问:
Collatz Conjecture: If a non-trivial cycle exists, would it follow that $2^{K+L} - 3^K < (10^{-5})3^K$
Does it follow from Lemma 8 in the paper by Simons and de Weger (2005) that if a non-trivial cycle exists:
$$2^{K+L} - 3^K < (10^{-5})3^K$$

Let:

$x\ge 1$ be an integer

$T(x) = \begin{cases}
\frac{1}{2}(3x + 1), && \text{if }x\text{ is odd}\\
\frac{1}{2}x, && \text{if }x\text{ is even}\\
\end{cases}$

$T_1(x) = T(x)$

$T_{c+1}(x) = T(T_c(x))$

$\delta = \dfrac{\log 3}{\log 2}$

A non-trivial cycle is any case where $x > 2$ and there exists integer $c > 0$ where $T_c(x) = x$

$[x_1, x_2, \dots, x_{c-1}]$ be the distinct numbers which make up a non-trivial cycle with:

$K = $ the number of odd integers

$L = $ the number of even integers

$K+L = c-1$

Here's my thinking:

(1)  From Lemma 8 on page 9:

$$K + L < 1.000001\delta K$$

(2)  $2^{K+L} - 3^K < 2^{1.000001\delta K} - 2^{\delta K} = 2^{\delta K}\left(2^{.000001\delta K}  - 1\right) < 3^K(3^{.000001K} - 1) < 3^K(1.00001 - 1) = (10^{-5})3^K$

Am I understanding the implications of the lemma correctly?  Did I make a mistake?

Update:  The last part is wrong.  For larger $K$, it is incorrect.

Body:
You could use this instead: If a non-trivial m-cycle exists,

$$K+L-K\log_23<\frac{m}{X_0\log2}$$
The current $X_0=2,95...10^{20}$, so
$$(K+L)\log2-K\log3<\frac{m}{2,95\cdot 10^{20}}$$
$$\log \frac{2^{K+L}}{3^K}<\frac{m}{2,95\cdot 10^{20}}$$
$$\frac{2^{K+L}}{3^K}-1<e^\frac{m}{2,95\cdot 10^{20}}-1$$
or
$$2^{K+L}-3^K<3^K(e^\frac{m}{2,95\cdot 10^{20}}-1)$$


问:
Calculate the Flux through S where $\vec F=y\vec j-z\vec k, \text{enclosed by}\quad y=x^2+z^2, \space y\in[0,1], \quad x^2+z^2 =1$
I am struggling with Flux integrals and would like someone to check what I am not doing right.

I am to evaluate

$$\int\int_S \vec F \cdot d\vec S $$
where $\vec F=y\vec j-z\vec k, \text{enclosed by}\space S : y=x^2+z^2, \space y\in[0,1], \quad x^2+z^2 =1$

So I am thinking that I can use

$$\int\int_S \vec F \cdot (\vec r_x \times\vec r_z)d S $$

where the parametarization of $S$ is obtained by
$$x=x, y=x^2+z^2, z=z \Rightarrow \vec r(x,z)=x\vec i+(x^2+z^2)\vec j +z\vec k $$

I believe that this leads to

$$\vec F \cdot (\vec r_x \times\vec r_z) = -y-2z^2$$

and all I have to do is evaluate the double integral

$$-\int \int _D y+2z^2 dA$$

Using the cylindrical method I have

$$-\int_0^{2\pi} \int_0^{1} r^3(1+2\cos^2\theta) dr d\theta = -\pi $$

but the answer is supposedly zero.

Either I made some calculation mistake or there is something that I fundamentally got wrong . . . may I have some assistance?

答:
$\vec F=y\vec j-z\vec k, \text{enclosed by} \ y=x^2+z^2, \ x^2+z^2 =1, \ y\in[0,1]$

It is easier to see this by divergence theorem.

The surface is bound by paraboloid $y = x^2+z^2$ and cylinder $x^2+z^2 = 1$ for $y \in (0,1)$. As the surface is open at $y=0$, we first close the surface by adding a disc at $y = 0$ as applying divergence theorem requires closed surface. We can separately find flux through the disc and subtract it later (which by the way is also zero but that is for later). Now applying divergence theorem,

$div{\vec{F}} = \nabla \cdot \vec{F} = \frac{\partial{(y)}}{{\partial y}} + \frac{\partial{(-z)}}{{\partial z}} = 0$

As the divergence of the vector field is $0$, the flux through the closed surface is zero. But we now need to subtract the flux through the disc at $y = 0$ which we added and is not part of the original surface.

The outward unit normal vector to the disc is $(0,-1,0)$ so $\vec{F} \cdot \hat{n} = -y \ $ but as $y = 0$, the flux through this surface is also zero.

So the surface integral is zero.

EDIT:
On your working,

First visualizing the region between $y=0$ and $y=1$, its inner surface is given by the equation of the paraboloid and outer surface is given by the equation of the cylinder. Now when you are finding the flux through the paraboloid surface, please note that out of surface here would mean pointing generally upward in $y$ direction. So while your working seems fine, you have taken normal vector pointing in downward $y$ direction which is not correct for this question.

If I parametrize it in cylindrical coordinates,

$r(\rho, \theta) = (\rho \cos\theta, \rho^2, \rho\sin\theta)$

$r'_{\rho} = (\cos\theta, 2\rho, \sin\theta)$

$r'_{\theta} = (-\rho\sin\theta, 0, \rho\cos\theta)$

$r'_{\theta} \times r'_{\rho} = (-2\rho^2\cos\theta, \rho, - 2\rho^2\sin\theta)$ (keeping $y$ component positive)

$I_1 = \displaystyle \int_0^{2\pi}\int_0^1 (0, \rho^2, -\rho\sin\theta) \cdot (-2\rho^2\cos\theta, \rho, - 2\rho^2\sin\theta) \ d\rho \ d\theta = \pi$

Now parametrizing cylindrical surface as

$r(y, \theta) = (\cos\theta, y, \sin\theta)$

you will get $r'_{y} \times r'_{\theta} = (\cos\theta, 0, \sin\theta)$

So $I_2 = \displaystyle \int_0^{2\pi}\int_0^1 (0, y, -\sin\theta) \cdot (\cos\theta, 0, \sin\theta) \ dy \ d\theta = - \pi$

$I_1 + I_2 = 0$.

ContentLicense: CC BY-SA 4.0


问:
Brownian motion exercise
Given $Z(t)=B(t)Y$, with $B(t)$ standard brownian motion, also $t \in (0,1)$ and $Y \sim N(0,1)$. $B(t)$ and $Y$ are independent.
Compute $E[Z(t)]$ , $Var[Z(t)]$ and $Cov[Z(t),Z(s)]$ .
Prove also that $Z(t)$ is a brownian motion.
It's easy to obtain
$$
E[Z(t)]=0 \ \ \ \ \ \ \ \ Var[Z(t)]=t
$$
Then I suppose that $t>s$ and compute

$$
Cov[Z(t),Z(s)]=E\big\{[B(t)Y][B(s)Y]\big\} =\\
=E\big\{Y^2B(s)[B(t)-B(s)]+Y^2B^2(s) \big\} = \\
E[Y^2B^2(s)]=s
$$
For a generic $s$ we get

$$
Cov[Z(t),Z(s)]=min\{t,s\}
$$

Here comes the doubt. To prove that Z(t) is also a brownian motion I have to prove the independence of the increments.
For $t>s$
$$
E[Z(t)-Z(s)]=0 \\
Var[Z(t)-Z(s)]=s
$$

To prove that generic increments are mutually independent is sufficient to prove that $Cov[Z(t)-Z(s),Z(s)-Z(0)]=0$?
Thank you for the answer. :)

答:
You computed $\mathbb{E}(Z(t))$ , $Var(Z(t))$ and $Cov(Z(t),Z(s))$ correctly.

But what you've been asked to prove is not true, i.e. $Z$ is not a Brownian motion.
To see this, just note that $Z(t)$ is the product of two Normal random variables, so it cannot be normal.

Title: Brownian motion exercise


问:
Convergence of: $\sum_{n=1}^{\infty} \frac{3^n \cdot n^n}{(n!)^2}$
Convergence of:

$$\sum_{n=1}^{\infty} \frac{3^n \cdot n^n}{(n!)^2}$$

Tried ratio test but somehow it doesn't work for me.

答:
$$a_n=\frac{3^n \, n^n}{(n!)^2}\implies \log(a_n)=n\log(3)+n \log(n)-2\log(n!)$$ Use Stirling approximation
$$\log(a_n)=n (-\log (n)+2+\log (3))-\log (2 \pi  n)-\frac{1}{6
   n}+O\left(\frac{1}{n^3}\right)$$ Use it twice and continue with Taylor series
$$\log(a_{n+1})-\log(a_n)=(-\log (n)+1+\log (3))-\frac{3}{2 n}+\frac{5}{6
   n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1} } {a_n}=e^{ \log(a_{n+1})-\log(a_n)}=\frac{3 e}{n}-\frac{9 e}{2 n^2}+O\left(\frac{1}{n^3}\right)$$


问:
How to show Lipschitz gradient of Norm doesn't exist
How can I show that there is no Lipschitz gradient $\nabla f$ for any $C < \infty$ for the function $ f: \mathbb{R}^2 \rightarrow \mathbb{R}$

$$f(\text{x}) = {\Vert\text{x}\Vert}^{3/2}$$

Updated with gradient calculation below:

$$f(\text{x}) = ((\sum_{k=1}^2 x_k^2)^{1/2})^{3/2} =(\sum_{k=1}^2 x_k^2)^{3/4}$$

$$\begin{aligned} \nabla_j f(\text{x}) &= \partial/\partial x_j(\sum_{k=1}^2 x_k^2)^{3/4}\\
&= \frac{3}{4} (\sum_{k=1}^2 x_k^2)^{-1/4} (2x_j)\\
&=\frac{3}{2} x_j (\sum_{k=1}^2 x_k^2)^{-1/4}=\frac{3}{2}\frac{x_j}{\sqrt{\Vert x \Vert}}
\end{aligned}$$

Updated to add definition of Lipschitz Gradient for a continuously differentiable function $f$, where $ L \in [0, \infty)$

$$ \Vert\nabla f(\text{x}) - \nabla f(\text{y})\Vert \leq L \Vert \text{x} - \text{y} \Vert$$

答:
So you have

$$\nabla_j f(x)=\frac{3}{2}\frac{x_j}{\sqrt{\Vert x \Vert}}.$$

Taking $x=(t,0)$ with $t \gt 0$ you get

$$\nabla f(x)=\frac{3}{2}(\sqrt t, 0)$$ and the desired result as $t \mapsto \sqrt t$ is not Lipschitz. Last statement is a consequence of $\lim\limits_{t \to 0^+} \frac{\sqrt t }{t} = \infty$.


问:
Unity in the rings of matrices
Suppose we are given an arbitrary ring $R$. Then the set $M_n(R)$ of all square matrices  with elements from $R$, together with usual matrix addition and multiplication forms a ring. If R is a unitary ring then $M_n(R)$ too.

My question seems to be very trivial but is it possible that $M_n(R)$ has unity while $R$ hasn't?

Thanks in advance.

答:
Let
$$
\begin{bmatrix}a & b\\c& d\end{bmatrix}
$$
be the identity in $M_2(R)$. Then, for any $r$ in $R$, we have
$$
\begin{bmatrix}r & 0\\0& 0\end{bmatrix}=
\begin{bmatrix}r & 0\\0& 0\end{bmatrix}
\begin{bmatrix}a & b\\c& d\end{bmatrix}=
\begin{bmatrix}ra & rb\\0& 0\end{bmatrix}
$$
so $ra=r$, for all $r\in R$, and $a$ is a right unity in $R$. Similarly
$$
\begin{bmatrix}r & 0\\0& 0\end{bmatrix}=
\begin{bmatrix}a & b\\c& d\end{bmatrix}
\begin{bmatrix}r & 0\\0& 0\end{bmatrix}=
\begin{bmatrix}ar & 0\\cr& 0\end{bmatrix}
$$
so $a$ is also a left unity.

The example generalizes easily to any matrix size.

ContentLicense: CC BY-SA 3.0

Title: Unity in the rings of matrices


问:
Find a closed form to the solution of $\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-x}}}}}=x$
Hi I try to solve the following nested radical :

$$\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-x}}}}}=x$$

Miraculously the related polynomials is a quintic .More precisely :

$$ x^5 - x^4 - 4 x^3 + 3 x^2 + 3 x - 1=0$$

I know that we can reduce the quintic to a Bring quintic form and use Jacobi theta function .

My question :

Can we hope to see a closed form with radicals ?

Any helps is greatly appreciated

Thanks a lot for all your contributions.

答:
One of the solution that can be obtained easily is by substituting $x = 2\cos\theta$

steps as follows

$\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-x}}}} = x$ will become $\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-2\cos\theta}}}} = 2\cos\theta$

Now by applying Half angle cosine formula we can simplify as follows
$\sqrt{2+\sqrt{2+\sqrt{2-2\sin\frac{\theta}{2}}}} = 2\cos\theta$

$\sqrt{2+\sqrt{2+\sqrt{2-2\cos(\frac{\pi}{2} - \frac{\theta}{2})}}} = 2\cos\theta$

$\sqrt{2+\sqrt{2+2\sin(\frac{\pi}{4} - \frac{\theta}{4})}} = 2\cos\theta$

$\sqrt{2+\sqrt{2+2\cos(\frac{\pi}{2} -\frac{\pi}{4} + \frac{\theta}{4})}} = 2\cos\theta$

$\sqrt{2+2\cos(\frac{\pi}{4} -\frac{\pi}{8} + \frac{\theta}{8})} = 2\cos\theta$

$2\cos(\frac{\pi}{8} -\frac{\pi}{16} + \frac{\theta}{16}) = 2\cos\theta$

Now solving for $\theta$ will give rise to $\frac{\pi}{16} + \frac{\theta}{16} = \theta$

$\frac{16\theta}{16}- \frac{\theta}{16} = \frac{\pi}{16} $

$\theta = \frac{\pi}{15}$. Therefore $x$ becomes $2\cos\frac{\pi}{15}$ or $2\cos12^\circ$

As this is multiple of 3, we can calculate value of $2\cos12^\circ$ which we can represent as a finite nested radical as $\frac{1}{2}\times\sqrt{9+\sqrt5+\sqrt{(30-6\sqrt5)}}$

Refer here


问:
A is a matrix of integers , prove that A+I is invertible
Question:

$2 \le d \in \Bbb Z$

Let $A \in M_n(\Bbb Q) s.t$ All of it's elements are integers divisible by d.

Prove that $I+A$ is invertible.

What I thought:

I thought of using the determinant of A and extracting $d^2$ from it, but I can't use that on det(A+I)... tried also to calculate the determinant for n=2, and got a polynomial of 2nd degree (With no idea on how to prove that it doesn't equal 0).

答:
Hint: Can $-1$ be an eigenvalue for $A$?

