问:
What is the connection between sum of tuple products and permutations by cycle?
From my previous question Disjoint sets in a combinatoral sum (continued), let $X = \{x_1, \dots, x_n\}$ be a set and
$$f_X(m) = \sum_{X' \in \binom{X}{m}} \prod_{x \in X'} x$$

Let $S_k = \sum x_i^k$.  We can calculate with inclusion-exclusion:

$$f_X(2) = \frac{1}{2!} (S_1^2 - S_2)$$

$$f_X(3) = \frac{1}{3!} (S_1^3 - 3 S_2 S_1 + 2 S_3) $$

and so on.

In the previous answer I discovered a connection between permutations by cycle type and the coefficients of the expression $f_X(m)$ (table at https://oeis.org/A181897), and came up with a tenuous explanation. Can someone confirm my answer and elucidate if my reasoning is correct? Also what is the connection to integer partitions? I am not that experienced with combinatorics or group theory so an answer that is as elementary as possible (possibly at the expense of generality) is appreciated.
I'm looking for references to this problem which has surely been explored in the past.

答:
To find a closed form for $f_X(m)$ in terms of the $S_\ell$ we first
require the exponential formula for the cycle index of the unlabeled set
operator

$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}.$$

Let  $A$ be  a  generating  function in  some number  of
variables and  let $B =  c_B X_B$  a contributing monomial  term where
$c_B$ is the leading coefficient  (positive) and $X_B$ the product
of the variables to their respective powers, so that $A= \sum_{B\in A}
c_B  X_B$. We  then have  from  first principles  that the  generating
function of sets drawn from $A$ containing $m$ elements is

$$[z^m] \prod_{B\in A} (1 + z X_B)^{c_B}.$$

Manipulate this to obtain

$$[z^m] \prod_{B\in A} \exp \log (1 + z X_B)^{c_B}
\\ = [z^m] \prod_{B\in A}
\exp \left(- c_B \log \frac{1}{1 + z X_B} \right)
\\ = [z^m] \exp
\sum_{B\in A} \left(- c_B \log \frac{1}{1 + z X_B} \right)
\\ = [z^m] \exp
\sum_{B\in A} \left(- c_B \sum_{\ell\ge 1} (-1)^\ell X_B^\ell
\frac{z^\ell}{\ell} \right)
\\ = [z^m] \exp \left( \sum_{\ell\ge 1} (-1)^{\ell-1}
\left(\sum_{B\in A} c_B X_B^\ell\right)
\frac{z^\ell}{\ell} \right).$$

But $\sum_{B\in A} c_B X_B^\ell$ is by definition the Polya  substitution
applied to $A$ through the cycle index variable $a_\ell$ and we have
proved the exponential formula for the unlabeled set operator, which says
that

$$Z(P_m) = [z^m] \exp
\left(\sum_{\ell\ge 1} (-1)^{\ell-1}
 a_\ell \frac{z^\ell}{\ell}\right).$$

Now we have from the definition applying PET that

$$f_X(m) = Z(P_m; x_1+x_2+\cdots+x_n).$$

Hence

$$f_X(m) = [z^m] \exp
\left(\sum_{\ell\ge 1} (-1)^{\ell-1}
 S_\ell \frac{z^\ell}{\ell}\right)
\\ = [z^m] \prod_{\ell\ge 1}
\exp \left( (-1)^{\ell-1}
 S_\ell \frac{z^\ell}{\ell}\right)
\\ = [z^m] \prod_{\ell\ge 1}
\sum_{q\ge 0} \frac{1}{q!} (-1)^{q(\ell-1)}
S_\ell^q \frac{z^{\ell q}}{\ell^q}.$$

We want to  expand this product. We are interested  in the coefficient
on $[z^m]$  so we consider  integer partitions $\lambda\vdash  m$. Let
the  partition be  $1^{q_1} 2^{q_2}  3^{q_3} \cdots$  where all  but a
finite number of exponents are zero. We now obtain

$$[z^m] \sum_{\lambda\vdash m} \prod_{\ell\ge 1}
\frac{1}{q_\ell!}
(-1)^{(\ell-1)q_\ell} S_\ell^{q_\ell}
\frac{z^{\ell q_\ell}}{\ell^{q_\ell}}.$$

Since $\sum_{\ell\ge 1} \ell q_\ell = m$ this becomes

$$\sum_{\lambda\vdash m} \prod_{\ell\ge 1}
\frac{1}{q_\ell!}
(-1)^{(\ell-1)q_\ell} S_\ell^{q_\ell}
\frac{1}{\ell^{q_\ell}}
\\ = (-1)^m \sum_{\lambda\vdash m}
(-1)^{\sum_{\ell\ge 1} q_\ell}
\prod_{\ell\ge 1} S_\ell^{q_\ell}
\frac{1}{q_\ell! \times \ell^{q_\ell}}.$$

We thus get the closed form

$$\bbox[5px,border:2px solid #00A000]{
f_X(m) = \frac{(-1)^m}{m!} \sum_{\lambda\vdash m}
(-1)^{\sum_{\ell\ge 1} q_\ell}
\left(\prod_{\ell\ge 1} S_\ell^{q_\ell} \right)
\left(m! \prod_{\ell\ge 1}
\frac{1}{q_\ell! \times \ell^{q_\ell}}\right).}$$

This yields for example

$$f_X(4) =
\frac{1}{4!}
\left({S_{{1}}}^{4}-6\,{S_{{1}}}^{2}S_{{2}}
+8\,S_{{1}}S_{{3}}+3\,{S_{{2}}}^{2}-6\,S_{{4}}\right)$$

and

$$f_X(5) =
\frac{1}{5!}
\left({S_{{1}}}^{5}-10\,{S_{{1}}}^{3}S_{{2}}+20\,{S_{{1}}}^{2}S_{{3}}
+15\,S_{{1}}{S_{{2}}}^{2}-30\,S_{{1}}S_{{4}}
-20\,S_{{2}}S_{{3}}+24\,S_{{5}}\right).$$

We now show that

$$m! \prod_{\ell\ge 1}
\frac{1}{q_\ell! \times \ell^{q_\ell}}$$

counts the number of permutations with cycle structure $\lambda.$

First, selecting the values to go on the cycles yields the multinomial
coefficient

$$\frac{m!}{\prod_{\ell\ge 1} (\ell!)^{q_\ell}}.$$

A set of $\ell$ values gives $\frac{\ell!}{\ell}$ cycles:

$$\prod_{\ell\ge 1} \left( \frac{\ell!}{\ell} \right)^{q_\ell}.$$

Any permutation of the cycles of length $\ell$ yields the same
permutation:

$$\prod_{\ell\ge 1} \frac{1}{q_\ell!}.$$

Multiply these to obtain

$$\frac{m!}{\prod_{\ell\ge 1} (\ell!)^{q_\ell}}
\prod_{\ell\ge 1} \left( \frac{\ell!}{\ell} \right)^{q_\ell}
\prod_{\ell\ge 1} \frac{1}{q_\ell!}
= m! \prod_{\ell\ge 1} \frac{1}{q_\ell! \times\ell^{q_\ell}}.$$

This is the claim and concludes the argument.

As an addendum  we have by inspection for  the  boxed closed form that
it is  given by a substitution  into $Z(Q_m)$, the cycle  index of the
symmetric group (the variable $S$ is in use already), which is

$$Z(Q_m; a_\ell = (-1)^{\ell -1} S_\ell).$$

With $Z(Q_m)$  being the average  of all $m!$  permutations factorized
into cycles, where $a_\ell$ stands for a cycle of $\ell$ elements we have

$$Z(Q_m) =
\frac{1}{m!} \sum_{\lambda\vdash m}
\left(\prod_{\ell\ge 1} a_\ell^{q_\ell} \right)
\left(m! \prod_{\ell\ge 1}
\frac{1}{q_\ell! \times \ell^{q_\ell}}\right).$$

Now  put $a_\ell = (-1)^{\ell-1}  S_\ell$ to  get the  boxed form.  The
substitution $a_\ell := (-1)^{\ell-1} a_\ell$  converts  the  unlabeled
multiset operator into the unlabeled  set operator through their cycle
indices.

There is also some Maple code for  the curious who would want to study
these polynomials and verify the correctness of the closed form.

```
with(combinat);

fX1 :=
proc(m, n)
local t;

    coeff(expand(mul(1+u*x[t], t=1..n)), u, m);
end;

fX2 :=
proc(m)
    local res, part, mset, ent;

    res := 0;

    part :=  firstpart(m);

    while type(part, list) do
        mset := convert(part, `multiset`);

        res := res +
        (-1)^add(ent[2], ent in mset)
        * mul(S[ent[1]]^ent[2], ent in mset)
        * m!/mul(ent[2]!*ent[1]^ent[2], ent in mset);

        part := nextpart(part);
    od;

    res*(-1)^m/m!;
end;

fX :=
proc(m, n)
local l, sl, t;

    sl := [seq(S[l] = add(x[t]^l, t=1..n), l=1..m)];
    expand(subs(sl, fX2(m)));
end;

pet_cycleind_symm :=
proc(n)
local l;
option remember;

    if n=0 then return 1; fi;

    expand(1/n*add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;

fX2A :=
proc(m)
local sl, l;

    sl := [seq(a[l] = (-1)^(l-1) * S[l], l=1..m)];
    subs(sl, pet_cycleind_symm(m));
end;
```

Body:
To find a closed form for $f_X(m)$ in terms of the $S_\ell$ we first
require the exponential formula for the cycle index of the unlabeled set
operator

$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\textsc{SET}.$$

Let  $A$ be  a  generating  function in  some number  of
variables and  let $B =  c_B X_B$  a contributing monomial  term where
$c_B$ is the leading coefficient  (positive) and $X_B$ the product
of the variables to their respective powers, so that $A= \sum_{B\in A}
c_B  X_B$. We  then have  from  first principles  that the  generating
function of sets drawn from $A$ containing $m$ elements is

$$[z^m] \prod_{B\in A} (1 + z X_B)^{c_B}.$$

Manipulate this to obtain

$$[z^m] \prod_{B\in A} \exp \log (1 + z X_B)^{c_B}
\\ = [z^m] \prod_{B\in A}
\exp \left(- c_B \log \frac{1}{1 + z X_B} \right)
\\ = [z^m] \exp
\sum_{B\in A} \left(- c_B \log \frac{1}{1 + z X_B} \right)
\\ = [z^m] \exp
\sum_{B\in A} \left(- c_B \sum_{\ell\ge 1} (-1)^\ell X_B^\ell
\frac{z^\ell}{\ell} \right)
\\ = [z^m] \exp \left( \sum_{\ell\ge 1} (-1)^{\ell-1}
\left(\sum_{B\in A} c_B X_B^\ell\right)
\frac{z^\ell}{\ell} \right).$$

But $\sum_{B\in A} c_B X_B^\ell$ is by definition the Polya  substitution
applied to $A$ through the cycle index variable $a_\ell$ and we have
proved the exponential formula for the unlabeled set operator, which says
that

$$Z(P_m) = [z^m] \exp
\left(\sum_{\ell\ge 1} (-1)^{\ell-1}
 a_\ell \frac{z^\ell}{\ell}\right).$$

Now we have from the definition applying PET that

$$f_X(m) = Z(P_m; x_1+x_2+\cdots+x_n).$$

Hence

$$f_X(m) = [z^m] \exp
\left(\sum_{\ell\ge 1} (-1)^{\ell-1}
 S_\ell \frac{z^\ell}{\ell}\right)
\\ = [z^m] \prod_{\ell\ge 1}
\exp \left( (-1)^{\ell-1}
 S_\ell \frac{z^\ell}{\ell}\right)
\\ = [z^m] \prod_{\ell\ge 1}
\sum_{q\ge 0} \frac{1}{q!} (-1)^{q(\ell-1)}
S_\ell^q \frac{z^{\ell q}}{\ell^q}.$$

We want to  expand this product. We are interested  in the coefficient
on $[z^m]$  so we consider  integer partitions $\lambda\vdash  m$. Let
the  partition be  $1^{q_1} 2^{q_2}  3^{q_3} \cdots$  where all  but a
finite number of exponents are zero. We now obtain

$$[z^m] \sum_{\lambda\vdash m} \prod_{\ell\ge 1}
\frac{1}{q_\ell!}
(-1)^{(\ell-1)q_\ell} S_\ell^{q_\ell}
\frac{z^{\ell q_\ell}}{\ell^{q_\ell}}.$$

Since $\sum_{\ell\ge 1} \ell q_\ell = m$ this becomes

$$\sum_{\lambda\vdash m} \prod_{\ell\ge 1}
\frac{1}{q_\ell!}
(-1)^{(\ell-1)q_\ell} S_\ell^{q_\ell}
\frac{1}{\ell^{q_\ell}}
\\ = (-1)^m \sum_{\lambda\vdash m}
(-1)^{\sum_{\ell\ge 1} q_\ell}
\prod_{\ell\ge 1} S_\ell^{q_\ell}
\frac{1}{q_\ell! \times \ell^{q_\ell}}.$$

We thus get the closed form

$$\bbox[5px,border:2px solid #00A000]{
f_X(m) = \frac{(-1)^m}{m!} \sum_{\lambda\vdash m}
(-1)^{\sum_{\ell\ge 1} q_\ell}
\left(\prod_{\ell\ge 1} S_\ell^{q_\ell} \right)
\left(m! \prod_{\ell\ge 1}
\frac{1}{q_\ell! \times \ell^{q_\ell}}\right).}$$

This yields for example

$$f_X(4) =
\frac{1}{4!}
\left({S_{{1}}}^{4}-6\,{S_{{1}}}^{2}S_{{2}}
+8\,S_{{1}}S_{{3}}+3\,{S_{{2}}}^{2}-6\,S_{{4}}\right)$$

and

$$f_X(5) =
\frac{1}{5!}
\left({S_{{1}}}^{5}-10\,{S_{{1}}}^{3}S_{{2}}+20\,{S_{{1}}}^{2}S_{{3}}
+15\,S_{{1}}{S_{{2}}}^{2}-30\,S_{{1}}S_{{4}}
-20\,S_{{2}}S_{{3}}+24\,S_{{5}}\right).$$

We now show that

$$m! \prod_{\ell\ge 1}
\frac{1}{q_\ell! \times \ell^{q_\ell}}$$

counts the number of permutations with cycle structure $\lambda.$

First, selecting the values to go on the cycles yields the multinomial
coefficient

$$\frac{m!}{\prod_{\ell\ge 1} (\ell!)^{q_\ell}}.$$

A set of $\ell$ values gives $\frac{\ell!}{\ell}$ cycles:

$$\prod_{\ell\ge 1} \left( \frac{\ell!}{\ell} \right)^{q_\ell}.$$

Any permutation of the cycles of length $\ell$ yields the same
permutation:

$$\prod_{\ell\ge 1} \frac{1}{q_\ell!}.$$

Multiply these to obtain

$$\frac{m!}{\prod_{\ell\ge 1} (\ell!)^{q_\ell}}
\prod_{\ell\ge 1} \left( \frac{\ell!}{\ell} \right)^{q_\ell}
\prod_{\ell\ge 1} \frac{1}{q_\ell!}
= m! \prod_{\ell\ge 1} \frac{1}{q_\ell! \times\ell^{q_\ell}}.$$

This is the claim and concludes the argument.

