问:
Vector potentials of vector potentials
In $\mathbb{R}^{3}$, we have that $\mathbb{div}(A_0)=0 \implies A_0=\mathbb{curl}(A_1)$
for some vector field $A_1$

We can also, using the Lorentz Gauge, choose $A_1$ such that $\mathbb{div}(A_1)=0$, which made me wonder;

What happens if you repeat this process, producing $A_2$ a divergence free potential for $A_1$ and so on. Does this terminate to a fixed point? Does it enter a loop?

答:
Following the prescription of this question, given a vector field $\mathbf A_0$ such that $\nabla \cdot \mathbf A_0=0$, we construct a sequence $\mathbf{A}_0, \mathbf A_1, \mathbf A_2, \ldots$ such that
$$
\begin{array}{cc}
\nabla \cdot \mathbf A_1 = 0, & \nabla \times \mathbf A_1= \mathbf A_0\\
\nabla \cdot \mathbf A_2 = 0, & \nabla \times \mathbf A_2= \mathbf A_1\\
\nabla \cdot \mathbf A_3 = 0, & \nabla \times \mathbf A_3= \mathbf A_2\\
\vdots & \vdots
\end{array}$$
Taking the curl of the right column, and using the formula for the double curl, we see that
$$
-\nabla^2 \mathbf A_{n+2} = \mathbf A_n, \quad n=0, 1, 2, \ldots,$$
where $\nabla^2$ denotes the vector Laplacian. This is the Poisson equation and it can be solved by convolution against  the Newtonian potential;
$$\tag{*}
\mathbf A_{n+2}(x)=\int_{\mathbb R^3} \frac{ \mathbf A_n(x-y)}{4\pi \lvert y\rvert}\, dy.$$
Remark. The vector field $\mathbf A_{n+2}$ given by (*) is divergence-free if $\mathbf A_n$ is. The proof is immediate, just take the divergence termwise and interchange differentiation and integration. Thus, (*) has the correct gauge.

To gather information about the limit as $n\to \infty$, (*) is impractical. However, taking the Fourier transform, defined as
$$
\hat{f}(k)=\int_{\mathbb R^3} f(x)e^{-i x\cdot k}\, dx, $$
we see that
$$\tag{**}
\hat{\mathbf A}_{n+2}(k)=\frac{\hat{\mathbf A}_n(k)}{\lvert k\rvert^2}.$$
We conclude that
$$
\hat{\mathbf A}_n(k)=\begin{cases} \frac{\hat{\mathbf A}_0(k)}{\lvert k\rvert^m}, & n=2m, \\
\frac{\hat{\mathbf A}_1(k)}{\lvert k\rvert^m}, & n=2m+1.
\end{cases}$$
In particular:

the sequence never reaches a fixed point, except for the trivial cases in which $\mathbf A_0=\mathbf 0$ or $\mathbf A_1 =\mathbf 0$.

if $\hat{\mathbf A}_0$ is supported in $\lvert k\rvert <1$, then the sequence $\hat{\mathbf A}_n$ for even $n$ blows up. If it is supported in $\lvert k\rvert >1$, it tends to zero. Similar considerations for $\hat{\mathbf A}_n$ for odd $n$.

Title: Vector potentials of vector potentials


问:
How to show that $a$ can be divided by $6$ if and only if it can be divided by both $2$ and $3$?
Prove that for: $a \in\mathbb Z$, $a$ is divisible by 2 and $a$ is divisible by 3 if and only if $a$ is divisible by 6.

EDIT: Sorry, I wasn't aware of how exactly this site worked. This is pretty much a question from my Discrete Math course that I was having a little trouble understanding. Apologies for not being clear about the background and only posting a question with no context.

ContentLicense: CC BY-SA 3.0

Body: Hint: $\ \dfrac{a}6\, =\, \dfrac{a}2\, -\, \dfrac{a} 3$


问:
To which subspace does $v_4$ belong?
Let $h\in\mathbb{R}, h\neq 0$ and three vectors
$$
\begin{pmatrix}
1\\ 0\\ 0\\ 1
\end{pmatrix},\quad \begin{pmatrix}
2\\ h\\ h\\ 2
\end{pmatrix}\quad\begin{pmatrix}
1\\ 1+h\\ 2h+1\\ 1
\end{pmatrix}.
$$
Let $$v_4=\begin{pmatrix}
1\\ 0\\ 0\\ 0
\end{pmatrix} \quad\mbox{ and }\quad A=\begin{pmatrix}
1& 2& 1\\
0& h& 1+h\\
0& h & 2h+1\\
1& 2& 1
\end{pmatrix}.$$
If C(A) is the column space of $A$ and $N(A)$ is the null space of $A$, to which of these belongs $v_4$?

My solution: Now, since $h\neq 0$, thus $v_1, v_2, v_3$ are linearly independent, so
$$
C(A) =\left\{
\begin{pmatrix}
1\\ 0\\ 0\\ 1
\end{pmatrix}, \begin{pmatrix}
2\\ h\\ h\\ 2
\end{pmatrix}\begin{pmatrix}
1\\ 1+h\\ 2h+1\\ 1
\end{pmatrix}\right\},
$$
while dim$N(A) = 3-{\rm rank}(A)=0$, do that $N(A) =\{0\}$ and then $v_4$ does belong nither to $C(A)$ nor $N(A)$.

Is my solution right? Could anyone please help me?
Thank you in advance!

答: Concerning the fact that $v_4\notin C(A)$, you did not justify it. You can say that, for instance, since $h\ne0$, the set $\{v_1,v_2,v_3,v_4\}$ is linearly independent. And $v_4\notin N(A)$ because $N(A)$ is a set of vectors of $\Bbb R^3$, whereas $v_4\in\Bbb R^4$.

ContentLicense: CC BY-SA 4.0

Title: To which subspace does $v_4$ belong?


问:
Differentiate an equation
I'm learning about implicit differentiation and i'm stuck at understanding the concept of "differentiating an equation".

If I define $ f(x) = x^2$ and $g_1(y) = 9$ and $g_2(y) = 9 + y^2$

Then I define 2 equations:

$(1)$ $f(x) = g_1(y)$ (i.e. $x^2 = 9$)

Then I can't differentiate both side of this equation with respect to x (ortherwise I will get: $2x = 0$)

$(2)$ $f(x) = g_2(y)$ (i.e. $x^2 = 9 + y^2$)

Then I can differentiate both side of this equation with respect to $x$ to get: $ 2.x = 2.y.\frac{dy}{dx}$

Could you please explain me the difference between the 2 cases ? And in general, in which condition could I differentiate both side of an equation concerning more than 1 variable ?

Thank you very much for your help!

答:
When we write $x^2=9$, $x$ is an unknown, rather than a variable. By contrast, the expression $\frac{d}{dx}(x^2)=2x$ can be viewed as a shorthand for

Let $f$ be the function defined by $f(x)=x^2$ for all $x$. The function $f'$, evaluated at the arbitrary point $x$, is equal to $2x$, i.e. $f'(x)=2x.$

Here, $x$ is simply a dummy variable used to illustrate what the derivative of the function is. It makes no difference whether we use $x$ or $y$ or any other  variable. So $x$ is being used in two different ways here, hence why contradictory results appear.

By contrast, the second example, $x^2=9+y^2$, does show a relationship between the variables $x$ and $y$. Again, this could be expressed more formally as

Let $y$ be an arbitrary function satisfying $y(x)^2+9=x^2$ for all $x$.

When we differentiate both sides with respect to $x$, what we are doing is showing that $y'(x)$ must satisfy the following property:

$2x=2y(x)y'(x) \, .$

Even though we might not have an explicit formula about $y$, implicit differentiation allows us to make meaningful statements about this function.

Title: Differentiate an equation


问:
Is the height associated to a degree zero divisor always bounded?
Let $X$ be a smooth, projective, geometrically irreducible curve defined over a number field, and let $D$ a divisor on $X$. To these data, we can associate a height function on the $\Bbb{\overline{Q}}$-rational points of $X$
$$
\operatorname{ht}_D: X(\Bbb{\overline{Q}}) \rightarrow \Bbb{R}
$$ as follows. Write $D=D_1-
D_2$, with $D_1,D_2$ very ample divisors on $X$, and let $\phi_{D_1},\phi_{D_2}: X \hookrightarrow \Bbb{P}^n$ be embeddings determined by $D_1,D_2$. Then
$$
\operatorname{ht}_D(P):=\operatorname{ht}_{D_1}(P)-\operatorname{ht}_{D_2}(P):=\operatorname{h}(\phi_{D_1}(P))-\operatorname{h}(\phi_{D_2}(P))
$$
where $\operatorname{h}: \Bbb{P}^n \rightarrow \Bbb{R}$ is the usual height function on projective space. One can check that, up to a bounded function, $\operatorname{ht}_D$ does not depend on the choices of $D_1,D_2$.

Now I know that if $\deg(D) >0$, the height function $\operatorname{ht}_D$ is bounded from below, and if $\deg(D) < 0$, then $\operatorname{ht}_D$ is bounded from above.

$\textbf{Question:}$ What happens if $\deg(D)=0$? Is there a curve $X$ as above and a divisor $D$, with $\deg(D)=0$ such that $\operatorname{ht}_D$ is not bounded from below/above?

答:
The question as to whether $\operatorname{ht}_D$ for a degree zero divisor $D$ is bounded depends on the class of $D$ in the Picard group of $X$.

Theorem: Let $X$ be a smooth, projective curve over a number field, and $D$ a divisor on $X$. Then $\operatorname{ht}_D$ is bounded, if and only if $D$ is a torsion element of the Picard group $\operatorname{Pic}(X)$ of $X$.

More generally, the theorem is true for an arbitrary smooth, projective variety.

The "if" part is clear, and the only if part is proved in

J.-P. Serre, Lectures on the Mordell-Weil Theorem, Theorem 3.11.

The most important fact for the proof seems to be that the Jacobian of $X$ is principally polarised, i.e. it is isomorphic to its dual abelian variety.

The above theorem is somewhat interesting in my opinion, partly because it resembles the well-known result that on an abelian variety $A$ a point is a torsion point, if and only if its Néron-Tate height (associated to an ample, symmetric divisor) vanishes.


问:
Prove that if $A$ is diagonal, than $\operatorname{Adj}(A)$ is diagonal as well
I'm not sure if my proof is formal enough, and I'd appreciate a review or if someone could write his/her proof of the statement. Here's how my proof goes:

Given a diagonal matrix $A \in M_n(F)$, we can say that $a_{ij}$, $1\le i\le n$ and $1\le j\le n$, being an entry of $A$, equals to $0$ if $i\ne j$. According to the definition of $\operatorname{Adj}(A)$, we can find it using the following formula:
$$[\operatorname{Adj}(A)]_{ij}=(−1)^i(−1)^j|A_{ji}|$$
$|A_{ji}|$ being the determinant of the minor matrix $A_{ji}$. Since only $a_{ii}$ can differ (either be an opening entry or $0$) from $0$, $\forall 1\le i\le n$, while all of the other entries are $0$, achieving $A_{ji}$ ($j\ne i$) requires removing column $i$ and row $j$ - and thus removing a column with the only potential opening entry at $a_{ii}$. This leaves $A_{ji}$ with a zeroes row where $a_{ii}$ was removed - and as known, a square matrix with a row/column of zeroes is not invertible $\iff |A_{ji}| = 0$. Which means:
$$[\operatorname{Adj}(A)]_{ij}=(−1)^i(−1)^j\cdot0=0\quad\forall i\ne j$$
just as required.

答:
Let $$A= \begin{pmatrix}A_1 & 0 & \cdots & 0 \\ 0 & A_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & A_n  \end{pmatrix}$$

Now, $\text{adj} A$ is the transpose of the cofactor matrix of $A$. If you carefully see, the cofactor of each element in the diagonal of $A$ is nothing but the product of the remaining entries taken with sign, and the cofactor of all zero entries (other than diagonal entries), will be $0$, because you could find one column or a row containing all $0$'s in it. Hence, it will be a diagonal matrix. Transpose of a diagonal matrix is again a diagonal matrix. So, done!