Alternatively: What is the determinant of $A+I$ modulo $d$?


问:
Decomposing the element of a group with elements with coprime order
I have the following exercise of group theory, it is supposed to be easy but I don't know how to show the existence of the elements that the problem is asking for. Can you give me some hints?

If $o(g)$ is the order of the element $g$, consider: $o(a) = mn$ with $\gcd (m, n) = 1$, show that $a = gh = hg$ for some $g, h \in G$ with $o(g) = m$ and $o(h) = n$.

答:
If $o(g)$ is the order of the element $g$, consider: $o(a) = mn$ with $\gcd (m, n) = 1$, show that $a = gh = hg$ for some $g, h \in G$ with $o(g) = m$ and $o(h) = n$.

By Euclid's algorithm there are integers $s,t$ such that $$sm+tn=1.$$
Then $a=a^1=...$ Can you finish the proof now?


问:
What can replace "dimension $0$" to measure discreteness of varieties over $ \mathbb{R} $?
Consider the algebraic set $Z(x^2 + y^2) \subseteq \mathbb{R}[x, y]$. Set $A = \mathbb{R}[x, y]/(x^2 + y^2)$. In this ring, we have a chain of prime ideals $(0) \subseteq (x,y) $: the ring has dimension $1$. Hence the algebraic set has dimension $1$ when viewed as the (closed points of the) scheme $ \mathrm{Spec}\ A$. By Definition 2.8.1 on p.50 of Bochnak/Coste/Roy, this is also the dimension of the set as a real algebraic variety.

Question. As a scheme, the algebraic set has dimension 1. On the other hand, it is a real algebraic variety consisting of a single real closed point. Is there an algebraic way of defining a `dimension' so that this real algebraic variety has dimension $0$?

I fully expect the answer might be 'no', as the polynomial we look at here inherently has 'dimension 1-ness' built into it intrinsically (it is a conic) and I think we get dim = 1 here because we are in some sense detecting an infinitsimal tangent vector pointing in the imaginary direction(?).

Some context: I am studying a large family of complicated polynomials defined over $\mathbb{Q}$ and I am trying to detect whether given polynomials in the family have discrete zeros over $\mathbb{R}^n$; to do this, I was using Macaulay2 to compute the dimension of the hypersurface generated by each polynomial and looking for polynomials where this quantity was zero. Of course, this did not work as there were many polynomials with discrete zeros but generating an ideal with positive Krull dimension! I chose the example above as it is the simplest polynomial I could think of which has this "defect" in the hopes that the answers would be enlightening when dealing with the more complicated case.

Related: Two-variable polynomials over $\mathbb{Q}$ with finitely many roots in $\mathbb{R}^2$ (but I am interested in checking for discreteness of roots, not finiteness; I am also interested in studying the embedding of the real points of the varieties inside the complex points, so I'm more interested in algebraic results as opposed to the analytic ones in the answers to that question)

答:
Let $I:=(f_1,..,f_l)$ where $f_i\in A:=\mathbb{R}[x_1,..,x_n]$ let $ X:=\mathrm{Spec}(A/I)$. Let $F:=f_1^2+\cdots + f_l^2$ and let $Y:=\mathrm{Spec}(A/(F))$.
It follows $Y$ is a "hypersurface" in real affine space $\mathbb{A}^n_\mathbb{R}:=\mathrm{Spec}(A)$.

Lemma. Since $(F) \subseteq I$ there is a canonical map of affine schemes $\phi: X \rightarrow Y$ inducing a 1-1 correspondence at the level of $\mathbb{R}$-rational points:

$$ X(\mathbb{R}) \cong Y(\mathbb{R}).$$

Proof:Let $f^*: Spec(\mathbb{R}) \rightarrow X$ be an $\mathbb{R}$-rational point. It follows the corresponding map of $\mathbb{R}$-algebras $f$ satisfies $f(x_i):=a_i\in \mathbb{R}$ with $f_j(a_1,..,a_n)=0$ for all $j=1,..,l$. Hence it follows $F(a_1,..,a_n)=0$, and $f$ defines an $\mathbb{R}$-rational point of $Y$. Conversely if $f^*: Spec(\mathbb{R}) \rightarrow Y$ is an $\mathbb{R}$-rational point, it follows $f(x_i):=a_i$ and $F(a_1,..,a_n):= \sum_j f_j(a_1,..,a_n)^2=0$, hence $f_j(a_1,..,a_n)=0$ for all $j$. Hence $f$ defines an $\mathbb{R}$-rational point of $X$. This is a 1-1 correspondence and the Lemma is proved. QED.

Note 1. By the Lemma it follows the two schemes $X,Y$ have the same $\mathbb{R}$-rational points. A hypersurface in affine $n$-space should have dimension $n-1$, but the ideal $I$ is an arbitrary ideal in $A$. Hence if you are studying the $\mathbb{R}$-rational points $X(\mathbb{R})$, you may realize $X(\mathbb{R})$ as a hypersurface for any $X$.

Note 2. The map $\phi: X \rightarrow Y$ is a closed immersion, since $X$ may be defined using an ideal $\overline{I}\subseteq A/(F)$.

Note 3. This type of result holds for any subfield $\mathbb{Q} \subseteq K \subseteq \mathbb{R}$.

Example: If $I:=(x_1-a_1,..,x_n-a_n)$ is a maximal ideal, it follows $X:=\mathrm{Spec}(A/I)$ has one $\mathbb{R}$-rational point: The point $a:=(a_1,..,a_n)$. If $F:=(x_1-a_1)^2+\cdots + (x_n-a_n)^2$
it follows $Y:=\mathrm{Spec}(A/(F))$ is a hyper surface in affine $n$-space.

Lemma. $Y$ has $a$ as the only $\mathbb{R}$-rational point.

Proof. Let us repeat the proof: Assume $\phi:\mathrm{Spec}(\mathbb{R}) \rightarrow Y$ is a $\mathbb{R}$-rational point. we get a corresponding map of $\mathbb{R}$-algebras

$$ f: A/(F) \rightarrow \mathbb{R}$$

with $f(x_i)=u_i \in \mathbb{R}$. Since $f(F)=0$ we get

$$ f(F):=f(u_1,..,u_n):=(u_1-a_1)^2+\cdots + (u_n-a_n)^2=0$$

hence $u_i=a_i$ for all $i$, and it follows $(u_i)=a$ is the only $\mathbb{R}$-rational point. QED

Example 1. Let $D \subseteq \mathbb{R}$ be a sub-ring and let $I:=\{f_1,..,f_l\} \subseteq B:=D[x_0,..,x_n]$ be an ideal generated by a set of homogeneous polynomials of the same degree $deg(f_i)=d$. Let $A:=V(I)\subseteq \mathbb{P}^n_D$ be the projective scheme defined by $I$. Let $F:= \sum_j f_j^2 \in A$ and let $H:=V(I)$. It follows $A(D) \cong H(D)$.

In your case $a:=(0,0)$ has maximal ideal $I=(x,y)$ and corresponding hypersurface $F:=x^2+y^2$. The ring $B:=\mathbb{R}[x,y]/(x^2+y^2)$ has one maximal ideal with $\mathbb{R}$ as residue field. It has many maximal ideals with with $\mathbb{C}$ as residue field:

"Question. As a scheme, the algebraic set has dimension 1. On the other hand, it is a real algebraic variety consisting of a single real closed point. Is there an algebraic way of defining a `dimension' so that this real algebraic variety has dimension 0?"

For any non-zero complex number $w \in \mathbb{C}$ it follows the map $f: B\rightarrow \mathbb{C}$ defined by
$f(x)=iw, f(y)=w$ is a well defined surjection with $I(w):=\mathrm{ker}(f) \subseteq B$ a maximal ideal. Hence the ring $B$ has many maximal ideals with $\mathbb{C}$ as residue field.

What you are asking for is a formula detecting how many rational points a real algebraic variety $X$ has, and I believe this is an open problem in diophantine geometry in general.


问:
Convergence of Sum of a series
$$\frac{(1+r)^{N+1}-(1+r)-rN}{r^2(1+r)^N}$$
I'm calculating the sum of a special series and finally get a general formula for the series. $r$ is a constant in the formula. And I'm curious whether this series will be convergent or not. Could someone explain how to calculate the sum when $N$ is towards infinity? Thank you very much.

答:
Welcome to MSE!

If you write $\alpha = 1+r$ to simplify notation you get

$$
\frac{1}{(\alpha-1)^2}
\left (
\alpha - \frac{\alpha}{\alpha^n} - \frac{\alpha n}{\alpha^n} + \frac{n}{\alpha^n}
\right )
$$

But we know that

$\frac{\alpha}{\alpha^n} \to 0$ when $|\alpha| > 1$

$\frac{\alpha n}{\alpha^n} \to 0$ when $|\alpha| > 1$

$\frac{n}{\alpha^n} \to 0$ when $|\alpha| > 1$

This is basically because $\alpha^n$ grows exponentially in $n$, which dominates any linear (or constant) growth. You can use L'hospital if you want to verify this yourself.

Then, applying these limits, we see that your limit converges if and only if $|\alpha| > 1$, and then it converges to

$$\frac{\alpha}{(\alpha - 1)^2}$$

I'll leave it to you to convert this back into a formula in terms of $r$.

If you like, you can code this up quite quickly in
sage and convince yourself that this formula is correct. Checking with $n \approx 500$ should be good enough to see the convergence for most values of $\alpha$.

I hope this helps ^_^

Title: Convergence of Sum of a series


问:
Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $\frac{a+b+c+ab+ac+bc}{1+abc}$ is a real number.

Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $$x = \frac{a+b+c+ab+ac+bc}{1+abc}$$ is a real number.

I wanted to calculate $2 \cdot Im(x) = x- \overline x$ and show that it's equal to zero:

$$\frac{a+b+c+ab+ac+bc}{1+abc} - \frac{\overline{a}+\overline{b}+\overline{c}+\overline{ab}+\overline{ac}+\overline{bc}}{1+\overline{abc}} =$$ $$=a+b+c+ab+ac+bc-\overline{a}-\overline{b}-\overline{c}-\overline{ab}-\overline{ac}-\overline{bc} = $$ $$=2(Im(a) + Im(b)+Im(c)+Im(ab)+Im(ac)+Im(bc))=$$$$=2(Im(a+b+c+ab+bc+ac))$$But I have no idea what to do next or what am I missing.

答:
Consider $(a+b+c+ab+ac+bc)(1+\overline {abc})$.

When you multiply out remember that $a\overline {a}=1$.


问:
A Geometric Description of Injective Modules
I've found that when studying commutative algebra, thinking of things in terms of their algebro-geometric interpretation helps them stick as well as motivates otherwise odd and abstract concepts.

(For example, thinking of localization as referring to open sets, completions to infinitesimal neighborhoods, primary decomposition to irreducible components, etc.)

I'd like to do the same for concepts from homological algebra. I'm dimly aware that projective modules correspond geometrically to vector bundles (i.e. to the condition of being locally free) at least in nice cases, and flatness seems to have a deeply geometric interpretation as meaning something difficult to describe about "smoothly varying families". Is there a similarly geometric way to think about injective modules? If not, is there some intuition to why dualizing the concept of a projective module breaks the geometric correspondence?

If you're aware of other nice correspondences between geometric ideas and homological algebra constructions, I'd also love to hear them

答:
The analogous of a module $M$ over a ring $A$ is a (quasi coherent) sheaf of modules over a scheme $X$. Indeed, in the case $X=\operatorname{Spec} A$ is an affine variety, the two notions coincide. When

First point: how one can think of a sheaf $F$ over $X$? Well, you can take the associated étalé space and think of it as a space $p: E \to X$, such that

$$F(U) \simeq \{e \in E: p(e) \in U\}$$

Note that sections are $R$-modules in this case. In other words, you can always think a sheaf as a sheaf of sections. A cool way to think about sections are a sort of "generalized functions" that can take more than one value. For example, take the function $x \mapsto x^2$ over $\mathbb{C}$: the local sections in a neighborhood of 1 are exactly the two functions $\sqrt{x}, -\sqrt{x}$.

Another set of geometric examples of sheaf - despite less general and less common in algebraic geometry - is the sheaf of solutions to a given constraint (as a differential equation) : locally it can have a lot of solutions, but it is not necessarily true that they glue together to a global solution. This point of view is useful because is more common to have an action of a ring $R$ on the space of solutions. For example, if they have some symmetry under the action of a group and stable for multiplication by a real constant, the ring $\mathbb{R}[G]$ is acting on the space of solutions; however in this situation it is much more natural to have a ""continous"" group of symmetries (like rotations) , and one considers the space of solutions as a module over the associated lie algebra.

Note that the example of "generalized functions" above is a special case of the "solutions to a constraint" case: sections of the function $x^2 \mapsto x$ are exactly the complex functions that satisfy $f(x)^2 = x$.

Second point: what does it mean to be an injective sheaf? The most "neat" consequence is that its cohomology groups are all zero (except the zero-th one which coincide with the module).

Intuitively, I think as "having zero cohomology" like being without holes of any dimension. A concrete reincarnation of this intuition is the fact that an injective sheaf is "flasque", or " flabby": this means that any local section extends to a global section (read: any local solution extends tk a global solution!).

Now we see a completely different approach, which is equivalent to be injective but not pictorial as the previous.

Recall the criterion for which is enough to check that every morphism $I \to M$ extends to a morphism $R \to M$, where $I$ is an ideal of $\mathbb{R}$. Suppose the ring is noetherian, so that the ideal $I$ is generated by a finite number of elements $f_1, \ldots, f_n$. This means we have an exact sequence

$$ 0 \to K \to R^n \to I \to 0 $$

And that a map from $I$ to $M$ is the same as choosing n elements $m_1, \ldots, m_n\in M $ satisfying the relations prescribed by $K$. Finding an extension to $R$ means to find an element $m \in M$ such that $f_j m = m_j$. Let us look at an easy example: if you take an ideal generated by one element $(f) $, for any $m_1 \in M$ you can find an element $m$ such that $fm = m_1$. In other words, you have "" $ m =m_1/f$, that is the module is divisible. For PID like $\mathbb{Z}$ this is equivalent to be injective. In general, you can always find a "common divisor" given that the elements satisfy the obvious relations they should satisfy. If for example $R= \mathbb{C}[x, y]$ and you want an element $m$ such that $xm = m_1, ym= m_2$, it's necessary that $m_1, m_2$ satisfy $ym_1 = xym = xm_2$.

In terms of sheaves, this is not much more significant. Given an ideal $I=(f_1, \ldots, f_r) $, it means that if some global sections $m_1, \ldots, m_r$ satisfies the relations analogous to the relations satisfied by $f_1, \ldots, f_r$, you can always find a global section $m$ such that $f_j m = m_j$.


