e^{\frac{a}{x}}-1}$?
I am asking, why does
$$\sum_{n=1}^{∞}e^{-\frac{an}{x}}=\frac{1}{e^{\frac{a}{x}}-1}$$
(where 'a' is just an arbitrary constant)
Is it to do with power series expansion of the terms, or something else?

答: Fact: for $x$ such that $|x| < 1$, one has $\sum_{n=1}^\infty x^n = \frac{1}{1-x} - 1$. Now use this formula with $\mathrm{e}^{-\frac{a}{x}}$ in place of $x$.


问:
how to give elements of $A × B$ and $A ∩ B$
$A$ = {$3n+1: n∈Z$} and $B$ = {$12m+7: m∈Z$}

I have proven that $A⊄B$ by showing that $3n+1$ $⇒$ $3n+1+7-7$ ⇒ $3n-6+7$ ⇒ $3(n-2)+7$ ⇒ $4((3(n-2))+7)/4$ ⇒ $12(n-2)(1/4) +7$ ⇒ $12m-7$, where $m = (n-2)(1/4)$, and this proves that $A⊄B$, since $m$ will not always give an integer value.

I have also proved that $B⊆A$ by showing that $12m+7$ ⇒ $12m+7-6+6$ ⇒ $12m+6+1$ ⇒ $3(4m+2)+1$ ⇒ $3n+1$, where $n = (4m+2)$.

If my proofs for the above statements can be improved, let me know.

Now, I am trying to work on giving two concrete elements of $A × B$ and giving two concrete elements of $A ∩ B$.

$A × B$ is defined as $:=$ {$(a,b): a∈A$ and $b∈B$}

$A ∩ B$ is defined as $=$ {$x: (x∈A)$ and $(x∈B)$}

I am not exactly sure how to give concrete elements on $A × B$ and $A ∩ B$ or what it means to give concrete elements. If someone can explain this to me, I would really appreciate it.

Could I say that a concrete element of $A×B$ is $(1,6)$ and $(3, 10)$? I say that $b=6$ and $b=10$ because $(n−2)(1/4)$ will give an integer with those values of $b$? I am not sure what makes an element "concrete" and if I am understanding this correctly.

Could I say that a concrete element of $A∩B$ is $x=10$ or $x=6$ or $x=0$ or $x=14$. I am basically giving out numbers that $(n−2)(1/4)$ will yield an integer value for, since $B$ is contained in $A$, so all values of $B$ will work for $A$ but not the other way around?

Thanks for any answers given!

答:
Giving a concrete element of $A \times B$ amounts to finding a concrete element $a$ of $A$, and a concrete element $b$ of $B$, and then writing $(a, b)$.

Similarly, giving a concrete element of $A \cap B$ amounts to finding a concrete element $a$ of $A$ which also happens to be an element of $B$.

Also, you are unfortunately mistaken that $A \subseteq B$. We have $1 \in A$ but $1 \notin B$. Perhaps you were trying to prove the other direction?


问:
Using binomial theorem to express $ \frac{1}{\sqrt{5}} \ [((\sqrt{5}/2)+(1/2))^{10} - (-(\sqrt{5}/2)+(1/2))^{10}]$ as a single finite series
I am trying to write
$$\frac1{\sqrt5}\left[(\frac{\sqrt5}2+\frac12)^{10}
- (-\frac{\sqrt5}2+\frac12)^{10}\right]
$$
as a single finite series of the form $\sum^{10}_{j=0}a_j$, where $a_j$ depends on $j$.

So far I have expanded the term $(\frac{\sqrt5}2+\frac12)^{10}$ and the term $(-\frac{\sqrt5}2+\frac12)^{10}$ using the binomial theorem.

For the term  $(\frac{\sqrt5}2+\frac12)^{10}$, I got $\sum^{10}_{j=0}\binom{10}{m}(\frac12)^m(\frac{\sqrt5}2)^{10-m}$.

For the term $(-\frac{\sqrt5}2+\frac12)^{10}$, I got $\sum^{10}_{j=0}\binom{10}{m}(\frac12)^m(-\frac{\sqrt5}2)^{10-m}$.

I am not sure how to combine the terms to write them under one series.

This is what I have written so far:
$$\frac1{\sqrt5}\left[(\frac{\sqrt5}2+\frac12)^{10}
- (-\frac{\sqrt5}2+\frac12)^{10}\right]
$$

$=$$\sum^{10}_{j=0}\binom{10}{m}(\frac12)^m(\frac{\sqrt5}2)^{10-m}$$ - $ $\sum^{10}_{j=0}\binom{10}{m}(\frac12)^m(-\frac{\sqrt5}2)^{10-m}$

$=$ $\sum _{m=0}^{10}\binom{10}{m}\:\left(\left(\frac{\sqrt{5}}{2}\right)^m\left(\frac{1}{2}\right)^{10-m}-\left(\left(-\frac{\sqrt{5}}{2}\right)^m\left(\frac{1}{2}\right)^{10-m}\right)\right)$

$=\sum _{m=0}^{10}\binom{10}{m}\:\left(\frac{1}{2}\right)^{10-m}\left(\left(\frac{\sqrt{5}}{2}\right)^m-\left(\left(-\frac{\sqrt{5}}{2}\right)^m\right)\right)$

The goal of the excersise is to show that $\sum^{10}_{j=0}a_j$ will give the same terms as $\frac1{\sqrt5} = \left[(\frac{\sqrt5}2+\frac12)^{10} - (-\frac{\sqrt5}2+\frac12)^{10}\right]\\$

which will give same terms as
$\sum^5_{j=1}\binom{10}{ 2j-1}\frac{5^{j-1}}{2^9}
$

So I need help in writting  $\sum^{10}_{j=0}a_j$, and so far I have  $\sum^{10}_{j=0}a_j$

$=\sum _{m=0}^{10}\binom{10}{m}\:\left(\frac{1}{2}\right)^{10-m}\left(\left(\frac{\sqrt{5}}{2}\right)^m-\left(\left(-\frac{\sqrt{5}}{2}\right)^m\right)\right)$

The goal of the final exercise is to show that these terms yield the same terms as the series $\sum^5_{j=1}\binom{10}{2j-1}\frac{5^{j-1}}{2^9}$. Or in other words, show that

$$\frac1{\sqrt5}
\left[(\frac{\sqrt5}2+\frac12)^{10} - (-\frac{\sqrt5}2+\frac12)^{10}\right]\\
= \sum^5_{j=1}\binom{10}{ 2j-1}\frac{5^{j-1}}{2^9}
$$

To do so, I need help in combining the expansion of the terms under one series in the form $\sum^{10}_{j=0} a_j$, where $a_j$ depends on $j$.

I mainly need someone to check if $\sum^{10}_{j=0}a_j$ $=\sum _{m=0}^{10}\binom{10}{m}\:\left(\frac{1}{2}\right)^{10-m}\left(\left(\frac{\sqrt{5}}{2}\right)^m-\left(\left(-\frac{\sqrt{5}}{2}\right)^m\right)\right)$ is correct or not. I can expand the terms myself and I know that they cancel out. I am also not supposed to use any calculator computations, so I am not looking for the sum of the series, just the series itself.

Thank you for any solutions and answers or any sort of help provided.