问:
show this $\lceil \frac{n}{1-a_{n}}\rceil =n+1$
let $a_{n}$ be squence such $a_{1}=2-\dfrac{\pi}{2}$, and
$$a_{n+1}=2a_{n}+\dfrac{a_{n}-1}{n},n\in N^{+}$$
show that
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$

I try:since
$$f(1)=\lceil \dfrac{1}{\dfrac{\pi}{2}-1}\rceil=2$$
and $$a_{2}=2a_{1}+a_{1}-1=3a_{1}-1=5-\dfrac{3}{2}\pi$$
so
$$f(2)=\lceil \dfrac{2}{\dfrac{3\pi}{2}-4}\rceil=3$$
and so on ,if we want prove

$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
or
$$n<\dfrac{n}{1-a_{n}}\le n+1$$
or
$$0<a_{n}\le\dfrac{1}{n+1}$$
if we use induction to prove it.then
$$a_{n+1}=(2+\dfrac{1}{n})a_{n}-\dfrac{1}{n}\le (2+\dfrac{1}{n})\cdot\dfrac{1}{n+1}-\dfrac{1}{n}=\dfrac{1}{n+1}$$
is wrong,becasue we want to prove $a_{n+1}\le\dfrac{1}{n+2}$

答:
You can convert the recurrence equation into a differential equation by multiplying it with $x^n$, summing over $n=1,2,3,...$ and defining $f(x)=\sum_{n=0}^\infty a_{n+1} x^n$, which gives the first order DE $$\sum_{n=1}^\infty a_{n+1}x^n = 2\sum_{n=1}^\infty a_n x^n + \sum_{n=1}^\infty \frac{a_n}{n} x^n - \sum_{n=1}^\infty \frac{x^n}{n} \\
\Rightarrow \quad f(x)-a_1=2x f(x) + \sum_{n=1}^\infty \frac{a_n}{n} x^n + \log(1-x) \\
\Rightarrow \quad (1-2x)f'(x) - 3f(x) + \frac{1}{1-x} = 0$$
that can be solved in full generality. You can use that formula or also just check that $$f(x)=-\frac{2\arctan\left(\sqrt{1-2x}\right)}{\left(1-2x\right)^{3/2}} + \frac{2}{1-2x} + \frac{\pi-c}{(1-2x)^{3/2}}$$
solves the ODE where $c$ is an arbitrary constant. The initial condition $$-\frac{\pi}{2}+2 + \pi -c=f(0)=a_1=2-\frac{\pi}{2}$$ implies $c=\pi$ and it is easy to see that this very special case removes the singularity at $x=1/2$. This singularity would lead to the violation of the inequality $a_n \leq \frac{1}{n+1}$ as the coefficients $a_{n+1}$ would then obey $\sim (-2)^n\binom{-3/2}{n}$ asymptotically. Now we express $f(x)$ as an integral and use the geometric series $$\frac{2}{1-2x} - \frac{2\arctan\left(\sqrt{1-2x}\right)}{\left(1-2x\right)^{3/2}} = \int_0^1 \frac{2t^2}{1+(1-2x)t^2} \, {\rm d}t = \sum_{k=0}^\infty x^k \, \underbrace{2^{k+1} \int_0^1 {\rm d}t \, \frac{t^{2k+2}}{(1+t^2)^{k+1}}}_{a_{k+1}}$$

where $0<x<1/2$ was assumed. Interchanging summation and integration is allowed by dom. convergence. Hence it needs to be shown that $$0\leq \frac{1}{n+1} - 2^n \int_0^1 {\rm d}x \, \frac{x^{2n}}{(1+x^2)^{n}} = \int_0^1 {\rm d}x \left( x^n - \frac{2^n x^{2n}}{(1+x^2)^{n}} \right) \\ = \int_0^1 {\rm d}x \, \frac{x^n}{(1+x^2)^n} \left\{ (1+x^2)^n - (1+1)^n x^n \right\} \, .$$ From this it suffices to show

$$0\leq (1+x^2)^n - (1+1)^n x^n \\
= \sum_{k=0}^n \binom{n}{k} \left(x^{2k} - x^n\right)
= \sum_{0 \leq k < n/2} \binom{n}{k} (x^{2k}-x^n) + \sum_{n/2 < k \leq n} \binom{n}{k} (x^{2k} - x^n) \\
= \sum_{0 \leq k < n/2} \binom{n}{k} (x^{2k}-x^n) + \sum_{0 \leq k < n/2} \binom{n}{n-k} (x^{2(n-k)} - x^n)
=\sum_{0 \leq k < n/2} \binom{n}{k} \left( x^k - x^{n-k} \right)^2 \, .$$


问:
$f(S)^c \subseteq f(S^c)$, $\forall S\subseteq A$, if and only if $f\colon A\to B$ is surjective
Let $f:A \rightarrow B$.

The complements indicated below are taken within $A$ or $B$.

I need to prove that
$f(S)^c \subseteq f(S^c)$, $\forall S \subseteq A$, if and only if $f$ is surjective.

So I need to prove $f(A) = B$ right?

How can I prove this?

答:
(I assume it is OK to post a full solution to what was long ago probably a homework question.)

Let me provide a complete (but perhaps overly long-winded) proof which does not use separate $\Rightarrow$ and $\Leftarrow$ parts, but only equivalences.

The most complex part seems to be $f[S]^c \subseteq f[S^c]$, so let's try to simplify that: for any $S \subseteq A$,
$$
\begin{align}
& f[S]^c \subseteq f[S^c] \\
\equiv & \;\;\;\;\;\text{"definition of $\subseteq$"} \\
& \langle \forall y :: y \in f[S]^c \Rightarrow y \in f[S^c] \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $^c$, which means $B$-complement here"} \\
& \langle \forall y :: y \in B \land y \not\in f[S] \Rightarrow y \in f[S^c] \rangle \\
\equiv & \;\;\;\;\;\text{"logic: rearrange to bring both occurrences of $f[\cdot]$ together"} \\
& \langle \forall y :: y \in B \Rightarrow y \in f[S] \lor y \in f[S^c] \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\cup$ -- since we know distribution properties of $\cdot[\cdot]$"} \\
& \langle \forall y :: y \in B \Rightarrow y \in f[S] \cup f[S^c] \rangle \\
(*) \; \equiv & \;\;\;\;\;\text{"$f[\cdot]$ distributes over $\cup$"} \\
& \langle \forall y :: y \in B \Rightarrow y \in f[S \cup S^c] \rangle \\
\equiv & \;\;\;\;\;\text{"set theory: basic property of $^c$, which here means $A$-complement"} \\
& \langle \forall y :: y \in B \Rightarrow y \in f[A] \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\subseteq$"} \\
& B \subseteq f[A] \\
\equiv & \;\;\;\;\;\text{"using $f[S] \subseteq B$ for any $S$, since the range of $f$ is $B$"} \\
& B = f[A] \\
\equiv & \;\;\;\;\;\text{"one of the definitions of surjectivity, using $f : A \to B$"} \\
& f \textrm{ is surjective} \\
\end{align}
$$

Now formally wrapping up, we have
$$
\begin{align}
& \langle \forall S :: f[S]^c \subseteq f[S^c] \rangle \\
\equiv & \;\;\;\;\;\text{"by the above calculation"} \\
& \langle \forall S :: f \textrm{ is surjective} \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify: $S$ does not occur inside $\forall S$"} \\
& f \textrm{ is surjective} \\
\end{align}
$$
which proves the statement in question.

The key step was $(*)$, and this is the most 'creative' part in an otherwise fairly mechanical proof, provided one is familiar with logic and the definitions and basic properties of set theory and functions.

ContentLicense: CC BY-SA 2.5


问:
How many fours can be formed from eight sports teams?
Suppose we have eight teams, let them be A, B, C, D, E, F, G and H. Two teams can form a pair, e.g. AB (BA is exactly the same pair). Then, pairs can make up into fours. For instance, AB, CD, EF and GH is an example of a four. Again, the order of pairs in the four does not matter.

The question is: how many different fours can be formed?

At school, we were solving problems like "In how many ways can you arrange 8 different objects (no identical ones)?" It's 8! = 40320. Here, however, you may arrange/combine them in any way you wish. In the described above situation with sports teams, you cannot use any letter more than once. E.g. BC and CD cannot coexist in the same four.

Basically, the answer is not that important for me; I just wonder which formula should I use or where I can read some information or maybe watch any Youtube videos about combinatorics problems where certain objects don't 'like' the other ones (combinations have restrictions).

答:
You are asking for the number of fours, not pairs.

So imagine that there are two numbered courts , you can send the players into the two courts in $\dfrac{8!}{4!4!} = 70$ ways

But since we aren't interested in which court number a player goes to, the required answer is $\dfrac{70}{2!} = 35$ ways

Answer modified according to OP's clarification

OP has clarified that by "fours" is meant four pairs including all $8$ teams.

Then, by the same logic of first imagining $4$ courts, division can be into $\dfrac{8!}{2!2!2!2!} \div4! = 105$ fours


答:
If you suppress a row, the corresponding column becomes all zeroes. If you suppress this column, you get a diagonal submatrix, and if you suppress another, the minor is perforce zero.

$$\begin{pmatrix}a&0&0&0\\0&b&0&0\\0&0&c&0\\0&0&0&d\end{pmatrix}\to
\begin{pmatrix}a&0&\color{red}0&0\\0&b&\color{red}0&0\\0&0&\color{red}0&d\end{pmatrix}\to
\begin{cases}\begin{pmatrix}a&0&0\\0&b&0\\0&0&d\end{pmatrix}\\\\\begin{pmatrix}a&0&\color{red}0\\0&b&\color{red}0\\0&0&\color{red}0\end{pmatrix}\end{cases}$$

Formally:

If we suppress the row $r$ and the column $c$,
$$M_{i\ne r,j\ne c}=(j\ne r) M_{i,j}.$$

Title: Prove that if $A$ is diagonal, than $\operatorname{Adj}(A)$ is diagonal as well

Body:
I'm not sure if my proof is formal enough, and I'd appreciate a review or if someone could write his/her proof of the statement. Here's how my proof goes:

Given a diagonal matrix $A \in M_n(F)$, we can say that $a_{ij}$, $1\le i\le n$ and $1\le j\le n$, being an entry of $A$, equals to $0$ if $i\ne j$. According to the definition of $\operatorname{Adj}(A)$, we can find it using the following formula:
$$[\operatorname{Adj}(A)]_{ij}=(−1)^i(−1)^j|A_{ji}|$$
$|A_{ji}|$ being the determinant of the minor matrix $A_{ji}$. Since only $a_{ii}$ can differ (either be an opening entry or $0$) from $0$, $\forall 1\le i\le n$, while all of the other entries are $0$, achieving $A_{ji}$ ($j\ne i$) requires removing column $i$ and row $j$ - and thus removing a column with the only potential opening entry at $a_{ii}$. This leaves $A_{ji}$ with a zeroes row where $a_{ii}$ was removed - and as known, a square matrix with a row/column of zeroes is not invertible $\iff |A_{ji}| = 0$. Which means:
$$[\operatorname{Adj}(A)]_{ij}=(−1)^i(−1)^j\cdot0=0\quad\forall i\ne j$$
just as required.


答:
There are $7$ possible choices for $A$'s partner. Now let $X$ be the next unpaired player (i.e. $X$ is $B$ unless $B$ was paired with $A$, in which case $X$ is $C$). There are $5$ choices for $X$'s partner. Now let $Y$ be the next unpaired player. There are $3$ choices for $Y$'s partner.

So the answer is $7\times 5\times 3=105$.

Title: How many fours can be formed from eight sports teams?

Body:
Suppose we have eight teams, let them be A, B, C, D, E, F, G and H. Two teams can form a pair, e.g. AB (BA is exactly the same pair). Then, pairs can make up into fours. For instance, AB, CD, EF and GH is an example of a four. Again, the order of pairs in the four does not matter.

The question is: how many different fours can be formed?

At school, we were solving problems like "In how many ways can you arrange 8 different objects (no identical ones)?" It's 8! = 40320. Here, however, you may arrange/combine them in any way you wish. In the described above situation with sports teams, you cannot use any letter more than once. E.g. BC and CD cannot coexist in the same four.

Basically, the answer is not that important for me; I just wonder which formula should I use or where I can read some information or maybe watch any Youtube videos about combinatorics problems where certain objects don't 'like' the other ones (combinations have restrictions).


问:
Bilinear forms on C[0,1]
Let $C[0,1]$ be the vector space of real-valued continuous functions on $[0,1]$. Then
$$B(f,g) = \int_0^1{f(x)g(x)\, dx}$$
is a bilinear form on $C[0,1]$. More generally, if $k:[0,1]^2\rightarrow \mathbb{R}$ is continuous, then
$$B_k(f,g) = \int_{[0,1]^2}{f(x)g(y)k(x,y)\,dx\, dy}$$
is a bilinear form.

In Keith Conrad's notes on bilinear forms, he asks if there is a $k$ corresponding to the first example. My intuition first suggested the characteristic function on the diagonal, but that is not continuous. Surely it must be something like this, but I can't think of what it should be.

There is another question where he asks us to find conditions on $k$ under which $B_k$ will be a symmetric bilinear form. I sketched a proof that its both necessary and sufficient for $k(x,y) = k(y,x)$, but my proof is not pretty. Is there a natural proof?

答:
For the first question: we would have for any $f$ and $g$ continuous that $\int_0^1f(x)\left(g(x)-\int_0^1k(x,y)g(y)dy\right)dx$. This implies that $g(x)=\int_0^1k(x,y)g(y)dy$ for all $g$ continuous. By Stone-Weierstrass theorem, $k$ is approximable by finite sums of terms of the form $a(x)b(y)$, hence the identity from $[0,1]$ to $[0,1]$ would be compact: it's not possible.

In $B_k(g,f)$, switch the role of $x$ and $y$: this gives for each $f$ and $g$ continuous,
$$\int_{[0,1]^2}f(x)g(y)(k(x,y)-k(y,x))dxdy=0.$$
Take $F(x,y)$ continuous on $[0,1]$ and approximate by linear combinations $\sum a_i(x)b_i(y)$.

Title: Bilinear forms on C[0,1]


问:
Why is it necessary to find extreme values of L-functions?
Apparently I have been exploring Riemann Zeta Function and have lately come across $L$-functions. After going through few papers, I realised that many mathematicians are giving their full time knowing the extreme values of $L$-functions. I need to know the following:

What is the motivation to study these areas?

What information do we get after finding central values of these functions?

Thank you in advance.

答:
The OP probably meant the local extrema of the $\Bbb{R\to R}$ function $$\Xi(t)=(1/2+it)(1/2-it)\pi^{-(1/2+it)/2}\Gamma((1/2+it)/2)\zeta(1/2+it)$$ obtained from the functional equation.

Under the RH $$\Xi(t)=\Xi(0)\lim_{K\to \infty}\prod_{k=1}^K (1-\frac{t}{\Im(\rho_k)})^{e_k}(1+\frac{t}{\Im(\rho_k)})^{e_k}$$ where $\rho_k$ are the non-trivial zeros of $\zeta$ and $e_k$ their multiplicity.

$f_K(t)=\prod_{k=1}^K (1-\frac{t}{\Im(\rho_k)})^{e_k}(1+\frac{t}{\Im(\rho_k)})^{e_k}$ is a real polynomial whose all roots are real, its derivative $f_K'$ has at least one zero between each consecutive zero of $f_K$ (at the maximum of $|f_K|$ between those consecutive zeros) so we obtain $2K-1+\sum_{k=1}^K 2(e_k-1)$ zeros for $f_K'$,  thus we have found them all.

Since the zeros of $f_K'$ converge locally uniformly to those of $\Xi'$ we know all the zeros of $\Xi'$,

and hence if the RH is true then there is exactly one local extremum of $\Xi$ between two consecutive zeros.