问:
Orthogonal on an ellipse
Consider two points $a \in \mathbb{R}^2$ and $b \in \mathbb{R}^2$ on the boundary of an ellipse such that $a^\top \Sigma^{-1} a = b^\top \Sigma^{-1} b$. I am trying to understand which points on the boundary of this ellipse satisfy $a^\top \Sigma^{-1} b = 0$. One possibility for this to happen is when $a \perp v_1$ and $b \perp v_2$, where $v_1$ and $v_2$ are eigenvectors of $\Sigma$, then $a^\top \Sigma^{-1} b = 0$. It is also clear that if $a$ and $b$ are valid, then $(-a, b), (a, -b) \text{ and } (-a, -b)$ are also valid solutions. Is it possible to find other points on the boundary of the ellipse such $a^\top \Sigma^{-1} b = 0$ or show that no other points besides these four exist, where this condition is satisfied ?

答:
I assume that $\Sigma$ is symmetric postive definite. Let $R$ be the rotation $$R = \begin{pmatrix}0 & -1 \\[1 ex] 1 & 0 \end{pmatrix}.$$ Take any vector $b \neq 0$ and $$a = \Sigma ^{1/2} R \Sigma ^{-1/2} b.$$
Then $$a^{\mathrm t} \Sigma^{-1} a = b^{\mathrm t} \Sigma^{-1} b$$ and $$a^{\textrm t}\Sigma^{-1}b=0.$$

Title: Orthogonal on an ellipse


问:
Generalized Gronwall inequality for singular differential inequality
Consider a differential inequality of the form
$$
  y'(t) \leq K(t)y(t) + f(t), \: y(0) = a < \infty, \; t\in(0,1]
$$
Here $K(t)>0$ is decreasing function of $t$ such that
$$
  K(t) \leq C\frac{\vert\ln t\vert}{t}, C>0.
$$

Is there a generalization of Gronwall inequality to study this kind problems?

答:
Let $t_0\in (0,1]$ and set $\kappa (t)=\int_{t_0}^tK(u)\,\mathrm d u$, $t\geq t_0$. Then, for all $t\geq t_0$,

\begin{align*}
y'(t)\leq K(t)y(t)+f(t)&\iff y'(t)e^{-\kappa (t)}-K(t)y(t)e^{-\kappa (t)}\leq f(t)e^{-\kappa (t)}\\
&\iff y(t)e^{-\kappa (t)}\leq y(t_0) +\int_{t_0}^t f(u)e^{-\kappa (u)\,\mathrm d u}\\
&\iff y(t)\leq e^{\kappa (t)}y(t_0)+e^{\kappa (t)}\int_{t_0}^tf(u)e^{-\kappa (u)}.
\end{align*}


Title: Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $\frac{a+b+c+ab+ac+bc}{1+abc}$ is a real number.

Body:
Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $$x = \frac{a+b+c+ab+ac+bc}{1+abc}$$ is a real number.

I wanted to calculate $2 \cdot Im(x) = x- \overline x$ and show that it's equal to zero:

$$\frac{a+b+c+ab+ac+bc}{1+abc} - \frac{\overline{a}+\overline{b}+\overline{c}+\overline{ab}+\overline{ac}+\overline{bc}}{1+\overline{abc}} =$$ $$=a+b+c+ab+ac+bc-\overline{a}-\overline{b}-\overline{c}-\overline{ab}-\overline{ac}-\overline{bc} = $$ $$=2(Im(a) + Im(b)+Im(c)+Im(ab)+Im(ac)+Im(bc))=$$$$=2(Im(a+b+c+ab+bc+ac))$$But I have no idea what to do next or what am I missing.

Body:
$$x = \frac{(a+1)(b+1)(c+1)-(1+abc)}{1+abc} = \frac{(a+1)(b+1)(c+1)}{1+abc}-1$$

and using the property $a \bar{a} = |a|^2 = 1$ repeatedly:

$$\frac{(a+1)(b+1)(c+1)}{1+abc} = \frac{(1+1/a)(1+1/b)(1+1/c)}{1/abc+1} = \frac{(1+\bar{a})(1+\bar{b})(1+\bar{c})}{\overline{abc}+1}$$

so $x = \bar{x}$ (why?) and hence $x$ is real.


问:
Stokes theorem on calculating the integral
Using Stokes theorem calculate the integral:$\int_C x^2y^2 dx+dy+zdz$ where C is curve bounded with $x^2+y^2=2x, y\ge 0, z=1$ oriented from $A(2,0,1)$ to $B(0,0,1)$.
So using Stokes formula I should bring this down to solving surface integral od second kind.
After finding partial derivates od $P=x^2y^2, Q=1, R=z$, I need to calculate normal vector, and then scalar product of field F and normal n. But not sure how to use given orientation, what is it telling me?

答:
Vector field is $\vec{F} = (x^2y^2,1,z)$. The curve is bounded by $x^2+y^2=2x, y\ge 0, z=1$ and you are going from point $A(2,0,1)$ to $B(0,0,1)$.

You can rewrite the surface as $(x-1)^2+y^2 = 1$ which is a cylinder of radius $1$ with center at $(1,0,z)$. Now coming to the curve, it is a semicircle on this surface from point $A$ to point $B$ at $z=1$. Please note you must go anti-clockwise from $A$ to $B$ as $y \geq 0$.

You can parametrize your curve as

$r(t) = (2\cos^2 t, 2\cos t \sin t, 1) $ or $(1+\cos2t, \sin2t,1), 0 \leq t \leq \frac{\pi}{2}$.

$r'(t) = (-2\sin2t, 2\cos 2t,0)$

$\vec{F} \cdot r'(t) = (\sin^2 2t (1+\cos 2t)^2,1,1) \cdot (-2\sin2t, 2\cos2t,0)$

$ = -2\sin^32t(1+\cos 2t)^2 + 2\cos2t$

Line integral is $ \ \displaystyle \int_0^{\pi/2} \vec{F} \cdot r'(t) \ dt$


问:
Logical systems that are complete but not sound
I was wondering, are there any commonly used logics(with both notions of deductions and of semantics) that are complete but not sound? I'm looking for an example that has actually proven useful to logicians. Thank you for your wisdom!

Sincerely,

VIen

答:
For anyone that comes across this:

A good example in an exam would be the calculus consisting only of the rule:

$\emptyset \vdash_{R_{1}} F$

Any rule can be derived from it (making it complete), because F can be arbitrarily chosen. It is clearly not sound, since every formula can be derived from it (including false ones). It must follow for the calculus to be sound that every formula derived from some arbitrary set $\mathbf{M}$ under the rule $R_{1}$ must be a tautology:

$M \vdash_{K}F \implies M\models F$ (Definition of soundness for a calculus K, where M is some arbitrary set and F is some formula.)

In this case we have $K = \{R_{1}\}$. And therefore:

$\emptyset  \vdash_{R_{1}}F \implies \emptyset \models F \implies F \ is \ a \ tautology$

Since not all formulae derived from the rule are tautologies the calculus cannot be sound.


问:
Affine cone is a closed set
I have seen in the notes of algebraic geometry that if I have a reducible projective algebraic set, i.e given $X$ projective algebraic set, $X=X_1\cup X_2$, $X_i\subsetneqq X$, $X_i$ closed, then $C(X)=C(X_1)\cup C(X_2)$, $C(X_i)\subsetneqq C(X)$ closed. Why $C(X_i)$ is closed?

答: If $I$ is a homogenous ideal in $k[T_0,...,T_n]$ it gives a closed set of the projective space of dimension $n$, but you can forget that it is homogenous, and it will correspond to a closed set of the affine space of dimension $n+1$, this is the cone.

Title: Affine cone is a closed set


问:
How to find instances when $d(a,b) = p^2$ for $p$ a prime.
Suppose I have a dimension formula (for a Lie algebra representation) given by
$\mathrm{dim}_{a,b}  = {(a+1)(b+1)(a+b+2) \over 2}$. I now would like to find pairs $(a,b)$ where $\dim_{a,b} = p^2$ for $p$ a prime? What are some techniques for accomplishing this? Should I try coding it up in python?</p>

答:
Suppose that
$$(a+1)(b+1)(a+b+2)=2p^2,$$
with $p$ prime and $a,b \geq 0$.  Let's take $a \geq b$ without loss of generality.  If $a,b \geq 1$, then each term of the left-hand side above is bounded below by $2$.  Thus each contains a prime, and we may match up factors in descending pairs
$$(a+b+2)=p,\quad a+1=p, \quad b+1=2.$$
This contradicts that $(a+1)+(b+1)=(a+b+2)$, so we may assume $b=0$.  Our dimension equation now reads
$$(a+1)(a+2)=2p^2.$$
The prime $p$ cannot appear in the factorizations of both $a+1$ and $a+2$, so $p^2$ divides one of them.  The other factor is then bounded above by $2$, which bounds the whole left-hand side above by $6$.  Yet $2p^2 \geq 8$, a contradiction.</p>


问:
How to use central difference formula if step size is not constant
In the problem I am solving, table values are given for function and it says to find $f''(0.5)$ using second order central difference formula. I know the formula which is
$\frac{f(x+\triangle x)-2f(x)+f(x-\triangle x)}{{\triangle x}^2}$. But the problem is, table values are as follows(I am listing values around $0.5$ only).

$f(0.48)=1.336, f(0.5)=1.405,f(0.51)=1.481$

I am confused what value should I use as $\triangle x$.