答:
It’s not so bad if you do the simplifications in a better order:

$$\begin{align*}
&\frac1{\sqrt5}\left(\left(\frac{\sqrt5}2+\frac12\right)^{10}-\left(-\frac{\sqrt5}2+\frac12\right)^{10}\right)\\
&\qquad=\frac1{2^{10}\sqrt5}\left(\sum_{k=0}^{10}\binom{10}k5^{k/2}-\sum_{k=0}^{10}(-1)^k\binom{10}k5^{k/2}\right)\\
&\qquad=\frac1{2^{10}\sqrt5}\sum_{k=0}^{10}\binom{10}k5^{k/2}\left(1-(-1)^k\right)\\
&\qquad=\frac1{2^9\sqrt5}\sum_{k=0}^4\binom{10}{2k-1}5^{k+\frac12}\\
&\qquad=\frac1{2^9}\sum_{k=0}^4\binom{10}{2k+1}5^k\\
&\qquad=\frac1{2^9}\sum_{k=1}^5\binom{10}{2k-1}5^{k-1}
\end{align*}$$

The first step factors out the denominator and applies the binomial theorem to $\left(\sqrt5+1\right)^{10}$ and $\left(-\sqrt5+1\right)^{10}$; the third recognizes that the terms with even $k$ cancel out and keeps only the terms with odd $k$; and the last simply shifts the index by $1$.


问:
Show that if $p$ is prime then $p\neq (a/b)^2$ for any integers $a$ and $b$
Question:

Show that if $p$ is prime then $p\neq (a/b)^2$ for any integers $a$ and $b$

答:
Let's use a proof by contradiction.

Assume for the sake of contradiction that there exists a prime $p$ such that $p=\left(\frac{a}{b}\right)^2=\frac{a^2}{b^2}$. This implies that $\frac{a^2}{b^2}$ is an integer, and since $a,b$ are integers, so is $\frac{a}{b}$. Let $\frac{a}{b}=k$. We know that $p=k^2$. However, we have a contradiction because both $k$ and $k^2$ divide $p$ (unless $k=k^2$, in which case $p=1$, which is not prime). We're done! $\blacksquare$


问:
Bezier curves algorithm complex yet simple
I've got a path that is made up of multiple points - i.e. 0,0  0,50  50,50  70,20

If I just draw this line on the screen it looks quite harsh as it puts a sharp angle at the joining of each point.

Hence I was wondering what the bezier curves algorithm/method would look like which I could call that automatically change the sharp angles to "tight" curves?

I wouldn't want the curve to be too big or generally effect the fall of main path, just soften the join out. If you have a look at the below here is a quick sample I have put together. The line on the left is what I have now, the one in the middle is the line I want.

The image on the right represents what I think I need the algorithm to do. Essentially, I add an additional point to each arc that makes up the join at a point which 10% away from the join, then I remove the join point and adjust the handles so that they are where the point was (not in the diagram they are slightly apart, this is just so that you can see). This is what I need to be able to do.

答: It seems like you answered your own question in the last paragraph. You create the two new control points at each corner, and then replace the corner with a quadratic Bezier curve determined by the two new control points and the original corner point. So, it is not clear to me what you are asking. The de Casteljau algorithm can be used to render the curves.</p>

ContentLicense: CC BY-SA 2.5


问:
Subtracting roots of unity. Specifically $\omega^3 - \omega^2$
This is question that came up in one of the past papers I have been doing for my exams. Its says that if $\omega=\cos(\pi/5)+i\sin(\pi/5)$. What is $\omega^3-\omega^2$. I can find $\omega^3$ and $\omega^2$ by De Moivre's Theorem. But I cant make much headway into how to subtract these?
The answer they give is $2cos(\dfrac{3\pi}{5})$

Please, some help will be greatly appreciated. I cant seem to find help over the internet either.

Body:
$$\omega=\cos(\pi/5)+i\sin(\pi/5)$$
$$\omega^3-\omega^2=\cos(3\pi/5)+i\sin(3\pi/5)-(\cos(2\pi/5)+i\sin(2\pi/5))$$
$$\omega^3-\omega^2=\cos(3\pi/5)-\cos(2\pi/5)+i(\sin(3\pi/5)-\sin(2\pi/5))$$
$$\omega^3-\omega^2=\cos(3\pi/5)-\cos(2\pi/5)+i(\sin(3\pi/5)-\sin(2\pi/5))$$
$$\omega^3-\omega^2=-2\sin(\pi/2)\cdot\sin(\pi/10)-2i\cos(\pi/2)\cdot\sin(\pi/10)$$
$$\omega^3-\omega^2=-2\sin(\pi/10)$$
$$\omega^3-\omega^2=-2\cdot\dfrac{\sqrt{5}-1}{4}$$
$$\omega^3-\omega^2=\dfrac{1-\sqrt{5}}{2}$$

use some trig. formula:
$\cos C-\cos D=2\cdot \sin\dfrac{C+D}{2}\cdot \sin \dfrac{D-C}{2}$

and $\sin C-\sin D=2\cdot \cos\dfrac{C+D}{2}\cdot \sin \dfrac{C-D}{2}$

and $\sin (\pi/10)=\dfrac{\sqrt5-1}{4}$


问:
Divisibility of Chebyshev Polynomials
I was trying to solve a problem involving an Insect crawling on the Cartesian/Coordinate Plane.

We have an insect on the origin of the coordinate plane, who remembers a particular angle $\theta.$ We can command the insect to move in terms of a given string of commands with $\text{R's}$ and $\text{M's}.$

The command $R$ is for "Rotate an angle $\theta$ anticlockwise", the command $M$ for "Move $1$ Step forward in whatever angle you are facing"

For example, if the insect remembers the angle $\theta=90^{\circ},$ and performs the string MRMRMRM, then it will trace a square path, and will finally get back to the origin.

I need to prove that if $\theta$ is an angle such that all of the numbers;
$$\cos{\theta}$$
$$\cos(2\theta)$$
$$\cos(3\theta)$$
$$\vdots$$

are irrational, then there is no such string of $\text{M's}$ and $\text{R's}.$

But this is not really question I want help on, I have an argument that can reduce this to proving an easier fact.
We let $T_k$ denotes the $k^\text{th}$ Chebyshev Polynomial $\big($so that $T_k(\cos{\theta})=\cos{(k\theta)}\big),$ and let $n$ be a natural number.

Lemma 1: We need to prove that, for all nonnegative integers $a_0,a_1,\dots,a_n,$ there exists some polynomial $p(x),$ and integers $a, b,$ and a Chebyshev Polynomial $T_m$ such that;
$$p(x)\big(a_n T_n(x)+a_{n-1}T_{n-1}(x)+\cdots+a_0T_0(x)\big)=aT_{m}(x)+b.$$

This is what I need help on. I have tried proving this by Strong Induction, but it didn't help much :(

Can you help me prove Lemma 1, since I can manage the rest. If you think it is not true, you can tell me about it. In that case you can try to answer the original question.

Thank You and Regards,

MathEagle

答:
For example, $P(z) = 1 + z + z^2 + z^4 + z^6 + z^7 + z^8$
is irreducible over the rationals, and has four roots on the unit circle and four not.
Because of the roots off the unit circle, it does not divide any cyclotomic polynomial, and its roots can't be roots of unity.  If $r$ is one of the roots on the unit circle, $\theta = \log(r)/i$ has the property that none of $\cos(k\theta)$ are rational.  But with this $\theta$ the insect would come back to the origin for the string MRMRMRRMRRMRMRM.