It works the same way for other L-functions.


问:
A proof about Boundary Asymptotics of the Cauchy Transform using the Dominated Convergence Theorem
In this paper, on page 8, the authors make the remark that, for a complex, finite borel measure $\mu$ on the circle/unit interval, and for the Cauchy Transform: $$K\left\{ \mu\right\} \left(z\right)\overset{\textrm{def}}{=}\int_{0}^{1}\frac{d\mu\left(t\right)}{1-e^{-2\pi it}z},\textrm{ }\forall\left|z\right|<1$$ one can use the Dominated Convergence Theorem to show that: $$\lim_{r\uparrow1}\left(1-r\right)K\left\{ \mu\right\} \left(re^{2\pi it}\right)=\mu\left(\left\{ t\right\} \right)$$ I would like to see this argument written out in full detail (no handwaving, no skipped steps, all objects explicitly constructed whenever possible, no skipped computations), seeing as it is directly related to what I'm researching, and I want to be able to have a baseline proof that I can refer to without needing to worry about whether or not I might have made a mistake in writing up said baseline proof.

答:
Write $\mu=h|\mu|$ for the polar decomposition of $\mu$ with a measurable function $h\colon [0,1]\to \mathbb{T}$. You want to evaluate the limit
$$\lim_{r\nearrow 1}\int_0^1\frac{1-r}{1-re^{2\pi i(t-s)}}h(s)\,d|\mu|(s).$$
By the reverse triangle inequality, $|1-re^{2\pi i(t-s)}|\geq 1-r$. Thus the absolute value of the integrand is bounded by $1$, which is integrable since $\mu$ is finite.

For $s=t$, the integrand is $1$. For $s\neq t$, the numerator of the integrand tends to $0$ as $r\nearrow 1$, while the denominator converges to $1-e^{2\pi i(t-s)}\neq 0$. Altogether
$$
\lim_{r\nearrow 1}\frac{1-r}{1-re^{2\pi i(t-s)}}h(s)=1_{\{t\}}(s)h(s).
$$
Hence
$$
\lim_{r\nearrow 1}\int_0^1\frac{1-r}{1-re^{2\pi i(t-s)}}\,h(s)d|\mu|(s)=\int_0^1 1_{\{t\}}(s)\,h(s)d|\mu|(s)=\mu(\{t\})
$$
by the dominated convergence theorem.


问:
Absolute value of sub-gaussian
Let $\eta_1 , \eta_2, \cdots, \eta_t$ be 1-subgaussian independent random variables with mean 0(but not necessarily identical).

Now we know several nice equations about $|\sum_{s=1}^t \eta_s|$.

How about $\sum_{s=1}^t |\eta_s|$?

Is there any hints about $\mathbb{E}|\eta_s|$?

I believe $\sum_{s=1}^t |\eta_s|\leq O(t \log t)$ with high probability. Is it true?

(Surprisingly, there is no 'sub-gaussian' tag.......)

答:
Any 1-subgaussian random variable $X$ satisfies that $\|X\|_p \leq C \sqrt{p}$ for an absolute constant $C$ [1, Proposition 2.5.2].
Therefore, $E |X| = \|X\|_1 \leq C $.

Yes, it is true. Actually something stronger is true. Following the definitions in [1], if $X$ is $\sigma$-subgaussian ,i.e., $P(|X| \geq t) \leq 2 \exp(-ct^2/)$, then $|X|$ is also $\sigma$-subgaussian.
Let $Z = |X|$. Using the centering lemma [1, Lemma 2.6.8], $\|Z - \mathbb E Z\|_{\psi_2} \leq K\|Z\|_{\psi_2} \leq K$ for an absolute constant $K$.

Applying generalized Hoeffdings's inequality[1, Theorem 2.6.2], we get that with probability $1 - \delta$, $|\sum_{i=1}^n (Z_i - \mathbb E Z)| = O(\sqrt{n \log(1/\delta)})$.

Using part 1 and triangle inequality, we get that w.p. $1 - \delta$, $\sum_{i=1}^n |X_i| = O(n) + O(\sqrt{n \log(1/\delta)})$

P.S.: If you don't want logarithmic dependence on $\delta$, then what you require is satisfied by random variables with much relaxed assumptions, say, unit variance.

[1]: High-Dimensional Probability, Roman Vershynin. https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.html

Title: Absolute value of sub-gaussian


问:
Estimation of conditional probability
The joint frequency distribution of two discrete random variables,  and , is given in the following table

a) Can anybody help me how to calculate the conditional expectation of  for a given value of : (|=). For example, similar to the formula in we have: (|=1)=Σ⋅(=|=1). Find (|=1) and

b) Calculate (|=3)

答:
It is very easy. Fixed a value of $X=x$ in the above table you will see which are the values of $Y$ (in this case for any value of $X$  you have $Y=1,2,3,4$ but it could be not necessary so, in general) and all you have to do is to calulate the conditional probability with the usual formula

Example, fix $X=3$ and you get

$$(Y|X=3)=\begin{cases}
2/31,  & \text{if $Y=1$} \\
5/31,  & \text{if $Y=2$} \\
20/31,  & \text{if $Y=3$} \\
4/31,  & \text{if $Y=4$}
\end{cases}$$

The expectation follows as a consequence being

$$E(Y|X=x)=\Sigma_y yp(y|x)$$

Title: Estimation of conditional probability


问:
A frog from k lilypad either jumps to k+1 or k+2 lilypad. Find the probability that it reaches pad 7 if it is initially at pad 1.
Lilypads numbered $1,2,3,\cdots$ lie in a row in a pond. A frog is initially on pad $1$. From any pad $k$, the frog will jump to either pad number $k+1$ or $k+2$ with equal probability of ${1\over 2}$. The probability that the frog visits pad $7$ is ${m\over n}$ where ($m,n$) = $1$, Evaluate $n-m$.

Well I wanted to know as to how to attack this problem. I mean Probability = Number of favorable outcomes / Total Outcomes. What is the total outcomes here ?? What are the favorable outcomes ?? How to count them ??

答:
We can enumerate the number of pads the frog leaps over to get to pad 7:

Case 1$$1+1+1+1+1+1$$
1 permutation, $p=(\frac{1}{2})^6$

Case 2
$$1+1+1+1+2$$
5 permutations, $p=5\times(\frac{1}{2})^5$

Case 3
$$1+1+2+2$$

6 permutations, $p=6\times(\frac{1}{2})^4$

Case 4
$$2+2+2$$

1 permutations, $p=(\frac{1}{2})^3$

Total $=\frac{43}{64}$


问:
Numerically solve differential equation that has no solution
I am not really experienced in the field of numerical differential equations. I am only studying existence and uniqueness results about PDEs at the moment. Now I was wondering what would happen, if one tries to solve a PDE numerically, although the PDE 1) has no solution or 2) the solution is not unique?

The Navier-Stokes-Equation in 3 dimensions comes to mind. How do physicists approach this equation? Do they work with numerical solutions although there is no proof, that there is a solution at all?

I really appreciate your help!

Best,

Luke

答:
I am engineer and I made quite a lot of numerical simulations on PDEs. My experience does not result form mathematical insight, but by practical experince.

Engineers are interested in dayly work problems. We can assume that some kind of solutions exists: No such thin like a Dirac impuls somewhere. You need experience, some kind of feeling, which problems tend to fail and which are not problematic. The usual necessary mesh play an important role.

If I am unsure whether the solution is correct I tend to change something slightly: The geometry, the mesh, the boundary conditions. It may be happen that I got a very different solution. That means I am on a wrong way.

There are other situations where I know that I solutions seems to exist, lets say turbulent hydrodynamics. But I know that cannot come close to that solution with my algorithms. Then it is a useful idea to calibrate the solution against a real measuring. Then you may change something (say optimzing a geometry) an need only to take into account the changes against the calibrated situation.


问:
Can someone give me a natural language example of the following in-equivalence?
$$\forall x \exists y (F(x) \leftrightarrow K(x,y)) \not\equiv \forall x (F(x) \leftrightarrow \exists y K(x,y)) $$

The statement on the right is not same as the statement on the left and I understand it by finding counter models to the equivalence. But can some one give me a natural language example where this in-equivalence is more intuitive ?

For counter model:

https://www.umsu.de/trees/#%E2%88%80x%E2%88%83y(Fx%20%E2%86%94%20Kxy)%20%20%E2%86%92%20%E2%88%80x%20(Fx%20%E2%86%94%20%E2%88%83yKxy)

答: I'll give an example where the left-hand part is true, but the right-hand part isn't. Suppose a club needs face ID. For each person $x$, there is a face $y$ such that that person may enter the club ($F(x)$) iff their ID has that face on it ($K(x,\,y)$); this is, of course, their own face. On the other hand, it's not true that each person can enter iff some face appears on their ID; it has to be their face.


答:
I might be wrong. It's happened before (once).

$P($ reaches pad $7)$ = $1 - P(6 \to 8 \ | $ reaches pad $6)$

$\color{red}{= 1 - \frac12P(\text{ reaches pad 6 })}$ [because assuming it reaches pad $6,$ it has $\frac12$ chance of landing on $8$.]

$= 1 - \frac12(\ 1 - P(5\to7\ |$ reaches pad $5)\ )$

$\color{red}{ = 1 - \frac12(\ 1 - \frac12 P( \text{reaches pad } 5)\ )}$

$= 1 - \frac12(\ 1 - \frac12(\ 1 - P(4 \to 6|$ reaches pad $4\ )\ )\ )$

$\color{red}{= 1 - \frac12(\ 1 - \frac12(\ 1 - \frac12P(\text{reaches pad 4})\ )\ )}$

$ = ...$

$ \color{red}{= 1 - \frac12\left(\ 1 - \frac12\left(\ 1 - \frac12\left(\ 1 - \frac12\left(\ 1 - \frac12\left(\ 1 - \frac12P\left(\text{reaches pad 1}\ \right)\ \right)\ \right)\ \right)\ \right)\ \right)}$

$ = 1 - \frac12\left(\ 1 - \frac12\left(\ 1 - \frac12\left(\ 1 - \frac12\left(\ 1 - \frac12\left(\ 1 - \frac12 \times 1\ \right)\ \right)\ \right)\ \right)\ \right) $

$ = \frac{43}{64}$.

Since $(n,m) = (64,43) = 1,\ n - m = 64 - 43 = 21.$

Title: A frog from k lilypad either jumps to k+1 or k+2 lilypad. Find the probability that it reaches pad 7 if it is initially at pad 1.

Body:
Lilypads numbered $1,2,3,\cdots$ lie in a row in a pond. A frog is initially on pad $1$. From any pad $k$, the frog will jump to either pad number $k+1$ or $k+2$ with equal probability of ${1\over 2}$. The probability that the frog visits pad $7$ is ${m\over n}$ where ($m,n$) = $1$, Evaluate $n-m$.

Well I wanted to know as to how to attack this problem. I mean Probability = Number of favorable outcomes / Total Outcomes. What is the total outcomes here ?? What are the favorable outcomes ?? How to count them ??


问:
Suppose $|A|=n, f:A\to A$ is injective $\implies \exists k\in [n]:f^k(x)=x$

My Attempt

Case 1

Suppose $f:A \to A$ is an identity function. then $k=1\in [n]:f(x)=x$.

Case 2

Suppose $f:A \to A$ is an injective function and non identity function.

so, there exists $x\in A$ such that $f(x) \neq x.$
So, there are $n-1$ possibilities for $f(x).$
if $f(f(x))=x, $ then $k=2.$
so on. I am not able to complete the proof. Is my method correct?

答:
For fixed $a$ we have a chain $$a,f(a), f^2(a),...f^{n}(a)$$ $n+1$ elements from the set with $n$ elements, thus at least two of them are the same, so there are $0\leq i<j\leq n$ such that $$f^i(a) = f^j(a) $$
so $$f^i(a) = f^i(f^{j-i}(a))$$

thus, since $f^i$ is also injective, we have $$a= f^{j-i}(a)$$ and thus $k:=j-i$ does the work.


答:
Define a graph $G$ with vertices from the set $A$ and two vertices being connected iff one is image of the other.

Clearly each vertex has degree at most 2.

If all vertices have degree $2$ then this graph is not forest, so exist cycle and thus a conclusion.

If some vertex $a$ has degree $1$ then $f(a)=a$ and we are done again.

Title: Suppose $|A|=n, f:A\to A$ is injective $\implies \exists k\in [n]:f^k(x)=x$

Body:
My Attempt

Case 1

Suppose $f:A \to A$ is an identity function. then $k=1\in [n]:f(x)=x$.

Case 2

Suppose $f:A \to A$ is an injective function and non identity function.

so, there exists $x\in A$ such that $f(x) \neq x.$
So, there are $n-1$ possibilities for $f(x).$
if $f(f(x))=x, $ then $k=2.$
so on. I am not able to complete the proof. Is my method correct?


问:
$f(x)$ is continuous in $[0,\infty)$ and $f(0)=1 , \lim_{x\to \infty}f(x)=-3$

$f(x)$ is continuous in $[0,\infty)$ and $f(0)=1 , \lim_{x\to \infty}f(x)=-3$ , Prove exist $c \in (0,\infty)$ such that $f(c)=\frac {c-1}{c+1}$

Hello everyone.
I'm posting this because I'm not sure if I proved it right and I have a feeling I assumed something wrong.. anyways my try :

We can see that $-1<\frac {c-1}{c+1}<1$ because $c \in [0,\infty)$.

since $"f(\infty)"=-3<-1<\frac {c-1}{c+1}<1=f(0)$ , we know that $\frac {c-1}{c+1}$ is between $f(0)$ and $f(\infty)$.

by Intermediate value theorem we get that exist $c\in [0,\infty)$ such that $f(c)=\frac {c-1}{c+1}$ as required.

As you can see I'm pretty sure I can't write $f(\infty)$ and there is the mistake but I don't know how to write it in different way.