答:
Apply Taylor expansion to a general linear combination with coefficient sum zero, so that the constant terms cancel from the start.
\begin{align}
af(x+Δx)&−(a+b)f(x)+bf(x−2Δx)
\\
&=(a-2b)f'(x)Δx+\frac12(a+4b)f''(x)Δx^2+\frac16(a-8b)f'''(x)Δx^3+...
\end{align}
To remove the linear/first derivative term one needs $a=2b$, and the third order term does not cancel. The approximation formula is then
$$
f''(x)\approx\frac{2f(x+Δx)-3f(x)+f(x-2Δx)}{3Δx^2}
$$
Inserting values gives

```<code>(1.336-3*1.405+2*1.481)/(3*0.01**2) = 276.66666
</code>```

Body:
Apply Taylor expansion to a general linear combination with coefficient sum zero, so that the constant terms cancel from the start.
\begin{align}
af(x+Δx)&−(a+b)f(x)+bf(x−2Δx)
\\
&=(a-2b)f'(x)Δx+\frac12(a+4b)f''(x)Δx^2+\frac16(a-8b)f'''(x)Δx^3+...
\end{align}
To remove the linear/first derivative term one needs $a=2b$, and the third order term does not cancel. The approximation formula is then
$$
f''(x)\approx\frac{2f(x+Δx)-3f(x)+f(x-2Δx)}{3Δx^2}
$$
Inserting values gives

(1.336-3*1.405+2*1.481)/(3*0.01**2) = 276.66666


问:
Finding isomorphism question
By given the next structures: $M_1=\langle\mathbb{N},<\rangle$, $M_2=\langle\mathbb{Z},<\rangle$, $M_3=\langle\mathbb{Q},<\rangle$, $M_4=\langle\mathbb{R},<\rangle$, $M_5=\langle\mathbb{R}^{+},<\rangle$,$M_6=\langle(0,1)\cup (1,2),<\rangle$, I need to find isomorphism between the models.

I have tried looking for the same cardinality of the models separately. Therefore, there is a chance for isomorphism between $M_1, M_2, \ M_3$, or between $M_4, M_5, \ M_6$. Now I find it hard to contradict isomorphism since I have to find a sentence that one satisfies and the other doesn't, for each couple of models. I believe finding a sentence isn't the fastest way to contradict isomorphism, so I will be happy to hear from you a piece of advice on this topic, and your full way of solving this question.

答:
Hint:

Welcome to MSE!

Hint:

To distinguish $M_1$, $M_2$, and $M_3$ you should try to find some simple features that intuitively make them different. Then you can formalize these features to find a sentence that distinguishes them.

For instance, notice $\mathbb{N}$ has a least element (either $0$ or $1$, depending on your definition of $\mathbb{N}$). This distinguishes it from both $\mathbb{Z}$ and $\mathbb{Q}$. Can you make this precise (by finding a sentence describing the existence of a least element)?

To distinguish $\mathbb{Z}$ and $\mathbb{Q}$, you might think about the "next integer". Is there such a thing as a "next rational"? Can you make this precise?

As for $M_4$, $M_5$, and $M_6$, there's some finesse needed.

First, does your $\mathbb{R}^+$ include $0$? If yes, then it has a least element. Can you show that it isn't isomorphic to either of the others (a similar argument to before will work). If $\mathbb{R}^+$ does not include $0$, then actually $M_4$ and $M_5$ are isomorphic! Can you find an explicit isomorphism? What are some nice monotone functions on $\mathbb{R}$ whose output is always positive?

$M_6$ is very subtle indeed. It turns out you won't be able to find a (first order) sentence which distinguishes it from $M_4$. This is because any two "dense linear orders without endpoints" agree on all first order sentences.

Of course, intuitively $M_6$ is "disconected" whereas $M_4$ and $M_5$ should be "connected". We can make this precise by thinking about second order sentences. In particular, notice every bounded subset of $M_4$ has a supremum. Can you find a (necessarily higher order) formula expressing this fact? Can you show that $M_6$ doesn't satisfy this property?

I hope this helps ^_^

Title: Finding isomorphism question


问:
Does $\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{\lceil n/2\rceil}\frac{1}{j}$ exist?
Does $$\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{\lceil n/2\rceil}\frac{1}{j}$$ exist?
To show this, consider $$0\leq \lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{\lceil n/2\rceil}\frac{1}{j}\leq \lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{n}\frac{1}{j}.$$ Does $$\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{n}\frac{1}{j}$$ equal to 0?

Body:
Cesàro Mean. If $a_n\to a$, then $\frac{1}{n}(a_1+\cdots+a_n)\to a$.

So $\frac{1}{n}\to 0$ and hence $\frac{1}{n}\sum_{j=1}^n\frac{1}{j}\to 0$.  Since
$$
0<\frac{1}{n}\sum_{j=1}^{[n/2]}\frac{1}{j}<\frac{1}{n}\sum_{j=1}^n\frac{1}{j}
$$
then
$$
\frac{1}{n}\sum_{j=1}^{[n/2]}\frac{1}{j}\to 0,
$$
as well.


问:
Examples of system of one dimensional first order hyperbolic PDEs with source term
I am writing an algorithm to solve a system of one- dimensional first order hyperbolic PDEs with constant coefficient matrices and I require some example problems which are already solved analytically or numerically with initial and boundary conditions.

The system of PDEs is of the form:
$$A\dot{U} + BU^{'} + F(U) = 0$$
Where, $A$, and $B$ are constant matrices and $F(U)$ is a linear or nonlinear function of $U$.

$\dot{\begin{Bmatrix} \\ \end{Bmatrix}}$ represents the derivative w.r.t. time ($t$), and $\begin{Bmatrix} \\ \end{Bmatrix}^{'}$ represents the derivative w.r.t. spatial variable ($s$).

Example:

One of such problems is the transmission equation given by:

$$\begin{bmatrix}L & 0\\0 & C\end{bmatrix}\dot{\begin{Bmatrix}I \\V\end{Bmatrix}} + \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\begin{Bmatrix}I \\V\end{Bmatrix}^{'} + \begin{Bmatrix} RI \\GV \end{Bmatrix} = \begin{Bmatrix}0 \\0 \end{Bmatrix}$$

$I$ represents current, and $V$ represents voltage.

Where, $L = 4$, $R = 4$, $C = 1$, and $G = 1$

The initial conditions are given by:

$$I(s, 0) = I_0$$
$$V(s, 0) = V_0$$

An analytical solution is available for the above example problem, when $RC = LG$

If anyone has knowledge of similar system of equations (with 3 or more variables if possible) with initial and boundary conditions please let me know. I have searched online and could not find another example.

答:
The scalar case $$a\dot u + b u' + c u =0$$ has analytical closed-form solutions deduced from the method of characteristics. It's much more difficult to do so in the case of systems $$
A\dot{U} + B U' + F(U) = 0
$$ where $F(U)$ is nonzero. Nevertheless, some boundary-value problems can be solved using the Fourier method in the case where $F(U) = CU$ is a linear function. I'd suggest to analyse the scalar case in Fourier space before moving to other systems.

This type of system represents lossy wave propagation. As for examples, I can suggest the case of the damped wave equation (telegraph equation), of linear viscoelastic waves (Maxwell model), or of linear poroelasticity (Biot equations). You might find some other example in the literature related to this post.


问:
Real function with intermediate value property and one more property as set $\{x:f(x)=r\}$ is always closed for every rational $r$, is continuous.
This is exercise $19$ from fourth chapter of baby Rudin.

Suppose $f$ is a real function with domain $R^1$ which has IVP i.e. if $f(a) < c < f(b)$, then $f(x)=c$ for some $x$ between $a$ and $b$. Suppose also for every rational $r$, that the set of all $x$ with $f(x)=r$ is closed. Prove that $f$ is continuous.

I use the hint given by author and got the answer. You can look here.

When I was thinking about this problem, I drew some graphs. One of the graph I draw has puzzled me. It seems to satisfy both the hypotheses, but fails to be continuous.
The graph is as follows.

Graph seems to satisfy IVP and also any horizontal line $y=k$ will hit graph at most twice, so second hypotheses also seems to satisfy. But $f$ is not continuous at origin. What I am messing up here. Thanks.

答: Your example does not satisfy IVP for $[-x_{0},x_{0}]$ and $c=2$ if $x_{0}$ is small enough.


答: It does not satisfy the intermediate value property: $f(0)=3$ and $f(-1)<2$, but there is no $a\in(-1,0)$ such that $f(a)=2$.

Title: Real function with intermediate value property and one more property as set $\{x:f(x)=r\}$ is always closed for every rational $r$, is continuous.

Body:
This is exercise $19$ from fourth chapter of baby Rudin.

Suppose $f$ is a real function with domain $R^1$ which has IVP i.e. if $f(a) < c < f(b)$, then $f(x)=c$ for some $x$ between $a$ and $b$. Suppose also for every rational $r$, that the set of all $x$ with $f(x)=r$ is closed. Prove that $f$ is continuous.

I use the hint given by author and got the answer. You can look here.

When I was thinking about this problem, I drew some graphs. One of the graph I draw has puzzled me. It seems to satisfy both the hypotheses, but fails to be continuous.
The graph is as follows.

Graph seems to satisfy IVP and also any horizontal line $y=k$ will hit graph at most twice, so second hypotheses also seems to satisfy. But $f$ is not continuous at origin. What I am messing up here. Thanks.


问:
Why is $PGL_2(5)\cong S_5$?
Why is $PGL_2(5)\cong S_5$? And is there a set of 5 elements on which $PGL_2(5)$ acts?</p>

答:
As David Speyer explains,

there are 15 involutions of $P^1(\mathbb F_5)$ without fixed points (one might call them «synthemes»). Of these 15 involutions 10 («skew crosses») lie in $PGL_2(\mathbb F_5)$ and 5 («true crosses») don't. The action of $PGL_2(\mathbb F_5)$ on the latter ones gives the isomorphism $PGL_2(\mathbb F_5)\to S_5$.

Title: Why is $PGL_2(5)\cong S_5$?


问:
Exhibit the complete table of multiplication for $(\mathbb{Z} / 15 \mathbb{Z})^*$, and use this to solve $7X \equiv 3 \pmod{15}$.
I have no idea how to start this question? I know how to build a multiplication table for $(\mathbb{Z} / 15 \mathbb{Z})$, but not for $(\mathbb{Z} / 15 \mathbb{Z})^*$? Can anyone help?

答:
For the first part, use that the group of units functor respects products, so that we get $\Bbb Z_2\times\Bbb Z_4$.  That is we exploit the fact that $\Bbb Z_{15}\cong\Bbb Z_3\times\Bbb Z_5$.  It should be fairly routine (if not a little tedious, since it will contain $64$ entries) to write out the table.

For the second part, note that $-2$ is the multiplicative inverse to $7\bmod15$, since indeed $-2\cdot7+1\cdot15=1$.  Thus we get $x\equiv-6\equiv9\bmod15$.


问:
probability problem/combinations
Confused on how to do this.

4 people play a game of cards. These cards are marked with three out of five symbols (A,B,C,D, and E). Each person gets to create one custom card of their preference with any three symbols they like. What is the probability that exactly 2 people from the 4 players end up having the same custom card with the same symbols? (Assume you cannot repeat the same symbol on a card, for instance, you cannot have a card with the symbol combination ACA, DBB, or EEE).

My attempt:
$$P(X=2) = \frac{10 \times10 \times \dbinom{4}{2}^2}{\dbinom{5}{3}^4} = 0.06$$

Denominator because multiplication rule of how many ways 4 people can choose a custom card (3 unique symbols chosen from 5). Numerator: If two people have the same custom card, there are 6 ways to do it hence 4 choose 2, and then the remaining possibilities for the other 2 people is 10 times 10. Is my logic/understanding here ok?

答:
Not quite.  Since exactly two of the people must have the same symbols on their card, there are not $\binom{5}{3} =  10$ possible choices for the symbols on the cards of the remaining people since they must be distinct from the choice made by the people who chose the same three symbols and from each other.

There are $\binom{4}{2}$ ways to select the two people who have the same set of symbols on their card, $\binom{5}{3}$ ways to choose which three of the five symbols appear on the cards of the people who have the same symbols on their card, $10 - 1 = 9$ ways to select the symbols which appear on the card of the younger of the two remaining people, and $9 - 1 = 8$ ways to pick the symbols on the card of the remaining person.  Hence, the number of favorable cases is
$$\binom{4}{2}\binom{5}{3} \cdot 9 \cdot 8$$
Therefore, the probability that exactly two of the four people will select the same three symbols for their custom cards is
$$\Pr(X = 2) = \frac{\dbinom{4}{2}\dbinom{5}{3} \cdot 9 \cdot 8}{\dbinom{5}{3}^4}$$

Title: probability problem/combinations


问:
Find the matrix representation $T:R^2\rightarrow P^2$ if the basis is given
Find the matrix representation $T:R^2\rightarrow P^2$ if the basis is given $B_{R^2}=\{[2,0],[0,-2]\}$,  $B_{R^2}=\{2t^2-1,t+3,2\}$ and $T\{[a,b]\}=(a^2+b^2)t^2+(a-c)t+ab$
Please help me,but i didnt know how to do. Help me!!

I know how to find:

$T\{B_{R^2}\}=T\{[2,0]\}=(a^2+b^2)t^2+(a-c)t+ab$ and $T\{B_{R^2}\}=T\{[0,-2]\}=(a^2+b^2)t^2+(a-c)t+ab$ ?????

$T\{B_{P^2}\}=((2t-1)^2+(t+3)^2)t^2+(2t^2-1-2)t+(2t^2-1)(t+3)=4t^6-3t^4+10t^3+16t^2-4t-3$

but how to continue

答:
Observe that $\text{dim}(\mathbb R^2)=2$ and $\text{dim}(\mathbb R_{\le 2}[x])=3\implies$ the representative matrix of the map is such that $\mathcal M(T)\in \operatorname{M}(3\times 2,\mathbb R)$.

Let $T$ be your map and $\mathcal B$ and $\mathcal D$ the bases of the two vector spaces, then
$$(2,0)\overset{T}{\mapsto} 4t^2+2t=2(2t^2-1)+2(t-3)+4\cdot2$$
$$(0,-2)\overset{T}{\mapsto} 4t^2+2t=T((2,0))\implies$$
$$\mathcal M(T)=\begin{pmatrix}2 &&2\\2 && 2\\4&&4 \end{pmatrix}.$$
Tell me if the text of the problem is fine.


问:
On normal forms of group elements