As an addendum  we have by inspection for  the  boxed closed form that
it is  given by a substitution  into $Z(Q_m)$, the cycle  index of the
symmetric group (the variable $S$ is in use already), which is

$$Z(Q_m; a_\ell = (-1)^{\ell -1} S_\ell).$$

With $Z(Q_m)$  being the average  of all $m!$  permutations factorized
into cycles, where $a_\ell$ stands for a cycle of $\ell$ elements we have

$$Z(Q_m) =
\frac{1}{m!} \sum_{\lambda\vdash m}
\left(\prod_{\ell\ge 1} a_\ell^{q_\ell} \right)
\left(m! \prod_{\ell\ge 1}
\frac{1}{q_\ell! \times \ell^{q_\ell}}\right).$$

Now  put $a_\ell = (-1)^{\ell-1}  S_\ell$ to  get the  boxed form.  The
substitution $a_\ell := (-1)^{\ell-1} a_\ell$  converts  the  unlabeled
multiset operator into the unlabeled  set operator through their cycle
indices.

There is also some Maple code for  the curious who would want to study
these polynomials and verify the correctness of the closed form.

with(combinat);

fX1 :=
proc(m, n)
local t;

    coeff(expand(mul(1+u*x[t], t=1..n)), u, m);
end;

fX2 :=
proc(m)
    local res, part, mset, ent;

    res := 0;

    part :=  firstpart(m);

    while type(part, list) do
        mset := convert(part, `multiset`);

        res := res +
        (-1)^add(ent[2], ent in mset)
        * mul(S[ent[1]]^ent[2], ent in mset)
        * m!/mul(ent[2]!*ent[1]^ent[2], ent in mset);

        part := nextpart(part);
    od;

    res*(-1)^m/m!;
end;

fX :=
proc(m, n)
local l, sl, t;

    sl := [seq(S[l] = add(x[t]^l, t=1..n), l=1..m)];
    expand(subs(sl, fX2(m)));
end;

pet_cycleind_symm :=
proc(n)
local l;
option remember;

    if n=0 then return 1; fi;

    expand(1/n*add(a[l]*pet_cycleind_symm(n-l), l=1..n));
end;

fX2A :=
proc(m)
local sl, l;

    sl := [seq(a[l] = (-1)^(l-1) * S[l], l=1..m)];
    subs(sl, pet_cycleind_symm(m));
end;


问:
Proof of $(ab)^{-1} = a^{-1}b^{-1}$
I wonder if you would consider this proof of my hypothesis of

$(ab)^{-1} = a^{-1}b^{-1}$ with $a,b\ne0$ correct.

By definition of the inverse of $ab$: $$(ab)^{-1}(ab)=1$$

Multiplying the equation with $a^{-1}$ and $b^{-1}$ gives $$(ab)^{-1}aba^{-1}b^{-1} = 1a^{-1}b^{-1}=a^{-1}b^{-1}$$ using property $1x=x$.
On the left hand side I rearrange the terms using the property $xy=yx$ to $$(ab)^{-1}aa^{-1}bb^{-1}= a^{-1}b^{-1}$$ which gives by using again the definition of the inverse $$(ab)^{-1}1\cdot1= a^{-1}b^{-1}$$
Using the property $1x=x$ on the left twice I have the answer.

答:
Yes, this is correct. You could even do this without assuming commutativity, but then you would get
$$(ab)^{-1} = b^{-1}a^{-1}.$$

ContentLicense: CC BY-SA 4.0

Title: Proof of $(ab)^{-1} = a^{-1}b^{-1}$


答:
You are correct, but this can be written more concisely as
$(ab)a^{-1}b^{-1}=a(ba^{-1})b^{-1}=a(a^{-1}b)b^{-1}=(aa^{-1})(bb^{-1})=1\cdot1=1$

The $a^{-1}b^{-1}(ab)$ case is similar.

Title: Proof of $(ab)^{-1} = a^{-1}b^{-1}$

Body:
I wonder if you would consider this proof of my hypothesis of

$(ab)^{-1} = a^{-1}b^{-1}$ with $a,b\ne0$ correct.

By definition of the inverse of $ab$: $$(ab)^{-1}(ab)=1$$

Multiplying the equation with $a^{-1}$ and $b^{-1}$ gives $$(ab)^{-1}aba^{-1}b^{-1} = 1a^{-1}b^{-1}=a^{-1}b^{-1}$$ using property $1x=x$.
On the left hand side I rearrange the terms using the property $xy=yx$ to $$(ab)^{-1}aa^{-1}bb^{-1}= a^{-1}b^{-1}$$ which gives by using again the definition of the inverse $$(ab)^{-1}1\cdot1= a^{-1}b^{-1}$$
Using the property $1x=x$ on the left twice I have the answer.


问:
If $a,b >0$ and $a+b \sqrt{d} = \left(\alpha+\beta \sqrt{d} \right)^k$ with $k>1$ then $|\alpha|<a$ and $0<|\beta|<b$
I'm trying to show that if $d$ is a positive, squarefree integer with $d \not \equiv 1 (\operatorname{mod} 4)$ and $a,b, \alpha, \beta$ are integers with $a,b >0 $ and such that $a+b\sqrt{d}$ and $\alpha+\beta\sqrt{d}$ are units in $\mathbb{Q}(\sqrt{d})$ and satisfy
$$a+b \sqrt{d} = \left(\alpha+\beta \sqrt{d} \right)^k$$
for some integer $k>1$, then $|\alpha|<a$ and $0<|\beta|<b$.

I have no idea from where to start, I've shown that it is true for $k=2$ looking for an inductive proof, buy I got stuck really fast so any help would be really appreciated.

答:
Clearly $a+b\sqrt d>1$. It follows that $|\alpha+\beta\sqrt{d}|>1$ also. Therefore $\alpha$ and $\beta$ have the same sign. If that sign is negative, then $k$ must be even, and we might as well replace $\alpha$ and $\beta$ with their negatives. So without loss of generality $\alpha>0,\beta>0$.

Write $u=\alpha+\beta\sqrt d$, $v=\alpha-\beta\sqrt d$. We know that $v=\pm 1/u$, but either sign is possible. Anyway $|v|<1$. By conjugation we see that $v^k=a-b\sqrt{d}$. Therefore
$$
2a=u^k+v^k>u^k-1,\qquad 2\alpha=u+v\le u+1.
$$
But $u\ge1+\sqrt2$ (with equality possible when $d=2$), so elementary tools give the inequality $u^k\ge u^2>u+2$. This implies
$$
2a>u^k-1>u+1\ge 2\alpha.
$$
I suspect that the claim $b>\beta$ can be proven with a similar method. $2b\sqrt d=u^k-v^k$ etc.


问:
Here's a fun math/Zoom question for you all:
If I have 6 people in a group and each participant needs to meet each other in pairs of 2, how could I arrange Zoom Breakout rooms so that no one meets another twice? (3 Breakout rooms consisting of 2 people per session)

Here are the possible pairs I have (the letters represent initials of participants):
K x R, K x J, K x C, K x N, K x Z, R x J, R x C, R x N, R x Z, J x C, J x N, J x Z, C x N, C x Z, N x Z

So with this in mind, it should break down into 5 total sessions with 3 breakout rooms in each session and 2 participants per breakout room. I just can't seem to figure out how to configure it without duplicate people being in a session.

答:
What you're describing is the $1$-factorization (i.e., splitting into $1$-factors, which are perfect matchings) of a complete graph on $n$ vertices.

From the Wikipedia article on graph factorization:

One method for constructing a $1$-factorization of a complete graph on an even number of vertices involves placing all but one of the vertices on a circle, forming a regular polygon, with the remaining vertex at the centre of the circle. With this arrangement of vertices, one way of constructing a $1$-factor of the graph is to choose an edge $e$ from the centre to a single polygon vertex together with all possible edges that lie on lines perpendicular to $e$. The $1$-factors that can be constructed in this way form a $1$-factorization of the graph.

In other words, make a diagram like so, then consider its five rotations (rotate just the edges; keep the labels fixed):

$\hskip{3cm}$


问:
Additive order of modular product ring elements
Let
$$
R = \mathbb{Z}_m\times\cdots\times\mathbb{Z}_m = \times_{i=1}^\ell\mathbb{Z}_m
$$
I'm trying to find the additive order of elements of $R$.

The additive order $\text{ord}(a)$ of $a\in\mathbb{Z}_m$ is the smallest integer $k\in\mathbb{N}$ such that
$$
k\cdot a\equiv0\;\bmod m.
$$
And resp. for $r\in R$
$$
\text{ord}(r)=\text{min}\{k\in\mathbb{N}:\forall i:k\cdot r_i\equiv0\;\bmod m\}.
$$

So far I know that for $a\in\mathbb{Z}_m$
$$
\text{ord}(a) = \frac{\text{lcm}(a, m)}{a}
$$
so I would expect
$$
\text{ord}(r) = \text{lcm}\left(\frac{\text{lcm}(r_1, m)}{r_1},\ldots,\frac{\text{lcm}(r_\ell, m)}{r_\ell}\right)
$$
for $r=(r_1,\ldots,r_\ell)\in R$.

But here I'm already stuck. Is there a way to simplify this any further?

答:
Since you're only concerned with additive order, we can disregard the ring structure and the question becomes that is the order of an element from the direct product of groups. In that context, we have that the order of an element from the direct product of groups is the lcm of the order of each coordinate, which is exactly as you suggest. This can be seen fairly easily:

Suppose $g= (g_1,...,g_\ell)$ is an element of some direct product of $\ell$ groups. Let $L$ be the lcm of the orders of $g_1,...,g_\ell$. Then clearly
$$
L(g_1,...,g_\ell) = (Lg_1,...,Lg_\ell) = 1
$$
So the order of $g$ is divisible by $L$. Suppose $g$ has order $k$ strictly less than $L$, then $kg_i = 1$ for all $g_1,...,g_\ell$, but for at least one such $g_i$, $k$ is not a multiple of its order so we have contraction. Thus $g$ has order $L$ and we can apply the result you invoked to get the order of the element in this specific case.

Also this is pretty, minor, but usually the direct product of rings, groups, etc. when you have a set of them is denoted as
$$
\prod_{i=  1}^\ell\mathbb{Z}_m
$$
instead of the $\times_{i = 1}^\ell \mathbb{Z}_m$, which may help you find answers to this kind of question more easily in the future.


问:
Solution to second order homogeneous ODE
I have the 2nd order homogeneous ODE:
$$y(x)''+\frac{2}{x}y'(x)-By=0 $$
Where $B$ is a constant. Since I am a bit rusty with analytical solutions, I plugged it in MatLab's symbolic solver which found the following solution.
$$y(x)=C_1\frac{e^{\sqrt{B}x}}{x}-C_2\frac{e^{-\sqrt{B}x}}{2\sqrt{B}x}$$
Unofrtunately, the constant B takes both positive and negative values, but since the ODE is derived from a physical problem I'd like the solution to always be real.
Any suggestion/comment is highly appreciated.

答:
Hint Try the substitution
\begin{equation}
y(x) = \frac{z(x)}{x}
\end{equation}


问:
Find the maximum of $(x_1\cdots x_n)^2$ subject to $x_1^2+\ldots +x_n^2=1$
Find the maximum of $(x_1\cdots x_n)^2$ subject to $x_1^2+\ldots +x_n^2=1$ und show that $$\left (\prod_{k=1}^na_k\right )^{\frac{1}{n}}\leq \frac{1}{n}\sum_{k=1}^na_k$$

For the first part I applied the method of Lagrange multipliers.

We have the function $f(x_1, \ldots , x_n)=(x_1\cdot \ldots \cdot x_n)^2$ and the constraint $g(x,y,z)=x_1^2+ \ldots + x_n^2-1=0$.

The Lagrange function is \begin{equation*}L(x_1, \ldots , x_n  ,\lambda )=f(x_1, \ldots , x_n)+\lambda g(x_1, \ldots , x_n)=(x_1\cdot \ldots \cdot x_n)^2+\lambda \left (x_1^2+ \ldots + x_n^2-1\right )=\left (\prod_{j=1}^nx_j\right )^2+\lambda \left (\sum_{j=1}^nx_j^2-1\right )\end{equation*}
We calculate the partial derivatives of $L$ :
\begin{align*}&\frac{\partial}{\partial{x_i}}(x_1, \ldots , x_n  ,\lambda )=2\left (\prod_{j=1}^nx_j\right )\cdot \left (\prod_{j=1, j\neq i}^nx_j\right ) +2\lambda x_i  \\ & \frac{\partial}{\partial{\lambda }}L(x_1, \ldots , x_n  ,\lambda )=\sum_{j=1}^nx_j^2-1 \end{align*} with $1\leq i\leq n$.

To get the extrema we set the partial derivatives equal to zero.

Then we get the following system:
\begin{align*}&2\left (\prod_{j=1}^nx_j\right )\cdot \left (\prod_{j=1, j\neq i}^nx_j\right ) +2\lambda x_i =0 \Rightarrow x_i\cdot \prod_{j=1, j\neq i}^nx_j^2 +\lambda x_i =0 \Rightarrow x_i\cdot \left (\prod_{j=1, j\neq i}^nx_j^2 +\lambda \right ) =0 \\ & \sum_{j=1}^nx_j^2-1=0  \end{align*}

How can we continue?

答:
Obviously, $x_i \ne 0$. Multiply each of your first $n$ equations by $x_i$ and add them up:
$$
 n\prod_{j=1}^nx_j^2+\lambda\sum_{i=1}^nx_i^2 = 0,
$$
$$
 n\prod_{j=1}^nx_j^2+\lambda= 0.
$$
It follows that
$$
\lambda = -n\prod_{j=1}^nx_j^2, \quad \quad x_i = \pm \frac{1}{\sqrt{n}}.
$$


问:
How can one compute the determinant of a tridiagonal matrix when using integers?
My question is similar to How to compute the determinant of a tridiagonal matrix with constant diagonals?
However, all of the options seem to include computations that are not permitted for integers. I am trying to find the determinant for a matrix in integers, more specifically, this matrix:

$X = \begin{bmatrix}
 2 & -1 & 0 & 0 & \cdots & 0 & 0 \\
 -1 & 2 & -1 & 0 & \ddots & 0 & 0 \\
 0 & -1 & 2 & -1 & \ddots & 0 & 0  \\
 \vdots & \ddots & \ddots & \ddots & \ddots & \vdots & \vdots \\
 0 & 0 & 0 & \cdots & -1 & 2 & -1 \\
 0 & 0 & 0 & \cdots & 0 & -1 & 2 \\
\end{bmatrix} \in \mathbb{Z}^{n,n}$

Computing the determinant for smaller $n$ with the Rule of Sarrus gave me the suspicion that it is always $n + 1$, however, how do I show this for larger $n$ as well?