答:
Put $g(x) = f(x) - \frac{x-1}{x+1}$. Then $g$ is continuous on $(-1, \infty)$, $g(0)=2$ and $\lim_{x \to \infty} g(x) = -4$. In particular, there exists $B >0$ such that $g(B)<0$. Applying the intermediate value theorem to $g$, we get $c \
\in (0, B)$ such that $g(c)=0$.


问:
Is it true that $\Big[(ak + \frac{b}{2})\cos(\frac{t}{2})\Big] + \Big[(bk - \frac{a}{2})\sin(\frac{t}{2})\Big] = k\cos(\frac{t}{2})$?
Is it true that
$$
\left[\left(ak + \frac{b}{2}\right)\cos\left(\frac{t}{2}\right)\right] + \left[\left(bk - \frac{a}{2}\right)\sin\left(\frac{t}{2}\right)\right] = k\cos\left(\frac{t}{2}\right)
$$
If yes, how to prove it?
Is there a hidden trigonometric identity behind it?

Question taken MIT 18.03 (end of problem 1 page 2)

答:
Generally, it's trivially false for $k=0$ and $a,b\neq 0$, and there's no reason why it would be true for other values. For instance for $t=0$, it gives $ak+b/2 = k$.

But in your exercise, you know that $(ak + \frac{b}{2})\cos(\frac{t}{2}) + (bk - \frac{a}{2})\sin \frac{t}{2} = k\cos \frac{t}{2}$, and the point is to find a relation between $a,b,k$ such that this is true.

Since $(t\mapsto \cos \frac{t}{2}, t\mapsto \sin \frac{t}{2})$ is linearly indepedent, you get that :

$ak+b/2 = k$ and $bk-a/2 = 0$.


答:
1) No. If such a $k$ existed, denote $M:=\sup |k|$. Then
$$
\big|\int_0^1 f^2\Big|=\Big|\iint f(x)f(y)k(x,y)dxdy\big|\leq \iint|f(x)||f(y)|Mdxdy=M\left(\int_0^1 |f|\right)^2
$$
for every continuous function $f$ on $[0,1]$. Then for each $n\geq 1$, consider the continuous function $f_n$ defined by $f_n(x):=\frac{1}{\sqrt{x}}$ on $[1/n,1]$ and $f_n(x):=\sqrt{n}$ on $[0,1/n]$. We get
$$
\log n=\int_{1/n}^1\frac{1}{x}dx\leq \int_0^1f_n^2\leq M\left(\int_0^1 |f_n|\right)^2\leq M\left(\int_0^1 \frac{1}{\sqrt{x}}dx\right)^2=4M.
$$
Let $n$ tend to $+\infty$ to obtain the desired contradiction.

2) If $B_k(f,g)=B_k(g,f)$, set $u(x,y):=k(x,y)-k(y,x)$. You get, as observed by Davide Giraudo,
$$
\iint_{[0,1]^2} f(x)g(y)u(x,y)dxdy=0
$$
for every $f,g$ continuous on $[0,1]$. From here, there are several ways to conclude that $u=0$. Here are two possible ones.

a) Applying the latter to $f(x)=x^n$ and $g(y)=y^m$, we get, by linearity, $\iint p(x,y)u(x,y)dxdy=0$ for every polynomial in two variables $p(x,y)$. By Stone-Weierstrass, $u$ can be uniformly approximated by such polynomials. So $\iint u^2=0$ whence $u^2=0$ since $u^2$ is continuous and nonnegative.

b) Approximating uniformly characteristic functions of intervals by continuous functions, we get $\iint 1_I(x)1_J(y)u(x,y)dxdy=\iint 1_{I\times J}\cdot u=0$ for every intervals $I,J$ in $[0,1]$. If $u$ was not $0$, by continuity, there would exist non trivial intervals $I,J$ such that $u\geq \alpha>0$ on $I\times J$ (or $u\leq -\alpha<0$). Then $0=\iint 1_{I\times J}\cdot u\geq \alpha \lambda(I)\lambda(J)>0$ (or $<0$). Contradiction.

Title: Bilinear forms on C[0,1]

Body:
Let $C[0,1]$ be the vector space of real-valued continuous functions on $[0,1]$. Then
$$B(f,g) = \int_0^1{f(x)g(x)\, dx}$$
is a bilinear form on $C[0,1]$. More generally, if $k:[0,1]^2\rightarrow \mathbb{R}$ is continuous, then
$$B_k(f,g) = \int_{[0,1]^2}{f(x)g(y)k(x,y)\,dx\, dy}$$
is a bilinear form.

In Keith Conrad's notes on bilinear forms, he asks if there is a $k$ corresponding to the first example. My intuition first suggested the characteristic function on the diagonal, but that is not continuous. Surely it must be something like this, but I can't think of what it should be.

There is another question where he asks us to find conditions on $k$ under which $B_k$ will be a symmetric bilinear form. I sketched a proof that its both necessary and sufficient for $k(x,y) = k(y,x)$, but my proof is not pretty. Is there a natural proof?


答:
It might help to model the process as a Markov Chain (though this isn't necessary).

Let $p_n$ be the probability that the frog hits lilypad $n$. Then $p_n=\frac{1}{2}\left(p_{n-1}+p_{n-2}\right)$, and $p_1=1$, $p_2=\frac{1}{2}$. At this point, you can either pump the recursion for $n=7$, or solve the relation to get $p_n=\frac{2}{3}+\frac{1}{3}\left(-\frac{1}{2}\right)^{n-1}$. Either way, you get $p_7=\frac{43}{64}$.


问:
Coefficients of stationary process
Problem

Given an equation $x_t -2.5x_{t-1} + x_{t-2} = u_t$ where $u_t$ is white noise.

Does this equation have a solution of form $ x_t = \dots + \alpha_{-1}y_{t+1} + \alpha_0 u_t + \alpha_{1}y_{t-1} + \dots$ ?

If does, find $\alpha_{-1}, \alpha_{0},\alpha_{1}$

My Ideas

If we solve the characteristic polynomial correspond $ b^t -2.5b^{t-1} + b^{t-2} = 0$ we get that $b = 1/2, 2, 0$. According to some (unfortunately, I don't remember the its name) when we have both $|b|$ both $<1$ and $>1$< out equation does have solution of form $\dots + \alpha_{-1}y_{t+1} + \alpha_0 u_t + \alpha_{1}y_{t-1} + \dots$, however I do not understand how to evaluate $\alpha_{-1}, \alpha_{0},\alpha_{1}$.

答:
The idea with characteristic polynomial is correct, however we need some other approaches to evaluate the coefficients.

Let us use the following operators:
$$ L:\quad L\cdot y_t = y_{t-1}, \quad L^2 y_t = y_{t-2},\quad ...$$
$$F: \quad L^{-1} = F$$
Then
$$u_t =  x_t -2.5x_{t-1} + x_{t-2} =  x_t -2.5L\cdot x_t + L^2 x_t =$$
$$= x_t(L-2)(L-\frac{1}{2}) = (x_t - 2Lx_t)(1-\frac{L}{2}) $$
Note, $x_t = Fx_{t-1}$. From that follows
$$ -2x_{t-1} = u_t \cdot \frac{1}{1-\frac{F}{2}} \cdot \frac{1}{1-\frac{L}{2}} = $$
$$ = u_t \sum_{i=0}^\infty \bigg(\frac{F}{2}\bigg)^i  \sum_{i=0}^\infty \bigg(\frac{L}{2}\bigg)^i$$
Note, $F^m\cdot L^m = 1$ and $F^m\cdot L^{m+k} = L^k$. Then we need to regroup the summands in the resulting sum (after the product) such that $L^0, L^1, L^2$ are left.

$$ = \dots +  u_tL^2\sum_{i=1}^\infty \bigg(\frac{1}{4}\bigg)^i + u_tL^1\sum_{i=1}^\infty \bigg(\frac{1}{4}\bigg)^i + u_0/2L + u_tL^0\sum_{i=0}^\infty \bigg(\frac{1}{4}\bigg)^i + \dots = $$
$$ = \dots + \frac{1}{3}u_{t-2} + \frac{2}{3}u_{t-1} + \frac{4}{3}u_{t}+\dots$$
Hence,
$$\alpha_0 = - \frac{1}{3},$$
$$\alpha_{-1} = - \frac{2}{3},$$
$$\alpha_1 = - \frac{1}{6}$$

Title: Coefficients of stationary process


答: That equation is true iff $$\left( k(a-1) +\frac b2 \right) \cos (t/2) = \left(\frac a2 -bk\right) \sin(t/2) \\ \iff \tan(t/2) = \frac{k(a-1) +\frac b2}{\frac a2 -bk} $$ The RHS is just a constant. If this were to hold true for all $t$, then that would imply that $\tan x$ is a constant function.

Title: Is it true that $\Big[(ak + \frac{b}{2})\cos(\frac{t}{2})\Big] + \Big[(bk - \frac{a}{2})\sin(\frac{t}{2})\Big] = k\cos(\frac{t}{2})$?

Body:
Is it true that
$$
\left[\left(ak + \frac{b}{2}\right)\cos\left(\frac{t}{2}\right)\right] + \left[\left(bk - \frac{a}{2}\right)\sin\left(\frac{t}{2}\right)\right] = k\cos\left(\frac{t}{2}\right)
$$
If yes, how to prove it?
Is there a hidden trigonometric identity behind it?

Question taken MIT 18.03 (end of problem 1 page 2)


问:
Long run percentage of customers who wait for a bus less than x units of time if customers arrive according to a homogenous Poisson process?
Assume that customers arrive to a bus stop according to a homogenous Poisson process with rate $\alpha$ and that the arrival process of buses is an independent renewal process with interarrival distribution $F$. What is the long run percentage of customers who wait for a bus less than $x$ units of time?

I tried to solve this problem by considering a renewal reward process. In the case that $x$ is very large so that it imposes no restriction we may proceed as follows to solve a similar problem: we say that a new cycle of length $T$ begins each time a bus arrives. Further, we assume customers pay us money at a rate of $1$ per time unit while they wait for a bus. Then our "reward rate" at any time corresponds to the number of people who are waiting at that time. So, by the renewal reward theorem we have that: average number waiting = $E[R]/E[T]$, where $R$ is the total reward during a cycle. In that case, $E[R] = 1/2E[NT] = 1/2\alpha E[T^2]$, where $N$ is the number of arrival in a cycle, which follows from the definition of a Poisson process and since the points of the Poisson process are uniformly distributed in a cycle conditional on the total number of points in that cycle.

But I don't know how to define the reward with the restriction imposed by $x$. I tried the following conditional expectation: $E[R|T,N]=N x1_{T>x}/2+NT1_{T\leq x}/2$, but got stuck. Can someone please help me?

答:
First recall the renewal reward theorem:

$E[R(t)]/t \rightarrow E[R]/E[T]$,

where $E[R]$ is the expected reward during a cycle and $E[T]$ is the expected length of a cycle. We denote $W_i$ the waiting time for customer $i$, $Z_i = 1$ if $W_i < x$, $Z_i = 0$ if $W_i \geq x$ and $N$ denotes the number of customers during a cycle. Then the proportion $\pi$ of customers who waited less than $x$ time units during a cycle with $N$ customers becomes:

$\pi = E[Z_1+...Z_N]/E[N]$.

We have that $E[Z_1+...Z_N|T=t, N=n]=\frac{nx1_{t>x}}{t}+n1_{t \leq x}=n\min(x, t)/t$, considering Poisson points are uniformly distributed conditional on the number of points in the cycle, so that the first equality follows because $x/t$ is the probability that a customer wait less than $x$ time units. Now, because $E[N]=\alpha E[T]$, we have that:

$\pi = E[Z_1+...Z_N]/E[N] = \frac{\alpha E[\min(x, T)]}{\alpha E[T]} = \frac{\int_0^xP(T>t)dt}{E[T]}$.

The rationale for the last equality is explained in this post.


问:
Topological manifolds (dimension)
I am taking an introductory course to topology and the professor defined a topological manifold of dimension $n$ if it is hausdorff and if for every point $x$ there exists an open set $U$ around $x$ such that $U$ is homeomorphic to $\mathbb{R}^n$. My question is: By the way she defined it, one could have (apriori) a topological space $X$ being a manifold of dimensions $n$ and $m$? Meaning I could find open sets $U$ and $V$ around every $x$ and $U$ is homeomorphic to $\mathbb{R}^n$ and $V$ is homeomorphic to $\mathbb{R}^m$, and this would make the notion of dimension not well defined. My guess is that since it is called "topological manifold of dimension $n$" what I described probably cant happen, but I dont see why not. </p>

答:
You are correct.

There are several ways to show that it is impossible, and it basically boils down to the fact that no open subset of ${\bf R}^n$ is homeomorphic to an open subset of ${\bf R}^m$ if $n<m$ (as $U\cap V$ would be in your case). You can assume without loss of generality that the sets in question are connected.

One way to show it is to use the fact that you can find an $(n-1)$-dimensional sphere which disconnects any open subset of ${\bf R}^n$, but it is a general fact that if a compact set disconnects ${\bf R}^m$, then it must admit a non-nullhomotopic map onto $S^{m-1}$ (in fact, it is an equivalent condition), while on the other hand, no proper compact subset of $S^{m-1}$ admits such a map.

So, summing it all up, a set homeomorphic to $S^{n-1}$ can't disconnect an open subset of ${\bf R}^m$ if $m>n$ and we're done.

(The results used can be shown by combinatorial or algebraic topology, but are not simple enough to show here.)

Title: Topological manifolds (dimension)


问:
Permutations in lexicographic order
A permutation is a bijection from a set to itself.

$$\pi: \{1,\ldots , n\} \mapsto \{1,\ldots , n\}$$

One way to get permutations in lexicographic order is based on the algorithm successor which finds each time the next permutation.

This procedure works as follows:

We start from the end of the list and we find the first number that is smaller from its next one, say $x$.

Then we start again from the end of the list and we find the first number that is bigger than $x$, say $y$.