I was reading about finitely presented groups. Some finitely presented groups (like the braid groups) have normal forms, which means that every element of that group has a unique representation of sorts. I wanted a few examples of finitely presented groups without normal forms. I was wondering if we have a finitely presented group, can we prove that the given group does not have a normal form?

答:
You can define a normal form for any finitely generated group.

Suppose $G$ is generated by the finite set $X$ and let $A = X \cup X^{-1}$. As your representative of the group element $g$, choose  the least word in the alphabet $A$ that represents $g$ under some well-ordering of the set of words in $A$.

An example of such a well-ordering is a lenlex ordering. For that, you choose some arbitrary total ordering $<$ of $A$, and then, for word $v,w \in A$,  define $v<w$ if either $v$ is shorter than $w$, or if $v$ and $w$ have the same length but $v$ precedes $w$ in the lexicographical (i.e. dictionary) ordering of words.

The problem with this is that in general (i.e. for groups with unsolvable word problem) there is no algorithm for computing this normal form.

For a normal form to be useful, it needs to satisfy two properties. Firstly, it should be easy to recognize when a given word is in normal form. Secondly, there should be an algorithm, preferably reasonably efficient, for computing the normal form equivalent of an arbitrary word.

The normal forms that you read about, such as for braid groups and polycyclic groups, satisfy these two properties.

Title: On normal forms of group elements


问:
Asking for help in proving a result related to Mangoldt function in number theory
I am trying exercises of ch-4 of Apostol Introduction to analytic number theory and in 1 question I am struck. This question uses following condition

Can it be proved that $\sum_{n\leq x}\frac {\Lambda(n) } { n} $ = log (x) +O(1) ?

Due to Mangoldt function sum will be calculated only on prime values but I am not able to get desired result. Can you please help?

答:
Using the fact that
$$n!=\prod_{p\;\mathrm{prime}} p^{j(n,p)}=\prod_{\substack{p\;\mathrm{prime}\\p\leq n}} p^{j(n,p)}\quad\mathrm{where}\quad j(n,p)=\sum_{m=1}^{\infty} \bigg\lfloor \frac{n}{p^m}\bigg\rfloor$$
Therefore we obtain
$$\sum_{n=2}^{\lfloor x\rfloor} \ln(n)=\ln\Bigg(\prod_{\substack{p\;\mathrm{prime}\\p\leq x}} p^{j(\lfloor x\rfloor,p)}\Bigg)=\sum_{\substack{p\;\mathrm{prime}\\p\leq x}} j(\lfloor x\rfloor,p)\ln(p)=\sum_{\substack{p\;\mathrm{prime}\\p^m\leq x}} \bigg\lfloor \frac{x}{p^m}\bigg\rfloor\ln(p)=\sum_{n\leq x} \bigg\lfloor \frac{x}{n}\bigg\rfloor\Lambda(n)$$
since $\ln(1)=0$ and for any real number $x$ and any positive integer $n$, we have $\lfloor\lfloor x\rfloor/n\rfloor=\lfloor x/n\rfloor$. Now we use the equivalent statement of the Prime Number Theorem, namely the asymptotic formula for Chebyshev's $\psi$-function
$$\psi(x)=\sum_{n\leq x} \Lambda(n)=O(x)$$
If we remove the floor function in the last sum, we have
\begin{align*}\begin{split} &\sum_{n\leq x} \frac{x}{n}\Lambda(n)<\sum_{n\leq x} \bigg\lfloor \frac{x}{n}\bigg\rfloor\Lambda(n)+\sum_{n\leq x}\Lambda(n)=\sum_{n\leq x} \ln(n)+O(x)\\ &\Leftrightarrow \sum_{n\leq x} \frac{x}{n}\Lambda(n)=\sum_{n\leq x} \ln(n)+O(x) \end{split}\end{align*}
since for any real number $x$, we have $x<\lfloor x\rfloor+1$. Now we use the asymptotic formula for $\sum_{n\leq x} \ln(n)=x\ln(x)+O(x)$, therefore we obtain
$$\sum_{n\leq x} \frac{x}{n}\Lambda(n)=x\ln(x)+O(x)$$
Dividing by $x$ on both sides and it follows.


问:
Determine the partial derivative of $\frac{\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}}{\cos ^2\left(x^2y^2\right)}$.

Determine the partial derivative of $f(x)$:
$$f(x,y)=\frac{\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}}{\cos ^2\left(x^2y^2\right)}$$

Here's what I have so far:

\begin{align}
\frac{\partial \:}{\partial \:x}\left(\frac{\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}}{\cos ^2\left(x^2y^2\right)}\right) &= \frac{\frac{\partial \:}{\partial \:x}\left(\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}\right)\cos ^2\left(x^2y^2\right)-\frac{\partial \:}{\partial \:x}\left(\cos ^2\left(x^2y^2\right)\right)\left(\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}\right)}{\left(\cos ^2\left(x^2y^2\right)\right)^2} \\
&= \frac{\left(\frac{2\tan \left(\frac{1}{x^2+1}\right)}{x}-\frac{2x\ln \left(y^2x^2\right)\sec ^2\left(\frac{1}{x^2+1}\right)}{\left(x^2+1\right)^2}+\frac{2x}{3\left(x^2+y^2\right)^{\frac{2}{3}}}\right)\cos ^2\left(x^2y^2\right)-\left(-2y^2x\sin \left(2y^2x^2\right)\right)\left(\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}\right)}{\left(\cos ^2\left(x^2y^2\right)\right)^2} \\
&= \frac{\cos ^2\left(x^2y^2\right)\left(\frac{2\tan \left(\frac{1}{x^2+1}\right)}{x}-\frac{2x\ln \left(x^2y^2\right)\sec ^2\left(\frac{1}{x^2+1}\right)}{\left(x^2+1\right)^2}+\frac{2x}{3\left(x^2+y^2\right)^{\frac{2}{3}}}\right)+2xy^2\sin \left(2x^2y^2\right)\left(\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}\right)}{\cos ^4\left(x^2y^2\right)}
\end{align}

However, I am not sure if this answer is correct since I get a different answer on emathhelp website.

答:
The only issue is that  $ \frac{\partial}{\partial x} \cos^2(x^2 y^2)$ should be equal to

$$- 4 x y^2 \cos(x^2 y^2) \sin(x^2 y^2) $$

and with that correction it is good.


问:
Is polynomial algebra the only minimal dense subalgebra of $C^{\infty}(U)$
Let $U$ be connected open set of $\mathbb{R}^{n}$. Consider $C^{\infty}(U)$ with sup norm, we said $A\subset C^{\infty}(U)$ is a minimal dense subalgebra of $C^{\infty}(U)$ if and only if any subalgebra contain in $A$ is not dense in $C^{\infty}(U)$. I want to ask are the polynomial algebra of $U$, either with coefficients in $\mathbb{Q}$ or $\mathbb{R}$\ $\mathbb{Q}$, the only two minimal dense subalgebra of $C^{\infty}(U)$?

答: For proper U, polynomials in even powers or trigonometric polynomials i.e. span of ${e^{ikt}}$ will do. Also, if $U=[a,b]\subset (0,1)$ then polynomials with integer coefficients are good.


问:
Distinct composition series of $Z_{30}$
Given two composition series for a group G:
$$\{e \} \triangleleft G_1 \triangleleft ... \triangleleft G_k \triangleleft G $$
$$\{e \} \triangleleft H_1 \triangleleft ... \triangleleft H_k \triangleleft H $$

If we say they are distinct when $G_i \ncong H_i$ for at least one i. How many distinct composition series does $Z_{30}$ have?

I am a bit confused with this question I initially thought it was linked to the Jordan-Holder theorem and so there would only be 1, but then I realized that this only tells us that series only have the same composition factors up to isomorphism.

I just wanted to check that I am right and this doesn't hold for individual $G_i \cong H_i$ too?

If this is true does anyone have any hints for an easy way to work this out or is it just a case of trying to find all the normal subgroups of $Z_{30}$, and would Sylow theorems possibly help to find these?

答:
$Z_{30}={0,1,2,...29}$ (addition $\bmod 30$ here):

$\langle2\rangle= {0,2,4,...28}$

$\langle3\rangle= {0,3,6,...,27}$ and

$\langle5\rangle$, $\langle6\rangle$, $\langle10\rangle$, $\langle15\rangle$ are subgroups of $Z_{30}$ other then ${0}$ and $Z_{30}$.

Therefore, composition series are

${0}⊂\langle15\rangle⊂\langle5\rangle⊂G$

${0}⊂\langle15\rangle⊂\langle3\rangle⊂G$

${0}⊂\langle10\rangle⊂\langle5\rangle⊂G$

${0}⊂\langle10\rangle⊂\langle2\rangle⊂G$

${0}⊂\langle6\rangle⊂\langle3\rangle⊂G$

${0}⊂\langle6\rangle⊂\langle2\rangle⊂G$

for the first series is $\langle5\rangle/\langle15\rangle$
and $|\langle5\rangle/\langle15\rangle|= |\langle5\rangle|/|\langle15\rangle|=6/2=3$ which is prime no.

therefore $\langle5\rangle/\langle15\rangle$ is simple
similarly, we can show that all factor group $G_i/G_{i+1}$ are simple.

Title: Distinct composition series of $Z_{30}$


问:
Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $\frac{a+b+c+ab+ac+bc}{1+abc}$ is a real number.

Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $$x = \frac{a+b+c+ab+ac+bc}{1+abc}$$ is a real number.

I wanted to calculate $2 \cdot Im(x) = x- \overline x$ and show that it's equal to zero:

$$\frac{a+b+c+ab+ac+bc}{1+abc} - \frac{\overline{a}+\overline{b}+\overline{c}+\overline{ab}+\overline{ac}+\overline{bc}}{1+\overline{abc}} =$$ $$=a+b+c+ab+ac+bc-\overline{a}-\overline{b}-\overline{c}-\overline{ab}-\overline{ac}-\overline{bc} = $$ $$=2(Im(a) + Im(b)+Im(c)+Im(ab)+Im(ac)+Im(bc))=$$$$=2(Im(a+b+c+ab+bc+ac))$$But I have no idea what to do next or what am I missing.

答:
$$x=\frac{a+b+c+ab+bc+ca}{1+abc}$$
$$=\frac{abc}{abc}\cdot\frac{a+b+c+ab+bc+ca}{1+abc}$$
$$=\frac{\frac{a+b+c+ab+bc+ca}{abc}}{\frac{1+abc}{abc}}$$
$$=\frac{\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}}{\frac{1}{abc}+1}$$
$$=\frac{\overline{a}+\overline{b}+\overline{c}+\overline{ab}+\overline{bc}+\overline{ca}}{1+\overline{abc}}$$
$$\overline{\left(\frac{a+b+c+ab+bc+ca}{1+abc}\right)}=\overline{x}$$

using if $|a|=1$ then $a\overline{a}=1$, and if $x=\overline{x}$, then $x\in\mathbb{R}$.


问:
$\lim_{x \to 0} \frac{f(\sin \frac{\pi + x}{3}) + f(\cos \frac{\pi + x}{3})}{e^x -1} = 1$, Prove that $f(x_i)=0$ for some $x_i\in [0,1]$

Let $f \colon [0,1] \to \mathbb{R}$ be a continuous function such that:
$$
\lim_{x\to 0} \frac{f(\sin \frac{\pi + x}{3}) + f(\cos \frac{\pi + x}{3})}{e^x -1} = 1
$$
Prove that there is an $x_i \in [0,1]$ that $f(x_i)= 0.$

I know that it means that the sum of function values with trigonometric functions behave like $e^x -1$ near $0$ and $e^x - 1$ near $0$ behaves like $x$, but what do I do with those trig functions?

答: As in the limit the denominator is zero thus the numerator must be zero otherwise the limit won't exist so $ f\left(\dfrac{\sqrt{3}}{2} \right) + f\left( \dfrac{1}{2} \right) = 0 $ And as $[0,1]$ is complete and $f$ is continuous the result follows from IVT


问:
Find all primes p such that p!+p is a perfect square
The question is that,
Find all primes $p$ such that $p!+p$ is a perfect square.

By using Wilson's Theorem I got,
$$p!+p=p((p-1)!+1)=p^{2}k$$
If $p^{2}k$ is a perfect square, then k must be a perfect square.

I was unable to proceed after this. I also tried another approach

We can see that for p=2,3 the given expression is a perfect square. Let $p>3$. If $p!+p=k^{2}$, then it is easy to see that k must be odd. Also, if k is odd, we know that $k^{2}\equiv 1\pmod{8}$, which implies that $p\equiv 1\pmod{8}$, since $8|p!$
 Similarly, we also get $p\equiv 1\pmod{3}$. Thus, we get
$$p\equiv 1\pmod{24}$$Unfortunately, I was unable to capitalze from here as well

Could someone please complete my solutions or provide an alternate solution.

Thanks a lot for your help!! :)

答:
Suppose there is such a prime greater than $3$.

Let $n$ be any natural number less than $p$, then $p$ is a square modulo $n$. As you have observed, $p\equiv 1 \pmod 8$ and therefore, by quadratic reciprocity, any odd $n$ is a square modulo $p$ and by the second supplement $2$ is a square modulo $p$.

Therefore all natural numbers less than $p$ are squares modulo $p$. This contradicts the fact that only half the numbers less than $p$ are quadratic residues.


答:
Hint:

$$1+\frac12+\frac13+\frac14+\frac15+\frac16+\frac17+\cdots<1+\frac12+\frac12+\frac14+\frac14+\frac14+\frac14+\cdots$$

shows that the sum of the $2^m-1$ first terms is bounded by $m$.

With $n=2^m-1$, you can conclude.

Title: Does $\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{\lceil n/2\rceil}\frac{1}{j}$ exist?

Body:
Does $$\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{\lceil n/2\rceil}\frac{1}{j}$$ exist?
To show this, consider $$0\leq \lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{\lceil n/2\rceil}\frac{1}{j}\leq \lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{n}\frac{1}{j}.$$ Does $$\lim_{n\to\infty}\frac{1}{n}\sum_{j=1}^{n}\frac{1}{j}$$ equal to 0?


问:
Multidimensional integral of symmetric function
It holds that
$$
\int_0^\infty \int_0^\infty f(x_1,x_2) dx_1 dx_2 = 2 \int_0^\infty \int_0^{x_2} f(x_1,x_2) dx_1 dx_2
$$
when $f$ is symmetric, or, in other words, it is invariant under permuting the variables.

Does it hold for some $C$ that
$$
\int_0^\infty \cdots \int_0^{\infty}f(x_1, \cdots, x_n) dx_1 \cdots dx_n = C \int_0^\infty\int_0^{x_n}\cdots\int_0^{x_n}f(x_1, \cdots, x_n) dx_1 \cdots dx_n
$$
where $n$ is the number of integrals, again for $f$ being symmetric. If so, how does $C$ depend on $n$?

答:
In your exact formulation of the question, there is no such constant $C$. This can be seen by trying the pair of functions

$$f_1(x,y,z)=\mathrm{e}^{-x^2-y^2-z^2} \\ f_2(x,y,z)=\mathrm{e}^{-(x-1)^2-y^2-z^2}$$
The reasoning behind the first equation
$$
\int_0^\infty \int_0^\infty f(x_1,x_2) dx_1 dx_2 = 2 \int_0^\infty \int_0^{x_2} f(x_1,x_2) dx_1 dx_2
$$

is that the RHS is equal to the sum

$$\iint_{x_1 \leq x_2} f + \iint_{x_1 \geq x_2} f .$$

In the general case with $n$ variables, there are $n!$ permutations of the variables, with $x_1 \leq x_2 \leq \cdots \leq x_n$ being one of them. Thus, the true analogue of your 2-d observation is

$$\int_{\mathbf{R}^n} f(x_1,x_2,\dots,x_n) \mathrm{d} \mathbf{x} = n! \int_0^\infty \int_0^{x_n} \cdots \int_{0}^{x_2}  f(x_1,x_2,\dots, x_n)\mathrm{d}x_1 \mathrm{d} x_2 \dots \mathrm{d} x_n. $$


问:
Questions about epimorphisms and projectives in functor categories