答:
While I don't quite understand why you only want to use integers, you can prove this by induction. Let $d_n$ be the determinant of your matrix of size $n \times n$. Then $d_1 = 2$, $d_2 = 3$. So assume that $d_k = k+1$ for all $k < n$. By Laplace expansion and the induction hypothesis, we get
$$d_n = 2d_{n-1} - d_{n-2} = 2n - (n-1) = n+1$$
and we are done.


答:
Hint: $\quad\DeclareMathOperator{\lcm}{lcm}\dfrac{\lcm(r_i,m)}{r_i}=\dfrac{m}{\gcd(r_i,m)}$, so
$$\lcm\biggl(\frac{\lcm(r_1,m)}{r_1},\dots,\frac{\lcm(r_\ell,m)}{r_\ell}\biggr)=\lcm\biggl(\frac{m}{\gcd(r_1,m)},\dots,\frac{m}{\gcd(r_\ell,m)}\biggr)=\frac{m}{\gcd(r_1,\dots,r_\ell,m)}. $$
Added: I  used the formula
$$\operatorname{lcm}\Bigl(\frac m r,\frac ns\Bigr)=\frac{\operatorname{lcm}(m,n)}{\gcd(r,s)} $$
(There's an analogous formula for the $\gcd$ of two rational numbers.)

Title: Additive order of modular product ring elements

Body:
Let
$$
R = \mathbb{Z}_m\times\cdots\times\mathbb{Z}_m = \times_{i=1}^\ell\mathbb{Z}_m
$$
I'm trying to find the additive order of elements of $R$.

The additive order $\text{ord}(a)$ of $a\in\mathbb{Z}_m$ is the smallest integer $k\in\mathbb{N}$ such that
$$
k\cdot a\equiv0\;\bmod m.
$$
And resp. for $r\in R$
$$
\text{ord}(r)=\text{min}\{k\in\mathbb{N}:\forall i:k\cdot r_i\equiv0\;\bmod m\}.
$$

So far I know that for $a\in\mathbb{Z}_m$
$$
\text{ord}(a) = \frac{\text{lcm}(a, m)}{a}
$$
so I would expect
$$
\text{ord}(r) = \text{lcm}\left(\frac{\text{lcm}(r_1, m)}{r_1},\ldots,\frac{\text{lcm}(r_\ell, m)}{r_\ell}\right)
$$
for $r=(r_1,\ldots,r_\ell)\in R$.

But here I'm already stuck. Is there a way to simplify this any further?


答:
Pair 1
Pair 2
Pair 3

AB
CD
EF

AC
BE
DF

AD
CE
BF

AE
CF
DB

AF
CB
DE

I solved it by approaching it as you would sudoko

Title: Here's a fun math/Zoom question for you all:

Body:
If I have 6 people in a group and each participant needs to meet each other in pairs of 2, how could I arrange Zoom Breakout rooms so that no one meets another twice? (3 Breakout rooms consisting of 2 people per session)

Here are the possible pairs I have (the letters represent initials of participants):
K x R, K x J, K x C, K x N, K x Z, R x J, R x C, R x N, R x Z, J x C, J x N, J x Z, C x N, C x Z, N x Z

So with this in mind, it should break down into 5 total sessions with 3 breakout rooms in each session and 2 participants per breakout room. I just can't seem to figure out how to configure it without duplicate people being in a session.


问:
Showing that $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$ [Direct approach].
I want to check if my reasoning is correct and makes sense. I know other approaches, but I wonder if this one works also:

To show that $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$, it's enough to check that $x^2+(y^2-1)$ is irreducible in $\mathbb{Q}[y][x]$ as there is a ring isomorphism between $\mathbb{Q}[x,y]$ and $\mathbb{Q}[y][x]$.

As $x^2+y^2-1$ is a quadratic polynomial, then it's enough to show that $\pm\sqrt{y^2-1}$ does not belong to $\mathbb{Q}[y]$.

If $\sqrt{y^2-1}$ belongs to $\mathbb{Q}[y]$, then there exists $f(y)\in \mathbb{Q}[y]$ s.t.
$$\sqrt{y^2-1}=f(y)\text{ i.e. }y^2-1=f^2(y)\text{ }[*].$$

But, then $(y-f(y))(y+f(y))=1$. So, $y-f(y)=c$ for some $c\in\mathbb{Q}$ with $c\neq0$ i.e. $f(y)=y-c$.

From $[*]$, follows that $y^2-1=(y-c)^2$ i.e. $y=\frac{c^2-1}{2c}\in\mathbb{Q}$ which is a contradiction.

Thank you!

答:
You actually need to show that $1 - y^2$ is not a square in $\mathbb{Q}[y]$ (you need to pay attention to the sign here because you're not working over $\mathbb{C}$), which is even easier since its quadratic coefficient is negative. Of course it is still not a square over $\mathbb{C}$ and there are many ways to see this, e.g.

it has two distinct roots, or

by comparing degrees we see that a square root must be linear, and by comparing the quadratic and constant terms it must have the form $\pm iy \pm 1$, whose square has the wrong linear term.


答:
Lagrange multipliers are just asking (when gradients of both constraint and objective functions are nonzero) where they are parallel. This is simply, in the case that all partials are nonzero, asking for points where the ratios of the partials all come out the same.

If any element is zero, the product is zero. Otherwise, we can demand all $x_i > 0$ and continue. The $x_i$ partial of $P^2$ is $$2 \frac{P^2}{x_i}$$
where $P$ is the product of the $x_i$

The $x_i$ partial of the sum of squares is $$ 2 x_i$$

Each ratio is $$ \frac{P^2}{x_i^2}$$  Lagrange multipliers demand all these ratios equal, meaning all the $x_i^2$ must be equal. So, each $$ x_i = \frac{1}{\sqrt n} $$

Body:
Find the maximum of $(x_1\cdots x_n)^2$ subject to $x_1^2+\ldots +x_n^2=1$ und show that $$\left (\prod_{k=1}^na_k\right )^{\frac{1}{n}}\leq \frac{1}{n}\sum_{k=1}^na_k$$

For the first part I applied the method of Lagrange multipliers.

We have the function $f(x_1, \ldots , x_n)=(x_1\cdot \ldots \cdot x_n)^2$ and the constraint $g(x,y,z)=x_1^2+ \ldots + x_n^2-1=0$.

The Lagrange function is \begin{equation*}L(x_1, \ldots , x_n  ,\lambda )=f(x_1, \ldots , x_n)+\lambda g(x_1, \ldots , x_n)=(x_1\cdot \ldots \cdot x_n)^2+\lambda \left (x_1^2+ \ldots + x_n^2-1\right )=\left (\prod_{j=1}^nx_j\right )^2+\lambda \left (\sum_{j=1}^nx_j^2-1\right )\end{equation*}
We calculate the partial derivatives of $L$ :
\begin{align*}&\frac{\partial}{\partial{x_i}}(x_1, \ldots , x_n  ,\lambda )=2\left (\prod_{j=1}^nx_j\right )\cdot \left (\prod_{j=1, j\neq i}^nx_j\right ) +2\lambda x_i  \\ & \frac{\partial}{\partial{\lambda }}L(x_1, \ldots , x_n  ,\lambda )=\sum_{j=1}^nx_j^2-1 \end{align*} with $1\leq i\leq n$.

To get the extrema we set the partial derivatives equal to zero.

Then we get the following system:
\begin{align*}&2\left (\prod_{j=1}^nx_j\right )\cdot \left (\prod_{j=1, j\neq i}^nx_j\right ) +2\lambda x_i =0 \Rightarrow x_i\cdot \prod_{j=1, j\neq i}^nx_j^2 +\lambda x_i =0 \Rightarrow x_i\cdot \left (\prod_{j=1, j\neq i}^nx_j^2 +\lambda \right ) =0 \\ & \sum_{j=1}^nx_j^2-1=0  \end{align*}

How can we continue?


问:
Simplify derivative expression
Trying to derive how

$$f(x) = \frac{\mathrm{d}y}{\mathrm{d}x}e^{-cosx} + y\sin(x)e^{-cosx}$$

becomes

$$f(x) = (ye^{-cosx})'$$
Please kindly help.

Thank you

答:
It's the product rule in reverse.  $d(uv) = u\,dv + v\,du$.  Therefore, the reverse is also true. $u\,dv + v\,du = d(uv)$.  In this case, $u = y$ and $v = e^{-\cos(x)}$.

Although, technically, I'm not sure you could call it $f(x)$ since it also has other terms, unless they knew the other terms were expressible in terms of $x$.

Title: Simplify derivative expression


问:
Can I write $f[i-1] \neq value$ instead of $s=i-1 \Rightarrow f[s] \neq value$
I have got the following statement:

$$s=i-1 \Rightarrow f[s] \neq value$$.

This states that the element $f[i-1]$ does not equal $value$. Thats why I am wondering if I can use the following equivalence

$$s=i-1 \Rightarrow f[s] \neq value \Leftrightarrow f[i-1] \neq value$$

It feels right but I am not sure if it is formally correct. Thank you for any advice.

答:
Your question is a bit difficult to understand: it would be clearer if you stated the actual question in the body and not just in the title. Also, I don't know why you are using square brackets for function applications and I will use round brackets as is customary. Given that change of notation and looking at the question in your title:

$$s=i-1 \Rightarrow f(s) \neq value$$

is equivalent to:

$$s=i-1 \Rightarrow f(i - 1) \neq value$$

But that is only equivalent to $f(i - 1) \neq value$ in a context where you either know that $s = i - 1$ or you know that $f(i - 1) \neq value$.


问:
what is exactly analytic continuation of the product log function
When I solve in wolfram equation like this $xe^x=z$
they give me the solution $x=W_n(z)$

I know about $x=W_0(z) $ and $x=W_{1}(z)$ but for $n$ I searched in the internet but I didn't find anything can give me an expression about $x=W_n(z)$

Please can anyone help me and give me an explain $x=W_n(z)$ and give me an example how we can find $x=W_n(t)$ $t\in C$ or $t\in R$

答:
All that, and much more, is here:

Corless, R.M.; Gonnet, G.H.; Hare, D.E.G.; Jeffrey, D.J.; and Knuth, D.E. "On the Lambert W Function." Advances in Computational Mathematics, Vol. 5, (1996): 329-359.

ContentLicense: CC BY-SA 3.0


问:
Is saying "the rate of change is five-thirds meters per second" equivalent to saying "the rate of change is five meters every three seconds"?
I am a teaching assistant for a winter school course in basic algebra and I was explaining the meaning of the slope of a line to a student today using the following example: $$y = \frac{5}{3}x$$ where $y$ is measured in units of meters and $x$ is measured in units of time. I told them that the slope of the line is the rate of change of $y$ with respect to $x$, however, I then followed up by saying for this line in particular the rate of change is "five-thirds meters per second". This got me thinking, is this the same thing as saying "the rate of change is five meters every three seconds"? I concluded that these two statements do in fact mean the same thing since $\frac{5}{3} \cdot 3 = 5$ which precisely what you should have if you think about "your position" after $3$ seconds has elapsed.

I understand this is a little informal, but the concept to new students is important to me. Any feedback is welcome.

答: Yes, the two statements are equivalent. Since you can think of the unit for rate of change in this case as fraction with meters divided by seconds. So the measurement  $\frac{5}{3}$ mps is the same as $\frac{5 \text{meters}}{3 \text{seconds}}$, or 5 meters per 3 seconds as you said.


问:
Determinant of associated real matrix from a complex one
Given $X\in \operatorname{GL}(n, \mathbb{C}) $, let $X_r\in \operatorname{GL}(2n, \mathbb{R}) $ be the real matrix obtained by substituting to each complex entry $a+ib$ the matrix $\begin{pmatrix} a & & - b \\ b & & a \end{pmatrix}$.
Does someone can give me the details of the proof that $\det(X_r) =|\det(X)|^2$?
The only thing I have tried is to use the fact that the determinant is a wedge product but it seems not work.

答:
One may view this as a determinant of a block matrix. In general (cf. John Silvester, Determinants of Block Matrices), if $R$ is a commutative ring and $C$ is a commutative subring of $M_m(R)$, each matrix $Z\in M_n(C)$ can be viewed as a matrix $Y\in M_{mn}(R)$ and
$$
\det{}_RY=\det{}_R\left(\det{}_CZ\right).\tag{1}
$$
In your case, let $m=2,\, R=\mathbb R$ and $C=\left\{\pmatrix{a&-b\\ b&a}:a,b\in\mathbb R\right\}$. Since $f:a+ib\mapsto\pmatrix{a&-b\\ b&a}$ is a ring isomorphism between $\mathbb C$ and $C$, when it is applied entrywise to a matrix $X\in M_n(\mathbb C)$, we have
$$
\det{}_Cf(X)=f(\det{}_{\mathbb C}X).\tag{2}
$$
Now let $Z=f(X)\in M_n(C)$ and $Y=X_r\in M_{2n}(\mathbb R)$. By $(1)$ and $(2)$,
$$
\det{}_{\mathbb R}Y
=\det{}_{\mathbb R}\left(\det{}_CZ\right)
=\det{}_{\mathbb R}\left(\det{}_Cf(X)\right)
=\det{}_{\mathbb R}\left(f(\det{}_{\mathbb C}X)\right)
=|\det{}_{\mathbb C}X|^2.
$$


问:
Example of functions satisfying certain little-oh condition
Let $f:\mathbb{Z}_{>0} \rightarrow \mathbb{C}$ be a function. Now require $f$ to satisfy the condition
$$
|f(n)|\leq n^{o(1)}
$$
where $o(1)$ is any function $g(n)$ satisfy $\frac{g(n)}{1}\rightarrow 0$ as $n\rightarrow\infty$.

Can someone give some example of such $f(n)$ besides constant functions?

答: Other more interesting examples are the divisor function and $n/\phi(n)$ where $\phi$ is the Euler function. Not so interesting examples are constant functions.


问:
Find $\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx$ without resorting to series

Find $$\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx$$

Firstly, note that the upper limit makes this an improper integral.

The method I employed against this problem was integration by parts, as shown below:
$$\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx=2\int_0^1\text{artanh} x ~dx=2\left[x~\text{artanh}~x+\frac{1}{2}\ln\lvert 1-x^2\rvert\right]_0^1$$
However, I am unsure how to find the value that this converges to.