We replace $x$ and $y$.

We reverse the sublist that starts from the next position of (the new position) of $y$ till the end.

The algorithm is :

I haven't really understood the idea of that algorithm. Could you explain that further to me? Do we change the order of the elements at the end of the list to get the next permutation in lexicographic order?

$$$$

Other methods to get get permutations in lexicographic order is with rank and unrank.

These algorithms are:

Is the idea of these ones that we write all the permutations in lexicograhic order and enumerate them to get the rank and the unrank is, given a number (rank), that we can give the respective permutation?

Is that correct? Or is the idea of rank or unrank related to permutations elsewise?

$$$$

EDIT :

I am trying to understand the formula that is used in permLexRank(n,L).

The formula is $$\text{rank}(\pi,n)=(\pi(1)-1)(n-1)!+\text{rank}(\pi',n-1) \ \text{ with } \ \text{rank}([1],1)=0$$ where $\pi'$ is the permutation that we get if at $\pi$ we delete the first element and subtract all elements that are greater than that by $1$.

The idea for this formula is :  At the lexicographic order of permutations of $S=\{1,\ldots  n\}$ first there are $(n-1)!$ permutations that start by $1$ then there are $(n-1)!$ permutations that start by $2$, etc.

So it holds that $$(\pi(1)-1)(n-1)!\leq \text{rank}(\pi)\leq \pi(1)(n-1)!-1$$
If $r'$ is the rank of $\pi$ in the set of $(n-1)!$ permutations that start with $\pi(1)$, then $r'$ is the rank of $[\pi(2), \ldots , \pi(n)]$ if we conside it as a permutation of $\{1, \ldots , n\}\setminus \{\pi(1)\}$.

If we subtract each element of $[\pi(2), \ldots , \pi(n)]$ by $1$ we get a permutation $\pi'$ of $\{1, \ldots , n-1\}$ that has again rank $r'$.

I haven't really understood that. Could you explain that to me?

答:
When it comes to understanding the pseudocode, one of method is to run a non-trivial example to see how it works. Obviously, it is not a proof, but it is rather to see the mechanics. Here I believe, indexing starts with 0. Run the following steps on your favorite example and see what is happening (for instance [a, c, b] and n=3 or [z, b, a, k, d] and n=5 etc.)

When it comes to theoretical understanding step by step.

Firstly, you inspect index i. The idea is to find a place in your sequence such that all elements from i until n-2 are sorted in decreasing order. You can see that because you decrease i only when the element on right is smaller. You remember i as your pivot index. If i is at the beginning of the array, it means that the whole array is (almost) sorted.

Then you work with j index. It starts with n-1 element (the last one) but it goes all the way down and ensures that all elements from j until the end of array are smaller than pivot. Notice that j is always larger than i.

Then we do swap of L[j] and L[i]. Notice that usually j = i+1 (when not?)

Then it is convenient to realize that (n - (i+1))//2 + 1 is just halfway through between n and i+1. And these swaps in for loop is just reflecting the whole array around this axis (so changing the order, take [10, 3, 2, 1, 0] and n=5, i=1 (so k is running from 1 to 2 and swapping L[2]=3 and L[4]=0 and L[3]=2 and L[2]=1 to obtain [10, 0, 1, 2, 3].

Title: Permutations in lexicographic order


问:
What is covariance of two different wiener processes?
I wonder what is covariance of two different wiener processes, say $W_t$ and $V_t$:

$$Cov(W_t,V_t) = \text{ }?$$

If it was one process we could use the property that $\Delta = W_t - W_s \text{~} N(0, t-s)$ to get that

$$Cov(W_t,W_s) = Cov( W_s + (W_t - W_s) ,W_s) = s, \text{    } s \leq t$$

However, I don't know, whether it's possible to do it in the same way. Intuitively, the  $Cov(W_t,V_t)$ is zero.

答:
Indeed, it depends on the task. Usually, it is stated in the problem set whether the the processes are independent.

Roughly, if $X$ and $Y$ are independent, then $ \Bbb Corr(X,Y) = 0$ or
$$\Bbb Corr(X,Y) = \frac{ \Bbb Cov(X,Y)}{\sigma_X\sigma_Y} \implies \Bbb Cov(X,Y) = 0$$
If nothing about independence is said, we cannot say the $\Bbb Cov$ is zero.


问:
Finding an isomorphism between spans
The qustion is how to find an isomorphism between two spans: span{(1,0,0,0),(0,1,0,0)} and span{(3,2,1,0),(1,1,1,1)}. I can see they are both lineary independet so they form bases, but how do I find the isomorphism?</p>

答:
A linear mapping between the spans (subspaces) can be given by mapping a basis of span to the basis (generating system in general) of the other span, here

$\phi(1,0,0,0) = (3,2,0,0)$ and $\phi(0,1,0,0)=(1,1,1,1).$

Since both spans have the same dimension you already have an isomorphism.

Title: Finding an isomorphism between spans


问:
A way to find this shaded area without calculus?
This is a popular problem spreading around. Solve for the shaded reddish/orange area. (more precisely: the area in hex color #FF5600)

$ABCD$ is a square with a side of $10$, $APD$ and $CPD$ are semicircles, and $ADQB$ is a quarter circle. The problem is to find the shaded area $DPQ$.

I was able to solve it with coordinate geometry and calculus, and I verified the exact answer against a numerical calculation on Desmos.

Ultimately the result is 4 terms and not very complicated. So I was wondering: Is there was a way to solve this using trigonometry? Perhaps there is a way to decompose the shapes I am not seeing.

A couple of years ago there was a similar "Find the shaded area" problem for Chinese students. I was able to solve that without calculus, even though it was quite an involved calculation.

Disclosure: I run the YouTube channel MindYourDecisions. I plan to post a video on this topic. I'm okay posting only the calculus solution, but it would be nice to post one using only trigonometry as many have not taken calculus. I will give proper credit to anyone that helps, thanks!

Update: Thanks for everyone's help! I prepared a video for this and presented 3 methods of solving it (the short way like Achille Hui's answer, a slightly longer way like David K and Seyed's answer, and a third way using calculus). I thanked those people in the video on screen, see around 1:30 in this link: https://youtu.be/cPNdvdYn05c.

答: Tried BFI numerical sums, didn't get back to error again for the needed extra independent condition ..


问:
Differential Geometry Video Lectures
I know there's a similar question here, however since what I found there wasn't what I was looking for I thought on creating a new question. I'm studying Differential Geometry through Spivak's book "A Comprehensive Introduction to Differential Geometry Vol. 1" and I'm looking for video lectures on differential geometry at this level (introduction to the analysis on manifolds and so on).

Until now I didn't find any video lectures. The only ones I've founded related to that are the one in the question I've mentioned, however that's too advanced for now. There are any good video lectures on this introductory (but yet rigorous) Differential Geometry? And I've pointed the rigorous part because I've found some videos on youtube, however the ones I've found simply ignore rigorous aspects of the subject and do things too loosely (even without defining manifolds properly).

Thanks very much in advance and sorry if this question is somehow silly!

答: Differential Geometry for Computer Science is the best I can find.</p>

Title: Differential Geometry Video Lectures


问:
Disproving $ L^2(X, U(\mathbb{C},n)) = U(L^2(X,\mathbb{C}^n))$.
Here $X$ is some measurable space  and  $U(L^2(X,\mathbb{C}^n))$ denotes the Banach space of unitary automorphisms of the Hilbert space $L^2(X,\mathbb{C}^n)$.

Let $U(\mathbb{C},n)$ denote the unitary matrices $n\times n$, we can consider the space of $L^2$ functions  $\phi: X\to U(\mathbb{C},n)$, denoted as $L^2(X, U(\mathbb{C},n))$.
An element $\phi$ in the latter space defines an element in $U(L^2(X,\mathbb{C}^n))$ that sends $f\in L^2(X, \mathbb{C}^n)$ to $\phi\cdot f $ such that

$\phi \cdot f(x) = \phi(x) (f(x))$.

Since $\phi(x)$ is unitary, it preserves the norms hence $\phi\cdot f\in L^2(X,\mathbb{C}^n)$ (this is why I am taking the unitary group and not $GL$) and the operator norm is $1$.
Therefore we have an inclusion

$L^2(X, U(\mathbb{C},n))\to U(L^2(X,\mathbb{C}^n)) $

Is this map an isomorphism?
Can you provide an example of an element of $U(L^2(X,\mathbb{C}^n))$ which is not in the image of the above map?

I believe the answer to the first question is negative. The LHS is an Hilbert space while the RHS is a Banach (but I cannot disprove that it can't be given an Hilbert structure, e.g. is it reflexive? I don't know).

Also, I thought that $U(L^2(X,\mathbb{C}^n))$ is contractible (at least if $\dim L^2(X,\mathbb{C}^n) = \infty$)  by Kuipers' theorem, therefore another possible way of disproving it would be show that $L^2(X, U(\mathbb{C},n))$ has more connected components.

答:
There are several problems here. First, you need a measure to define $L^2(X,U(n))$, not just a measurable space. Moreover, $L^2(X,U(n))$ is not a Hilbert space or even a vector space - neither sums nor scalar multiples of unitaries are unitary in general. In fact, if the underlying measure is infinite, then $L^2(X,U(n))=\emptyset$. For similar reasons, $U(L^2(X,\mathbb{C}^n))$ is not a vector space.

Even if the underlying measure is finie, the map from the question is almost never surjective. To see this, note that a bounded linear operator of the form
$$
Tf(x)=\psi(x)f(x)
$$
with $\psi(x)\in\mathbb{C}$ for a.e. $x$ commutes with all elements of the image of $L^2(X,U(n))$, but the only operators commuting with all elements of $U(L^2(X,\mathbb{C}^n))$ are scalar multiples of the identity. So the map cannot be surjective if $\dim L^2(X)\geq 2$.


问:
How are integer sets of "all odd positive" or "all even negative" or "all negative" described?
The infinite set of all positive integers, i.e., evens and odds combined (sometimes including, sometimes without, $0$), is called the "natural numbers", denoted by $\mathbb{N}$. What about with a specific half of them omitted, or all of them converted to their additive inverse? And does this work the same way for finitely-defined sets?

Slightly off-topic but how might "all complex numbers" or "all rational numbers" (two cases also with special abbreviations) or "all natural numbers" or cetera be described in terms of a subset of "all numbers" (i.e. all numerical values conceivable/possible) without using their respective abbreviations?

答:
How are integer sets of "all odd positive" or "all even negative" or "all negative" described?

To answer the question in the title, recall how can we define a set.

We can define a set by extension, listing all its elements (note that this way of defining a set doesn’t contemplate the order and the number of times the elements are listed). This method works fine for finite sets (and “small” sets) since you have to list each element, one by one.

Also, you can define a set by stating one or more conditions that are satisfied by the elements of that set (and only by them). And this is a good approach to accomplish what you want.

If you want the set of all integers that are odd and positive, then you want a set of integers $z$ that satisfies the property of “$z$ is odd and positive”. And you can represent the set as follows

$$\{n \in \mathbb{Z} : n > 0 \wedge \exists k \in \mathbb{Z} : n = 2k + 1\}.$$

Equivalently, you can define the other two sets in you question as

$$\{n \in \mathbb{Z} : z < 0 \wedge \exists k \in \mathbb{Z} : n = 2k\} \text{ and } \{n \in \mathbb{Z} : n < 0\}.$$

What about with a specific half of them omitted, or all of them converted to their additive inverse?

Suppose that you have the set $X = \{1,2,3\}.$ If you want the set of the additive inverses of the elements of $X,$ you can do it as follows (similar to the other three above)

$$\{-x : x \in X\} = \{-1,-2,-3\}.$$

If by finitely-define sets you mean sets described by extension, then the answer is yes as the example above shows.

For the set of all numbers there is no such structure.

But you can describe, for example, the set off all rational numbers and complex numbers without using its abbreviation. Just recall what is a rational number and what is a complex number. By using the notation above you can described this sets as follows

$$\{x : \exists a \in \mathbb{Z}, b \in \mathbb{N} : x = \frac{a}{b}\} \text{ and } \{x : \exists a,b \in \mathbb{R} : x = a + bi\}.$$


问:
Is there any existing standard name for primes of the following kind in mathematics?

Given a prime $p$ all primes $q$ such that $$q \bmod p = 1\text{ and }2^{(q-1)/p}  \bmod q = 1$$Do primes $q$ have any standard name ?

答:
I don't think that there is a special name for primes like these. Note however that the condition
$$2^{(q-1)/p} \equiv 1 \pmod q$$
means that the order of $2$ in $(\mathbb Z/q\mathbb Z)^* = \mathbb Z/(q-1)\mathbb Z$ is a proper divisor of $q-1$. In other words, your condition means that $2$ does not generate $(\mathbb Z/q\mathbb Z)^*$.

This, for instance, excludes all Fermat primes: if $q = 2^{2^n} + 1$, then necessarily $p = 2$, so that $1 < 2^{(q-1)/p} = 2^{2^{n-1}} < 2^{2^n}$. In particular

$$2^{(q-1)/p} = 2^{2^{n-1}} \not\equiv 1 \pmod{2^{2^n}+1}.$$

So Fermat primes never satisfy what you want. However, this does not mean that if $q$ is not Fermat, then $q$ satisfies your condition ($q = 19$ is a counterexample).

You can have a look at the the table on Wikipedia, which lists generators of $(\mathbb Z/n \mathbb Z)^*$. Just have a look at the rows where $n$ is prime and $2$ is not a generator.


问:
Lambda application terms MN where M is not a Lambda abstraction term?
I believe such terms are syntactically legal in (untyped) lambda calculus (see below), but the motivation for including them is not very clear to this newbie (e.g. because MN has no potential for reducibility). Perhaps I am hung up on the fact that while computation is about reduction, we are permitted to introduce irreducible expressions. I see that MN might reduce in a broader context, e.g. $(\lambda x. PMN)$. Perhaps this is  why it is permitted. I guess this freedom is not necessary (?) but adds some additional expressive power?