Suppose $I$ is a small category, $R$ is a ring and $_R\mathrm{Mod}$ is the category of left $R$-modules. How do I show that the category $[I,~_R\mathrm{Mod}]$ of all functors from $I$ to $_R\mathrm{Mod}$ is an abelian category with enough projectives and injectives?

It's my first time verifying the axioms of an abelian category and looking at projectives/injectives in a context different than that of modules. I think I can handle the abelian part on my own. However, in order to get a feel for the projectives I would like to ask a few questions:

A natural transformation $F\stackrel{\alpha}{\Rightarrow} G$ with all $\alpha_i~(i\in I)$ epimorphisms is an epimorphism in the functor category. Are there any others?

Are the functors $P:I\to~_R\mathrm{Mod}$, with all $P_i$ projective, projective, is this condition even necessary or sufficient? Is there a description of the projectives?

Given a functor $F$, is there a simple description of a projective $P$ and an epimorphism $P\Rightarrow F\Rightarrow 0$?

Even the simplest of examples $I\equiv\bullet\rightarrow\bullet$ i don't understand.

答:
Let $\mathcal{R} = [\mathcal{I}, R\text{-}\mathbf{Mod}]$, let $\mathcal{A} = [\mathcal{I}, \mathbf{Ab}]$, and let $\mathcal{S} = [\mathcal{I}, \mathbf{Set}]$. First, observe the following: there is an evident forgetful functor $\mathrm{Hom}_R (R, -) : \mathcal{R} \to \mathcal{A}$, and it has both a left adjoint $R \otimes_\mathbb{Z} {-} : \mathcal{A} \to \mathcal{R}$ and a right adjoint $\mathrm{Hom}_\mathbb{Z} (R, -) : \mathcal{A} \to \mathcal{R}$, induced by the corresponding functors when $\mathcal{I}$ is the terminal category. There is also an adjunction
$$F \dashv U : \mathcal{A} \to \mathcal{S}$$
that corresponds to the usual free–forgetful adjunction.

A well-known fact about functor categories $[\mathcal{I}, \mathcal{C}]$ is that the monomorphisms and epimorphisms are precisely the componentwise ones when $\mathcal{C}$ has kernel pairs and cokernel pairs. (Use the fact that a morphism is a monomorphism if and only if its kernel pair is trivial, and the fact that kernel pairs in $[\mathcal{I}, \mathcal{C}]$ are computed componentwise when $\mathcal{C}$ has kernel pairs.) This is true in particular for $\mathcal{R}$, $\mathcal{S}$, and $\mathcal{A}$. Thus we deduce that $\mathrm{Hom}_R (R, -)$ and $U$ preserve monomorphisms and epimorphisms.

It is straightforward to show that $\mathcal{S}$ has enough projectives: indeed, every representable presheaf is projective (because they are free!), and every presheaf is a quotient of a coproduct of representable presheaves in a canonical way. Since $U$ preserves epimorphisms, we can use the adjunction $F \dashv U$ to deduce that $\mathcal{A}$ has enough projectives, and then we use $R \otimes_\mathbb{Z} {-} \dashv \mathrm{Hom}_R (R, -)$ to deduce that $\mathcal{R}$ has enough projectives. Note that componentwise projective diagrams need not be projective; see here for the case where $\mathcal{I} = \{ \cdots \to \bullet \to \bullet \to \bullet \to \cdots \}$.

For injectives, we have to do something non-trivial. I appeal to the following result:

Theorem. If $\mathcal{E}$ is a Grothendieck topos, then the category of internal abelian groups in $\mathcal{E}$ has enough injectives.

Of course, $\mathcal{S}$ is a Grothendieck topos, so we deduce $\mathcal{A}$ has enough injectives. Since $\mathrm{Hom}_R (R, -)$ preserves monomorphisms, we can use the adjunction $\mathrm{Hom}_R (R, -) \dashv \mathrm{Hom}_\mathbb{Z} (R, -)$ to deduce that $\mathcal{R}$ also has enough injectives. (An alternative approach would be to show that $\mathcal{R}$ is a Grothendieck abelian category; this is not too difficult since $\mathcal{R}$ is a locally finitely presentable category.)


问:
Show $\sum\limits_{i=0}^{k-1} \frac{r^i}{k-i}$ for $r \in (0,1)$ goes to 0 as $k\to \infty$
I wish to show that the following sequence $\alpha_k$ which is an infinite summation goes to 0 as $k\to\infty$

Let $$\alpha_k = \sum\limits_{i=0}^{k-1} \frac{r^i}{k-i}\quad r \in (0,1)$$
Then show that the sequence $\{\alpha_k\}$ converges to 0

Each term is clearly 0 as $k\to \infty$, but i'm unable to upperbound the infinite sum.

答: $0 \leq \alpha_k \leq \sum\limits_{i=0}^{m} \frac {r^{i}} {k-i}+\sum\limits_{i=m+1}^{k-1} \frac {r^{i}} {k-i}$ for any $m<k$. The second term is bounded by $\sum\limits_{i=m+1}^{k-1}  {r^{i}} $. It is easy to see that $\sum\limits_{i=m+1}^{\infty}  {r^{i}}  \to 0$ as $m \to \infty$ (by writing down the geometric sum explcitly). So choose and fix $m$ so that the second term is less than $\epsilon$. Now let $k \to \infty $ in the first term (with $m$ fixed).


问:
What's the necessary and sufficient condition of $p^m|F_n$?
Let {$F_n$} be Fibonacci Series defined by $F_{n+2}=F_{n+1}+F_n$ with $F_1=F_2=1$.I have proved that $3^m|F_n \Leftrightarrow 4\cdot 3^{m-1}|n$ for another problem.Then I began to think about the necessary and sufficient condition of $p^m|F_n$,where $p$ is a prime number and $m$ is a positive integer.By analyzing the first sixty terms of {$F_n$},I suppose that

where $\tau (p)$ is the least positive integer s.t. $p|F_{\tau(p)}$.

I think we could prove this with induction through several steps:

①Prove that $(F_m,F_n)=F_{(m,n)}$,where $(a,b)$ is the greastest common divisor of $a$ and $b$.

②Strengthen the proposition by proving $p^m||F_{\tau (p) \cdot p^{m-1}}$ $\ (m \geqslant 3 \ $when $p=2)$.

③Prove the proposition when $p=2$.

④Prove $p||F_{\tau (p)}$.

⑤Prove that $p^{m+1}|F_n \Leftrightarrow \tau (p) \cdot p^m|n $ ,and that $p^{m+1}||F_{\tau (p) \cdot p^m}$ ,when $p^{m}|F_n \Leftrightarrow \tau (p) \cdot p^{m-1}|n $ ,and $p^{m}||F_{\tau (p) \cdot p^{m-1}}$$\ \ $ for all odd primes $p$.

For ①,it's easy to prove $(F_n,F_{n+1})=1$ with the fact that $(F_1,F_2)=1$.And we have
$$\begin{align}F_m&=F_{m-1}+F_{m-2}=F_2F_{m-1}+F_1F_{m-2}=F_2(F_{m-2}+F_{m-3})+F_1F_{m-2}\\&=(F_2+F_1)F_{m-2}+F_2F_{m-3}=F_3F_{m-2}+F_2F_{m-2}\\&=\cdots\\&=F_nF_{m-n+1}+F_{n-1}F_{m-n}\end{align}$$
when $m>n$.Thus we have $(F_m,F_n)=(F_{m-n},F_n)$ when $m>n$.So by Euclid algotithm,we have $(F_m,F_n)=F_{(m,n)}$.

For ③,it's easy to prove that $2|F_n\Leftrightarrow3|n\ $,$4|F_n\Leftrightarrow6|n\ $,$8|F_n\Leftrightarrow6|n\ $,and $8||F_6$.When $2^m|F_n\Leftrightarrow3\cdot2^{m-2}|n$ and $2^{m}||F_{3\cdot2^{m-1}}$ $(m>2)$,we have
$$\begin{align}\frac{F_{3\cdot2^{m-1}}}{F_{3\cdot2^{m-2}}}&=\frac{(\frac{1+\sqrt{5}}{2})^{3\cdot2^{m-1}}-(\frac{1-\sqrt{5}}{2})^{3\cdot2^{m-1}}}{(\frac{1+\sqrt{5}}{2})^{3\cdot2^{m-2}}-(\frac{1-\sqrt{5}}{2})^{3\cdot2^{m-2}}}=(\frac{1+\sqrt{5}}{2})^{3\cdot2^{m-2}}+(\frac{1-\sqrt{5}}{2})^{3\cdot2^{m-2}}.\end{align}$$
Let {$Z_n$} be a sequence where $Z_1=1$,$Z_2=3$ and $Z_{n+2}=Z_{n+1}+Z_{n}$.Then we have $Z_n=(\frac{1+\sqrt{5}}{2})^n+(\frac{1-\sqrt{5}}{2})^n$,and
$$\begin{align}&Z_1\equiv1,Z_2\equiv1,Z_3\equiv0,Z_4\equiv1,Z_5\equiv1,Z_6\equiv0,\cdots(mod2)\\
&Z_1\equiv1,Z_2\equiv3,Z_3\equiv0,Z_4\equiv3,Z_5\equiv3,Z_6\equiv2,Z_7\equiv1,Z_8\equiv3,\cdots(mod4)
\end{align}$$
which tolds us that if $6|n$,then $2||Z_n$.So we have
$$\begin{align}2||Z_{3\cdot2^{m-2}}=\frac{F_{3\cdot2^{m-1}}}{F_{3\cdot2^{m-2}}}.\end{align}$$
And $2^{m}||F_{3\cdot2^{m-1}}\ \ $,thus $2^{m+1}||F_{3\cdot2^{m}}$.Obviously,the multiple of $2^{m+1}$ is the multiple of $2^{m}$.So if $2^{m+1}|F_n$,then $2^{m}|F_n$.So we have $3\cdot2^{m-2}|n$.Let $n=3k\cdot2^{m-2}$,where k is a positive integer.Then with ①,we have $(F_{3\cdot2^{m-1}}\ ,\ F_{3k\cdot2^{m-2}}\ )=F_{3\cdot2^{m-2}\ \ (2,k)}$.If $2\nmid k$,then $(F_{3\cdot2^{m-1}}\ ,\ F_{3k\cdot2^{m-2}}\ )=F_{3\cdot2^{m-2}}\ \ $,which means $2^m||F_{3k\cdot2^{m-2}}\ \ $;if $2|k$,then $(F_{3\cdot2^{m-1}}\ ,\ F_{3k\cdot2^{m-2}}\ )=F_{3\cdot2^{m-1}}\ \ $,which means $2^{m+1}|F_{3k\cdot2^{m-2}}\ \ $.So $2^{m+1}|F_n\Leftrightarrow3\cdot2^{m-1}|n$.By the induction principle,we've achieved ③.

For ⑤,similar to ③,we have only to prove $p^{m+1}||F_{\tau (p)\cdot p^{m}}$ ,when $p^{m}|F_n \Leftrightarrow \tau (p) \cdot p^{m-1}|n $ ,and $p^{m}||F_{\tau (p) \cdot p^{m-1}}\ $.But actually I can only do this when $p$ is a certain prime such as $3,5$.I can't find a way to prove it for a general $p\ $(constructing a sequence is not suitable to an odd prime ,or my poor knowledge of sequence can't support me to step further).When $p=3$,I proved it like that:

It's easy to know $\tau (3)=4$.And it's also easy to prove that $3|F_n\Leftrightarrow4|n$,and $3||F_4$.If $3^{m}|F_n\Leftrightarrow4\cdot3^{m-1}|n$ and $3^m||F_{4\cdot3^{m-1}}\ $,then: we have
$$\begin{align}\frac{F_{4\cdot3^{m}}}{F_{4\cdot3^{m-1}}}&=\frac{(\frac{1+\sqrt{5}}{2})^{4\cdot3^m}-(\frac{1-\sqrt{5}}{2})^{4\cdot3^m}}{(\frac{1+\sqrt{5}}{2})^{4\cdot3^{m-1}}-(\frac{1-\sqrt{5}}{2})^{4\cdot3^{m-1}}}=(\frac{1+\sqrt{5}}{2})^{8\cdot3^{m-1}}+(\frac{1+\sqrt{5}}{2})^{4\cdot3^{m-1}}(\frac{1-\sqrt{5}}{2})^{4\cdot3^{m-1}}+(\frac{1-\sqrt{5}}{2})^{8\cdot3^{m-1}}\\
&=\frac{(1+\sqrt{5})^{8\cdot3^{m-1}}+(-4)^{4\cdot3^{m-1}}+(1-\sqrt{5})^{8\cdot3^{m-1}}}{2^{8\cdot3^{m-1}}}=\frac{2\sum_{t=0}^{4\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{2t}}\cdot5^t}+4^{4\cdot3^{m-1}}}{2^{8\cdot3^{m-1}}}.
\end{align}$$
And because
$$\begin{align}2\sum_{t=0}^{4\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{2t}}\cdot5^t}+4^{4\cdot3^{m-1}}&\equiv 2\sum_{t=0}^{2\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{4t}}}+2\sum_{t=0}^{2\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{4t+2}}\cdot2}+1\\&\equiv 2\sum_{t=0}^{4\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{2t}}}+2\sum_{t=0}^{2\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{4t+2}}}+1\\&\equiv 2^{8\cdot3^{m-1}}+2^{8\cdot3^{m-1}-1}-2^{4\cdot3^{m-1}}+1\\&\equiv 1+2-1+1 \equiv 0 (mod3)
\end{align}$$
and it's easy to prove that $2\sum_{t=0}^{4\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{2t}}\cdot5^t}+4^{4\cdot3^{m-1}}\equiv 3(mod9)$when $m=1$,and when $m>1$
$$\begin{align}2\sum_{t=0}^{4\cdot3^{m-1}}{{{8\cdot3^{m-1}\ }\choose{2t}}\cdot5^t}+4^{4\cdot3^{m-1}}&\equiv 2\Biggl( \sum_{t=0}^{2\cdot3^{m-1}}{\biggl( {{8\cdot3^{m-1}}\choose{2t}}\cdot5^t+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-2t}}\cdot5^{8\cdot3^{m-1}-2t} \biggl)}-{{8\cdot3^{m-1}}\choose{4\cdot3^{m-1}}}\cdot5^{2\cdot3^{m-1}} \Biggr)+1\end{align}$$
$$\begin{align}&\equiv 2\Biggl( \sum_{t=0}^{3^{m-2}}{\biggl( {{8\cdot3^{m-1}}\choose{12t}}+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t}} \biggl)}+\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+2}}\cdot5+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-2}}\cdot2 \biggl)}\\
&\ +\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+4}}\cdot7+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-4}}\cdot4 \biggl)}+\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+6}}\cdot8+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-6}}\cdot8 \biggl)}\\
&\ +\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+8}}\cdot4+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-8}}\cdot7 \biggl)}+\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+10}}\cdot2+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-10}}\cdot5 \biggl)}-{{8\cdot3^{m-1}}\choose{4\cdot3^{m-1}}} \Biggr)+1\end{align}$$
$$\begin{align}&\equiv 2\Biggl( \sum_{t=0}^{3^{m-2}}{\biggl( {{8\cdot3^{m-1}}\choose{12t}}+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t}} \biggl)}+\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+2}}\cdot8+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-2}}\cdot8 \biggl)}\\
&\ +\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+4}}+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-4}} \biggl)}+\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+6}}\cdot8+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-6}}\cdot8 \biggl)}\\
&\ +\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+8}}+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-8}} \biggl)}+\sum_{t=0}^{3^{m-2}-1}{\biggl( {{8\cdot3^{m-1}}\choose{12t+10}}\cdot8+{{8\cdot3^{m-1}}\choose{8\cdot3^{m-1}-12t-10}}\cdot8 \biggl)}-{{8\cdot3^{m-1}}\choose{4\cdot3^{m-1}}} \Biggr)+1\end{align}$$
$$\begin{align}
&\equiv 2\Biggl( \sum_{t=0}^{4\cdot3^{m-1}}{{{8\cdot3^{m-1}}\choose{2t}}}+7\sum_{t=0}^{2\cdot3^{m-1}}{{{8\cdot3^{m-1}}\choose{4t+2}}} \Biggl)+1\\
&\equiv 2(2^{8\cdot3{m-1}-1}+7(2^{8\cdot3{m-1}-2}-2^{4\cdot3{m-1}-1}))+1\\
&\equiv 2(5+7\times(7-5))+1 \equiv 3(mod9)
\end{align}$$
(I'm terribly sorry for my making it so ugly,but it seems it has been beyond its limit),we have $3||\frac{F_{4\cdot3^m}}{F_{4\cdot3^{m-1}}}$,thus $3^{m+1}||F_{4\cdot3^m}$.Then $(F_{4\cdot3^m},F_{4k\cdot3^{m-1}})=F_{4\cdot3^{m-1}\ \ (3,k)}$.Similar to ③,We have $3^{m+1}|F_n \Leftrightarrow 4\cdot3^m|n$.Then by induction principle,$3^m|F_n\Leftrightarrow4\cdot3^{m-1}|n$.

But I can't work out in this way for a general $p$.

And I have no idea about ④,which is the basic of the induction.

How could I prove or disprove my suppose?Is there any theory about this?Is there a formula about $\tau (p)$?

Appreciated for any suggestion.

And I'm sorry for my possibly strange expressions and wrong grammar for I'm still an English learner.

答:
You clearly put a lot of effort into writing up your question, but I hope you can understand why the sheer number of symbols you wrote is a nightmare to try to read. I will not try. Trying to prove that rule by induction using the explicit formula for Fibonacci numbers and all those binomial coefficients seems horrific to contemplate.