I then tried to find a series representation of the value of the integral using the Maclaurin expansion of $\text{artanh}~x$, which was successful; I found that the value of the integral is
$$2\sum_{r=0}^\infty\frac{(2r)!}{(2r+2)!}=2\sum_{r=0}^\infty\left(\frac{1}{2r+1}-\frac{1}{2r+2}\right)=2\sum_{r=0}^\infty\frac{(-1)^r}{r+1}$$
which I know is $2\ln2$, which is the answer.
However, I would like to know how to use a regular method of integration such as my first method and still calculate the limit it converges to.

Thank you for your help.

答:
Simplify the limits by integrating by parts as follows

\begin{align}
\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx
&= -2\int_0^1 \tanh^{-1}x\> d(1-x)\\
&= -2(1-x) \tanh^{-1}x\bigg|_0^1+2 \int_0^1 \frac{dx}{1+x}
=0+2\ln2
\end{align}


问:
Formula for radius of circles on vertices of regular polygon
TLDR: Given a square, a number of circles, a rotation angle, and some spacing factor, how can I determine the largest possible radius of circles arranged on the vertices in a regular polygon such that the circles do not exceed the bounds of the square?

I am trying to determine a formula or algorithm for calculating the radius of circles situated on the vertices of regular polygons with 1 to n sides.