For reference, recall that the class of $\lambda$-terms is defined inductively as follows:

Every variable is a $\lambda$-term

If M and N are $\lambda$-terms, then so is (MN)

If M is a $\lambda$-term and x is a variable, then $\lambda x.M$ is a $\lambda$-term.

答:
In the untyped $\lambda$-calculus, one can indeed write "meaningless" terms. Such terms include the application of a non-function to an argument and the application of a function to an argument over which the function is not defined.
When attempting to evaluate such terms, you end up getting stuck at a point where you have reached an expression that is not yet in normal form, but can neither be further reduced. (After our discussion in the comments, I have to add that I consider reduction only over well-scoped terms, since  it makes no sense to reduce open terms that have no meaning.)

I don't think historically there was motivation for allowing such terms. On the contrary, after the untyped $\lambda$-calculus, came the simply typed $\lambda$-calculus (STLC) that disallows such meaningless terms, but on the other hand it can be too restrictive. For example, the fixed-point combinator $\lambda f.(\lambda x. f\,(\lambda y. x\,x\,y))\,(\lambda x. f\,(\lambda y. x\,x\,y))$ that allows general recursion in the untyped $\lambda$-calculus, is no longer an acceptable term in a typed $\lambda$-calculus. In order to gain back recursion, you need to add it as a language primitive with new typing and evaluation rules for it (or use other techniques).

Another problem that simple types introduce is more practical rather than theoretical and involves code modularity. When coding, usually you want a single concept to be coded in a single place. This is not always possible in STLC (but is possible in System F, which is a more advanced typed calculus). Take for example the identity function $(\lambda x. x)$. In the untyped setting, you can assign this function to a variable named, say, "id" and then you can use id on any argument. In STLC, on the other hand, you would have to define separate "id" functions for applying them on integers or on booleans. These variations of "id" would only differ in type annotations and nowhere else: $(\lambda x : Int. x)$ and $(\lambda x : Bool. x)$. The "id" example is a very simple one, but things can get pretty messy for a bit more complex functions.

So altogether, I'd say there is no motivation to allow meaningless terms, but by disallowing them, you end up loosing more than what you wanted to get rid of. However, there are ways to gain back the extra losses, but then your language becomes more complex.


问:
What is the functional inverse (with respect to $h$ (!)) of $f^{\circ h}(x)={F(h) +x F(h-1) \over F(1+h) +x F(h) }$?
I've considered the fractional iteration of $f(x) = {1 \over 1+x} $ for which the general expression depending on the iteration-height parameter $h$ might be assumed by the formula
$$ f^{\circ h}(x) = f(x,h)={F(h) +x \cdot F(h-1) \over F(1+h) +x \cdot F(h)  }$$  where

$ \displaystyle \qquad \small F(h)=\operatorname{fibonacci}(h) = {\varphi^h - (1-\varphi)^h \over \sqrt5 }$ and

$ \displaystyle \qquad \small \varphi = {1+\sqrt5 \over 2} \sim 1.618... $

(Note, that for fractional iterates we need $x \in \mathbb C$)

Today I tried to find a formulation for the functional inverse with respect to h, but don't find a good starting point so:

Q: How would look a function $ h = \operatorname{hgh}(x_0,x_h) $ which would indicate the required iteration-height $h$ given $x_0$ and $x_h = f(x_0,h)$ ? Is there even a closed form for it (I'd call it still closed-form  if it possibly includes e.g. the Lambert-W function)

[update]: for more background see the older exercise of mine

答:
Ahh, the hints to the matrices and to the superfunction were useful. They forced me to look more seriously in my own exercise - and of course the Abel-function is the log of the Schröder-function and in the exercise I've already defined a very simple series for the Schröder-function.

Let's denote $ \small \beta = \varphi-1 \sim 0.618...$ and $ \small u =-\beta^2$ then according to my own formula 2.7 the (recentered) Schröder function is
$$ \sigma (t) = -\sqrt5 \sum_{k=1}^\infty (-{t \over \sqrt5})^k $$ and the inverse schröderfunction
$$ \sigma^{-1} (t) = \sqrt5 \sum_{k=1}^\infty ({t \over \sqrt5})^k $$
such that the continuous iterate is
$$ f(x,h) = \sigma^{-1} \left(u^h \cdot \sigma( x - \beta) \right) + \beta $$

The height-function $hgh(x_h,x_0)$ can be taken from the logarithm to the base $u$ of the ratio of the schröder-values as
$$ \operatorname{hgh}(x_h,x_0) = \log_u \left({ \sigma(x_h - \beta) \over \sigma(x_0-\beta) } \right) $$
and if we use the fact, that the Schröder-function is just a geometric series and can be expressed in a closed form we get finally:
$$ \operatorname{hgh}(x_h,x_0) = \log_u \left(
 {x_h-\beta \over x_0 - \beta } \cdot{  \sqrt5 + x_0-\beta
 \over \sqrt5 +  x_h-\beta } \right) $$

Because $u$ is negative we get complex values for fractional $h$ and this must then be handled by selection of the correct branch of the complex logarithm.

Also it must be handled, that the trajectory of iterations with real heights form a spiral through the complex plane with one round per 2 units in the height and thus the winding-number for heights $h \ge 2 $ must be reflected, say for the example $hgh(2/3,1) = 2$ we need to introduce that correction writing/branching of log
$$ \cdots = { \log \left({ \sigma(x_h - \beta) \over \sigma(x_0-\beta) } \right) + 2 \pi i\over \log u} $$ otherwise we get "the shortest distance" expressed by a complex height of $\small 0.171595 + 0.560130 i$ instead.

But that are details which I can now work on by myself. Thanks for the hints - sometimes one needs only an external impulse to recover a unremembered concept ...


问:
Map comparing Alexander-Spanier and Čech cohomology
I'm trying to follow the argument in Bredon's Sheaf Theory (page 29) supposedly constructing an isomorphism between the Alexander-Spanier cohomology $H_{AS}^n (X; G)$ and Čech cohomology $\check H^n(X; G)$ for a space $X$ and an abelian group $G$ thought of as a constant sheaf. Specifically, Bredon constructs a map on the cochain groups that descends to cohomology. I'm having trouble seeing why this map is an isomorphism on cochains as claimed; in fact I believe I have a counterexample.

Specifically, $C_{AS}^n(X; G)$ is defined to be $A/A_0$, where $A$ is the group of all (possibly discontinuous) functions $X^{n+1} \to G$ and $A_0$ is the subgroup of functions that vanish on some neighborhood of the diagonal $X \hookrightarrow X^{n+1}$. The Čech cochain group $\check{C}^n(X; G)$ is defined to be the direct limit $\varinjlim \check C^n_{\mathfrak{U}}$ over all open covers $\mathfrak{U}$ of $X$ ordered by refinement, where given a cover $\mathfrak{U}$,
$$\check{C}^n_{\mathfrak{U}} = \bigoplus_{U_0, \dots, U_n \in \mathfrak{U}} G(U_0 \cap \dots \cap U_n)$$
with the understanding that $G(\varnothing) = 0$.

The map $C_{AS} \to \check{C}$ uses that a cofinal set of covers is given by covers of the form $\mathfrak{U} = \{U_x | x \in X\}$ indexed by $X$ with $U_x \ni x$, and refinements given by $\mathfrak{V} = \{V_x\}$ with $x \in V_x \subseteq U_x$. Then given an $f : X^{n+1} \to G$, we get $f_\infty  = \varinjlim f_{\mathfrak{U}} \in \check{C}^n(X; G)$ where
$$f_{\mathfrak{U}}(U_{x_0} \cap \dots \cap U_{x_n}) = f(x_0, \dots, x_n) \; .$$
The book claims that the kernel of the map $f \mapsto f_\infty$ is supposed to be exactly $A_0$ (and has an associated claim about supports of $f$ and $f_\infty$ but I do not think this part is crucial).

I believe the following is a counterexample: Let $X = \{1/k | k \ge 1\} \cup \{0\}$, $G = \mathbb{Z}/2$ and
$$f(x_0, \dots, x_n) = \begin{cases}0 & \text{if some } x_i = 0 \text{ or } x_0 = \dots = x_n\\ 1 & \text{otherwise.}\end{cases}$$ Then $f \notin A_0$ since any neighborhood of the diagonal contains $U^{n+1}$ for some neighborhood $U$ of $0$. But to compute $f_\infty$ we can take a further (cofinal) subset of covers $\mathfrak{U}$ where $U_{1/k} = \{1/k\}$ for each $k$. For such a $\mathfrak{U}$ the only way $U_{x_0} \cap \dots \cap U_{x_n}$ can be nonempty is if all the $x_i$'s are equal or at most one $x_i \ne 0$. So $f_{\mathfrak{U}} = 0$ and hence $f_\infty = 0$.

Am I misinterpreting something? Or does this statement need more assumptions on $X$ (or $G$)? Note that the counterexample has $X$ a compact subset of $\mathbb{R}$, although not (locally) connected. Is it still true that the cohomology is isomorphic even if the chain groups are not?

答:
$\!$Hi Ronno, just found this old question, small world. A few comments:

I agree that Bredon's book is mistaken here.

Define a rigid open cover to be a collection of open sets $U_x \ni x$ for all $x \in X$, with the additional property that the set $\{U_x | x \in X\}$ is a locally finite cover of $X$. If $X$ is paracompact then rigid open covers are cofinal among all open covers, so rigid open covers can be used to compute Cech cohomology. As you pointed out in a comment to Alan Zhang, rigid open covers will not in general be cofinal among all open covers indexed by $X$, but this is not quite the property we need, anyway. It seems to me that Bredon's argument will work to prove an isomorphism between Cech and Alexander-Spanier cohomology on paracompact spaces, once we make the choice of computing Cech cohomology using rigid open covers.

The above is a bit unsatisfying since paracompactness shouldn't be needed to compare Cech and Alexander-Spanier cohomology. There's a surprising theorem of Dowker ("Homology groups of relations", Annals 1952) that says that Cech cohomology and Alexander-Spanier cohomology are isomorphic on any topological space whatsoever. Usually one expects at least some kind of point-set assumption for "reasonable" cohomology theories to coincide! Both Cech and Alexander-Spanier theories are "badly behaved" for spaces that are not paracompact Hausdorff, but apparently they are badly behaved in exactly the same way. I haven't tried digging into Dowker's argument, though. (Edited to add: Actually Dowker's theorem is quite pretty and it seems much more like the "right" approach than what Bredon does.)


问:
If $\phi\otimes\mathrm{Id}_\mathbb Q$ and $\phi\otimes\mathrm{Id}_{\mathbb F_p}$ are $\mathbb Z$-algebra isomorphisms, is $\phi$ an isomorphism?
I'm currently trying to understand a line in a paper that would follow easily if the answer to the following question was yes:

Let $R, S$ be (commutative, unital) rings and let $\phi\colon R\hookrightarrow S$. Suppose that, as $\mathbb Z$-algebras, $\phi\otimes\mathrm{Id}_\mathbb Q\colon R\otimes \mathbb Q\to S\otimes\mathbb Q$ and $\phi\otimes\mathrm{Id}_{\mathbb F_p}\colon R\otimes \mathbb F_p\to S\otimes\mathbb F_p$ are isomorphisms for all primes $p$.

Is $\phi$ an isomorphism?

In the setting I care about, I know that $R, S$ are torsion free integral domains. I also know that $S$ is a subring of $\mathbb Z[X_1, \ldots, X_n]$ for some $n$ and that $S\otimes \mathbb Q$ is finitely generated as a $\mathbb Q$-algebra.

I know that the general statement is false when $R, S$ are just $\mathbb Z$-modules: a counterexample is $R = 0$ and $S = \mathbb {Q/Z}$.

答:
For $R\subset S$ torsion-free abelian groups such that $S\otimes_\Bbb{Z} \Bbb{Q}=R\otimes_\Bbb{Z} \Bbb{Q}$.

For $a\in S$, if $a\not \in R$ then $a\Bbb{Z}\cap R=an \Bbb{Z}$ for some $n\ge 2$.

For $p|n$ then $an\ne 0 $ in $R/pR$ whereas $an=0\in S/pS$ so the map $R/pR\to R/(pS\cap R)\subset S/pS$ is not injective.


问:
Evaluating $\lim_{x \to 0} \dfrac{\arctan x}{x}$

Evaluate $$\lim_{x \to 0} \dfrac{\arctan x}{x}$$

I know that it's $1$ using L'Hopital's rule or numerically, but what about algebraically? Thanks in advance and I really appreciate it.

答:
If you do not want to use L'Hopital, or series expansion, then I suggest the following :

Notice that for any $ x\in\mathbb{R}^{*} $, we have : $$ \frac{\arctan{x}}{x}=\int_{0}^{1}{\frac{\mathrm{d}y}{1+x^{2}y^{2}}}=1-x^{2}\int_{0}^{1}{\frac{y^{2}}{1+x^{2}y^{2}}\,\mathrm{d}y} $$

Since, for all $ x\in\mathbb{R}^{*} $, $$ \left|x^{2}\int_{0}^{1}{\frac{y^{2}}{1+x^{2}y^{2}}\,\mathrm{d}y}\right|\leq x^{2}\int_{0}^{1}{y^{2}\,\mathrm{d}y}=\frac{x^{2}}{3}\underset{x\to 0}{\longrightarrow}0 $$

Then $ \lim\limits_{x\to 0}{x^{2}\int_{0}^{1}{\frac{y^{2}}{1+x^{2}y^{2}}\,\mathrm{d}y}}=0 $, which means : \begin{aligned} \lim_{x\to 0}{\frac{\arctan{x}}{x}}&=\lim_{x\to 0}{\left(1-x^{2}\int_{0}^{1}{\frac{y^{2}}{1+x^{2}y^{2}}\,\mathrm{d}y}\right)}\\ &=1-0\\ \lim_{x\to 0}{\frac{\arctan{x}}{x}}&=1\end{aligned}


问:
Genus of a graph and planarity
In 'Introduction to graph theory' by Trudeau it is proved the theorem

Every graph has a genus

by first embedding the graph onto a sphere , and then adding handles, say we use $k$ of them, where the graph has edge-crossing , this is the example used :

Then, he writes that using a continuous deformation of the surface (stretching,shrinking or distorting the surface) it is possible to preserve planarity while deforming the $k$-handles sphere into a $k$-hole torus. So, the genus will be at most $k$.

