问:
Interior of a set and first countability
Summary: Let $(X,\tau)$ be a topological space and $A\subseteq X$. Is the first countability of $X$ needed for proving that any point $x$ is in the interior of $A$ iff every sequence in $X$ converging to $x$ is eventually in $A$?

$\newcommand{\nat}{\mathbb{N}}$
I've proved that when $X$ is first-countable, any point is in the closure of $A$ iff there is a sequence in $A$ that converges to $x$, that is,
$$
\forall x \in X:(x \in \overline{A} \iff \exists(x_i)_{i\in\nat} \subseteq A: \lim_{i \rightarrow \infty}x_i = x)
$$
where I used the first accountability in the construction of such a sequence ($\Rightarrow$ direction).

But I spent almost a day to figure out whether I even need first accountability in proving that any point $x$ is in the interior of $A$ iff every sequence in $X$ converging to $x$ is eventually in $A$, or
$$
\forall x \in X:(x \in A^{\circ} \iff \forall(x_i)_{i\in\nat} \subseteq X: (\lim_{i \rightarrow \infty}x_i = x \Rightarrow \exists j \in \nat:\forall k \geq j:x_k \in A))
$$

In my attempt in proving it, I didn't use the first accountability in both directions. It was easy to show the sufficiency ($\Rightarrow$ direction) and it does not require the first countability. When showing the other direction, I tried to show the contrapositive of the proposition using proof by contradiction.
$$
\forall x \in X:(\exists(x_i)_{i\in\nat} \subseteq X: (\lim_{i \rightarrow \infty}x_i = x \land \forall j \in \nat:\exists k \geq j:x_k \not\in A) \Rightarrow x \not\in A^{\circ})
$$
Unlike the closure case, this form does not require the construction of the sequence because we can just assume it exists. If we suppose $x \in A^\circ$, we end up with a contradiction due to $\forall j \in \nat:\exists k \geq j:x_k \not\in A$.

Solved: I mistook the direction of the contrapositive of the necessity ($\Leftarrow$), which means I have to actually construct a sequence.
$$
\forall x \in X:(x \not\in A^{\circ} \Rightarrow \exists(x_i)_{i\in\nat} \subseteq X: (\lim_{i \rightarrow \infty}x_i = x \land \forall j \in \nat:\exists k \geq j:x_k \not\in A))
$$

答:
Here is an example showing that it can fail if $X$ is not first countable.

Let $X$ be an uncountable set, and let $p\in X$. Let

$$\tau=\{U\subseteq X:p\notin U\text{ or }X\setminus U\text{ is countable}\}\,;$$

$\tau$ is a topology on $X$ making $X$ a Lindelöf Hausdorff space. This topology is not first countable at $p$. (This is a good exercise.) Let $A=\{p\}$. Then $A$ is not open in $X$, but a sequence $\langle x_n:n\in\Bbb N\rangle$ in $X$ converges to $p$ if and only if there is an $m\in\Bbb N$ such that $x_n=p$ for all $n\ge m$, i.e., if and only if the sequence is eventually in $A$.


问:
for any polynomial $f(t)$,we have that $f(A)$ is the matrix representation of $f(T)$.
Let $A$ be the matrix representation of a linear transformation $T$ from $V$ to $V$. Prove that, for any polynomial $f(t)$,we have that $f(A)$ is the matrix representation of $f(T)$.

How can I start proving such a statement? Can I prove it without using induction.

Thanks.

答:
The short answer to your question seems to be:

Let $v \in V$.  Show that $f(A)(v) = f(T)(v)$.  Conclude that $f(A)$ and $f(T)$ are the same function.  Observe that $f(A)$ is a matrix, so $f(A)$ is the/a matrix representation of $f(T)$.  (Whether "the" or "a" depends on other facts you may or may not have.  If you don't have uniqueness of representation by some other means, you need to show uniqueness as well.)

ContentLicense: CC BY-SA 4.0


问:
Formulas's deduction from the generating function of Hermite polynomials
In the book "Essential Mathematical Methods for Physicists" comes the following problem that I am trying to solve:

at first I could see that the first formula that is given as an answer is wrong since the hermite polynomial $ H_4 (x) $ does not match with the known since:

$\begin{align*}
H_{2n}(x)&=(-1)^{n}\sum_{s=0}^{n}(-1)^{2s}(2x)^{2s}\cfrac{(2n)!}{(2s)!(n-s)!}\\
H_{2n}(x)&=(-1)^{n}\sum_{s=0}^{n}\left[(-1)^{2}\right]^s\left[(2x)^{2}\right]^s\cfrac{(2n)!}{(2s)!(n-s)!}\\
H_{2n}(x)&=(-1)^{n}\sum_{s=0}^{n}\left[1\right]^s\left[4x^2\right]^s\cfrac{(2n)!}{(2s)!(n-s)!}\\
H_{2n}(x)&=(-1)^{n}\sum_{s=0}^{n}\left[4x^2\right]^s\cfrac{(2n)!}{(2s)!(n-s)!}\\
H_{2(2)}(x)&=(-1)^{2}\sum_{s=0}^{2}\left(4x^2\right)^s\cfrac{(2\cdot 2)!}{(2s)!(2-s)!}\\
H_{4}(x)&=(1)\sum_{s=0}^{2}\left(4x^2\right)^s\cfrac{(4)!}{(2s)!(2-s)!}\\
H_{4}(x)&=\sum_{s=0}^{2}\left(4x^2\right)^s\cfrac{24}{(2s)!(2-s)!}\\
H_{4}(x)&=(4x^2)^{0}\cfrac{24}{(2\cdot 0)!(2-0)!}+(4x^2)^{1}\cfrac{24}{(2\cdot 1)!(2-1)!}+(4x^2)^{2}\cfrac{24}{(2\cdot 2)!(2-2)!}\\
H_{4}(x)&=(1)\cfrac{24}{(0)!(2)!}+(4x^2)\cfrac{24}{(2)!(1)!}+(16x^4)\cfrac{24}{(4)!(0)!}\\
H_{4}(x)&=(1)\cfrac{24}{(1)(2)}+(4x^2)\cfrac{24}{(2)(1)}+(16x^4)\cfrac{24}{(24)(1)}\\
H_{4}(x)&=12+(4x^2)(12)+(16x^4)\\
H_{4}(x)&=12+48x^2+16x^4\\
H_{4}(x)&=16x^4+48x^2+12 \\
\end{align*}$