I have tried the following with inconsistent results.

```<code>2 * a * tan(π / n) / x
</code>```

Where x is a sizing factor and a is the apothem. For example say I wanted the spacing between circles to be such that you could fit another circle on the midpoint of the side, x could be 4. Or if I wanted the circles to be touching x could be 1. But for n = 3 the results look quite different than say for 6 in that the spacing between the circles is too small/big respectively.

As for the apothem, it's not clear how I would calculate that.

For n = 1 or 2 I would have to handle it differently. For 1 I'd just want a circle in the centre of the available space, and for 2 they would be arranged in a line. Is there a more general formula I could use?

Something like the circles on the vertices of the hexagon in the image below, where I could adjust the radius of the circles as needed.

Further complication is that I calculate the coordinates of the vertices based on a given square and an angle to offset the resulting vertices, but I reduce the size of the resulting polygon based on the radius of the circles to make sure the resulting circles do not exceed the bounds of the square.

Here's a few imperfect drawings I made for 1–3:

The circles don't have to fill the square necessarily; I want to be able to adjust the spacing between them by reducing the radius.

I'm a bit stuck on a circular dependency (no pun intended) since I think I need the positions of the vertices to calculate the circle radius, but I need the circle radius to determine the positions of the vertices so that the circles do not exceed the size of the square.

答: Seems like you can always grow the circles on the vertices to have radii equal to half the length of a side of the polygon.</p>


答:
Is saying “the rate of change is five-thirds meters per second” equivalent to saying “the rate of change is five meters every three seconds”?

Yes. To make things even simpler, the following two notions are equivalent:

five-thirds meters per second

five meters every three seconds

In general, if $p$ and $q$ are two positive integers,

$p/q$ meters per second

$p$ meters every $q$ seconds

are equivalent. Essentailly, you are looking at the function $y=\frac{p}{q}x$ in two different way by using two different "time unit":

$\displaystyle y\ (\textrm{meters})=\frac{p}{q}\ (\textrm{meter}/\textrm{second})\cdot x(\textrm{seconds})$

$\displaystyle y\ (\textrm{meters})=p\
(\textrm{meter}/\color{red}{q\ \textrm{seconds}})\cdot x(\textrm{seconds})$

If you want to use the second interpretation, the "slope" needs to be rational. Imagine how you want to interpret $y=\pi x$.

Title: Is saying "the rate of change is five-thirds meters per second" equivalent to saying "the rate of change is five meters every three seconds"?

Body:
I am a teaching assistant for a winter school course in basic algebra and I was explaining the meaning of the slope of a line to a student today using the following example: $$y = \frac{5}{3}x$$ where $y$ is measured in units of meters and $x$ is measured in units of time. I told them that the slope of the line is the rate of change of $y$ with respect to $x$, however, I then followed up by saying for this line in particular the rate of change is "five-thirds meters per second". This got me thinking, is this the same thing as saying "the rate of change is five meters every three seconds"? I concluded that these two statements do in fact mean the same thing since $\frac{5}{3} \cdot 3 = 5$ which precisely what you should have if you think about "your position" after $3$ seconds has elapsed.

I understand this is a little informal, but the concept to new students is important to me. Any feedback is welcome.


答:
Just to give a different approach, first substitute $u={1+x\over1-x}$ and then integrate by parts:

$$u={1+x\over1-x}\implies x={u-1\over u+1}\implies dx={2du\over(u+1)^2}$$

so

$$\begin{align}
\int_0^1\ln\left(1+x\over1-x\right)\,dx
&=2\int_1^\infty{\ln u\over(u+1)^2}\,du\\
&={-2\ln u\over(u+1)}\Big|_1^\infty+2\int_1^\infty{1\over u(u+1)}\,du\\
&=(0-0)+2\int_1^\infty\left({1\over u}-{1\over u+1}\right)\,du\\
&=2\ln\left(u\over u+1\right)\Big|_1^\infty\\
&=2(\ln1-\ln(1/2))\\
&=2\ln2
\end{align}$$

Title: Find $\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx$ without resorting to series

Body:
Find $$\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx$$

Firstly, note that the upper limit makes this an improper integral.

The method I employed against this problem was integration by parts, as shown below:
$$\int_0^1\ln\left(\frac{1+x}{1-x}\right)~dx=2\int_0^1\text{artanh} x ~dx=2\left[x~\text{artanh}~x+\frac{1}{2}\ln\lvert 1-x^2\rvert\right]_0^1$$
However, I am unsure how to find the value that this converges to.

I then tried to find a series representation of the value of the integral using the Maclaurin expansion of $\text{artanh}~x$, which was successful; I found that the value of the integral is
$$2\sum_{r=0}^\infty\frac{(2r)!}{(2r+2)!}=2\sum_{r=0}^\infty\left(\frac{1}{2r+1}-\frac{1}{2r+2}\right)=2\sum_{r=0}^\infty\frac{(-1)^r}{r+1}$$
which I know is $2\ln2$, which is the answer.
However, I would like to know how to use a regular method of integration such as my first method and still calculate the limit it converges to.

Thank you for your help.


问:
Topology induced by Basis B
I have recently started studying topology on my own. I have a confusion regarding defination of topology of set X induced by basis $\mathfrak{B}$. ${\tau} = \{ U: x\in U \implies \exists B :x \in B \in \mathfrak B, B \subset U   \}$

Here defination of topology $\tau$ means i need to include all sets such that x belongs to some B in basis. It means i need to include B, B+atleast 1 element in X .... X

Does it mean finally $\tau$ becomes powerset of X? please correct if wrong

答:
Another way to decribe the same topology is $$\{O\mid \exists \mathfrak{B}' \subseteq \mathfrak{B}: O= \bigcup \mathfrak{B}'\}$$

which is the set of all unions of subfamilies of the base $\mathfrak{B}$. (Note that all $B \in \mathfrak{B}$ will always be in the generated topology and topologies are closed under unions so all these sets are for sure in that topology.)

If $O$ is such as in my description, clearly every $x \in O$ is in some $B \in \mathfrak{B}'$ by the definition of a union.

And if $U$ is open as in your description, pick $B_x \in \mathfrak{B}$ such that $x \in B_x \subseteq U$ for every $x \in U$. Then $U = \bigcup \{B_x\mid x \in U\}$ and so we can take $\mathfrak{B}' = \{B_x\mid B_x \in U\}$ for my description etc.

A simple example: $X = \Bbb R$ with base $\mathfrak{B}  =\{(x, \rightarrow): x \in \Bbb R\}$; all unions of base elements are also of the form $(x, \rightarrow)$, as is easilt checked plus $\Bbb R$ and $\emptyset$ (for the union of the empty subfamily $\mathfrak{B}'$; in your description this set is trivially open too as  there are no elements so the implication condition is always satisfied). So the topology is not the power set, and in fact a set of the form $\{x\}$ will only be open if it that set $\{x\}$ is already in the base $\mathfrak{B}$.

Title: Topology induced by Basis B


问:
Equivalent condition on symmetry of product of matrices
This is not homework, but rather a problem I found in the practice problems for application to a Master in Mathematics. I have been unsuccessful to solve it and would appreciate any hint. Here it is

Let $A,B,C,D$ be square matrices. Show that the following two conditions are equivalent:

$AB^T$ and $CD^T$ are symmetric and $AD^T-BC^T=Id;$

$B^TD$ and $A^TC$ are symmetric and $A^TD-C^TB=Id$.

I have tried to manipulate this equations and virtually get nowhere.

答:
Hint. The first condition says that
$$
\pmatrix{A&-B\\ -C&D}\pmatrix{D^T&B^T\\ C^T&A^T}=I.
$$


问:
Problems with logical conjunction
There is a problem. I have line like this:
$$
x \in A \backslash(B\backslash C) \leftrightarrow x\in(A\backslash B)\cup(A\cap C)
$$
and I tried to prove it from left, but stopped here:
$$
x\in A \backslash(B\backslash C)) \leftrightarrow x\in A \wedge x\notin (B\backslash C) \leftrightarrow
x\in A \wedge(x\notin B \wedge x\in C) \leftrightarrow ???
$$
and tried from right, bout stopped here:
$$
x\in(A\backslash B)\cup(A\cap C) \leftrightarrow
(x\in A \wedge x \notin B) \vee (x \in A \wedge x \in C) \leftrightarrow
x \in A \wedge (x\notin B \vee x\in C) \leftrightarrow ???
$$
So, to be honest, I'm kinda lost. I tried it two times, but still get different things from both sides. It is supposed to be "truth" and I need to show how to prove it. But I have no idea where is the mistake.

Also other question.
Is it correct to do something like this:
$$
(x\in A \wedge x \in B) \wedge x \notin C\leftrightarrow x \in A \wedge x \notin C \wedge x \in B \wedge x \notin C
$$
and if yes (or not) based on what? Because I tried to find something to base my prove on, but I can't find anything specific, like theorem or definition.
If something isn't clear, please say what, and I will try to clarify.

答: Hint: You are getting confused between set operators and logical operators when you write $x \in A \setminus (x \not\in B \land x \in C)$. At that point, try using the fact that $x \not\in B \setminus C$ is equivalent to $x \not\in B \lor x \in C$.

Title: Problems with logical conjunction


问:
Unique maximal ideal and ring epimorphism kernel with prime numbers equivalence
Let $A$ be a ring. Let $f: \mathbf{Z} \to A$ be a surjective ring homomorphism. Prove that $A$ has a unique maximal ideal iff there exists $n\in \mathbf{N}$ and $p\in\mathbf{N}$ a prime number such that $\ker f =(p^{n})$.

$(p^{n})$ denotes the ideal generated by $p^{n}$, I'm not sure if that's the standard notation.

We had to prove that in a recent exam and nobody could prove it. I haven't found it online neither. Thank you!

答:
Use the Chinese remainder theorem.  Every ideal in $\mathbb Z$ looks like $n\mathbb Z$ for a unique nonnegative integer $n$.  If you write $n = p_1^{e_1} \cdots p_s^{e_s}$ for primes $p_i$, then $n\mathbb Z = p_1^{e_1}\mathbb Z \cap \cdots \cap p_s^{e_s}\mathbb Z$.  The Chinese remainder theorem gives

$$\mathbb Z/n\mathbb Z \cong \mathbb Z/p_1^{e_1}\mathbb Z \times \cdots \times \mathbb Z/p_s^{e_s}\mathbb Z$$.

The rings $\mathbb Z/p_i^{e_i}\mathbb Z$ are local rings with unique maximal ideals $p_i\mathbb Z/p_i^{e_i}\mathbb Z$.  Now, use the fact that if $R_1, ... , R_s$ are commutative rings with identity, the maximal ideals of

$$R = R_1 \times \cdots \times R_s$$

are all of the form

$$R_1 \times \cdots \times R_{i-1} \times \mathfrak m \times R_{i+1} \times \cdots \times R_s$$

for some $1 \leq i \leq s$, where $\mathfrak m$ is a maximal ideal of $R_i$.  Thus $R$ cannot be a local ring unless $s=1$.


问:
Let $H$ be a subgroup of a group $G$ such that $x^2 \in H$ , $\forall x\in G$ . Prove that $H$ is a normal subgroup of $G$
Let $H$ be a subgroup of a group $G$ such that $x^2 \in H$, $\forall x\in G$. Prove that $H$ is a normal subgroup of $G$.

I have tried to using the definition but failed. Can someone help me please.

答:
$H$ is a normal subgroup of $G$ $\iff\forall~h\in H ~\forall~ g\in G:g^{-1}hg \in H$

$g^{-1}hg=g^{-1}g^{-1}ghg=(g^{-1})^2h^{-1}hghg=(g^{-1})^2h^{-1}(hg)^2\in H(hg\in G \to (hg)^2\in H)$  then  $$g^{-1}hg \in H$$


问:
Prove that if f $c$-concave function and $c(x,y) = |x-y|^2/2$, then $|x|^2/2 - f(x)$ is convex
Let $c$ be a symmetric function. A function $f:\mathbb R^d \to \mathbb R$ is $c$-concave if there exists a function $g$ such that
$$f = \inf_y \left\{ c(x,y) - g(y) \right\}.
$$

Now, let $c(x,y) = \frac{|x-y|^2}{2}$. How does one then prove that if $f$ is $c$-concave,
then $h(x) = \frac{|x|^2}{2}-f(x)$ is convex? I saw this assertion in a book, but I've not been able to prove it.

答:
Plugging in the formula for $f$, you get:
\begin{align}
h(x) &= \sup_y \left\{ 0.5x^2 - 0.5(x-y)^2 + g(y) \right\} \\
&= \sup_y \left\{ xy - 0.5y^2 + g(y) \right\}.
\end{align}
The supremum of affine functions is convex.


问:
Analogue of Fubini's theorem for product space between natural numbers and an arbitrary measure space
This is the last part of Bartle's "Elements of Integration" exercise 10.J.

Let $(X, \mathcal{A}, \nu)$ be a measure space, and consider the measure space $(\mathbb{N}, \mathcal{P}(\mathbb{N}), \mu)$, where $\mu$ is the counting measure. Let $(Z = \mathbb{N} \times X, \mathcal{B}, \pi)$ be the product of the beforementioned measure spaces.

If $f: Z \to \mathbb{R}$ is an integrable function, and $f_n$ are its $n$-sections ($f_n: X \to \mathbb{R}$, $f_n(x) = f(n, x))$ for each $n \in \mathbb{N}$. I need to show that
$$\sum_{n=1}^\infty \left(\int_X f_n d \nu\right) = \int_X \left(\sum_{n=1}^\infty f_n\right)d\nu$$

I tried finding an integrable function that dominates $\sum f_n$, and got nowhere. Also tried to prove first for nonnegative functions, and using the definition of the integral for arbitrary function, but to no avail. Could anyone give me some hint?

答: For non-negative integrable $f$ this follows by Monotone Convergence Theorem: $\sum\limits_{k=1}^{n} f_k$ increases to $\sum\limits_{k=1}^{\infty} f_k$ as $n$ increases to $\infty$. For the general case use the decomposition $f=f^{+}-f^{-}$.


问:
If $x$ is an ordinal, then so is $x\,\bigcup\,\{x\}$
I wish to show that $x$ is an ordinal, then so is $x\,\bigcup\,\{x\}$ (with $y<x$ for all $y\in x$).

I have seen a proof in Cameron's Sets, Logic and Categories (p.41), but I find it unsatisfactory. He writes:

The set $a = x\,\bigcup\,\{x\}$ (with order as specified in the the statement of the theorem) has as sections all the sections of $x$ and one additional one, namely $a_x$. But since all the elements of $x$ are smaller than $x$, we have $a_x = x$.

I don't see why it necessarily follows from this that $a_x = x$. By definition, $a_x = \{y\in a : y\subset x\}$, so $a_x\subseteq x$, but this is different than $a_x = x$.

答: Suppose $y\in x$.  Since $x$ is an ordinal, this implies $y\subset x$.  So, $y$ is an element of $a$ such that $y\subset x$, which by definition means that $y\in a_x$.


问:
Trying to understand the total derivative/differential.
I'm asking this question because I'm having problems understanting the definition of differential/total derivative in multivariable calculus, and in order to improve my understanding of it, I want to make sure I have the right intuition behind the definition.

Let $F:\mathbb{R}^n\to\mathbb{R}^m$ be a function and $a\in\mathbb{R}^n$. We say that $F$ is differentiable at $a$ iff there exists a linear transformation $L:\mathbb{R}^n\to\mathbb{R}^m$ such that

$$\lim_{x\to a}\frac{\|F(x)-(L(x-a)+F(a))\|_m}{\|x-a\|_n}=0.$$

In this case $L$ will be called the differential of $F$ at $a$ (as it is later proven that this linear transformation is unique) and will be denoted $DF_a$. As my proffesor explained, the differential is the best linear aproximation there is to $F$ close to $a$. Now, I understand that $G:\mathbb{R}^n\to\mathbb{R}^m$ defined as $G(x)=DF_a(x-a)+F(a)$ is moving the graph of the function $DF_a$ ($Gr(DF_a):=\{(x_1,...,x_n,DF_a^1(x_1,...,x_n),...,DF_a^m(x_1,...,x_n))|x_i\in\mathbb{R},i=1,...,n\}$, where each $DF_a^i:\mathbb{R}^n\to\mathbb{R}$ is the $i$-th component function of $DF_a$), to the point $(a_1,...,a_n,F_1(a_1,...,a_n),...,F_m(a_1,...,a_n))$, and in this sense $G(x)$ approaches $F(x)$ as $x$ goes to $a$. This is where the confusion starts (please correct me in anything if I'm wrong).

How is it justified that $G$ approaches $F$ as $x$ goes to $a$? Is it true that $\lim_{x\to a}\|F(x)-G(x)\|_m=0$? If so, why is that not the definition? How does dividing over something that goes to zero while preserving the existence and value of the limit guarantee that $G$ is the best linear approximation? As I've heard before, it is to be understood that $\|F(x)-G(x)\|_m$ goes to zero faster than $\|x-a\|_n$ does, as $x$ approaches $a$, but once again, how is this justified?