I'm not able to visualize the bold part, how can I be sure that planarity will be preserved? Should it be obvious? Is there a simple explanation to understand clearly why this happens ? (I'm totally a beginner in algebraic topology, just visualization of trasformations are used in the book).

Another proof of a theorem stated earlier in the book : "every planar graph has genus $0$ (it can be embedded on a sphere)" was obscure to me but then searching online I found the explanation through the stereographic projection and all my doubts were gone , does something similar exist? Or even a good intuition is difficult to set up ?

EDIT :

For clarity I add the definition of genus given in the book :

The genus of a graph, denoted $g$, is the subscript of the first surface among the family $S_0,S_1,S_2,\dots$ on which the graph can be drawn without edge crossing .

I also add the explanation of the author to pass from the $k$-handle sphere to the $k$-hole thorus :

Think of the surface consisting of $S_0$ with $k$ handles as made of extremely pliable rubber which we are free to stretch, shrink or otherwise distort as much as we want , provided that we don't tear it off of fold it back onto itself. With this understanding we see that the surface consisting of $S_0$ with $k$ handles can be 'continuously deformed' into $S_k$ and the vertices and edges of $G$ can be carried along with this continuos deformation. Thus $G$ can be drawn on $S_n$ without edge-crossings.

The bold part is obscure to me in the sense explained above.

答:
To be honest, I am not really sure what is your question. What do you mean by "every graph has a genus"? I'll try to answer by the way, since geometrically your question is clear.

For me, the genus is a well defined quantity associated to a (compact) surface. When you have a planar graph, you imagine to fill the faces among the edges and you get a very simple surface, something like a polygon in the plane. If you also fill the exterior " ghost" face you obtain the whole plane. The genus is computed now as $1-\chi/2$, where

$$\chi =  \#\{\text{filled faces}\} - \#\{\text{edges}\} +\#\{\text{vertices}\}$$

And we would be done, but the plane is not a compact surface so we don't know what this should be equal to. To deal with this technical issue, we add a point to infinity and we get a sphere. Now the theory tells us that the genus (which is secretly the number of holes) of the sphere is zero, so that we get for a planar graph

$$\#\{\text{filled faces}\} - \#\{\text{edges}\} +\{\text{vertices}\}= 2$$

You can also reformulate this, recalling that the faces we filled were the internal faces + one ghost face:
$$\#\{\text{ faces}\} = \#\{\text{edges}\} - \{\text{vertices}\}+1$$

Check this on small graphs! That's true for real :)

Some complications arise if there are intersections. The main idea in algebraic topology is that if we cut a topological space in simple bricks - like a surface in triangles - the geometric properties of the space become combinatorial properties of how our bricks are combined. You have just seen this above: the genus of the sphere reflects in an equation for edges, vertices and faces for the graph that "builds up" the sphere.

When you have intersection, the problem is that the graph is apparently not building up anything. The idea of your book is to create small bridges (handles) so that the graph does not intersect himself anymore: you will have a planar graph sitting inside a new surface, no longer a sphere, but who cares! We can carry on our argument above, and we only have to change the genus in the end: for a nice coincidence, it will be the number of bridges we added.

If you are not comfortable with deforming the surface into a k-hole torus while preserving planarity, there is no need to do that: the genus is a homotopy invariant. If you are convinced that - beside the graph story - you can deform a k-handle sphere into a k-hole torus, then you know that the genus of the k handle sphere is the same of the k-hole torus, that is k.

Please ask in the comments if you need further clarifications :)

Title: Genus of a graph and planarity


问:
Show groups of symmetries of a cube and a tetrahedron are not conjugate in isometry group.
I've shown that the symmetry group of a cube and a tetrahedron are both isomorphic to S4, but I am now trying to show that they are not conjugate when considered as subgroups of isometries of 3D space.

I cannot think of any kind of criteria that gives when two groups are not conjugate?

EDIT: I meant to say the rotations of a cube! So I've shown the full symmetry group of the tetrahedron is isomporphic to the rotations of a cube, however I need to show that they are not conjugate.

答:
Something's wrong with your first statement. If you consider orientation preserving 3D movements, the symmetry group of the cube is twice as big as that of the tetrahedron. And the same holds if you disregard orientation for both solids.
Note that taking $4$ out of $8$ vertices of a cube, you get a tetrahedron. Any symmetry of that tetrahedron induces a symmetry of the cube. However, the cube allows additional symmetries that map the tetrahedron to the "other" tetrahedron (the one with the other four vertices).
In other words, the fact that half the vertices of a cube form a tetrahedron shows that the tetrahedron group is a subgroup of index 2 in the cube group.

However, you do get $S_4$ if you count

arbitrary isomometries of the tetrahedron (that is, just permutations of the four vertices) and

orientation preserving isometries of the cube (that can be identified with permutations of the four spatial diagonals).

These two subgroups of the group of isometries cannot be conjugate to each other because the conjugate of an orientation preserving isometry is orientation preserving.


问:
Proof Linearity of Conditional expectation
How we can proof that:

$E[X - Y|W] = E[X|W]-E[Y|W]$

I try to use the definition of Conditional Expectation:

$E[X|Y=y]= \sum_x \cdot p(x|y)$ and then to substitute $X=A-B$

Is it the right way? And how we continue from here?

答:
You did not explicitate if your rv's are contintuous or not. Thus I assume they are continuous:

$$E[X-Y|W]=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}(x-y)f_{XY|W}(x,y|w)dxdy=$$

$$=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}xf_{XY|W}(x,y|w)dxdy-\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}yf_{XY|W}(x,y|w)dxdy=$$

$$=\int_{-\infty}^{+\infty}xf_{X|W}(x|w)dx-\int_{-\infty}^{+\infty}yf_{Y|W}(y|w)dy=E(X|W)-E(Y|W)$$


问:
Proving $3^n>n^4$, for all $n\geq 8$

Prove $3^n>n^4$, for all $n\geq 8$.

I solved this for 4 - 5 hours now, but I still don't get any ideas other than plugging in $n$. Please give me some hints. Any hints are fine. Please use induction!

答:
We have $3^8 \geqslant 8^{4}$.

Let's suppose the property is true for some $n\geqslant 8$.
Then $3^{n+1} \geqslant 3.n^4$, and $(n+1)^4 = n^4 + 4n^3 + 6n^2 + 4n +1$.

So we want to prove that $2n^4-4n^3-6n^2-4n-1\geqslant 0$. By dividing by $n^2$, we have to prove that $2n^2 \geqslant 4n + 6 +4/n +1/n^2$, or just $2n^2 \geqslant 4n + 12$, which is equivalent to $n^2 \geqslant 2n + 6$. And since $n^2 - 2n - 6 = (n-4)(n+2)$ is $\geqslant 0$ for $n\geqslant 8$.

This gives the answer.

Note that you can also do without induction by directly deriving $f:x\mapsto 3^x - x^4$ $4$ times.

Title: Proving $3^n>n^4$, for all $n\geq 8$


问:
Generation of Hermite polynomials with Gram-Schmidt procedure
I want to use the Gram-Schmidt procedure to generate the first three Hermite polynomials. Given the set of linearly independent vectors $\{1,x,x^2,...\}$ in the Hilbert space $L^2(R,e^{-x^2}dx)$, I apply the orthogonalisation procedure as follows:

$u_{0}(x)=1$

$u_{1}(x)=x-\frac{\langle u_{0}|x \rangle}{\langle u_{0}|u_{0}\rangle} \cdot u_{0}=x-\frac{\langle 1|x\rangle}{\langle 1|1\rangle}\cdot 1 =x $

$u_{2}(x)=x^2-\frac{\langle u_{0}|x^2 \rangle}{\langle u_{0}|u_{0}\rangle} \cdot u_{0}-\frac{\langle u_{1}|x^2 \rangle}{\langle u_{1}|u_{1}\rangle} \cdot u_{1}=x^2-\frac{\langle 1|x^2\rangle}{\langle 1|1\rangle}\cdot 1-\frac{\langle x|x^2\rangle}{\langle x|x\rangle}\cdot x =x^2 -\frac{1}{2} $

The inner product I use is the one of $L^2(R,e^{-x^2}dx)$, namely:

$\langle u|v \rangle:=\int_{R}\overline{u(x)}v(x)e^{-x^2}dx$

Then I divide the functions I found by their norm in order to get orthonomal vectors:

$\tilde{H_0}(x)=\frac{1}{\sqrt{\langle u_{0}|u_{0} \rangle}}u_{0}(x)=\frac{1}{\pi^{1/4}}$

$\tilde{H_1}(x)=\frac{1}{\sqrt{\langle u_{1}|u_{1} \rangle}}u_{1}(x)=\frac{\sqrt{2}}{\pi^{1/4}}x$

$\tilde{H_2}(x)=\frac{1}{\sqrt{\langle u_{2}|u_{2} \rangle}}u_{2}(x)=\frac{\sqrt{2}}{\pi^{1/4}}(x^2-\frac{1}{2})$

I can't understand why the coefficients I get are different from the ones that can be found on Wikipedia (https://en.wikipedia.org/wiki/Hermite_polynomials) or in every list of Hermite's polynomials. Where is the mistake?

Body:
$u_{0}(x)=1 \quad u_{1}(x)=x \quad $
Let $\mathrm{H}_{0}(x), \mathrm{H}_{1}(x)$ be first two Hermite polynomials.
If $\frac{A H^{\prime}_{n}(x)}{\sqrt{\int_{0}^{\prime} H_{n}(x) H_{n}^{\prime}(x) e^{-x^{2} }d x}}=H_{n}(x)$
then $\int_{-\infty}^{\infty} H_{n}(x) H_{n}(x) e^{-x^{2}} d x=2^{n} n ! \pi^{1 / 2}$

From here we can simply find out that
$A=\sqrt{2^{n} n ! \pi^{1 / 2}}$

Let $H_{0}^{\prime}(x)=u_{0}(x)=1$
$\int_{-\infty}^{\infty} e^{-x^{2}} d x=\sqrt{\pi}$

$H_{0}(x)=\sqrt{2^{0} 0! \sqrt{\pi}} \frac{H_{0}^{\prime}(x)}{\sqrt{\sqrt{\pi}}}$
$H_{0}(x)=1$
$\begin{aligned} H_{1}^{\prime}(x) &=u_{1}(x)-H_{0}(x) \int_{-\infty}^{\infty} H_{0}(x) u_{1}(x) e^{-x^{2}} d x \\ &=x-\int_{-\infty}^{\infty} x e^{-x^{2} d x} \end{aligned}$

$H_{1}^{\prime}(x)=x$
$\int_{-\infty}^{\infty} H_{1}^{\prime}(x) \quad H_{1}^{\prime}(x) e^{-x^{2}} d x=\int_{-\infty}^{\infty} x^{2} e^{-x^{2}} d x=\frac{\sqrt{\pi}}{2}$

$\begin{aligned} H_{1}(x) &=\sqrt{2^{\prime} 1 ! \sqrt{\pi}} \frac{H_{1}^{\prime}(x)}{\sqrt{\sqrt{\pi} / 2}} \\ &=2 x \end{aligned}$


问:
Finding a primitive root of a prime number
How would you find a primitive root of a prime number such as 761? How do you pick the primitive roots to test? Randomly?

Thanks

答:
Building on Vadim's answer, we can cut down on the required powers by assessing whether a residue is quadratic, without using the power calculation.  Thereby the highest power we would have to calculate is eliminated from having to use the direct calculation.

To test $2$, check the residue of the prime modulus $p$ you're considering $\bmod 8$.  $2$ would be a quadratic residue, therefore not primitive, if and only if $p\in \{1,7\}\bmod 8$.

To test an odd prime, use quadratic reciprocity to simplify.

To test a squarefree product (all other composites are trivial), count the number of its prime factors that turned out to be nonquadratic residues.  This count must be odd for a primitive root.

Let's try these methods with $761$:

$2$ is a quadratic residue because $761\equiv1\bmod 8$.  This kills $2$ as a potential primitive root.

$3$ gives, by QR, $(3|761)=(761|3)=(2|3)=-1$, so we have a nonquadratic residue -- cheers!  We would still have to test for a higher-power residue, and Vadim's calculation of $3^{152}$ reveals a fifth-power residue instead of a primitive root.

$5$ gives $(5|761)=(761|5)=+1$, a quadratic residue.  Move along.

$6=2×3$ is a squarefree product in which one factor ($3$) is nonquadratic and the other ($2$) is quadratic.  Thus an odd nonquadratic-residue count.  So $6$ is a nonquadratic residue and we test it for a higher power resudue.  Vadim's calculations of $6^{152}$ and $6^{40}$ eliminate these pitfalls and thus $6$ is rendered a primitive root.


问:
Find PACF for die's results sum.
This is a variation of one of my previous questions:

Problem:

A die was thrown three times. The results of the throws are $X_1,X_2,X_3$. Let $L = X_1 + X_2$, $S = X_2 + X_3$ and $R = X_1 + X_2 + X_3$.

Find $p \Bbb Corr(L,S; R)$.

Note, $p \Bbb Corr(X, Y ;Z)$ means a partial correlation between $X$ and $Y$, "cleared" of the $Z$ effect.

Trouble

I came up to two solutions which have different answer.