EDIT: $e^{i\theta} = r$ is an algebraic integer, and so is its complex conjugate $\overline{r}$.  Therefore $2 \cos(k \theta) = r^k + \overline{r}^k$ is also an
algebraic integer.  But the only algebraic integers that are rational numbers are (ordinary) integers.  So the only possible rational values for $\cos(k\theta)$ would be $0$, $\pm 1/2$ or $\pm 1$.  But that would make $\exp(ik\theta)$, and thus also $\exp(i\theta)$, a root of unity, which can't happen.

Title: Divisibility of Chebyshev Polynomials


问:
Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $
How to determine the range of the function

$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$

It it is straightforward to verify that $f(x)$ is even and

$$f(0)= \frac13, \>\>\>\>\> \lim_{x\to\pm \infty} f(x) \to -\infty$$

which implies $f(x) \in (-\infty,\frac13]$, i.e.

$$\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le \frac13
$$

and is visually confirmed below

However, it is not obvious algebraically that $f(x)$ monotonically decreases away from $x=0$. The standard derivative tests are not viable due to their rather complicated functional forms.

So, the question is how to prove the inequality  $f(x) \le \frac13$ with rigor.

Note that it is equivalent to proving

$$\cot \left(\frac{\pi(1+x)}{3+x^2}\right)
\cot \left(\frac{\pi(1-x)}{3+x^2}\right)\le \frac13
$$

答:
$$g(x)=\frac{(1+x)^2}{3+x^2}=1+2\frac{x-1}{3+x^2}$$

$$\text{$g$ is continuous for all real numbers and has a derivative} \\ \text{therefore we can find the max and min} \\ g'=2\frac{3-x^2+2x}{(3+x^2)^2}; \quad g'=0\Rightarrow (x-3)(x+1)=0 \\ \text{After that it's obvious that the range of $g$ is } [f(-1),f(3)]=[0,\frac{4}{3}] $$

$$\text{At $g(2)=1$ the $\tan$ function inside $f(g)$ for $\frac{\pi}{2}$ isn't defined}$$

$$f(g)=\tan\Big(g(x)\frac{\pi}{2}\Big)\cdot \tan\Big(g(-x)\frac{\pi}{2}\Big)$$
$$g\cdot\frac{\pi}{2} :\mathbb{R} \rightarrow[0,\frac{2}{3}\pi]$$

$$\text{And also, at }x\geq0,\;g(x)\cdot \frac{\pi}{2}\in[\frac{\pi}{6},\frac{2}{3}\pi] \\ \hspace{135px} g(-x)\cdot\frac{\pi}{2}\in[0,\frac{\pi}{2})$$
$$\text{And since when both $tan$ parts of the function when $g(x \geq 0)\cdot \frac{\pi}{2}\in (\frac{\pi}{2}, \frac{2}{3}\pi]$} \\ \text{are multiplied, we get a negative result,} \\ \text{Therefore we should check for $\forall x\;\;g(x\geq 0)\cdot \frac{\pi}{2}\in[\frac{\pi}{6}, \frac{\pi}{2})$}$$

$$\text{From here it should be simpler to continue, what you should end up} \\ \text{with is that at $\frac{\pi}{6}$ the $\tan$ functions take their maximal value when} \\ \text{$g(x)\frac{\pi}{2} = \frac{\pi}{6} \;$and multiplied together, the result will be $\frac{1}{3}$}$$


问:
Elements of order $d$ are all powers of each other in $U_p$ without assuming $U_p$ cyclic
Without knowing that $U_p$ is already cyclic, is there a good way to show that all elements of a given order $d \mid p-1$ are powers of each other? That is, if we are able to find an element of order $d$, then all elements of order $d$ are its powers. I am trying to show that there is only $\varphi(d)$ of these elements, since once we find one element of order $d$, say $a$, then all the others can be found via $a^k$, for $k$ coprime to $d$.

So far I've tried doing a proof by contradiction. If we have a second element of order $d$ that is not $a$, say $b$, then $a^xb^y$ must have order $d$ as well for all $x, y$. However, there doesn't seem to be enough elements to exceed the limit of $p-1$.

The only way I have managed to prove this is first doing it for divisors that are prime powers through using polynomials on $U_p$, and then using multiplicativeness of $\varphi$. I am wondering whether there is some other solution that is more straightforward.

答:
Assume that $a,b$ are elements of order $d$.

Then $a,a^{2},...,a^{d-1}, a^d=1$ are $d$ distinct roots of the polynomial $X^d-1$, and therefore are all roots of this polynomial.

Next, $b^d=1$ and hence, $b$ is a root of $X^d-1$. Therefore, by the above there exists some $1 \leq k \leq d$ such that $b=a^k$.

This proves that $b=a^k$.

Note

$$d=\mbox{ord}(b)=\mbox{ord}(a^k)=\frac{d}{\mbox{gcd(d,k)}} \Rightarrow \mbox{gcd(d,k)}=1$$

You can use this to prove the stronger statement:

Assume that there exists an element $a$ of order $d$, then $b$ is an element of order $d$ if and only if there exists some $1 \leq k \leq d$ with $\mbox{gcd(d,k)}=1$ such that
$$
b=a^k \,.
$$
This immediately gives that there are either $0$ or $\phi(d)$ elements of order $d$.

If you are familiar with the Fundamental Theorem of ALgebra, it implies that there exists at least one element of order $d$, so combined with the above you can deduce that there are exactly $\phi(d)$ elements of order $d$.


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

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

答:
We can do this by hand, too.

First, we should determine what the vertices $1, 2, \dots, 10$ are doing. The idea is that when we generate the Prüfer code, one entry at a time, we are simultaneously deleting leaves of the tree, one leaf at a time (and recording the parent of the deleted leaf). So vertex $1$ becomes a leaf after the last "$1$" entry in the Prüfer code: it is the lowest label leaf, so it immediately gets deleted, and the very next entry of the code tells us $1$ is adjacent to $2$. Similar reasoning tells us that $2$ is adjacent to $3$, $3$ to $4$ and so on. So vertices $1, 2, \dots, 10$ form a path.

All other vertices in the tree must be leaves. They are filled in to make sure that vertices $1, 2, \dots, 10$ have degrees $2, 3, \dots, 11$ respectively: one more than the number of times they appear in the Prüfer code. We add them as leaves on the vertices $1, 2, \dots, 10$ in the order that they appear in the Prüfer code: $11$ is adjacent to $1$, $12$ is adjacent to $2$, $13$ and $14$ are adjacent to $3$, $15$ through $17$ are adjacent to $4$, and so on.

Title: Finding the Tree for Prufer Code


问:
Convergence in measure and almost everywhere

In a finite measure space, let $\{f_{n}\}$ be a sequence of measurable functions. Show that $f_{n}  \rightarrow f$ in measure if and only if every subsequence $\{f_{n_{k}}\}$ contains a subsequence $\{f_{n_{k_{j}}}\}$, that converges almost everywhere to f.

I can prove from converge in measure to converge almost everywhere, but I don't know how to write it down for the other direction.