Here is what I consider the correct (conceptual) way to think about your question.  Like you, I will use the explicit formula
$$
F_n = \frac{\varphi^n - \overline{\varphi}^n}{\sqrt{5}}
$$
where $\varphi = (1 + \sqrt{5})/2$ and $\overline{\varphi} = (1-\sqrt{5})/2$. Because $\varphi^2 = \varphi + 1$ and $\overline{\varphi} = 1 - \varphi$,
$\varphi$ and $\overline{\varphi}$ lie in the ring
$$
R = \mathbf Z[\varphi] = \mathbf Z + \mathbf Z\varphi,
$$
which is analogous to the Gaussian integers $\mathbf Z[i] = \mathbf Z + \mathbf Z i$.  For a prime $p$ other than $5$ (the prime $5$ is tricky because of the $\sqrt{5}$ in the denominator of the formula for $F_n$), you are seeking for each $m \geq 1$ a description of all $n \geq 1$ such that $F_n \equiv 0 \bmod p^m\mathbf Z$. Let's work not in $\mathbf Z$, but in the larger ring $R$.  For congruences with ordinary integers we don't lose anything, since
$$
a \equiv b \bmod M\mathbf Z \Longleftrightarrow a \equiv b \bmod MR
$$
for $a, b, M \in \mathbf Z$. This is because the only fractions (rational numbers) in $R$ are ordinary integers, so $(a-b)/M$ is in $\mathbf Z$ if and only if $(a-b)/M$ is in $R$. Thanks to this equivalence, we can rewrite the condition $p^m \mid F_n$ for primes $p \not= 5$ as follows:
$$
p^m \mid F_n \text{ in } \mathbf Z \Longleftrightarrow F_n \equiv 0 \bmod p^m\mathbf Z \Longleftrightarrow F_n \equiv 0 \bmod p^mR \Longleftrightarrow \frac{\varphi^n - \overline{\varphi}^n}{\sqrt{5}} \equiv 0 \bmod p^mR.
$$
Since $p$ is a prime other than $5$, $\sqrt{5}$ is invertible in $R/p^mR$.  Therefore
$$
\frac{\varphi^n - \overline{\varphi}^n}{\sqrt{5}} \equiv 0 \bmod p^mR \Longleftrightarrow
\varphi^n \equiv \overline{\varphi}^n \bmod p^mR
\Longleftrightarrow
\varphi^{2n} \equiv (-1)^n \bmod p^mR,
$$
where I passed between the last two congruences by multiplying both sides by $\varphi$ and using the relation $\varphi\overline{\varphi} = -1$. Multiplying both sides by $(-1)^n$, we finally obtain
$$
p^m \mid F_n \text{ in } \mathbf Z \Longleftrightarrow \boxed{(-\varphi^2)^n \equiv 1 \bmod p^mR}.
$$
So your question is actually about computing the order of a number in modular arithmetic in $R$ modulo powers of $p$.  In particular, this explains why the set of all $n \geq 1$ such that $p^m \mid F_n$ is the multiples of the least such $n$.
This is a special case of a result in group theory: the powers that make something the identity in a finite group are multiples of the least such power.

The number $\varphi \bmod p^mR$ is invertible (with inverse $-\overline{\varphi} \bmod p^mR$), so it lies in the finite group of units $(R/p^mR)^\times$.  Therefore there will be an $n$ such that $p^m \mid F_n$, and the smallest such $n$ is the (multiplicative) order of $-\varphi^2 \bmod p^mR$.  By Lagrange's theorem from group theory, the order of an element of $(R/p^mR)^\times$ divides the size of the group $(R/p^mR)^\times$.  If you know some algebraic number theory (how $p$ splits in $R$, a quadratic ring of algebraic integers), then you can compute the size of that group:
$$
|(R/p^mR)^\times| =
\begin{cases}
(p^{m-1}(p-1))^2, & \text{ if } p \equiv 1, 4 \bmod 5, \\
p^{2(m-1)}(p^2-1), & \text{ if } p \equiv 2, 3 \bmod 5.
\end{cases}
$$
So your least $n$ will be a factor of this number when $p \not= 5$.

Taking $m = 1$, when you write the least $n$ such that $p \mid F_n$ as $\tau(p)$, we can say $\tau(p)$ is a factor of
$$
|(R/pR)^\times| =
\begin{cases}
(p-1)^2, & \text{ if } p \equiv 1, 4 \bmod 5, \\
p^2-1, & \text{ if } p \equiv 2, 3 \bmod 5.
\end{cases}
$$
Actually, we can do a bit better: $\tau(p) \mid (p-1)$ when $p \equiv 1, 4 \bmod 5$ and $\tau(p) \mid (p+1)$ when $p \equiv 2, 3 \bmod 5$.

Case 1: $p \equiv 1, 4 \bmod 5$.
Here $5$ is a square in $\mathbf Z/p\mathbf Z$, so we can make sense of  $\sqrt{5}$ and thus $\varphi$ and $\overline{\varphi} = -1/\varphi$ in $\mathbf Z/p\mathbf Z$. Therefore $p \mid F_n$ is equivalent to $(-\varphi^2)^n \equiv 1 \bmod p\mathbf Z$, so the least such $n$ divides $|(\mathbf Z/p\mathbf Z)^\times|$, meaning $\tau(p) \mid (p-1)$.

Case 2: $p \equiv 2, 3 \bmod 5$.
For $p=2$, $\tau(p) = 3 = p+1$ since $F_3 = 2$. Now take $p > 2$.
Then $5$ is not a square in $\mathbf Z/p\mathbf Z$ but it is a square in the field $\mathbf F_{p^2}$.  (This is related to $R/pR \cong \mathbf F_{p^2}$ for these $p$.) We can make sense of  $\sqrt{5}$, and thus $\varphi$ and $\overline{\varphi} = -1/\varphi$, in $\mathbf F_{p^2}$.
I claim $(-\varphi^2)^{p+1} = 1$ in $\mathbf F_{p^2}^\times$. The $p$th power map on $\mathbf F_{p^2}$ is the nontrivial field automorphism, so $\varphi^p = \overline{\varphi}$ in $\mathbf F_{p^2}$.  Therefore in $\mathbf F_{p^2}$,
$$
(-\varphi^2)^{p+1} = (-1)^{p+1}(\varphi^{p+1})^2 =  (-1)^{p+1}(\overline{\varphi}\varphi)^2 = (-1)^{p+1}(-1)^2 = (-1)^{p+1}.
$$
For odd $p$, $(-1)^{p+1} = 1$ since $p+1$ is even.

I don't think it is realistic to expect a formula for $\tau(p)$.

Note: The number $\tau(p)$, the least $n \geq 1$ such that $F_n \equiv 0 \bmod p$, is not the same thing as the period of the Fibonacci sequence mod $p$, which is called the $p$th Pisano period and denoted $\pi(p)$: this is the least $n \geq 1$ such that $F_n \equiv 0 \bmod p$ and $F_{n+1} \equiv 1 \bmod p$. You can have $F_n \equiv 0 \bmod p$ and $F_{n+1} \not\equiv 1 \bmod p$ sometimes.  Always $\tau(p) \mid \pi(p)$. For example, $\tau(3) = 4$ while $\pi(3) = 8$, $\tau(7) = 8$ while $\pi(7) = 16$, and $\tau(13) = 7$ while $\pi(13) = 28$.

Your proposed formula for the least $n$ such that $p^m \mid F_n$ depends on your Step 4: you think the first time $p \mid F_n$ for some $n$, $p$ must divide that $F_n$ exactly once. This is the premise behind the $-1$ in the exponent of your rule $p^m \mid F_n \Longleftrightarrow \tau(p) p^{m-1} \mid n$. In similar problems about orders of numbers in modular arithmetic in $\mathbf Z$, this kind of rule often works but has counterexamples. It can happen that the first time $p$ divides a term like that, it divides it more than once.

For example, suppose you seek a formula for the decimal period of $1/p^m$, where $p$ is a prime other than $2$ or $5$ and $m \geq 1$. Call this decimal period $D(1/p^m)$. When $p = 7$, we have initially
$$
D(1/7) = 6, \ D(1/49) = 6\cdot 7, \ D(1/7^3) = 6 \cdot 7^2, \ D(1/7^4) = 6 \cdot 7^3,
$$
and in general $D(1/7^m) = 6 \cdot 7^{m-1} = D(1/7)7^{m-1}$ for all $m \geq 1$.
When $p = 3$, we have initially
$$
D(1/3) = 1, \ D(1/9) = 1, \ D(1/27) = 3, \ D(1/3^4) = 9, \ D(1/3^5) = 27,
$$
and in general $D(1/3^m) = 3^{m-2} = D(1/9)3^{m-2}$ for all $m \geq 2$, and $D(1/3) = 1$. The pattern does not start at $m = 1$. It resembles your formula starting at $m=2$, and this is why the power of $3$ in the formula is $3^{m-2}$, not $3^{m-1}$.

This is closely related to orders of units in modular arithmetic: for primes $p$ not $2$ or $5$, the decimal period of $1/p^m$ is the order of $10 \bmod p^m$.  We have $$D(1/p^m) = D(1/p)p^{m-1} \text{ for all } m \geq 1
$$
provided $p$ divides $10^{p-1}-1$ exactly once. This happens
at $p = 7$, since $10^6 - 1 = 7(142857)$ and $7 \nmid 142857$.
It does not happen at $p = 3$, since $10^{p-1} - 1 = 10^2-1 = 3^2$ is divisible by $3$ twice, not just once. In general, if $p^K$ is the highest power of $p$ dividing $10^{p-1}-1$, then $D(1/p^m) = D(1/p^K)p^{m-K}$ for all $m \geq K$, so there is a nice $p$-power pattern only starting at $m = K$.

The next prime $p$ after $3$ where $10^{p-1}-1$ is divisible by $p^2$ is $p = 487$:
$$
D(1/487) = 486, \ D(1/487^2) = 486, \ D(1/487^3) = 486 \cdot 487, D(1/487^4) = 486 \cdot 487^2,
$$
and in general $D(1/487^m) = 486 \cdot 487^{m-2}$ for all $m \geq 2$, but not at $m=1$.  And the next prime after $487$ where $10^{p-1}-1$ is divisible by $p^2$ is
$p = 56598313$:
$$
D(1/p) = p-1, \ D(1/p^2) = p-1, \
D(1/p^3) = (p-1) \cdot p,
$$
and in general $D(1/56598313^m) = 56598312 \cdot 56598313^{m-2}$ for $m \geq 2$.

Primes $p$ for which $p^2 \mid (10^{p-1} - 1)$ are called Wieferich primes to base $10$.  Only three of them are known ($3$, $487$, and $56598313$).  Standard probabilistic heuristics suggest the number of such $p \leq x$ should grow at the rate $\log \log x$, which tends to $\infty$ with $x$ , but extremely slowly. Note $\log(\log(487)) \approx 1.82 \approx 2$ and $\log(\log(56598313 )) \approx 2.88 \approx 3$. While we expect there to be infinitely many Wieferich primes to base 10, as Nick Katz once wrote "no computer experiment will ever convince us of it".

Returning to your problem, about $p^m \mid F_n$, everything hinges on how highly divisible $F_n$ is by $p$ the first time $p \mid F_n$. That is, everything depends on how highly divisible $\tau(p)$ is by $p$. If $p$ divides $\tau(p)$ just once, then your proposed formula is correct and proving it by induction is a complete nightmare that I will not think about. I prove the conjecture at the end of this post with a parameter $K$ depending on how highly divisible $F_n$ is by powers of $p$ the first time $F_n$ is divisible by $p$.

Example.  Consider $3^m \mid F_n$, which is the same as $(-\varphi^2)^n \equiv 1 \bmod 3^mR$.  The Fibonacci sequence begins
$1, 1, 2, 3, 5, \ldots$, so the least $n$ such that $3 \mid F_n$ is $n = 4$. This says $-\varphi^2 \bmod 3R$ has order 4 in $(R/3R)^\times$, and indeed
$$
(-\varphi^2)^2 = (\varphi+1)^2 = \varphi^2 + 2\varphi + 1 = 3\varphi + 2 \equiv 2 \bmod 3R,
$$
so $(-\varphi^2)^4 \equiv 2^2 \equiv 1 \bmod 3R$. Since $4-1$ is divisible by $3$ exactly once, the order of $-\varphi^2 \bmod 3^mR$ is $4 \cdot 3^{m-1}$ for all $m \geq 1$.

If $p^2 \mid F_{\tau(p)}$ then your proposed formula needs to be modified in a manner analogous to the way I described for decimal periods of $1/p^k$ if $p^2 \mid (10^{p-1}-1)$.  For the period $\pi(p)$ of the Fibonacci sequence modulo $p$, there is a name for primes $p$ where $p^2 \mid F_{\pi(p)}$: Wall-Sun-Sun primes. A Wikipedia page about them is here.  Probabilistic heuristics suggest that there should be infinitely many Wall-Sun-Sun primes, but not a single example is known.  For similar reasons, it is natural to conjecture that $p^2 \mid F_{\tau(p)}$ for infinitely many $p$, and that the set of such $p$ should be extremely sparse.

For a prime $p$ and integer $a \not= \pm 1$ such that $p \nmid a$,
let $O_a(p^m)$ be the order of $a \bmod p^m$. Here are formulas expressing $O_a(p^m)$ in terms of $O_a(p)$.  Note $O_a(p) \mid (p-1)$.

Case 1: $p > 2$. Let the highest power of $p$ dividing $a^{O_a(p)} - 1$ be $p^K$.  Then for all $m \geq K$, $O_a(p^m) = O_a(p)p^{m-K}$. That is, $a^n \equiv 1 \bmod p^m$ if and only if $O_a(p)p^{m-K} \mid n$.  In the "generic" situation, $K = 1$ (that is, $p$ divides $a^r - 1$ just once) and $O_a(p^m) = O_a(p)p^{m-1}$ for all $m \geq 1$.

Case 2: $p = 2$. If $a \equiv 1 \bmod 4$, let the highest power of $2$ dividing $a - 1$ be $2^K$, so $K \geq 2$.  Then for $m \geq K$, $O_a(2^m) = 2^{m-K}$. If $a \equiv 3 \bmod 4$, let the highest power of $2$ dividing $a + 1$ be $2^K$, so $K \geq 2$.  Then for $m \geq K+1$ (so starting at $m=K+1$, not at $m=K$), $O_a(2^m) = 2^{m-K}$.

To apply these order results to your Fibonacci question, we will use $p$-adic numbers and $p$-adic square roots of 5.

Theorem. If $F/\mathbf Q_p$ is unramified, $p \not= 2$,  and $u \in F$ has $|u|_p = 1$ with $u$ not a root of unity, let $u \bmod p\mathcal O_F$ have order $r$ and $p^K$ for $K \geq 1$ be the highest power of $p$ dividing $u^r - 1$.
Then for each $m \geq K$, the least $n \geq 1$ such that $u^n \equiv 1 \bmod p^m\mathcal O_F$ is $rp^{m-K}$.

If $F/\mathbf Q_2$ is unramified and $u \in F$ has $|u|_2 = 1$ with $u$ not a root of unity, let $u \bmod 4\mathcal O_F$ have order $r$ and $2^K$ for $K \geq 2$ be the highest power of $2$ dividing $u^r - 1$.
Then for each $m \geq K$, the least $n \geq 1$ such that $u^n \equiv 1 \bmod 2^m\mathcal O_F$ is $r2^{m-K}$.

Proof. For $p \not= 2$, the $p$-adic exponential function on $p\mathcal O_F$  is an isometry, so $u^r  = e^{p^Kv}$ where $|v|_p = 1$. For $m \geq 1$, if $u^n \equiv 1 \bmod p^m\mathcal O_F$ then $u^n \equiv 1 \bmod p\mathcal O_F$, so $r \mid n$. Write $n = rN$. Then
$$
u^n = (u^r)^N = (e^{p^{K}v})^N = e^{p^KvN},
$$
so
$$
|u^n - 1|_p = |e^{p^KvN}-1|_p = |p^KvN|_p = \frac{1}{p^K}|N|_p.
$$
Therefore
$$
u^n \equiv 1 \bmod p^m\mathcal O_F \Longleftrightarrow |u^n - 1|_p \leq \frac{1}{p^m} \Longleftrightarrow |N|_p \leq \frac{1}{p^{m-K}},
$$
so the least such $N$ is $p^{m-K}$ (when $m \geq K$).  Therefore the least value of $n = rN$ is $rp^{m-K}$.

Now we consider $p=2$.
The $2$-adic exponential function on $4\mathcal O_F$  is an isometry, so $u^r  = e^{2^Kv}$ where $|v|_p = 1$. For $m \geq 2$, if $u^n \equiv 1 \bmod 2^m\mathcal O_F$ then $u^n \equiv 1 \bmod 4\mathcal O_F$, so $r \mid n$. Write $n = rN$. Then
$$
u^n = (u^r)^N = (e^{2^{K}v})^N = e^{2^KvN},
$$
so
$$
|u^n - 1|_2 = |e^{p^KvN}-1|_2 = |2^KvN|_2 = \frac{1}{2^K}|N|_2.
$$
Therefore
$$
u^n \equiv 1 \bmod 2^m\mathcal O_F \Longleftrightarrow |u^n - 1|_2 \leq \frac{1}{2^m} \Longleftrightarrow |N|_2 \leq \frac{1}{2^{m-K}},
$$
so the least such $N$ is $2^{m-K}$ (when $m \geq K$). Therefore the least value of $n = rN$ is $r2^{m-K}$.

QED