My solution $1$

The idea is here to "offset" the change in C by substracting it from A and B
$$p \Bbb Corr(L,S; R) = p\Bbb Corr(X_1 + X_2, X_2 + X_3; X_1 + X_2 + X_3) = $$ $$ = \Bbb Corr(-X_3, -X_1) = 0 \quad \text{due to independence}$$

My solution $2$

I found the following formula on the Internet:

$$ r_{AB|Z} = \frac{ r_{AB} - r_{AC}\cdot r_{BC}}{\sqrt{(1- r^2_{AC})(1- r^2_{BC})}}$$

Using this formula I obtain number $p \Bbb Corr(A,B; C) = -\frac{1}{2}$.

As you can see the results are different. Which approach is wrong then?

答:
If we clear variables $L$ and $S$ from $R$ such that
$$L = aR + \nu_1, \quad \Bbb Cov(R, \nu_1) = 0$$
$$S = bR + \nu_2, \quad \Bbb Cov(R, \nu_2) = 0$$
than $p \Bbb Corr(L,S; R) =\Bbb Corr(\nu_1, \nu_2) = \Bbb Corr(L - aR, S - bR) = \frac{\Bbb Cov(L - aR, S - bR)}{\sqrt{\Bbb Var(L - aR) \Bbb Var( S - bR)}}$

Now, the task is to find $a$ and $b$.

From condition $\Bbb Cov(R, \nu_1) = 0$ the expression for $a$ follows
$$ a = \frac{ \Bbb Cov(L,R)}{\Bbb Var(R)}$$
Same for $b$.
Using this idea, you should get $a=-\frac{1}{2}$, $b=-1$
The fact that $X_1, X_2, X_3$ are from discrete uniform distribution should allow you calculate these covariance and variances easily.
$$p \Bbb Corr(L,S; R) = \frac{2}{3}$$

Title: Find PACF for die's results sum.


问:
If $f$ is a real function, continuous at $a$ and $f(a) < M$, then there is an open interval $I$ containing a such that $f(x) < M$ for all $x \in I$.
I have problem regarding the If f is a real function, continuous at a and f(a) < M, then there is an open interval I contianing a such that f(x) < M for all x in I. answer. If I used $\epsilon =M-f(a)$ which is also $\epsilon >0$ and $ \exists$  $  \delta>0$ so there is  an open interval $I$ containing such that $f(x)<M$ for all $x \in I$. I think this is also correct but not sure.

Can anyone verify my answer?

$\underline{Edit}$

Now let $\epsilon = {M-f(a)}$, clearly $\epsilon >0$, and hence there exists an open interval $I=(a-\delta, a+\delta)$, such that for any $x\in I$, $|f(x)-f(a)|<\epsilon= {M-f(a)}$ holds.

It follows that $f(x)<M$ for all $x \in I$

答:
The condition that $f$ is continuous at $a$ indicates that
\begin{equation}
\lim_{x \to a} f\left(x\right) = f\left(a\right).
\end{equation}
In other words, we have the following proposition:
\begin{equation}
\forall \epsilon > 0, \exists \delta > 0, \forall x, 0 < \lvert x-a\rvert < \delta \longrightarrow \lvert f\left(x\right)-f\left(a\right)\rvert < \epsilon.
\end{equation}
And we have the proposition that
\begin{equation}
f\left(a\right) < M.
\end{equation}
Using the fact that $M - f\left(a\right) > 0$, we have
\begin{equation}
\exists \delta > 0, \forall x, 0 < \lvert x-a\rvert < \delta \longrightarrow \lvert f\left(x\right)-f\left(a\right)\rvert < M - f\left(a\right),
\end{equation}
which further indicates that
\begin{equation}
\exists \delta > 0, \forall x, 0 < \lvert x-a\rvert < \delta \longrightarrow f\left(x\right) < M.
\label{main}
\end{equation}
If there is no such open interval $I$ that $f\left(x\right) < M$ for all $x \in I$, then we have the following proposition:
\begin{equation}
\forall \delta > 0, \exists x, 0 < \lvert x-a \rvert < \delta \wedge f\left(x\right) \geq M,
\label{sub}
\end{equation}
which obviously contradicts our conclusion.


问:
What is derivative of $z^*z$ : I find two contradictory results
What is derivative of $z^*z$ : I find two contradictory results

we have $f(z)=z^*z=x^2+y^2=u+iv$

so we have $(du/dx)+i(dv/dx)=2x$.

But according to formula 30 and 31 of
http://www1.spms.ntu.edu.sg/~ydchong/teaching/06_complex_derivatives.pdf

we could compute in another way :
we have $(dv/dy)-i(du/dy)=0-i2y=-2iy$, which is not equal to previous result.

Are the formulas (30) and (31) of this reference wrong ?

Is there a good reference for computing a complex derivative ?

答:
While partial derivatives of complex functions are straightforward, the total derivative is trouble. Proceeding in a calculus-$101$ analogy, we write $$\frac{dw\left(z,\bar{z}\right)}{dz} = \lim_{\Delta z \rightarrow 0} \frac{\partial w}{\partial z} + \frac{\partial w}{\partial \bar{z}}\frac{\Delta\bar{z}}{\Delta z} \:,$$ where if $\Delta z=\left|\Delta z\right| e^{i\theta}$, then the ratio $\Delta\bar{z}/\Delta z$ becomes $e^{-2i\theta}$, which can have any phase $\theta$ as $\Delta z\rightarrow 0$. The best thing we can do about the total derivative is to restrict $w$ to have no explicit $\bar{z}$-dependence, eliminating the second term altogether. We therefore take the following two equations as criteria of the total derivative: \begin{equation}\label{totalderiv}\frac{dw}{dz} = \frac{\partial w}{\partial z} \hspace{2.54cm} \frac{\partial w}{\partial\bar{z}}=0\end{equation}

The Cauchy-Riemann conditions follow right from this. I think the PDF you linked made the same point. Not sure if this helps overall.


问:
Finding an isomorphism between spans
The qustion is how to find an isomorphism between two spans: span{(1,0,0,0),(0,1,0,0)} and span{(3,2,1,0),(1,1,1,1)}. I can see they are both lineary independet so they form bases, but how do I find the isomorphism?</p>

答: Hint:  Any two finite dimensional vector spaces of the same dimension are isomorphic.  Let $\{v_1,\dots,v_n\}$ and $\{w_1,\dots,w_n\}$ be bases.  You can map $v_i\mapsto w_i$, and extend linearly.  This defines an isomorphism.

Title: Finding an isomorphism between spans


问:
Proving that an ideal in a PID is maximal if and only if it is generated by an irreducible

Let $R$ be a PID (Principal Ideal Domain) and $x$ is an element R. Prove that the ideal $\langle x\rangle$ is maximal if and only if $x$ is irreducible.

Ok, so I know what an irreducible is. I'm thinking that this problem is asking us to set up a proof by contradiction but I can't see how. No one in my study group has any clue.

答:
Suppose that $R$ is a PID and that $\mathcal{I}$ is an an ideal.  Then $\mathcal{I}$ is maximal iff for any $x$ generating $\mathcal{I}$, $x$ is irreducible.

Proof: $\Rightarrow$:  Suppose $\mathcal{I}$ is maximal and that $\mathcal{I}$ is generated by $x$.  Write $x = ab$ for some $a,b\in R$.  Since $a|x$, $\mathcal{I}$ must be a subset of the ideal generated by $a$.  Were this inclusion to be proper, by the maximality of $\mathcal{I}$, we would have $R$ being generated by $a$.  This make $a$ a unit. By symmetry, $a$ or $b$ is a unit.  We conclude that $x$ is irreducible.

$\Leftarrow$:
Suppose that $x$ is irreducible.   If $x$ is not a unit, there is a maximal ideal $\mathcal{I}$ of $R$ with $x\in\mathcal{I}$.  Since $R$ is a PID, we can choose $y\in R$ so that $\mathcal{I}$ is generated by $y$.  Since $x\in \mathcal{I}$, $y|x$.  Write $x = ay$ for some $y\in R$.  Since $x$ is irreducible $a$ is a unit.  Hence $x$ and $y$ generate the same ideal; this ideal is maximal.


问:
Periodic points of image of Cantor set by ternary dilation
Let $C$ be the Cantor. The set is built up by cutting all intervals into thirds and removing the middle third step one by one and to infinity.
\begin{align*}
 C_0 = &  [0,1]\\
  C_1 =&  [0, \frac 13] \cup [\frac 23, 1]\\
  C_2 =&  [0, \frac 19]\cup [\frac 29, \frac 13]\cup[\frac 23,\frac 79]\cup [\frac 89, 1].\\
\vdots \\
C= & \cap_{i=0}^{\infty} C_i
\end{align*}
The Cantor set corresponds to the resulting "infinite number of times" ( observe that $T(C)=C$ ).

I wonder how to get prove that the periodic points of  $T(C)$ are dense in $C$, where
$T$ is the transformation $T: C \to C$ defined by $x \longmapsto 3x \quad \mathrm{mod}\, \,  1$.

A periodic point is a point $x_0$ such that $f^k(x_0)=x_0$ for some  integer $k \geq 1$.

So
\begin{align*}
T(x)= x \quad \mathrm{mod} \, 1    \\
2x -  \lfloor 2x \rfloor =& 0 \\
2x =  \lfloor 2x \rfloor &.
\end{align*}
How to finish this exercise.

Many thanks for your help.

答:
Notice that every $x$ in $[0, 1]$  may be expressed as
$$
  x=\sum_{n=1}^\infty  a_n3^{-n},
  $$
with $a_n\in \{0,1,2\}$, and that $x$ lies in the Cantor set iff it has an expression as above with every  $a_n\neq 1$.

It should be stressed that $x$ may have two different such expressions, e.g.
$$
  \frac 13 = 1\times 3^{-1} = 2\times 3^{-2} + 2\times 3^{-3} + 2\times 3^{-4} + \cdots
  $$
and that it is enough for one of these expressions to satisfy   $a_n\neq 1$ in order  to qualify.

This said,  observe that  ternary dilation corresponds to shifting the digits, namely
$$
  T\left(\sum_{n=1}^\infty  a_n3^{-n}\right) =
  \sum_{n=1}^\infty  a_{n+1}3^{-n}.
  $$

This implies that $x$ is periodic for $T$ if its  digits are periodic,  that is,  if there exists $p$ such that
$$
  a_{n+p}=a_n, \quad\forall n\in \mathbb N.
  $$

Given any $x$ in $C$ we may therefore approximate it by a periodoc point $y$ whose digits $a_n$ coincide with those of $x$ up
to a big enough $n$,  and are then repeated periodically.


问:
Length of a curve (?) tangent to a circle
I'm preparing for my (hopefully) future university's entrance exams, and one example question reads as follows:

In the cargo bed of a truck with width 2d a tube with radius R is placed as seen in the picture below. The cargo bed itself also has a height of R. A belt is attached over the tube, perpendicular to the length direction of the tube and the cargo bed, with its ends at points A and B, on the edges of the cargo bed. Which expression displays the length of the belt?

(A) $~2R \cdot \arcsin\left(\frac{R}{d}\right)+2\cdot\sqrt{d^2-R^2}$

(B) $~2R \cdot \arcsin\left(\frac{R}{d}\right)+2\cdot\sqrt{d^2+R^2}$

(C) $~2R \cdot \arccos\left(\frac{R}{d}\right)+2\cdot\sqrt{d^2-R^2}$

(D) $~2R \cdot \arccos\left(\frac{R}{d}\right)+2\cdot\sqrt{d^2+R^2}$

I quite frankly have no idea how to approach this problem, and it's the first real roadblock I've encountered on the example tests. Any help is much appreciated!

答:
Hint:

$1)$ Find $x$ using Pythagoras.

$2)$ Calculate $\theta$.

$3)$ Use the formula for arc length to calculate the length of the curved part in the middle.


问:
If every bounded subsequence converges, then the sequence converges
I came across the following statement in a true or false exercise:

If every bounded subsequence converges, then the sequence converges

The answer is false, but I am struggling to figure out a counterexample to the statement to show that it is false.

答:
[Converting my commment to an answer.]

A sequence with no bounded subsequences (e.g., $1,2,3,4,\ldots$) would satisfy this statement vacuously.

You could interweave this with a convergent sequence to get a non-vacuous counterexample, e.g. $1,0,2,0,3,0,4,0,5,0,\ldots$.


问:
A good practise book suggestion
I am a beginner in Special Functions. So I am looking for a Good Practise/Application  Book  about the  properties of Hypergoemetric Function  and its relations with orthogonal polynomials. I want to learn finding
derivative, contiguous relations, integral representation, differential equalities and transformations practisely. You can suggest a thesis or article
, pdf, too. Thanks for your comments.

答:
For various properties of  Hypergoemetric Function, you can try the following references:

$(1)~~$ "Special functions and their applications" by N. N. Lebedev
$~~~~~~~~~$ It is a great reference for special functions with many physics - engineering applications. Also it is one of the very few books that give guidance and insight in the more difficult elements of that derivation.

$(2)~~$ "Textbook of Ordinary Differential Equations" by C. R. Mondal

$~~~~~~~~~$ It contains a detailed discussion (with lot of examples) on some well-known special functions. Many of the standard concepts and methods which are useful in the study of special functions are discussed. The properties of special functions are derived from their differential equations and boundary conditions.

$(3)~~$ "Special functions for scientists and engineers" by W. W. Bell

$~~~~~~~~~$ It is a good book of special functions for people who are just starting to study them, there are several examples of their properties that are useful, the only thing that lacks the book is that it does not contain examples of where to occupy the special functions

Also you can find some other references from these links:

Reference Book on Special Functions

Book recommendation on special functions

Good reference for Special functions/no elementary functions.

Title: A good practise book suggestion


问:
Upper bound on largest eigenvalue of symmetric matrix
Let $A$ be a square matrix with all eigenvalues in the unit disk: $|\lambda(A)| \leq 1$ for all eigenvalues $\lambda(A)$.

I am wondering if all the eigenvalues of $A + A^T$ are in twice the unit disk:
$\lambda(A + A^T) \leq 2$?

答: Nope. Take the two by two identity matrix and add a 2 above the diagonal.</p>