答:
"Convergence almost everywhere implies convergence in measure (for a finite measure)" follows almost trivially from the following lemma (e.g., lemma 2.5.4 of the Ash and Dade text):

Suppose $\mu$ is a finite measure. Then $f_n \to f$ pointwise $\mu$-a.e. if and only if
$$
\forall \epsilon > 0, \lim_{n \to \infty} \mu( \bigcup_{n \geq k} B_{k \epsilon}) = 0,
$$
where $B_{k\epsilon}$ is the set where the convergence is violated by the amount $\epsilon$ after index $k$,
$$
B_{k \epsilon} := \{x:  |f_k(x) - f(x)| > \epsilon \}.
$$

(we actually only need the $\implies$ direction).

Now to prove convergence in measure, fix any $\epsilon > 0$, then
$$
\lim_{n \to \infty} \mu(\{x: |f_n(x) - f(x)| > \epsilon \}) =  \lim_{n \to \infty} \mu(B_{n\epsilon}) \leq  \lim_{n \to \infty} \mu(\bigcup_{n \geq k} B_{k \epsilon}) = 0
$$
since $B_{n \epsilon} \subset  \bigcup_{n \geq k} B_{k \epsilon}$. That's it!


答:
$$\overline{\omega}=\omega^9 \Rightarrow \overline{\omega^3}=\omega^{27}=\omega^7=-\omega^2 \,.$$

Thus

$$\omega^3-\omega^2= \omega^3+\overline{\omega^3}=2 Re(\omega^3) \,.$$

Title: Subtracting roots of unity. Specifically $\omega^3 - \omega^2$

Body:
This is question that came up in one of the past papers I have been doing for my exams. Its says that if $\omega=\cos(\pi/5)+i\sin(\pi/5)$. What is $\omega^3-\omega^2$. I can find $\omega^3$ and $\omega^2$ by De Moivre's Theorem. But I cant make much headway into how to subtract these?
The answer they give is $2cos(\dfrac{3\pi}{5})$

Please, some help will be greatly appreciated. I cant seem to find help over the internet either.


答:
Note that each fraction $\frac{a}{b}$ has an irreducible representation, that is, there exist $a_{1},b_{1}$ such that $\frac{a}{b} =\frac{a_{1}}{b_{1}}$ and $\gcd(a_{1},b_{1}) = 1$.

So if $p = (\frac{a}{b})^{2}$, then $p = (\frac{a_{1}}{b_{1}})^{2}=\frac{a_{1}^{2}}{b_{1}^{2}} \implies b_{1}^{2}p = a_1{}^{2} \implies p \mid a_{1}^{2} \implies p \mid a_{1}.$

Similarly $p \mid b_{1} \implies p \mid a_{1},b_{1} \implies (a_{1},b_{1})  \neq 1$, a contradiction.

Body:
Question:

Show that if $p$ is prime then $p\neq (a/b)^2$ for any integers $a$ and $b$


问:
in barycentric coordinates why does $[PBC] = x[ABC]$?
From the Euclidean Geometry in Mathematical Olympiads written by Even Chan, there is a chapter about barycentric coordinates. It is said in the chapter that

Barycentric coordinates are also sometimes called areal coordinates because if $P = (x, y, z)$, then the signed area $[PBC]$ is equal to $x[ABC]$ and so on. In other words, these coordinates can be viewed as $$P = (\frac{[PBC]}{[ABC]}, \frac{[PCA]}{[BCA]}, \frac{[PAB]}{[CAB]})$$

I found wikipedia about this but it's a bit too technical for me to understand. Could anyone put it in a simpler way? Thanks in advance.

答:
Using the barycentric area formula (listed in $p.120$ of Evan's book) which states that for $P_1, P_2, P_3$ be points with barycentric coordinates $P_i = (x_i, y_i, z_i)$ for $i = 1, 2, 3$. Then,
$$\frac{[P_1P_2P_3]}{[ABC]} =\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}$$
where $[PQR]$ denotes the signed area of $\Delta PQR$. We have

$$\frac{[PBC]}{[ABC]} =
\begin{vmatrix}
P_1&P_2&P_3\\
0&1&0\\
0&0&1\\
\end{vmatrix}=P_1$$

The same logic goes for $P_2, P_3$.


问:
What is causing this weird behaviour in Wolfram Alpha, when computing this infinite Sum involving the Zeta function?
This is the Sum:

$$\sum\limits_{n=1}^{\infty} \frac{\zeta(2n+1)-1}{\zeta(2n)-1}$$

Sum is divergent, but Wolfram acts weird and after a correct display of early values, it starts converging to a random value:
Wolfram Alpha (Link computes the Sum up to $n=40$, just to get a faster result);

Probably just a bug.

答:
There are obviously some numerical issues.

Consider
$$S_p=\sum\limits_{n=1}^{p} \frac{\zeta(2n+1)-1}{\zeta(2n)-1}$$ Up to $p=23$, we have with a quite good accuracy
$$S_p \sim \frac {2p-1}4$$ For $p=23$, the exact value is $11.2656$ while the approximation gives $11.25$. However for $p=24$, a CAS reports a sum equal to $13.2656$ while the approximation would give $11.75$.

Where the problem is that, numerically, computed separately with unlimited precision
$$\zeta(49) \sim 1+1.776\times 10^{-15}$$
$$\zeta(48) \sim 1+3.553\times 10^{-15}$$ which would give

$$\frac{\zeta(49)-1}{\zeta(48)-1} \sim 0.49999999941187591$$
Computing the above as a whole leads to the same result. But, nevertheless, we are very close to machine $\epsilon$.

Computed seperately, the summand is asymptotic to $0.5^-$.

We can also use the double inequality (have a look at @reuns comment here)

$$1+2^{-k}+\frac{3^{1-k}}{k-1} < \zeta(k) < 1+2^{-k}+\frac{2^{1-k}}{k-1}$$ which gives
$$2^{-2 n-1}+\frac{3^{-2 n}}{2 n}<\zeta(2n+1)-1 <2^{-2 n-1}+\frac{2^{-2 n}}{2n}$$
$$2^{-2 n}+\frac{3^{1-2 n}}{2 n-1}<\zeta(2n)-1<2^{-2 n}+\frac{2^{1-2 n}}{2 n-1}$$ from which
$$\frac{2^{-2 n-1}+\frac{3^{-2 n}}{2 n}}{2^{-2 n}+\frac{2^{1-2 n}}{2 n-1}}<\frac{\zeta(2n+1)-1 }{\zeta(2n)-1 }< \frac{2^{-2 n-1}+\frac{2^{-2 n-1}}{n}}{2^{-2 n}+\frac{3^{1-2 n}}{2 n-1}}$$

Using $p=100$ this gives $47.8309 < S_{100} <51.8495 $. Notice that the mean of these bounds is $49.84$ while $\frac {199}4=49.75$.

Moreover, in the attached Wolfram Alpha, $S_{40}$ is given as $\sim 19.73$ which does not show at all in the plot ! The above would give $18.2817 < S_{40} <21.3951 $ (mean=$19.84$) while $\frac {79}4=19.75$


问:
Is this a observation or a experiment? And does x affect x or y affect x?
A study measures the mathematical knowledge among students who go to a school where the noise level is high and in one where the noise level is low. Assume that variable X = 'mathematics knowledge' and variable Y = 'school. Which scenario corresponds to this?

I feel like the answer is Observation study where Y affects X. because we are observing the world as it is and not experimenting with it. And we are not attempt to influence them, and we can not determine causation ( only correlation). is this correct way of thinking?