I'm sorry if this is too much or if it seems as if I have not worked as much in trying to learn this, but I really have been trying a lot lately and I'm having a hard time in doing so.

Any help is greatly thanked for. Greetings.

答:
How is it justified that  approaches  as  goes to ?

Well, this is simply because
$$
\lim_{x\to a}G(x)=
\lim_{x\to a}(F(a)+DF_a(x-a))=\lim_{x\to a}F(a)+\lim_{x\to a}DF_a(x-a)
=F(a)+0=F(a)
$$


问:
Probability from density and distribution
I can't figure out $P(3 \le X)$ from the density and distribution functions.

$$ f(x) = \begin{cases} cx^2 & 0 \le x \le 6 \\ 0 & \text{elsewhere} \\ \end{cases} ,$$

from which I determined $c$ to be $\frac{1}{72}$, giving me

$$ f(x) = \begin{cases} \frac{x^2}{72} & 0 \le x \le 6 \\ 0 & \text{elsewhere} \\ \end{cases} ,$$

and the distribution function to be

$$ F(x) = \begin{cases} 0 & x \le 0 \\ \frac{x^3}{216} & 0 \le x \le 6 \\ 1 & 6 \le x \\ \end{cases} .$$

I feel like there's an obvious solution but I cannot seem to figure it out.

Also I apologize for the poor layout I am completely new to all this

Body: $P(3 \leq X)=1-P(X<3)=1-F_X(3)=1-\frac {3^{3}} {216}=\frac 7 8$


问:
What are the values of $a$ and $b$?
I got this question from our quiz wrong I wonder how my teacher got the correct answers shown. I tried solving the first question but didn't get the right answers. I know the formula for the surface of a right circular cone is $S=πr(r+\sqrt {r^2+h^2})$ but I still followed the given formula for the sake of trying to obtain what my teacher got.

My solution:

Let $x$ be $r$ and $4x$ be $h$. Then,

$$πx\sqrt {x^2+(4x)^2}=8$$
$$πx\sqrt {17x^2}=8$$
$$17π^2x^4=8^2$$
$$x^4=\frac{8^2}{17π^2}$$
$$(x^4)^\frac{1}{4}=(\frac{8^2}{17π^2})^\frac{1}{4}$$
$$x=±\frac{2\sqrt 2}{17^\frac{1}{4}π^\frac{1}{2}}$$
$$x=\frac{2\sqrt 2}{17^\frac{1}{4}π^\frac{1}{2}}$$
$$h=\frac{8\sqrt 2}{17^\frac{1}{4}π^\frac{1}{2}}$$
$$h^2=\frac{128}{17^\frac{1}{2}π}$$
$$a=128$$
$$b=17$$

答:
Your math is correct. Given $S = 8$, $h = 4r$, and $h^2 = \frac{a\sqrt{b}}{b\pi}$, we have:

$$ S = \pi r\sqrt{r^2 + h^2} $$

$$ = \pi \frac{\pi h}{4} \sqrt{\frac{1}{16}h^2 + h^2} $$

$$ = \pi \frac{h}{4} \sqrt{\frac{17h^2}{16}} $$

$$ = \frac{\pi h^2 \sqrt{17}}{16} $$

Substituting in $S = 8$, we have

$$ 8 = \frac{\pi h^2 \sqrt{17}}{16} $$

$$ \Rightarrow h^2 = \frac{128}{\sqrt{17}\pi} $$

$$ = \frac{128\sqrt{17}}{17\pi} $$

$$\Rightarrow a = 128, b = 17 $$

Now, if we take the "correct" solution with $a = 2$ and $b = 8$, we still have (from the algebra above):

$$ S = \frac{\pi h^2 \sqrt{17}}{16} $$

Substituting in $a = 2$ and $b = 8$, we have

$$ S = \frac{\pi \sqrt{17}}{16} \frac{2 \sqrt{8}}{8 \pi} $$

$$ = \frac{\sqrt{34}}{32} \neq 8 $$

Title: What are the values of $a$ and $b$?


答:
Begin by doing the sub $x \mapsto 1-x$. You have
$$\int_0^1 \log\left({1+x\over 1-x}\right) = \int_0^1 \log\left({1+(1-x)\over x}\right) = \int_0^1 \log(2-x)\,dx  - \int_0^1 \log(x)\,dx$$
Finish by using integration-by-parts.


问:
Definition of a probability measure via expected value
I have seen quite a few times doing something like this:

I have a probability space $(\Omega, \mathbb{P})$ where $\mathbb{P}$ has to be defined.

I would expect them do define it on the measurable sets: given $E\subset\Omega $,  $\mathbb{P}(E)=...$

Instead, they bring in something someone called "test functions". It works like this: for every test function $f$ (if I understood well, they usually request $f$ to be continuous), then

\begin{equation}
\mathbb{E}(f(X))=...
\tag{1}
\end{equation}

where $X$ is a random variable on $\Omega$ with distribution $\mathbb{P}$.

I have two issues with this approach:

If I'm free enough to choose $f$, then if $f={1_E} $ charcteristic function, I get that
$$ \mathbb{E}(f(X))=\int 1_E(x)d\mathbb{P}= \int_E d\mathbb{P} = \mathbb{P}(E) $$
What's the need for the definition  (1), then?

If $f$ has to be continuous, then I can't choose it characteristic. So I don't even see how the (1) definition even works.

Can somebody help or provide some reference about this approach?

答:
https://en.wikipedia.org/wiki/Riesz%E2%80%93Markov%E2%80%93Kakutani_representation_theorem
Allows you to define a linear funtional on a suitable space of continuous functions, and deduce that it must arise from integration against a measure.</p>


问:
Chain rule in airy case
I have the following differential operator
\begin{equation}
L_1:=\hbar\partial^{1}-\frac{1}{2} Ax_{2}x_{3}-\hbar -\frac{1}{2}C \partial^{2}\partial^{3}-\hbar D,
\end{equation}
where $\partial^{c}=\frac{\partial }{\partial x_c}$, A,B,C, D are constant.
$L_1$ annihilates a function say $f(x_1, x_2 , x_3) $. I want to define a new function say
$$F(x_1, x_2 , x_3 ,s)= f(\frac{x_1}{s}, \frac{x_2}{s},\frac{x_3}{s} )$$
So I claim that by chain rule there exist an operator in $\partial^c,\frac{\partial }{\partial s}$ that annihilates $F(x_1, x_2 , x_3 ,s)$. My question how to get the annihilator in this particular case ?

答:
It looks simple, let $x_i = s y_i$. Then $f(y_1, y_2, y_3) = F(s y_1, s y_2, s y_3, s)$. We use
\begin{equation}
\hbar\partial^{1}f(y_1, y_2, y_3)-\frac{1}{2} Ay_{2}y_{3}f(y_1, y_2, y_3)-\hbar f(y_1, y_2, y_3) -\frac{1}{2}C \partial^{2}\partial^{3}f(y_1, y_2, y_3)-\hbar D f(y_1, y_2, y_3) = 0
\end{equation}
Then use that $\partial^i f(y_1, y_2, y_3) = s \partial^i F(s y_1, s y_2, s y_3, s)$ and $\partial^i\partial^j f(y_1, y_2, y_3) = s^2\partial^i\partial^j F(s y_1, s y_2, s y_3, s)$. Finally, replace $s y_i$ by $x_i$ and we are left with
\begin{equation}
\hbar s \partial^{1}F-\frac{1}{2} A \frac{x_{2}x_{3}}{s^2}F-\hbar F -\frac{1}{2}C s^2 \partial^{2}\partial^{3}F-\hbar D F = 0
\end{equation}

Also note that being homogeneous of degree $0$, $F$ anihilates another operator:
\begin{equation}
x_1 \partial^1 + x_2 \partial^2 + x_3 \partial^3 + s \partial^4
\end{equation}

Title: Chain rule in airy case


问:
Is it possible to solve this specific modulo equation?
So I have this equation:

$$x_1 + x_2 + 20 + 10 + 12 = 384 \pmod{31}$$

Is there any way to calculate $x_1$ and $x_2$ without knowing one of them or is that impossible?

Thanks in advance.

答:
Note that your equation is equal to
$$x_1+x_2+42=384\pmod{31},$$
and then also equal to
$$x_1+x_2=342 \pmod{31}.$$
Since $342 = 1 \pmod{31},$ the equation you have to solve is
$$\boxed{x_1 + x_2 = 1 \pmod{31}}$$
and that's a very easier one to see clear solutions.


问:
Continuous function and Lebesgue measure
I've spent many time working on an interesting problem. However, I've not been able to solve it. The problem says: Build, if possible, a function function g ∈ L1 ([0, 1]) so that g is not bounded in any environment of any x ∈ [0,1] . In case the above is possible, deduct if it is possible to make the construction so that the exact value of ∫(between 0 and 1)gdx. I hope someone knows the answer</p>

答:
One such example could be $g: [0,1] \rightarrow \mathbb{R}$ defined as
$$g(x) := \begin{cases} q &, \text{ if $x=\frac{p}{q}$ with $p,q \in \mathbb{N}$ and $\operatorname{gcd}(p,q)=1$} \\ 0 &, \text{otherwise} \end{cases} $$

This is unbounded in the neighborhood of any $x \in [0,1]$, but since the rational numbers in $[0,1]$ form a set of Lebesgue-measure $0$, we get that
$$\int_0^1 g(x) \: dx = 0$$


答:
You can take a function $g$ with that property such that $\int_0^1g(x)dx=k$ for any $k\in\mathbb R$. Since every rational number can be expressed uniquely as $p/q$ such that $gcd(p,q)=1$, let $$g(x)=\begin{cases} q&\text{if}& x=p/q\\
k&\text{if}&x\in\mathbb [0,1]\setminus\mathbb Q
\end{cases}$$
Since $\lambda(\mathbb Q)=0$ you have that $\int_0^1g(x)dx=k$, now take $(a,b)\subset[0,1]$ and take $n_0\in\mathbb N$ such that $2^{n_0}< b-a$, so for any $n>n_0$ there exists $p_n$ such that $\frac{p_n}{2^n},\frac{p_n+1}{2^n}\in(a,b)$ since $p_n$ is odd or $p_n+1$ is odd you have that $gcd(p_n,2^n)=1$ or $gcd(p_n+1,2^n)=1$ so for any $n>n_0$ there exists $x\in(a,b)$ such that $g(x)=2^n$ and this clearly implies that $g$ is unbounded in $(a,b)$.

Title: Continuous function and Lebesgue measure

Body: I've spent many time working on an interesting problem. However, I've not been able to solve it. The problem says: Build, if possible, a function function g ∈ L1 ([0, 1]) so that g is not bounded in any environment of any x ∈ [0,1] . In case the above is possible, deduct if it is possible to make the construction so that the exact value of ∫(between 0 and 1)gdx. I hope someone knows the answer</p>


问:
Does the function have a positive lower bound?
Let $\zeta$ be an irrational number and $p, q, n$ be natural numbers. The function $g$ is defined as:

$$g_n(\zeta,p,q)=p^n|p\zeta-q|.$$

Does there exist a positive $r$ such that:

$$\forall p\in\mathbb{N}-\{0\},\ \forall q\in\mathbb{N},\ g_n(\zeta,p,q)\geq r?$$

Is the result relevant to $\zeta$ and $n$?

答:
There is no lower bound.

Consider an increasing sequence $e_i$ of positive integers and let $b$ be a positive integer. Then
$$\zeta = \sum_i b^{-e_i}$$
will be irrational if the $e_i$ are chosen not to result in a periodic tail for the base-$b$ expansion of $\zeta$.

But $$\min_q g_n(\zeta, b^{e_k}, q) = \sum_{i > k}b^{(n+1)e_k-e_i}$$
and no matter what $n, k, r$ we choose, there will be sequences with $e_{k+1}$ large enough to make that remainder smaller than $r$.


问:
Polynomial making every matrix diagonalisable
For $n \geq 2$ a fixed integer, how to prove that there is no polynomial $P\in \mathbb{C}[X]$ such that for every matrix $A \in \mathcal{M}_{n}(\mathbb{C})$ : $P(A)$ is diagonalisable?
Edit : $ P \neq 0$

答:
You need $p$ to be non-constant, otherwise $p=1$ is a counterexample.

If $p$ is non-constant, then $p'(z)\ne0$ for some $z$. Let $J$ be the nilpotent Jordan block of size $n$ and $A=zI_n+J$, i.e. let $A$ be the $n\times n$ Jordan block for the eigenvalue $z$. Then
$$
p(A)=p(z)I_n+p'(z)J+\frac{p''(z)}{2!}J^2+\cdots+\frac{p^{(n-1)}(z)}{(n-1)!}J^{n-1}
$$
is not diagonalisable because all eigenvalues of $p(A)$ are equal to $p(z)$ but $p(A)\ne p(z)I_n$.


问:
epsilon delta question limit questions
I have two quick questions. In calculus books when proving limits, I always see something like, let $\delta = \min(\delta_{1}, \delta_{2})$. Does this mean that we take a $\delta$ such that it contains both $\delta_{1}$ and $\delta_{2}$?</p>

答: $\delta_1$ and $\delta_2$ are just numbers. So min($\delta_1$, $\delta_2$) is just the minimum, i.e., min$(1,2) = 1$.</p>

Title: epsilon delta question limit questions


问:
application of maximal function inequality
Given the function

$$
u(x,t)=\frac{1}{2 \Vert x \Vert}\int_{\vert \Vert x\Vert -t \vert}^{\Vert x\Vert +t}{s h(s)ds}
$$

where $h \in C^2(\mathbb R)$, $h \in L^2(\mathbb{R}^3)$ and $x \in \mathbb{R}^3$.

We define the maximal function  $$(Mf)(t) = \sup_{r>0}{\frac{1}{2r}\int_{-r}^r}{\vert f(t-r) \vert}dr$$

Given the property $\Vert M(f)\Vert_{L^2(\mathbb{R}^n)} \le \Vert f\Vert_{L^2(\mathbb{R}^n)}$ for every $f \in L^2(\mathbb{R}^n)$ show that
$$
\int_{\mathbb{R}}{\Vert u(t,\cdot)\Vert_{L^{\infty}(\mathbb{R}^n)}^2} \le C \Vert h \Vert_{L^2(\mathbb{R}^n)}^2
$$

This almost looks like a direct application of the inequality for maximal functions but the integrand of $u$ is not quite right. How can I apply it here ?

Would appreciate any help/hints

答:
Several things should be clarified. It seems that there are many typos on your questions. If I change the definition of $u$, then I can show that the desired assertion(modified one) holds. Since it is too long to explain in the comment, I write it here. I'm not sure that why $s$ came in the definition of $u$. So I omit it.

Let $h\in L^2 (\mathbb{R})$. For $x\in \mathbb{R}^n$, we define
$$ u(x,t) = \begin{cases}
\frac{1}{2\Vert x \Vert} \int_{t-\Vert{x}\Vert}^{t+\Vert{x}\Vert} h(s) ds&\text{if } x\neq 0,\\
0 & \text{if } x=0.
\end{cases}
$$
Our definition is well-defined since $h$ is locally integrable. Then by definition of maximal function, we have
$$ |u(x,t)| \leq M(h)(t)\quad \text{a.e. on } x.$$
Hence it follows from $L^2$-boundedness of Hardy-Littlewood maximal function that
$$ \int_{-\infty}^\infty \Vert u(\cdot,t)\Vert_{L^\infty(\mathbb{R}^n)}^2 dt\leq \int_{-\infty}^\infty |M(h)(t)|^2 dt \leq C \int_{-\infty}^\infty |h(t)|^2 dt=C\Vert h \Vert_{L^2(\mathbb{R})}^2. $$

If you correct typos on your original question, let me know.


问:
Proof that a particular subset of $L^2[\Pi]$ is dense and first category set (Baire's category)
I hope I don't make too many mistakes since this is my first post.
I was trying to prove that $$M = \left[ f\in L^2[\Pi] : \sum_{n=-\infty}^{+\infty}\hat f(n) \quad converges \right]$$ is dense and first category set (Baire's category) in $L^2[\Pi]$.

With $ \ \Pi \ $ I mean the Torus  and with $\hat f(n)$ I mean the Fourier coefficients of $f\in L^2[\Pi]$: $$\hat f(n)=\frac{1}{2\pi}\int_{\Pi}f(t)e^{-int}dt$$

When I say a set $E$ is first cathegory set I mean it's equal to a countable union of sets $E_n$ such that $int(\bar E_n)=\emptyset$ fo every n.

I know that a similar proof about the divergence set of the Fourier series of a $f \in L^1[\Pi]$ has been done using the linear and bounded operators $$L_n=\sum_{m=-n}^{n}\hat f(m)=\frac{1}{2\pi}\int_{\Pi}f(t)D_n(t)dt$$ and calculating their norm in the dual space of $L^2[\Pi] \;$ ($D_n(t)=\sum_{m=-n}^{n}e^{int}$ is the Dirichlet kernel).

I struggled with these for a couple of days and I couldn't manage to get an answer. hope someone can help me :).

答:
Let me address only the non-trivial part of the question which is that $M$ is of first category.

Considering the map
$$
  T: (a_n)_{n\in \mathbb Z}\in  \ell ^1(\mathbb Z) \ \mapsto \ \sum_{n\in \mathbb Z}a_ne^{int} \in  L^2(\mathbb T),
  $$
it is clear that $M$ is precisely the range of $T$.  It is also clear that $T$ is bounded and  not surjective.  So the
result is a consequece of  the following more general result:

Theorem.  Let $X$ and $Y$ be Banach spaces and let $T:X\to Y$ be a non-surjective bounded map.  Then the range of $T$ is
of first category.

Proof.  Let $B_X$ be the closed unit ball of $X$.  Observing that
$$
  T(X) = \bigcup_{n\in \mathbb N} T(nB_X),
  $$
the proof will be concluded once we show that  the closure of each $T(nB_X)$ has empty interior, and it is clearly enough to consider only
the case $n=1$.

(If $X$ is reflexive this would be much easier because  then $B_X$ is weakly compact,  whence the same holds for
$T(B_X)$,  so $T(B_X)$  would be closed.  However,  since we are aiming at an application in which $X=\ell ^1$, we cannot assume
that $X$ is reflexive.)

So let us assume  by contradiction that $\overline{T(B_X)}$ has a nonempty interior.

It so happens that the standard
proof of the Open Mapping Theorem  (if $T:X\to Y$ is surjective then it is open),  e.g. Theorem III.12.1  in Conway's "A Course
in Functional Analysis",  start out  by using that $T$ is surjective to deduce that $\overline{T(B_X)}$ has a nonempty
interior.
After this crucial step, the surjectivity of $T$ is no longer needed and, with the sole information that
$\overline{T(B_X)}$ has a nonempty interior, it is proved that $T$ is open.

In other words, we can borrow the argument of that proof to deduce that our $T$ is open, but this is an absurd since we
are assuming that $T$ is not surjective.   This concludes the proof.  $\qquad \square$


问:
Why is $\operatorname{Hom}_\Bbb{Z}(A,A)$ a ring?
I am trying to prove and understand why the homomorphisms from an abelian group A to itself $\operatorname{Hom}_\Bbb{Z}(A,A)$ is a ring. Is my reasoning correct in the following proof?

Show the homomorphisms from an abelian group A to itself $\text{Hom}_{\mathbb{Z}}(A,A)$ form a ring.