We will prove your conjecture for $p \not= 5$ by applying this theorem to $F = \mathbf Q_p(\varphi) = \mathbf Q_p(\sqrt{5})$, which is unramified for $p \not= 5$, and $u = -\varphi^2$.

For $p \not= 5$,

$$
|F_n|_p = |\varphi^n - \overline{\varphi}^n|_p  = |(-\varphi^2)^n - 1|_p.
$$
Therefore the least $n$ such that $p \mid F_n$  is the order of $-\varphi^2 \bmod p\mathcal O_{\mathbf Q_p(\sqrt{5})}$, so $r = \tau(p)$ when $p \not= 2, 5$.

First suppose $p \not= 2$ (so $p \not= 2$ or $5$). Let $p^K$ for $K \geq 1$ be the highest power of $p$ dividing $\tau(p)$. The theorem tells us that for each $\boxed{m \geq K}$, the least $n \geq 1$ such that $(-\varphi^2)^n \equiv 1 \bmod p^m\mathcal O_{\mathbf Q_p(\sqrt{5})}$ is $\tau(p)p^{m-K}$, so the least $n \geq 1$ such that $F_n \equiv 0 \bmod p^m$ is $\tau(p)p^{m-K}$. When $K = 1$, meaning $p$ divides $\tau(p)$ just once, we have shown for each $m \geq 1$ that the least $n \geq 1$ such that $F_n \equiv 0 \bmod p^m$ is $\tau(p)p^{m-1}$.

Next suppose $p = 2$. When is $|(-\varphi^2)^n -  1|_2 \leq 1/4$ for the first time?  This is the same as $F_n$ being divisible by $4$ for the first time.  The initial even Fibonacci numbers are $F_3 = 2$ and $F_6 = 8$, so the least $n$ is $n=6$.  In terms of $\varphi$,
$$
(-\varphi^2)^3 = -8\varphi - 5 \not\equiv 1 \bmod 4, \ \
(-\varphi^2)^6 = 144 \varphi + 89 \equiv 1 \bmod 4.
$$
The highest power of $2$ dividing $(-\varphi^2)^6 - 1 = 144\varphi + 88$ is $8 = 2^3$, so $r = 6$ and $K = 3$. The theorem tells us that for each $m \geq 3$, the least $n \geq 1$ such that $(-\varphi^2)^n \equiv 1 \bmod 2^m$ is $6 \cdot 2^{m-3} = 3 \cdot 2^{m-2}$. So for $\boxed{m \geq 3}$, the least $n \geq 1$ such that $F_n \equiv 0 \bmod 2^m$ is $n = r \cdot 2^{m-K} = 3 \cdot 2^{m-2}$.
We can work out the cases $m = 1$ and $2$ directly: the least $n$ such that $F_n$ is even (equivalently, $(-\varphi^2)^n \equiv 1 \bmod 2$) is $n=3$ and the least $n$ such that $F_n \equiv 0 \bmod 4$ (equivalently, $(-\varphi^2)^n \equiv 1 \bmod 4$) is $n = 6$.


问:
Examples of almost compact spaces.
This question is a follow up question to this one

Definition 1. A non-compact Hausdorff topological space X is called almost compact if its Stone-Cech compactification coincides with its one point compactification.

The only two examples of almost compact spaces I know are from the book Pseudocompact topological spaces. M. Hrusak, A. Tamariz-Mascarua, M. Tkachenko. On the page 17 authors say that $[0,\omega_1)$ is almost compact and Mrowka-Isbell space $\Psi(\mathcal{A})$ is almost compact for some specific maximal almost disjoint family $\mathcal{A}\subset 2^\omega$.

I would like to know more on almost compact spaces, but I found almost nothing on this subject.

Questions:

What are other examples of almost compact spaces?

Is true that $\beta\mathbb{N}\setminus\{p\}$ is almost compact for $p\in\beta\mathbb{N}\setminus\mathbb{N}$ ?

Is it true that $X\setminus \{p\}$ is almost compact whenever $X$ is extremally disconnected and $p\in X$.

答:
Let $Y=\beta \Bbb N \setminus \{p\}$ for $p \in \Bbb N^\ast$.
Then theorem 6.4/6.7 from Gillman and Jerrison tells us that $\beta Y=\beta \Bbb N$ and another standard theorem tells us that the one-point compactification of $Y$ also equals $\beta \Bbb N$. So $Y$ is almost compact.

8L in that book also shows that $\Omega$ (the square of $\omega_1 +1$ with $(\omega_1, \omega_1)$ removed), is another example of an almost compact space, as is the Tychonoff plank, which is closely related. They are introduced in excercise 10R as spaces with a unique uniformity and the non-compact ones are characterised as those Tychonoff $X$ that have $|\beta X\setminus X| =1$.

Title: Examples of almost compact spaces.


问:
Are projection and norm enough to define an inner product?
Given an inner product, one can define a projection and a norm. Can we do the opposite?

That is, suppose we have:

a complex vector space V

a norm $|V|^2 : V \rightarrow \mathbb{R}$ such that:

is  posite definite

$|\alpha V|^2 = |\alpha|^2 |V|$ with $\alpha \in \mathbb{C}$

a family of projections $\forall v \in V \; \exists \mathbf{P}_v : V \rightarrow V$ such that:

$\mathbf{P}_v ^2 = \mathbf{P}_v$

$\mathbf{P}_v (\alpha w_1 + \beta w_2) = \alpha \mathbf{P}_v (w_1) + \beta \mathbf{P}_v (w_2)$

$\forall w \in V$ $\mathbf{P}_v(w)=\alpha v$ with $\alpha \in \mathbb{C}$

$\mathbf{P}_v(v)=v$

Is the map $\langle \cdot , \cdot \rangle : V \times V \rightarrow \mathbb{C}$ defined such that $\mathbf{P}_v(w)=\frac{\langle v, w \rangle}{|v|^2} v$ an inner product?

$\langle \cdot , \cdot \rangle$ is positive definite. $\mathbf{P}_v(v) = v$. $\langle v, v \rangle = |v|^2$. The norm induced by the map is the original norm which is positive definite.

$\langle \cdot , \cdot \rangle$ is linear in the second argument because the projection is linear.

Conjugate symmetry is the only thing that I am missing. Is it an extra requirement, or can it be derived by the previous?

答:
The family of projections with one-dimensional range is a red herring:
It is equivalent to require a family of linear functionals (if you assume the axiom of choice). In the case of a normed space, it is equivalent to require a family of bounded projections/bounded linear functionals by Hahn-Banach (again, if you assume the axiom of choice).

Since you require nothing on the family of functionals/projections with respect to $v$, there are plenty of functionals/projections violating the conjugate symmetry.

A prominent family of examples are semi-inner products, see https://en.wikipedia.org/wiki/L-semi-inner_product
(The examples given there are linear in the first argument instead of the second, but this is of course just a notational difference.)


问:
Complex Exponential Expansion
I'm currently working through a problem using exponentials with complex numbers!
I am simply wondering what the steps are to go from

$$\pi \cdot (e^{i\frac{\pi}{3}} - e^{i\pi})  = \frac{\pi}{2}\cdot(3 + i\sqrt{3})$$

If someone could point me in the direction of the correct expansion that would be very helpful!

Thank you

答:
Dear Sarah, first of all, you may cancel the factor of $\pi$ on both sides.

Second, $\exp(ix)=\cos(x)+i\sin(x)$ and
$$\cos(\pi/3)=\frac 12, \quad \sin(\pi/3) = \frac{\sqrt{3}}{2}, \quad e^{i\pi} = -1$$
so
$$e^{i\pi/3} - e^{i\pi} = \frac 12 + i \frac{\sqrt{3}}{2} + 1 = \frac 32 + i\frac{\sqrt{3}}{2} = \frac{1}{2} (3 +\sqrt{3}) $$
just like you wrote. If you need it, $\cos(\pi/3)$ is $\cos (60°)$ - thanks, Hans - which is equal to $1/2$ because of a simple picture of an equilateral triangle. $\sin(\pi/3)$ is calculated via $\sin^2x+\cos^2 x = 1$. For a derivation of the exponential of an imaginary number in terms of cosines and sines, see e.g.

Complex Exponents

Title: Complex Exponential Expansion


问:
Open annulus covers any orientable non simply-connected surface without boundary.

$\textbf{Problem:}$ Let $\Sigma$ be an orientable surface without
boundary, possibly non-compact, with the non-trivial fundamental
group. Then there is a covering map $p:\Bbb S^1\times \Bbb R\to
\Sigma$.

$\textbf{Attempt:}$ Let $\alpha\in \pi_1(\Sigma)$ be a non-trivial element and consider the covering $p:\Sigma_\alpha\to \Sigma$ corresponding to the subgroup $\langle \alpha\rangle$ of $\pi_1(\Sigma)$. Then, the fundamental group of the surface $\Sigma_\alpha$ is $\Bbb Z$. Note that $\Sigma_\alpha$ has no boundary, as $p$ is a local homeomorphism. Also, $\Sigma_\alpha$ can not be compact surface by classification theorem. So, $\Sigma_\alpha$ is an open surface. Now, every connected open surface with finitely generated fundamental group is the interior of some compact surface. Also, every manifold is homotopically equivalent to its interior. So, $\Sigma_\alpha$ can be open Möbius strip or open annulus.

$\bullet$  I don't know how to show open Möbius strip can not cover an
orientable surface. More generally, is it true that any non-orientable manifold can not cover an orientable manifold?

$\bullet$ Are the lines in the Attempt portion correct?

$\bullet$ Is there any quick proof of the original problem?

答:
Your attempt looks fine, although you may want to justify why $\langle\alpha\rangle \cong \mathbb{Z}$ (as opposed to $\mathbb{Z}_k$). It is true that any covering space of an orientable manifold is necessarily orientable.

Let $p : M \to N$ is a continuous covering map of topological manifolds. Equip $N$ with a smooth atlas $\{(U_{\alpha}, \varphi_{\alpha}\}_{\alpha \in A}$ such that $U_{\alpha}$ is a connected, evenly covered neighbourhood for all $\alpha \in A$. Denote the connected components of $p^{-1}(U_{\alpha})$ by $\{V_{\alpha}^i\}_{i \in I}$; note that $p|_{V_{\alpha}^i} : V_{\alpha}^i \to U_{\alpha}$ is a homeomorphism. Then the smooth atlas on $N$ induces a smooth atlas on $M$ given by $\{(V_{\alpha}^i, \varphi_{\alpha}\circ p|_{V_{\alpha}^i})\}_{\alpha \in A, i \in I}$; with respect to these smooth structures on $M$ and $N$, the map $p$ is smooth. Now suppose $V_{\alpha}^i\cap V_{\beta}^j \neq \emptyset$, then

$$(\varphi_{\alpha}\circ p|_{V_{\alpha}^i})\circ(\varphi_{\beta}\circ p|_{V_{\beta}^j})^{-1} = \varphi_{\alpha}\circ p|_{V_{\alpha}^i}\circ p|_{V_{\beta}^j}^{-1}\circ \varphi_{\beta}^{-1} = \varphi_{\alpha}\circ\varphi_{\beta}^{-1}.$$

In particular, the atlas on $M$ is orientable if and only if the atlas on $N$ is. Therefore, if $p : M \to N$ is a covering map and $N$ is orientable, then so is $M$. Note, the above assumes that $N$ admits a smooth structure (which is always the case in dimension two). The conclusion is also true without this hypothesis, but the proof relies on a different characterisation of orientability (for example, in terms of homology).


问:
Why does Friedberg say that the role of the determinant is less central than in former times?
I am taking a proof-based introductory course to Linear Algebra as an undergrad student of Mathematics and Computer Science. The author of my textbook (Friedberg's Linear Algebra, 4th Edition) says in the introduction to Chapter 4:

The determinant, which has played a prominent role in the theory of linear algebra, is a special scalar-valued function defined on the set of square matrices. Although it still has a place in the study of linear algebra and its applications, its role is less central than in former times.

He even sets up the chapter in such a way that you can skip going into detail and move on:

For the reader who prefers to treat determinants lightly, Section 4.4 contains the essential properties that are needed in later chapters.

Could anyone offer a didactic and simple explanation that refutes or asserts the author's statement?

答:
In my mathematics experience, determinants are an essential theoretical tool. It is true, you can almost never compute it in big matrices, but lot of stuff is not necessarily computable and still very useful: for example, cohomology or fundamental groups of spaces (invariants attached to topological spaces).

An example of application among others:

In multivariate analysis, the change of variables formula for integrals includes the determinant (isn't useful to change variables in integrals?)

Any matrix is diagonalizable (up to an infinitesimal perturbation). This has also a significance in applications. Indeed, the fact that a square matrix $A$ of size $n$ has distinct eigenvalues can be reformulated as its characteristic polynomial $p(A)= \det(A-\lambda I) $ having only single roots. Again, this can be reformulated as the resultant $r_A$ of $p(A) $ and $p'(A) $ being different from zero when calculated in zero. Since all these operations are polynomial in the entries of the matrix $A$, a constraint of the form "polynomial $\neq 0$ " is almost always true (its solutions are dense).

Given a finite dimensional representation of a group $\rho: G \to GL(n) $, the determinant is a one dimensional representation;

Given two square matrices $A, B$ of size $n$, such that $AB-BA $ is invertible and $A^2+AB+B^2=0$, then $n$ is a multiple of 3.

Given a vector bundle $E $ over a space $M$ , its determinant bundle is a line bundle (which are much easier to study).

In algebraic number theory, if you have a finite extension of number fields $L/K$, the norm $N_{L/K}$ of an element $x \in L$ is defined as the determinant of the multiplication action of $x$ as a $K$ -linear map. For example, if one takes the Gauss integers, one can show with a bit of theory that any prime number congruent to 1 modulo 4 can be written as $u^2+v^2$. Multiplicativity of the determinant allows to extend the result to any number that has prime numbers congruent to 1 modulo 4 in his decomposition.

The determinant of the Frobenius (I think?) matrices in representation theory yield the not so trivial factorization of polynomials

$$ X^3+Y^3+Z^3 -3XYZ = (X+Y+Z) (X+\omega Y+\omega^2 Z) (X+\omega^2 Y+ \omega Z) $$
Where $\omega$ is the third root of unity.

LaTex has a dedicated symbol $\det$ for the determinant. Look here: $\Gauss$. So gauss is less important than the determinant, and Gauss is important.

And there are much more. These are the first few examples that come to my mind. The fact that a number governes the invertibility of a matrix is a very peculiar fact; the absence of this criterion over other rings yield very serious problems.

Sincerely,

The "we love determinants" Association