None of the above

Observation study where X affects Y.

Experimental study where Y affects X.

Observation study where Y affects X.

Experimental study where X affects Y

答:
Your thinking is correct. There are two questions here we have to work through:

Does $X$ affect $Y$ or does $Y$ affect $X$?

Is this study an experiment or is it observational?

First, on the direction of the effect. The study wants to determine the relationship between the noise level in the school ($Y$) and the mathematics knowledge of the students ($X$). Either we have a study that wants to know whether the mathematics knowledge of the students affects how noisy the school is ($X$ affects $Y$), or we have a study that asks whether students at a noisier/quieter school perform better in mathematics ($Y$ affects $X$). Although both are plausible, the latter is much more likely the question the study is trying to answer, so we'll go with $Y$ affects $X$.

Then, we want to know if it's observational or experimental. An experiment is typically denoted through some aspect of randomness in the independent variable, which is $Y$ (school noisiness) in this case. We have two schools, where one is noisy and the other is quiet. We didn't randomly assign one school to the noisy condition and one to the quiet condition. There may be something else (a confounding variable) that affects both how noisy the school is and how well the students perform in math, like the behavior of the students, the rules in place at the school, etc. So, in this case, we have an observational study, where we simply observe that there is a difference in mathematics performance that's correlated with the noise levels at the school. That's why this type of study can only measure correlation, and not causation.


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

答: For a sort of a sledgehammer here, it falls out of the orbit-stabilizer theorem.  For the "size" of the orbit of any element is the index of the stabilizer: $[G:G_x]=|Gx|$.  But $[G:G_x]\le|G|$.  That'll crack this nut.


答:
Let
$v_p(k)$ be the exponent of
the largest power of $p$
that divides $k$
so that
$p^{v_p(k)} | k$
and
$p^{v_p(k)+1} \not\mid k$.

Then
if $m, a, b \ge 1$
then $ m |v_p((a/b)^m)$.

This problem is the case
$m = 2$.

Proof.

If
$ v_p((a/b)^m) = n$
where $n \ge 1$,
let
$v_p(a) = u$
and
$v_p(b) = v$
where $u, v \ge 0$.

Then
$v_p(a/b) = u-v$
so
$v_p((a/b)^m) = m(u-v)$
so
$n = m(u-v)
$.

Therefore
$m | n$.

Title: Show that if $p$ is prime then $p\neq (a/b)^2$ for any integers $a$ and $b$


答:
This depends on what you know about factorization.

Do you know every number has a unique prime factorization?  If you know that then you know that if $p = \frac{a^2}{b^2}$ and $p$ is an integer,  then $b^2$ divides evenly into $a^2$ and the primes that divide into $b$ also divide into $a$ but to higher power and the powers that divide into $b^2$ and $a^2$ are the same that divide into $b$ and $a$ but to exactly half the powers. So if $b^2$ divides evenly into $a^2$ then $b$ divides evenly into $a$ and $\frac ab$ is an integer.  SO $p = \frac{a^2}{b^2} = (\frac ab)^2$ and so $p$ is a perfect square. .... but $p$ is prime.  SO that's impossible.

But knowing the every number has a unique prime factorization is a pretty sophisticated assumption?  Maybe you aren't allowed to assume it.

BUt can you assume that $\frac ab$ can be written in "lowest terms"?  If $\frac ab$  is in lowest terms then $a$ and $b$ have no factors in common.  Does that mean that $a^2$ and $b^2$ have no factors in common?  That depends on what you know.  But if $a^2$ and $b^2$ have  no factors in common then $b^2$ can't divide evenly into $a^2$ (unless we assume $b = 1$) and we cant have $\frac {a^2}{b^2}= p$ an integer.

Is it possible that $\frac ab$ is not in "lowest terms"?  If $a,b$ have a factor $k$ in common then $\frac ab = \frac {\frac ak}{\frac bk}$ and we can do this for all factors.  Is there always a finite number of factors and we will eventually factor them all out until we get $\frac ab = \frac {a'}{b'}$ where $a', b'$ don't have factors in common?  THe only other option would be if $a$ and $b$ both have an infinite number of factors in common and that's just plain nutty.

But do we know this stuff?

If $a,b$ have no factors in common than neither to $a^2, b^2$ so we can't have $\frac {a^2}{b^2} = p$ as in integer.... unless..... is it possible that somehow squaring $a$ and squaring $b$ add a new factor out of nowhere.  That's kinda sorta what the ancient pythagorians wanted to sweep under the rug.

DO we know that's impossible?  Do we know Euclid's Lemma?  If we have $p = \frac {a^2}{b^2}$ then $pa^2 = b^2$ so $p$ divides evenly into $a^2$. But $p$ is prime and indivisible.  Does that mean we can't split up $p$ among the two copies of $a$ in $a\times a = a^2$?  Does that mean $p$ must divide evenly into $a$?

But if $p$ divides evenly into $a$ that would mean $p^2$ divides into $a^2$, wouldn't it?  And if $pb^2 = a^2$ and $p^2$ divides $a^2$ the we can divide both sides by $p^2$ and get that $p$ divides evenly into $b^2$.  But that means $p$ divides $b$ for the same reason and $a, b$ have $p$ as common factor....

SO....  what do we know about factorization?  Which of those observations are legitimate?  What can we conclude?


问:
Finding intersection on Heterogenous sets
I am having a rather simple problem that looks on paper that would involve set theory. I am aware of the basic set theory principles like Union, Intersection, Commutativness, etc., However, I would like to find the intersection of a heterogenous set (don't know if this is even a term). To get to the problem, here is what I am trying to solve.

There are four dimensions in my problem set where I need to find whether one Set with a maximum possibility of 4 dimensions would intersect with another Set with a similar dimension. To explain it with an example

I have a region dimension whose complete set includes all the states of the United States. So the entire set R will comprise

R = [Alabama, Alaska, Arizona ...., Wyoming]

I also have another dimension say car brands in USA. And let C be all the car brands. So,

C = [Audi, BMW, Ford, Toyota..., Chrysler]

Now I need to find the intersection of these two different dimensions. If it were just the intersection within the same dimension, its straight forward. Again, let me say the examples.

```<code>S1 ========= R1 = [California, Arizona] and C1 = [BMW, Chrysler]

vs

S2 ========= R2 = [Wyoming] and C2=[BMW, Tesla]
</code>```

In the above example S1 ∩ S2 is going to be empty because the regions are entirely different, so no further need to dive into the Car brands further.

However, in the below example

```<code>S1 ========= R1 = R (the entire region set) and C1=[Tesla]

vs

S2 ========= R2 = [California] and C2=[Tesla, Audi]
</code>```

Here, S1 ∩ S2 = [California] [Tesla]

Questions

How do I model this in a program?

Also what is the result of something like S1 U S2? How do I represent it?

Any pointers/pseudo code, mathematical formulas to attack this problem would be greatly helpful.

Please note that the number of dimensions used in the above example is 2. In the problem that I am trying to solve, the dimensions are 4 and would increase in future. What is the best way to solve this problem?

答:
Your description is a little unclear, but I think you're talking about the intersection of Cartesian products.

Specifically, you have two "starting sets" $R$ and $C$, and the sets whose intersections you're interested in taking are subsets of $R\times C$, that is, sets whose elements are pairs with first coordinate from $R$ and second coordinate from $C$.