For two homomorphisms $\delta_1\ \colon V\to V$ and $\delta_2\ \colon V\to V$ $\in \text{Hom}_{\mathbb{Z}}(A,A)$ define their additive and multiplicative binary operataions as $[\delta_1+\delta_2](x)=\delta_1(x)+\delta_2(x)$
and $\delta_1 * \delta_2=\delta_1 \circ \delta_2$ respectively.

Additive Identity This would be the zero map so fix $\delta_1\in \text{Hom}_\mathbb{Z}(A,A)$ then $[\delta_1 + 0]=\delta_1$ $[0+\delta_1]=\delta_1$

Additive Associativity Addition of functions is associative.

Inverse Suppose we have a homomorphism $\delta$ in $\text{Hom}_{\mathbb{Z}}(A,A)$. Define another homomorphism from A to itself as $\delta^{-1}(v):=(\delta(v))^{-1}$ which exists because the image of A under $\delta$ is a subgroup of A (well known theorem of Homomorphisms).

Now we must show that the composition of these functions yields the zero function which is the identity of the Abelian group.

Let $v$ be an arbitrary element in $A$. Then $\delta(v)+\delta^{-1}(v)=\delta(v)+\delta(v^{-1})=\delta(v+v^{-1})=\delta(0)=0$ with the last equlaity following from the fact that $\delta$ is a group homomorphism.

Multiplicative Associativity Composition of functions is associative.

Multiplicative Identity This would be the identity homomorphism so fix $\delta_1 \in \text{Hom}_{\mathbb{Z}}(A,A)$ then $(\delta_1 + 1)=(\delta_1 \circ 1)=\delta_1$. Similarly, $(1 + \delta_1)=(1 \circ \delta_1)=\delta_1$

Multiplicative Distributive Laws Fix three homomorphisms $\delta_1, \delta_2, \delta_3$. $\delta_1 * (\delta_2 + \delta_3) (x)= \delta_1 \circ (\delta_2 + \delta_3) (x) =\delta_1((\delta_2 + \delta_3)(x))=\delta_1(\delta_2(x) + \delta_3(x))=
\delta_1(\delta_2(x)) + \delta_1(\delta_3(x))=(\delta_1*\delta_2 + \delta_1*\delta_3)$

Using the fact that $\delta_1$ is a homomorphism.

答:
Just some feedback:

Additive identity: If $0_A \in A$ is the identity element of $A$ ($a+0_A=a$ for any $a \in A$) define $0 : A \to A$ by $(\forall a \in A) \ 0(a) = 0_A$. Then it is easy to check that $0 \in \operatorname{Hom}_\mathbb Z(A,A)$ and $f+0=f$ for any $f \in \operatorname{Hom}_\mathbb Z(A,A)$ since $(f+0)(a) = f(a)+0(a) = f(a) + 0_A = f(a)$ for any $a \in A$.

Additive inverse: instead of writting $a^{-1}$ for the inverse of $a$, write $-a$. So, the additive inverse of $f \in \operatorname{Hom}_\mathbb Z(A,A)$ is the map $-f : A \to A$ given by $(\forall a \in A) \ (-f)(a) = -f(a)$ (as above, check that $(f+(-f))(a)=0_A$ for any $a \in A$, which proves that $f+(-f)=0$).

Multiplicative identity: Consider the identity map $\operatorname{id}_A : A \to A$ given by $(\forall a \in A) \ \operatorname{id}_A(a) = a$. Then it is easy to check that $\operatorname{id}_A \in \operatorname{Hom}_\mathbb Z(A,A)$ and that $\operatorname{id}_A \circ\, f = f \circ \operatorname{id}_A = f$ for any $f \in \operatorname{Hom}_\mathbb Z(A,A)$.

Distributive laws: you also need to prove that $(\delta_1+\delta_2) \circ \delta_3 = \delta_1 \circ \delta_3 + \delta_2 \circ \delta_3$, which follows from the definition of the sum in $\operatorname{Hom}_\mathbb Z(A,A)$.


问:
Find the maximum of $(x_1\cdots x_n)^2$ subject to $x_1^2+\ldots +x_n^2=1$
Find the maximum of $(x_1\cdots x_n)^2$ subject to $x_1^2+\ldots +x_n^2=1$ und show that $$\left (\prod_{k=1}^na_k\right )^{\frac{1}{n}}\leq \frac{1}{n}\sum_{k=1}^na_k$$

For the first part I applied the method of Lagrange multipliers.

We have the function $f(x_1, \ldots , x_n)=(x_1\cdot \ldots \cdot x_n)^2$ and the constraint $g(x,y,z)=x_1^2+ \ldots + x_n^2-1=0$.

The Lagrange function is \begin{equation*}L(x_1, \ldots , x_n  ,\lambda )=f(x_1, \ldots , x_n)+\lambda g(x_1, \ldots , x_n)=(x_1\cdot \ldots \cdot x_n)^2+\lambda \left (x_1^2+ \ldots + x_n^2-1\right )=\left (\prod_{j=1}^nx_j\right )^2+\lambda \left (\sum_{j=1}^nx_j^2-1\right )\end{equation*}
We calculate the partial derivatives of $L$ :
\begin{align*}&\frac{\partial}{\partial{x_i}}(x_1, \ldots , x_n  ,\lambda )=2\left (\prod_{j=1}^nx_j\right )\cdot \left (\prod_{j=1, j\neq i}^nx_j\right ) +2\lambda x_i  \\ & \frac{\partial}{\partial{\lambda }}L(x_1, \ldots , x_n  ,\lambda )=\sum_{j=1}^nx_j^2-1 \end{align*} with $1\leq i\leq n$.

To get the extrema we set the partial derivatives equal to zero.

Then we get the following system:
\begin{align*}&2\left (\prod_{j=1}^nx_j\right )\cdot \left (\prod_{j=1, j\neq i}^nx_j\right ) +2\lambda x_i =0 \Rightarrow x_i\cdot \prod_{j=1, j\neq i}^nx_j^2 +\lambda x_i =0 \Rightarrow x_i\cdot \left (\prod_{j=1, j\neq i}^nx_j^2 +\lambda \right ) =0 \\ & \sum_{j=1}^nx_j^2-1=0  \end{align*}

How can we continue?

答:
Just to provide a non-calculus alternative.

We can show that the maximum of $(x_1x_2\cdots x_n)^2$ subject to $x_1^2 + \cdots +x_n^2 =1$ must occur when $x_1^2 = \cdots = x_n^2$ by using AM-GM for two non-negatives (which is relatively easy to show):

AM-GM for two non-neagives: If $A,B\ge 0$, then $(A+B)/2 \ge \sqrt{AB}$ and equality holds if and only if $A=B$. (Proof of this later, or try it yourself.)

Now we prove that this maximum occurs when $x_1^2 =\cdots = x_n^2$. Suppose to the contrary, say $x_1^2\neq x_2^2$. Denote the maximum value as $M$ where $$M = (x_1^2 x_2^2)(x_3\cdots x_n)^2.$$
Now $x_1^2 + x_2 ^2 +\cdots+ x_n^2 = 1$, so $x_1^2 + x_2 ^2 = 1-Q$ for some $Q$. I claim we can pick better values for $x_1$ and $x_2$ that improves the maximum. Indeed, set $y_1 = y_2 = \sqrt{\frac{1-Q}{2}}$. Then note $y_1 ^2 + y_2 ^2 = 1-Q$, so $$y_1^2 +y_2 ^2 + x_3 ^2 +\cdots + x_n^2 = 1.$$ And by AM-GM for two non-negatives, we have $$x_1^2x_2^2 < \left(\frac{x_1^2+x_2^2}{2}\right)^2 = \left(\frac{1-Q}{2}\right)^2 = y_1^2y_2^2,$$ and note this inequality is strict as $x_1^2\neq x_2^2$.

Hence $y_1^2y_2^2(x_3\cdots x_n)^2> M$, contradicting the maximality of $M$.

Thus we conclude $x_i^2$ are all the same values when achieving maximum and hence $x_i^2 = 1/n$.  $\blacksquare$

Proof of AM-GM for two non-negatives.

The inequality is straightforward as $0\le (\sqrt A - \sqrt B)^2 = A + B -2\sqrt{AB} $, so $ \sqrt{AB}\le (A+B)/2$. For the equality case, note $ \sqrt{AB} =  (A+B)/2$ if and only if $(\sqrt A - \sqrt B)^2 = 0 $, if and only if $A=B$. $\blacksquare$

Remark. Of course we will have to believe that $(x_1x_2\cdots x_n)^2$ will attain a maximum somewhere on the unit sphere to begin with, which is true as $f(x_i)=(x_1\cdots x_n)^2$ is continuous on the compact unit sphere (extreme value theorem). This will be trickier to justify, a non-analysis proof eludes me at the moment.

Remark 2. It isn't too surprising we involve AM-GM here, as you may observe that the corollary of this problem is to demonstrate the general AM-GM inequality for $n$ non-negatives.


问:
Possible values of $b$ for $x^2 + bx + 6 = 0,$ for integer solutions
The question is essentially in the title. The actual question is here(and also my working out) if there's any confusion in the title http://m.imgur.com/a/lgNzt

My method was to find the discriminant = 0 and then use trial and error since this question was multiple choice.

Is there a way of doing this properly, as in algebraically without trial and error?

答:
$$x^2 + bx + 6 = 0\implies b=-\bigg(\frac{x^2+6}{x}\bigg)
= \bigg(-x - \frac{6}{x}\bigg)\\
 \implies x\in\{\pm1, \pm2, \pm3 \pm6 \}
 \quad\land\quad  b\in\{\pm5, \pm7\}$$

$$x=\frac{-b\pm \sqrt{b^2-24}}{2}\implies \sqrt{b^2-24}\in\mathbb{N}
\implies b\in \{ \cdots, \pm5, \pm7\, \cdots \}$$

Testing these results (and far beyond) confirms only one solution:
$\space (b,x)=(-7,1)$


问:
Proof of Lemma 5.1.5.3 in Jacob Lurie's HTT.
I am currently trying to understand the following proof in Higher Topos Theory.

I am fine with almost all of the argument, except with the claim that $\mathcal{E}^1$ is a deformation retract of $\mathcal{E}$. While I feel like this is true, I am unable to prove it properly, i.e writing down a well-defined morphism of simplicial sets $r : \mathcal{E} \to \mathcal{E}^1$ along with a homotopy witnessing it as a deformation retract.

An object (i.e a $0$-simplex) of $\mathcal{E}^1$ is just a map $C \to X$ in $\mathcal{P}(S)$, while an object of $\mathcal{E}$ is

Either a couple $(v, C \to X)$ where $v$ is the cone point of $(\mathcal{C}_{/X})^\rhd$.

Either a couple $(B \to X, C \to B)$ where $B$ is in $\mathcal{C}$.

The "obvious way" to produce a map $C \to X$ out of this kind of data would be to send $(v, C \to X)$ to $C \to X$ and $(B \to X, C \to B)$ to a composition $C \to B \to X$. My issue with this approach is that I am using composition internal to $\mathcal{P}(S)$, which is only weakly defined, and I do not know how to produce a full map of simplicial sets using it. The same problem arises when trying to work out higher-dimensional simplices.

I can show that $\mathcal{E}^1 \to \mathcal{E}^0$ is a weak homotopy equivalence by viewing it as a pullback of the initial (hence left anodyne) map $\{\mathrm{id}_{\mathcal{P}(S)}\} \hookrightarrow \mathcal{P}(S)_{C/ }$ by the right fibration (hence proper map) $\mathcal{C}_{/X} \to \mathcal{C}$. I can not make this argument work for $\mathcal{E}^1 \to \mathcal{E}$, unless the map $(\mathcal{C}_{/X})^\rhd \to \mathcal{P}(S)$ is proprer. I have no idea wether this can be expected or not.

Using a similar kind of argument (pulling back the right anodyne map $\{v\} \to (\mathcal{C}_{/X})^\rhd$ along the left fibration $\mathcal{P}(S)_{C/ } \to \mathcal{P}(S)$), I can show that $\mathcal{E}^0$ indeed has the same homotopy type as $\mathcal{E}$ and that it is the homotopy type of $\mathrm{Map}_{\mathcal{P}(S)}(C, X)$ but this argument does not say that this equivalence is induced by $\mathcal{E}^0 \to \mathcal{E}$.

Is there a simple argument to show that $\mathcal{E}^1 \to \mathcal{E}$ is a weak homotopy equivalence, or even to show directly that $\mathcal{E}^0 \to \mathcal{E}$ is a weak homotopy equivalence?

Update: It seems to me that one can show directly that $\mathcal{E}^0$ is a deformation retract of $\mathcal{E}$, and that the homotopies do not use composition internal to $\mathcal{P}(S)$. However, I am not yet been able to write the homotopy either, and would appreciate any hint or construction that would help defining such a map.

答:
I think you are right in following intuitions at the zero level, but I strongly recommend you not to use simplex by simplex constructions and use abstract stuff to create maps, as you were doing.

My first intuition was the following (see HTT 1.2.2 for definitions):

$\mathcal{E}$ can be contracted to the cone point in the first component, yielding $\text{Hom}^L_{\mathcal{P}(S)}(C,X)$;

$\mathcal{E}_0$ can be contracted in the second component (because the overcategory has a final object) to yield exactly $\mathcal{E}^1 = \text{Hom}^R_{\mathcal{P}(S)}(C,X)$;

The very non obvious thing is then to find a deformation retraction from $\text{Hom}^L \to \text{Hom}^R$, which I am not quite sure it exist; so we should change approach. On the other side, we can contract the second component on $\{1_C\}$ because it is initial, yielding a deformation retraction

$$ \mathcal{E} \to (\mathcal{C}_{/X})^{\triangleright} \times_{\mathcal{P}(S)} \{1_C\} $$

But the cone point is sent to $X \in \mathcal{P}(S)$ and will never contribute to the pullback, so the latter is equal to $\mathcal{E}^1$! This happens unless $X=C$, but in this case the inclusion $\mathcal{C}_{/C} \to (\mathcal{C}_{/C})^{\triangleright}$ is a deformation retraction by sending the cone point to $1_C$.

I don't know which lemma to invoke here so we just verify on simplices: an $n$ simplex $\Delta^n \to (\mathcal{C}_{/X})^{\triangleright} \times_{\mathcal{P}(S)} \{1_C\}$ is the same as a map $\phi : \Delta^{k+1} \to \mathcal{C}$ for some $-1 \le k \le n$  so that
$$\phi(k+1) = X, \ \ \  \phi|_{\{0, \ldots, k\}} * X : \Delta^n \to \mathcal{C} $$
is constantly $C$, where $X$ above denotes the constant $=X$ simplex $\Delta^{n-k-1} \to \mathcal{P}(S)$. In order for this to happen, we must have $n-k-1 = -1$ because $X \neq C$, i.e. $n=k$. But $n$ simplices $\phi: \Delta^{n+1} \to \mathcal{C}$ such that
$$\phi(n+1) = X, \ \ \ \phi|_{\{0, \ldots, n\}} = C $$
are exactly the simplices coming from $\mathcal{C}_{/X} \times_{\mathcal{P}(S)} \{1_C\}$.

Remark. I implicitly used above that maps $\Delta^n \to A^{\triangleright}$ are the same as maps $\phi:\Delta^k \to A$ with $-1 \le k \le n$, and under a simplicial map $F: A^{\triangleright} \to B$ such simplex is sent to $ F( \phi)* F(v)* \ldots * F(v)$; here $v$ is the cone point and $F(v)$ is repeated $n-k-1$ times. This is just another way of stating that

$$ (A^{\triangleright})_n = (A * \Delta^0 ) = A_n \cup (\Delta^0)_n \cup \bigcup_{i+j = n-1} A_i \times (\Delta^0)_j $$

Using the convenient convention that $\Delta^{-1} = \emptyset$, so that $A_{-1}$ has just one element (the empty map).


问:
When can we specify an everywhere orthogonal grid on a surface?
Is there a classification for surfaces on which we can specify a grid that forms right-angles at every intersection without singularities? For example, we can do this on the plane, cylinder, torus, and mobius-strip, but not the cone nor sphere which (I think) will always have singularities when we try. It seems related to the idea of a developable surface but not quite.

Do such surfaces have any special importance for this fact? I understand that coordinates are, by nature, quite arbitrary, but this is a question about the existence of coordinates with a certain property, so perhaps it can be more fundamentally related to the topology of the surface. Is there a formalization of this concept in terms of a manifold's atlas? Or perhaps something about an everywhere diagonal metric-tensor? (Just spitballing).

Edit: The reason the cone fails is obvious: it isn't a regular surface / manifold, i.e. it has a "sharp point." Meanwhile, the reason the sphere fails might be a result of the Hairy-Ball Theorem? I'm beginning to think that finding an orthogonal grid doesn't make a manifold special. What would make a manifold special is proving that no such grid can exist, which may require a Hairy-Ball-like theorem. Still open to answers!

答:
Here's an answer for finite type oriented surfaces by which I mean any surface $S$ that is diffeomorphic to an oriented surface $F$ which is closed --- meaning compact, connected, and empty boundary --- minus a finite subset $A \subset F$. (I suspect something similar works in the nonorientable case, but I think I"ll restrict my answer to the orientable case)

First, if $A = \emptyset$ and hence $S=F$ is closed then a necessary and sufficient condition is that the genus equals $1$.

I claim that if $A \ne \emptyset$ then $S$ has such a grid as you ask for.

First let me describe all genus $0$ examples, i.e. all examples where $F$ is a sphere minus $1$ or more points. If $A$ has just $1$ point then $S=F-A$ is diffeomorphic to $\mathbb R^2$, in which the horizontal/vertical lines give such a grid, as depicted in your question. If $A$ has $k \ge 2$ points then $S=F-A$ is homeomorphic to $\mathbb R^2$ minus $k-1$ points, and again the horizontal/vertical lines (punctured at $k-1$ points) gives such a grid.

Also one can do all genus $1$ examples, i.e. where $F$ is a torus, because $F$ itself has such a grid, and then you can remove any finite subset of points.

You can see from these examples that if the genus is fixed and if you can describe a grid on the once punctured surface of genus $g$, then you obtain a grid on a genus $g$ surface punctured at any nonempty finite set.

So, what does one do if $F$ is the closed surface of genus $2$, and $S$ is $F$ minus $1$ point? In this case, consider a regular octagon in the plane, and glue every side with the opposite parallel side, using a gluing map identifies each opposite pair of sides by a Euclidean translation. The quotient under this gluing is a closed surface of genus $2$. All 8 vertices of the octagon are identified to a single point of $F$ which we take to be $A$. Now puncture $F$ at that one point to give the surface $S=F-A$, so we can think of $S$ as the quotient of the octagon with its 8 vertices removed. What one observes is that if one restricts the horizontal/vertical lines of $\mathbb R^2$ to the octagon with vertices removed, and then maps those lines to the quotient surface $S=F-A$, the result is a grid as you desire.