In fact, even more specifically you're looking at "subproducts" of $R\times C$ - subsets of $R\times C$ which have the form $R'\times C'$ for some $R'\subseteq R$ and $C'\subseteq C$. (Not every subset of $R\times C$ will be of this form: consider e.g. something of the form $\{(a,c), (a,d), (b,c)\}$.) Intersections of Cartesian products turn out to be pretty easy to calculate: we always have $$(A\times B)\cap (A'\times B')=(A\cap A')\times (B\cap B').$$

(Note that unions are a bit more annoying: $(A\times B)\cup(A'\times B')$ is not the same thing as $(A\cup A')\times(B\cup B')$.)


问:
Area of the section enclosed by the curve
Curve is formed by the intersection of the surface:

$\displaystyle \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0$

with the plane $2x + 2y + z = 1$

What is the area enclosed by this curve?
Eliminating z from both equations, I get an implicit equation relating $x$ and y, with terms in $x^2, y^2, xy, x,$ and $y$. But I don't know where to go from there.

答:
Two given surfaces are
$ \ \displaystyle \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \ $ and
$ \ 2x+2y+z=1$

We need to find area enclosed by the intersection curve. The projection of the enclosed area in XY plane is ellipse
$x^2+y^2-\frac{x}{2}-\frac{y}{2}+ \frac{3xy}{2} = 0$.

As the enclosed area is in plane $z = 1- 2x - 2y$, the area element can be expressed in terms of the area element of its projection in $XY$ plane,

$dS = \sqrt{1+2^2+2^2} \ dA \implies dS = 3 \ dA$

Now I will suggest you to plot the ellipse $x^2+y^2-\frac{x}{2}-\frac{y}{2}+ \frac{3xy}{2} = 0$ in a tool and see. It is apparent that to make it easier to work, we either rotate the ellipse clockwise or anticlockwise by $\frac{\pi}{4}$. If we rotate it anticlockwise, we will get an ellipse with major axis parallel to $x-$ axis and minor axis along $y-$axis. A combination of rotation and shifting of origin will convert it into an ellipse centered at the origin and major and minor axes along coordinate axes. We can first rotate by substituting,

$x = \frac{X+Y}{\sqrt2}, y = \frac{Y-X}{\sqrt2}$. So the equation of ellipse becomes,

$\displaystyle \frac{X^2} {(\sqrt{2/7})^2} + \frac{(Y - \sqrt2 / 7)^2}{(\sqrt2 / 7)^2} = 1$

Now you can directly apply the formula for area of ellipse which is $A = \pi ab = \displaystyle \pi \frac{\sqrt2}{\sqrt7} \frac{\sqrt2}{7} = \frac{2 \pi}{7 \sqrt7}$

So the area enclosed by the curve is $\displaystyle \frac{6 \pi}{7 \sqrt7}$.

If you want to find the area by integral, use polar coordinates with the following transformation -

$X = \sqrt{\frac{2}{7}} \ r \cos\theta, Y = \frac{\sqrt2}{7} (1 + r \sin\theta)$

which transforms the ellipse into a unit circle centered at origin.

Now find the Jacobian of transformation and integrate with limits $0 \leq r \leq 1, 0 \leq \theta \leq 2\pi$.


问:
The Direct Sum in The Definition of Tensor Algebra
In the definition of the tensor algebra associated with the vector space $V$ over a field $\Bbbk$,
$$
T(V)= \bigoplus_{k=0}^{\infty} V^{\otimes k}
$$
writing it all out we get
$$
T(V)= \Bbbk \oplus V \oplus (V \otimes V) \oplus (V \otimes V \otimes V) \oplus \cdots \oplus (V \otimes \cdots \otimes V) \oplus \cdots.
$$

In this definition, the direct sum is between different objects, namely $\Bbbk \oplus V \oplus (V\otimes V) \oplus \cdots$ etc. However, I cannot find any sources explaining how this would work, could someone help me with this definition?

My sources are https://en.wikipedia.org/wiki/Tensor_algebra and "An Introduction to Clifford Algebras and Spinors by Jayme Vazz, Roldão da Rocha."

答:
Given a family of $k$-vector spaces $\{V_i\}_{i \in I}$, the direct sum $$V := \bigoplus_{i \in I} V_i$$ is defined to be the set of all functions $\alpha : I \to \bigcup_{i \in I} V_i$ such that $\alpha_i := \alpha(i) \in V_i$ for all $i \in I$, and that $\{i \in I : \alpha_i \neq 0\}$ is a finite set. Note that each $V_j$ can be naturally embedded in $V$ by identifying $v_j \in V_j$ with the function $v \in V$ such that $v_i = 0$ for all $i \neq j$. Thus, we write $\alpha \in V$ as a finite sum $\alpha = \sum_{i \in I} \alpha_i$.

Therefore, for example, $1 + v \otimes v + v \otimes v \otimes v \otimes v \in T(V)$ for $v \in V$.


问:
Unbounded convex hull is possible?
I'm reading this lecture note for additional study.(https://people.orie.cornell.edu/dpw/orie6300/fall2008/Lectures/lec05.pdf)

For Q the convex hull of a finite number of vectors v1, v2, . . . , vk, Q is a polytope.
An unbounded polyhedra is not a polytope

Think of unbounded polyhedra.

The convex hull should be also unbounded? (Since the polyhedra is also a convex?)

Or in this case does it means that the convex hull doesn't exists?

答: Assume everything is finite dimensional. Looking at the lecture note, your question is probably related to the Weyl-Minkowski theorem for the representation of a polyhedron. Namely, if the polyhedron is closed and bounded, then it is the convex hull of its extreme points. If the polyhedron is unbounded, then it is the Minkowski sum of the convex hull of its extreme points and conic hull of its extreme rays. See this What is Weyl-Minkowski theorem?</p>

Title: Unbounded convex hull is possible?


问:
Problem related to group action of $GL_2(\mathbb Z)$ on $\mathbb Z^2$

Let $G = GL_2(\mathbb Z)$ and $X = \mathbb Z^2$ with elements of $X$ written as column vectors.

(a) Show that $\phi: G\to \text{Perm}(X)$ given by $$\phi(g)\left(\begin{matrix} m \\ n\end{matrix}\right) = g\left(\begin{matrix} m \\ n\end{matrix}\right)$$ is a group action.

(b) Show that the orbits of this action are the same as the equivalence classes under the relation $(m,n)\sim (a,b)$ if $\gcd(m,n) = \gcd(a,b)$.

(c) Let $(m,n)\in\mathbb Z^2$. Show that there exists $g\in G$ such that $$g\ \text{stab}_G\left(\begin{matrix} m \\ n\end{matrix}\right)g^{-1} = \left\{\left(\begin{matrix} 1 & a \\ 0 & b\end{matrix}\right) : a\in\mathbb Z, b=\pm 1\right\}$$

Before anything, it's important to note that for all $g\in G$, $\det g = \pm 1$.

I was able to do (a). $\left(\begin{matrix} 1 & 0 \\ 0& 1\end{matrix}\right) \in G$ is the identity element $e$ for this group action, and $g\cdot (h\cdot x) = (gh)\cdot x$ for all $x\in X$ follows from associativity of matrix multiplication.

I tried (b). Firstly, $\sim$ is indeed an equivalence relation - this is easy to check. Now, I considered arbitrary $x\in X$ and looked at $G\cdot x = \{g\cdot x: g\in G\}$, the orbit of $x\in X$. I presume if I start working with arbitrary matrix entries at this point, things will get messy, so there must hopefully be a nicer way. Also, I noted that if $(m,n)\sim (a,b)$ then there exist integers $a_1,b_1,m_1,n_1$ such that $mm_1 + nn_1 = aa_1 + bb_1$. What do I do next? Don't have much clue about (c) either.

I'd appreciate any hints or solutions that would help since I have been working on this for over a week now without much progress.

P.S. $\text{Perm}(X)$ denotes the set (actually a group under composition itself) of all bijections $X\to X$.

答:
Hint for part (b): let $d$ be the greatest common divisor of $m$ and $n$, which is also the greatest common divisor of $a$ and $b$.  Try showing there is a $g \in \operatorname{GL}_2(\mathbb Z)$ such that

$$g \begin{pmatrix} m \\  n \end{pmatrix} = \begin{pmatrix} d \\ 0 \end{pmatrix}. \tag{$\ast$}$$

By the same reasoning, there will exist another matrix $h \in \operatorname{GL}_2(\mathbb Z)$ such that

$$h \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} d \\ 0 \end{pmatrix}.$$

It follows that $h^{-1}g$ will send $\begin{pmatrix} m \\ n \end{pmatrix}$ to $\begin{pmatrix} a \\ b \end{pmatrix} $, so they will be in the same orbit.

How do you know that such a matrix $g$ exists?  You know that $d$ is a divisor of $m$ and $n$, so $\frac{m}{d}$ and $\frac{n}{d}$ are integers.  You also know that there exist integers $x$ and $y$ such that $xm + ny = d$.  Now you can take

$$g = \begin{pmatrix} x & y \\ ? & ? \end{pmatrix}$$

so that the top row of $\ast$ comes out correctly.  What should go in the blanks '?' so that the determinant of $g$ is $1$ and the bottom row of $\ast$ comes out correctly?

The above reasoning shows that if $(a,b)$ and $(m,n)$ have the same GCD $d$, then they are in the same orbit, since they are both in the orbit of $(d,0)$.  You still need to show conversely that if $g \in \operatorname{GL}_2(\mathbb Z)$, and $g.(m,n) = (a,b)$, then $(m,n)$ and $(a,b)$ have the same GCD.

Hint for part (c): in general, if $G$ is a group acting on a set $X$, then for $g \in G$ and $x \in X$, we can consider the stabilizers $H_x$ and $H_{g.x}$ of the elements $x$ and $g.x$ of $X$, respectively.  It's quite easy to see that these subgroups of $G$ are related by conjugation: $gH_x g^{-1} = H_{g.x}$.

So the question is, what element of $\mathbb Z^2$ is

$$\{ \begin{pmatrix}1 & a \\ 0 & b \end{pmatrix} a \in \mathbb Z, b = \pm 1 \}$$ the stabilizer of?


问:
Why is the arccos of a negative ratio the obtuse angle?
This is a repost of this question since it was not answered and here it was not answered clearly.

if we can no longer think about triangles when in any quadrant other than the first, how else can I visualize or think about the angle and it's sin,cos,tan ?

Thank you in advance

答:
$\arccos:[-1,1] \to [0, \pi]$

If $x\in (0,1)$ then $\arccos x \in \big(0, \frac{\pi}{2}\big)$

If $x\in (-1,0)$ then $\arccos x \in \big(\frac{\pi}{2}, \pi \big)$

In words.

The arccos of a positive number less than 1 is an angle in the first
quadrant, an acute angle.

The arccos of a negative number greater than than -1 is an angle in the second
quadrant, an obtuse angle.


问:
Multivariable Taylor and partial derivatives
Any idea how to prove this? I know it's Taylor related, and the Taylor expansion of $(u^3-v)$ it's $(u^3-v)$ because it's a polynomial. But I don't how what to do with the sin...

Let $f(u,v)=(u^3-v)\cdot\sin(u+v) \Longrightarrow \dfrac{\partial^n f}{\partial u^j \partial v^{n-j}}(0,0)=0 \;\; \; \forall \; n\in \mathbb{N} \; \;\text{odd}.$

答: $f$ is an even function, so all the terms of odd degree in its Taylor expansion vanish.


问:
Solution Verification: area of sinx and sin2x
Let $A$ be the area of region $R$ bounded by $y = \sin(x)$ and $y = \sin(2x)$ over $[0,\pi]$. Calculate the area in this square $[0,\pi]*[-1,1]$ minus the region outside of $R$. Now this was very confusing to figure out, and it appears that the area of $[0,\pi]*[-1,1]$ is a rectangle instead of a square. With that in mind, I calculated the area between the 2 functions $y=\sin(x)$ and $y=\sin(2x)$. The final answer was $2(\pi-1)$. Objections? Any mistakes?

答:
The area between $\sin(x)$ and $\sin(2x)$ is

$$ A = \int_0^{\pi} {|\sin(2x) - \sin(x)| dx} . $$

To calculate $A$, we first need to determine where the two curves intersect so we can remove the absolute value and integrate. Setting the two functions equal, we have

$$ \sin(2x) = \sin(x) $$
$$ 2\sin(x)\cos(x) = \sin(x) $$
$$ 2\cos(x) = 1 $$
$$ \cos(x) = \frac{1}{2} .$$

On the interval in the question, $[0,\pi]$, this equation has exactly two solutions: $x = 0$ and $x = \pi / 3$. Dividing up our interval into two parts, we can test a random point to determine that $\sin(2x) \ge \sin(x)$ when $x \in [0,\pi / 3]$ and $\sin(x) \ge \sin(2x)$ when $x \in [\pi / 3, \pi]$.

Then, the area between the two curves is

$$ A = \int_0^{\pi} {|\sin(2x) - \sin(x)|  \ dx} $$
$$ = \int_0^{\pi / 3} {\sin(2x) - \sin(x)  \ dx} + \int_{\pi / 3}^{\pi} {\sin(x) - \sin(2x) \ dx} $$
$$ = \left[ -\frac{1}{2}\cos(2x) + \cos(x) \right]_0^{\pi / 3} + \left[ -\cos(x) + \frac{1}{2}\cos(2x) \right]_{\pi / 3}^{\pi} $$
$$ = \frac{5}{2} . $$

Then, the area in the region $[0, \pi] \times [-1, 1]$ outside of $R$ is $ 2\pi - 5/2$.


问:
A mathematical expression that gives 0 if variable is different from constant and 1 if variable same as constant
Example,
∆ = any mathematical expression;
c = constant;
x = variable;

c∆x=1 (if c = x)
c∆x=0 (if c ≠ x)

am just asking?

答: You can use an Iverson bracket and write $[\Delta=c]$, though you should explain it the first time that you use it.


问:
When throwing a die $3$ times, what is the probability of the product of the three results being a multiple of $9$?
When throwing a die $3$ times, find the probability of the product of the three results being a multiple of $9$.

I tried drawing a table for the possible outcomes of the first TWO dice. Then I tried considering the cases for each when the third die is rolled - as you can imagine not very efficient. Perhaps I could use $x+y+z = 9k$ as a divisibility test, where $k$ is an integer. I'm pretty sure this would involve stars and bars, but I'm not sure how to implement it in this case. Would I have to solve for $k$?