And then this generalizes to any genus: take a regular planar $4g$-gon, remove its vertices, and glue opposite side pairs; the horizontal\vertical grid on the plane descends to a grid such as you ask for, on the genus $g$ surface puncture at $1$ point.


问:
The course of the function $f(x) = \sin^3 (x) + \cos^3 (x)$.
First, sorry for my english, hope you understand.

so I'm examining the course of this function $f(x) = \sin^3 (x) + \cos^3 (x)$ where I need to find everything I can about this function (e.g. definition field, range of values, inflection points, asymptots, global and local max/min, etc.). Now I'm struggling with 3 tasks.

Prove there are no limits in $+\infty$ by showing there are different limits of $f(\pi/2+2k\pi)$ and $f(2k\pi)$ where $k$ goes to $+\infty$. Is this right?
 Are the results of the limits different?

How to get inflection points? I know I get them by solving the 2nd derivative of $f(x)$ which I think I have but I don't know how to get the exact values.
graph and inflex points

Find intersection of $f(x)$ with axes. I got to the point where I have two equations and I need to prove why is it like that.

Thank you in advance

edit: Thank you all for helping me with this problem. I've used @Robert Lee solution in the end.

答:
(1) the idea is right, note that $\sin(k\pi) = 0$ and $\cos(k\pi) = (-1)^k$

(2) Note
$$
\begin{split}
f'(x) &= 3\sin^2 x \cos x - 3\cos^2x\sin x
       = 3 \sin x \cos x (\sin x - \cos x)\\
f''(x) &= \ldots = -3 (\cos x - \sin x)^2(\cos x + \sin x)\\
\end{split}
$$
and now solving $f''(x)=0$ is trivial.

(3) your argument makes sense, if you are asking something else, please clarify it


问:
$P(A − B) − \varnothing ⊆ P(A) − P(B)$
Prove that if $A$ and $B$ be sets. Then $P(A − B) − \{\varnothing\} ⊆ P(A) − P(B)$.

If there was no empty set, I know the rules such as $\{x\}$ is in $P(A)$ implies $x$ in $A$. But I don't know when there is an empty set.

答:
Always type-check when writing set operations (especially involving $P$ or $\varnothing$): When you wrote $P(A-B)-\varnothing$, this means the set containing subsets of $A-B$ but not elements of $\varnothing$. You probably mean the set containing subsets of $A-B$ that is not $\varnothing$, which should be $P(A-B)-\{\varnothing\}$.

Now by definition, we need to check for every $X\in P(A-B)-\{\varnothing\}$, that $X\in P(A)-P(B)$. By definition again, we need to check that if $X\subseteq A-B$ and $X\ne\varnothing$, then $X\subseteq A$ and $X\not\subseteq B$.

To check $X\subseteq A$, note that $X\subseteq A-B\subseteq A$. Then use transitivity of $\subseteq$ to conclude that $X\subseteq A$.

To check that $X\not\subseteq B$, we will use proof by contradiciton. If $X\subseteq B$, then since $X\subseteq A-B$ as well, we know that $X\subseteq (B)\cap(A-B)=\varnothing$. However $X\subseteq\varnothing$ implies $X=\varnothing$, contradicting that $X\ne \varnothing$. Thus by contradiction, we know that $X\not\subseteq B$.

QED

Okay some more formal nonsense: Note that there is the edge case where $A\subseteq B$, or even rather $A\subsetneq B$. The desired inclusion still holds in this case, even though it has completely lost meaning. In particular, $P(A-B)=\{\varnothing\}$, so there would be no such $X\in P(A-B)-\{\varnothing\}$. However since there is no such $X$, it is still valid to say "every such $X$ satisfies ...".

Title: $P(A − B) − \varnothing ⊆ P(A) − P(B)$


答: No, we choose the smaller one. $\delta, \delta_1, \delta_2$ represent various numerical values.</p>

Title: epsilon delta question limit questions

Body: I have two quick questions. In calculus books when proving limits, I always see something like, let $\delta = \min(\delta_{1}, \delta_{2})$. Does this mean that we take a $\delta$ such that it contains both $\delta_{1}$ and $\delta_{2}$?</p>


问:
Can two elliptic isometries generate a free group?
Standard applications of the ping-pong lemma can be used to show that two hyperbolic isometries or two parabolic isometries of $\mathbb{H}^2$ generate a free group. (Assuming they have disjoint fixed points and after passing to high enough powers, of course.)

I'm wondering if it's ever possible for two elliptic isometries to generate a free group (rank $2$). Clearly they would have to be irrational rotations about different fixed points. The action of each rotation on the boundary of $\mathbb{H}^2$ has dense orbits, so the standard ping-pong argument doesn't go through.

If they don't generate a free group, is there a (preferably geometric) way to see where a relation would come from?

答:
Here is an indirect proof. Let $G$ denote $SL(2, {\mathbb R})$. Then given any nonempty reduced word $w$ in the alphabet $\{x_1^{\pm 1}, x_2^{\pm 1}\}$, we have a map
$$
P_w: G\times G\to G
$$
sending $(g_1, g_2)$ to $w(g_1, g_2)$ (you substitute $g_i$ for $x_i$, $i=1, 2$). The map $P_w$ is polynomial in terms of the matrix coefficients of $g_1, g_2$. Since $P_w^{-1}(\{\pm 1\})\ne G\times G$ (as $G$ contains rank 2 free subgroups), it follows that each $X_w=P_w^{-1}(\{\pm 1\})$ is a closed subset with empty interior in $G\times G$. Thus, by Baire's theorem, the union
$$
X:= \bigcup_{w} X_w
$$
has empty interior in $G\times G$. (The union is taken over all nonempty reduced words $w$.) The subset $E\subset G$ consisting of matrices projecting to nontrivial elliptic elements of $PSL(2, {\mathbb R})$ is open (and, of course, nonempty). It follows that $E\times E$ has nonempty intersection with
$$
G\times G \setminus X.
$$
Every $(g_1, g_2)\in E\times E \setminus X$ will project to a pair of elliptic isometries of ${\mathbb H}^2$ generating a free subgroup of rank 2 in $PSL(2, {\mathbb R})$.


问:
Find all functions $f(x^2-y^2)=f(xy)(f(x)+f(y))$
Find all functions $f:\mathbb R\to\mathbb R$ such that, for all $x,y\in\mathbb R$, we have
$$f(x^2-y^2)=f(xy)(f(x)+f(y)).$$

Both $f(x)=0$ and $f(x)=1/2$ are solutions.

Swapping $x,y$ in the equation implies that $f$ is even. Plugging in $x=y=0$ implies that $f(0)=0$ or $f(0)=1/2$.

If $f(0)=0$, then plugging in $y=0$ implies that $f(x^2)=0$. So $f(t)=0$ for $t\geq0$, and $f$ even means that $f=0$ is the only solution in this case.

I'm struggling though with the case $f(0)=1/2$.

答:
Plugging in $y=0$ gives $f(x^2)=f(0)(f(x)+f(0))$, which, after plugging in $f(0)=\frac 1 2$, gives $$f(x^2)=\frac 1 2 f(x) + \frac 1 4$$
on the other hand, plugging in $x=y$ gives $f(0)=f(x^2)(f(x)+f(x))$, which is $$\frac 1 2=2f(x)f(x^2)$$
Eliminating $f(x^2)$ in those two equations provides $$2(f(x))^2+f(x)-1=0$$
which implies either $f(x)=\frac 1 2$ or $f(x)=-1$. Since $f(0)=\frac 1 2$, we can safely conclude that $f=\frac 1 2$ is the solution.


答:
$P(A)$ means the set of all subsets of $A$.

$A-B$ means the set of all elements in $A$ that are not in $B$.

Stop and first think about what any set minus $\{\varnothing\}$ would be.

Now, take something, call it $S$, that is in the left hand side. If the left hand side is the empty set, you are finished, so we can assume that there is something in the left hand side. That means $S$ is a subset of the set of elements that are in $A$, but not in $B$. That set is contained in $P(A)$, but it cannot be contained in $P(B)$. If $S$ were contained in $P(B)$, then all the elements of $S$ would be elements of $B$. But, that would mean $S$ would be a subset of $B$. If you are not convinced, take a specific $x$ that is an element of $S$. $S$ cannot be empty because the definition of the left hand side has excluded $\varnothing$. Convince yourself that $x$ has to be an element of $A$ but it cannot be an element of $B$.  Once you are convinced of that, you should see that the entire set $S$ is contained in $P(A)$, but not in $P(B)$.

Title: $P(A − B) − \varnothing ⊆ P(A) − P(B)$

Body:
Prove that if $A$ and $B$ be sets. Then $P(A − B) − \{\varnothing\} ⊆ P(A) − P(B)$.

If there was no empty set, I know the rules such as $\{x\}$ is in $P(A)$ implies $x$ in $A$. But I don't know when there is an empty set.


问:
Probability from 2 continuous random variables with joint density
How do you calculate the probability from two continuous random variables (X and Y) with joint density?

$$\begin{align}
\mathrm{i.}~&~\mathsf P(0 \leqslant X \leqslant 1/2, 1/2 \leqslant Y \leqslant 1)
\\[1ex]
\mathrm{ii.}~&~\mathsf P(0 \leqslant X \leqslant 1/4)
\\[2ex]
f(x,y)&=\begin{cases}2/3\,(2x+y) &\mbox{for}& 0 ⩽ x ⩽ 1, 0 ⩽ y ⩽ 1\\ 0 &&\mbox{elsewhere}\end{cases}\end{align}$$

答: Sometimes figures are worth 1000 words:

Body: Sometimes figures are worth 1000 words:


答:
Element chase.

Let $X \in P(A-B) - \emptyset$.  Then $X \subset A-B$ and $X \ne \emptyset$.  So $X$ consists entirely of elements within $A$ and has no elements of $B$.

So $X$ is a subset of $A$.  As $X$ has elements that are not in $B$, $X$ is not a subset of $B$. (Note if $X = \emptyset$ we would have $X$ is a subset of $B$ but because $X$ isn't empty it is not.)

So  $X$ is a subset of $A$.  So $X \in P(A)$.  And $X$ is not a subset of $B$ so $X \not \in P(B)$.  So $X\in P(A) -P(B)$ and $ P(A-B) - \emptyset\subset P(A) -P(B)$.

That's it.

....

Note the other direction does not hold.  A subset $K$ of $A$  that is not a subset of $B$ could contain an element that is in $A$ but not $B$, and an element that is in $A$ and $B$.  But then $K$ will contain elements of $B$ so it will not be a subset of $A-B$.

Title: $P(A − B) − \varnothing ⊆ P(A) − P(B)$


问:
If $(G,*)$ is a group and $a_1,a_2,\ldots,a_n \in G$, prove that$(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$.
If $(G,*)$ is a group and $a_1,a_2,\ldots,a_n \in G$, prove that
$(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$.

Attempt:
For the simplify and the convenience,
write $a*b=ab$.

I'll prove it by induction on $n$.

Let $a_1,a_2,\ldots,a_n \in G$.
Let $P(n)$ be the statement:
$(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$.

Base Case:
Clearly $P(1)$ is true since $(a_1)^{-1} = a_1^{-1}$. (Is it enough? What if I take $P(1)$ and $P(2)$ for the base case?)

Inductive Step:
Assume that $P(k)$ is true. That is,
$(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$.
To show that $P(k+1)$ is also true, here it is:
\begin{align*}
(a_1a_2\ldots a_{k+1})^{-1} &=
((a_1a_2\ldots a_k)a_{k+1})^{-1} \\
&= (a_{k+1})^{-1}(a_1a_2\ldots a_k)^{-1} \\
&= a_{k+1}^{-1}(a_k^{-1}a_{k-1}^{-1}\ldots a_1^{-1}) \\
&=a_{k+1}^{-1}a_k^{-1}a_{k-1}^{-1}\ldots a_1^{-1}
\end{align*}
as desired.

Hence, $(a_1*a_2*\ldots*a_n)^{-1} = a_n^{-1}*\ldots*a_1^{-1}$ for all
$a_1,a_2,\ldots,a_n \in G$. $\Box$

Is the approach above true?

答: Yes, this looks good with $P(2)$ which you use implicitly in your inductive step, if you're really trying to be exacting in the amount of detail, you may want to be explicit in you use of the base case and the inductive hypothesis in your inductive step, but I think almost anyone who read this would have no problem with it.


问:
Relation of surface area of a sphere to its volume.
As I was working on a problem I noticed that the formula for the surface area of a sphere is the derivative of a sphere's volume with respect to $r$. Is this purely coincidental or is there a reason for this?

答:
Not a coincidence. The same holds for circles - the circumference is the derivative of the area.

Start with a sphere of radius $r$. If the skin has thickness $dr$, the volume of the skin will be approximately $A.dr$, where $A$ is the surface area.

However, it will also be $V(r + dr) - V(r)$, so $$A\approx \frac{V(r+dr) - V(r)}{dr}$$

As $dr$ gets tiny, this gives $A=V^\prime(r)$.


问:
find the value of p to get converges (Improper Integrals)
find for which values of p the integral converges

$$\int_0^\infty \frac{\text{d}x}{x^p}\text{}$$

I just use the definition of the improper integrals and to separate into two parts.

one is from 0~1 and one is from 1~inf , after that , I cant find any real number which is suitable for

this question in order to take that two parts both Converges..

how could I do that?

btw.
$$\int_1^\infty \frac{\text{d}x}{x^p}\text{}$$ p>1---Converges
$$\int_0^1 \frac{\text{d}x}{x^p}\text{}$$ p<1---Converges

答:
You want a real number $p$ such that both
$$
\int_0^1\frac{1}{x^p}\,dx\quad\textrm{and}\quad \int_1^\infty\frac{1}{x^p}\,dx\tag{0}
$$
are convergent. As you have observed, such $p$ does not exist.

Notes.
On $(0,\infty)$,
$$
\int \frac{1}{x^p} dx = \frac{1}{1-p}x^{1-p}+C,\quad p\ne 1\tag{1}
$$
$$
\int \frac1x dx = \ln x+C\quad \tag{2}
$$

So you see when $p=1$, neither integral in $(0)$ converges.

If $p<1$, then $1-p>0$, $\int_1^\infty\frac{1}{x^p}\,dx$ diverges.

If $p>1$, then $1-p<0$, $\int_0^1\frac{1}{x^p}\,dx$ diverges.


问:
Time-invariant systems?
I am reading the paper http://web.mit.edu/2.14/www/Handouts/Convolution.pdf because I want to learn about convolutions.
With
$$
\delta_T (t) = \begin{cases}
      0 & t \leq 0 \\
      1/T & 0 < t \leq T
   \end{cases}
$$
the following is said:

We now assume that the system response to $\delta_T (t)$ is a known
function and is designated $h_T (t)$. Then if the system is linear and
time-invariant, the response to a delayed unit pulse, occurring at time nT, is simply a delayed version of the pulse response: $y_n (t) =
> h_T (t- nT)$

The linear system with the output function $y_n (t)$ is time-invariant. I think I get the definition of being time-invariant(not directly depend on time, possible indirectly from the input function). However, I still have no idea why $y_n (t)$ should be equal to $h_T (t - nT)$ -- like where does the $- nT$ part come from?

If you wanna take a look, it is formula 12 in the paper :))

答: The system is time-invariant, i.e., if the input is delayed by $\tau$, the output is also delayed by $\tau$, nothing else changes. So, if the response of the system to $\delta(t)$ is $h(t)$, then its response to $\delta(t-nT)$, which is $\delta(t)$ shifted to the right by $nT$, is simply $h(t-nT)$, which is $h(t)$ shifted to the right by $nT$.

Title: Time-invariant systems?


问:
Generating function question about arranging n objects with limitations
Generating functions question: There are n objects - rings, earring and bracelets. How many ways are there to arrange these objects, as the amount of earring is even and there are at most 4 bracelets.

[This is a question we must solve with generating functions].

Here's what I started:

There are $n$ objects, at most 4 bracelets. There are $2k$ bracelets when $2k \leq n$. So there are $n-2k-4$ rings at most.

I defined the following generating function: $$A(x)=(x^0+x^1+...+x^{n-2k-4})(x^0+x^2+...+x^{2k})(x^0+x^1+x^2+x^3+x^4)={1-x^{n-2k-3}\over 1-x}\cdot {1-x^{2k+1}\over 1-x^2}\cdot {1-x^5 \over 1-x}$$

So I tried simplfying it... and.... well it didn't go well.

Is there another way to solve this? Is my direction even correct?

Thanks for any input!

答:
You shouldn’t be building $n$ into your generating function: you actually want the coefficient of $x^n$ in the (correct) generating function, which is

$$A(x)=\left(\sum_{k\ge 0}x^k\right)\left(\sum_{k\ge 0}x^{2k}\right)\left(x^0+x^1+x^2+x^3+x^4\right)\;.$$

The first factor is for rings, of which you may have any number, the second is for earrings, of which you must have an even number, and the last is for bracelets, of which you must have at most $4$. Thus,

$$A(x)=\frac1{1-x}\cdot\frac1{1-x^2}\cdot\frac{1-x^5}{1-x}=\frac{1-x^5}{(1-x)^3(1+x)}\;.$$

Divide out, expand the remaining rational function into partial fractions, convert to power series, and find the coefficient of $x^n$.