I also worked out the boundaries but I'm not sure if that will help either: $9≤x+y+z≤216$.

Do I need to use stars and bars for this question?

答:
I do not agree with this solution.

I obtained the result $\frac{7}{27}$

In fact, the probability to have a product divisible by 9 means that, among 3 die's rolls, at least 2 must be #$3$ or #$6$, thus

$$\binom{3}{2}\left(\frac{1}{3}\right)^2\cdot\frac{2}{3}+\left(\frac{1}{3}\right)^3=\frac{7}{27}$$


问:
There is a polynomial $g(x)\in\mathbb{R}[x_1,..., x_n]$ such that $f(X) = g(a_0,..., a_{n−1})$ where $a_i$ is the coefficient of $λ^i$
Let $f: \mathfrak{gl}(n,\mathbb{R})\to \mathbb{R}$ be a polynomial such that $f(AXA^{-1})=f(X)$ for all $A\in GL(n,\mathbb{R})$. Prove that there is a polynomial $g(x)\in\mathbb{R}[x_1,..., x_n]$ such that $f(X) = g(a_0, . . . , a_{n−1})$ where $a_i$ is the coefficient of $λ^i$ in the characteristic polynomial of $X$.

I have been thinking about this problem for a long time and I still can't think of anything to do, I would like someone to give me a suggestion or tell me how I can easily solve this problem.

答:
Assuming that $f$ is a polynomial mapping, this statement is closely related to the theory of invariant polynomials. If instead of $\mathfrak{gl}_n$ we were looking at a (split) semisimple $\mathfrak g$, this would be a classical description of the $\mathfrak g$-invariant elements of the symmetric algebra $S(\mathfrak g)$ often credited to Chevalley (cf. MO 182871) which further would be shown to be the elements of the symmetric algebra $S(\mathfrak h)$ invariant under the Weyl group: and in the case of $\mathfrak{sl}_n$, this are the polynomials in $n-1$ variables which are invariant under permutation of the variables, i.e. the symmetric polynomials, i.e. what comes out here modulo the centre. (And Harish-Chandra linked that further to the centre of the universal enveloping algebra of $\mathfrak g$, but that's not important right now.)

But here's a handwavy proof for the case at hand. First notice that via conjugation it suffices to see how $f$ looks like on matrices in Jordan normal form. Then notice that e.g. the Jordan block

$\pmatrix{\lambda &1\\0&\lambda}$ via conjugation with $A:= \pmatrix{\varepsilon &0\\0&1}$ is brought to $\pmatrix{\lambda &\varepsilon\\0&\lambda}$, so by letting $\varepsilon \to 0$ via continuity (polynomial mapping!) we can further restrict to defining $f$ on diagonal matrices. (There's a complication here because over $\mathbb R$, we cannot even assume that every eigenvalue $\lambda \in \mathbb R$, so there are some technicalities to be filled in. However, one can prove quite generally for any Jordan-Chevalley decomposition $s+n$ that polynomial mappings are $0$ on the nilpotent part $n$.)

So now we assume our polynomial mapping $f$ is defined on the diagonal matrices, i.e. it is basically a polynomial in $n$ variables (one for each entry of the diagonal). But via conjugation with permutation matrices, the entries get permuted, i.e. our polynomial has to be invariant under all permutations of the $n$ variables. (Note that this is exactly the same as looking at $S(\mathfrak h)^W$, with $\mathfrak h$ here being the diagonal matrices, as in the general theory.) By definition, these are the symmetric polynomials, and it's known at least since Newton that these are all polynomials in the elementary symmetric polynomials, which evaluated on the eigenvalues of our matrix are also just the coefficients occuring in the characteristic polynomial.


问:
Continuous surjection between $\mathbb R^3 \setminus \mathbb S^2$ and $\mathbb R^2 \setminus \{0\}$
In my class, teacher says that there is a continuous surjection between $\mathbb R^3 \setminus \mathbb S^2$ and $\mathbb R^2 \setminus \{0\}$. I tried to find an example at home. I tried taking center of sphere at origin and then joining the lines any point on $\mathbb R^3\setminus \mathbb S^2$ and south pole $(0,0,-1)$ and taking the point at which the line cuts $\mathbb R^2$ but I stuck at $z$-axis as all points on $z$-axis goes to origin. Please help me. Thanks in advance.

答:
Notice that $\mathbb{R}^3\backslash S^2$ is not connected: there is the "inside" $B$ of the sphere and the "outside" $X$. We handle them separately.

First, $X\approx Z:=\mathbb{R}^3\backslash0$ by essentially a deformation retraction $f$. Let $Y=\{v\in X:\|v\|\le2\}$. For $v\in Y$, let $f(v)=(\|v\|-1)v$. This amounts to "stretching" $Y$ so that its distance-2 boundary stays fixed and its distance-1 boundary is drawn to the origin. For $v\in X\backslash Y$, let $f(v)=v$. These agree on the mutual boundary of $X$ and $X\backslash Y$, so they join continuously and surject onto $Z$.

The circle group $S^1\cong\mathbb{R}/2\pi\mathbb{Z}$ acts on $Z$ by rotating about the $z$-axis. Quotient by this action; this is a continuous map onto a space homeomorphic to the upper-half-(complex )plane $\mathbb{H}$ union the nonzero reals (notably, excluding 0). Symbolically, this is a continuous map from $X$ to $\mathbb{H}\sqcup(\mathbb{R}\backslash0)$.

Notice that all we have left to account for is the lower-half-plane (that is, excluding all of $\mathbb{R}$)! For $(a,b,c)\in B$, project onto the open unit disk $\mathbb{D}\subset\mathbb{C}$ via $(a,b,c)\mapsto a+bi$. Now, consider the (negative of the) standard conformal map (a biholomorphism) $-g:\mathbb{D}\to\mathbb{H}$ sending $z\mapsto-\frac{z-i}{z+i}$. This makes for a surjection from $B$ to $-\mathbb{H}$.

Combining these maps gives image $\mathbb{H}\sqcup(\mathbb{R}\backslash0)\sqcup-\mathbb{H}=\mathbb{C}\backslash0$.

To double-check continuity, open sets in the image are either entirely in one half-plane, or contain part of the real line.

In the former case, if the half-plane is upper then the preimage is just the original set revolved about the $z$-axis, and if the half-plane is lower then the preimage is the preimage under a biholomorphism (hence a bicontinuous map), then the preimage under a projection.

In the latter case, split the open set into that strictly below the real axis, and that weakly above it. The part above revolves about the $z$-axis, containing part of the axis, ensuring openness. The part below is still just the preimage of an open set under a bicontinuous map, then the preimage under a projection.

EDIT: Here's a "cleaner" solution (doesn't involve any conformal maps, or use $B$ at all). We can actually show that the result is true for $\mathbb{R}^3\backslash(S^2\sqcup B)$ with not much more work. Repeat the above solution up until we arrive at the surjection from $X$ to $\mathbb{H}\sqcup(\mathbb{R}\backslash0)$. Take this and identify the positive and negative reals, that is, $x$ and $-x$ for $x\in\mathbb{R}^+$. Quotienting gives a space homeomorphic to $\mathbb{C}\backslash0$. Now map $B$ via your favorite continuous map (easiest being any constant map).