But the real $H_4(x)$ is $H_4(x)=16x^4-48x^2+12$ so the $H_{2n}$ formula is wrong.
But trying to find errors in the formula for $ H_ {2n + 1} $ I did not find any for the polynomials 1,3,5 so I think that this formula is correct.

If the generating function of the hermite polynomials is $g (x, t) = e^{-t^2 + 2tx} = \sum_{n = 0} ^ {\infty}H_n (x) \cfrac{t ^ n}{n!} $

How can the formula for $H_{2n + 1}$ be derived?

I tried using the formula $H_n(x)=\sum_{s=0}^{[n/2]}(-1)^s\cfrac{n!}{(n-2s)!s!}(2x)^{n-2s}$ (
which is just the equation 13.40 that mentions the problem) substituting in the value of 2n + 1 but couldn't deduce anything, as there were terms that I couldn't adjust to look like the one the book asks for an answer. Any help is really appreciated!

答:
For $H_{2n}(x)$, the upper limit of their equation 13.40 is $\lceil (2n)/2\rceil=n$ and so the sum is $$H_{2n}(x)=\sum_{s=0}^n (-1)^s \frac{(2n)!}{(2n-2s)!s!}(2x)^{2n-2s}.$$ This is a descending sum, since it starts with highest power $(2x)^{2n}$ and terminates with $(2x)^0=1$. To make it ascending, we relabel $s\to n-s$ and so obtain

$$H_{2n}(x)=\sum_{s=0}^n (-1)^{n-s} \frac{(2n)!}{(2s)!(n-s)!}(2x)^{2s}.$$ This disagrees with the stated expansion in that it has $(-1)^s$ instead of $(-1)^{2s}$. Given that the latter would equal $1$ for all integers, though, I'm inclined to chalk that up to a typo. (By contrast, the case for odd $n$ does seem correct.)


问:
Is it true that primitive idempotents of $kS_n \otimes_k kS_m$ are just tensor of idempotents of $kS_n$ and $kS_m$?
I was working over a general field $k$ of characteristic $0$. For now, just think $k$ as the field of complex number.

Let $n$ and $m$ be some natural numbers. Let $kS_n$ be the group algebra of $S_n$, the symmetric group over $n$ objects. Let $A := kS_n \otimes_k kS_m$ be the tensor product of $k$-space equipped with a pointwise multiplication, i.e.
$$(a \otimes b)( c \otimes d) = ac \otimes bd.$$
Now, $A$ is a $k$-algebra.

My question is: is it true that any primitive idempotent of $kS_n \otimes_k kS_m$ is of the form $e_1 \otimes e_2$ where  $e_1$ and $e_2$ are primitive idempotents of $kS_n$ and $kS_m$ respectively?

Some work I have done: for the case when $n = 2 $ and $m=2$, my computation shows that it is true. Let $S_2 = \{1, u\}$ and
$$v = a(1 \otimes 1) + b(1 \otimes u) + c(u \otimes 1) +d(u \otimes u)$$
where $a,b,c,d \in k$. You get all the four primitive idempotents of $A$ by setting $(a,b,c,d)$ to $(x,x,x,x),(x,x,-x,-x),(x,-x,x,-x)$ and $(x,-x,-x,x)$ where $x = \frac{1}{4}$. I proved the completeness by computing out all the idempotents in $A$ which is a long process involving solving a system of equations. Note that all the four idempotents are precisely the tensor product of primitive idempotents of $kS_2$, which are either $\frac{1+u}{2}$ or $\frac{1-u}{2}$. So it is true for this case. I did not try larger case because it turns extreamly hard to determing all idempotents of $A$ as $n$ and $m$ grows larger.

答:
This is false for all $n, m \ge 3$.

As a general fact, the irreducible representations of $S_n$ over any field of characteristic $0$ are in fact already defined over $\mathbb{Q}$. This implies that $k[S_n]$ is a finite direct product of matrix algebras $M_{d_i}(k)$ over $k$ (rather than over division rings over $k$). Since $M_{d_1}(k) \otimes M_{d_2}(k) \cong M_{d_1 d_2}(k)$ the same is true for $k[S_n] \otimes k[S_m]$.

Hence an idempotent in $k[S_n] \otimes k[S_m]$ is exactly a choice of idempotent in each simple factor $M_{d_1 d_2}(k)$, and in particular is primitive iff it is a primitive idempotent in exactly one such simple factor. So the problem reduces to the corresponding problem for matrix algebras, namely:

Is a primitive idempotent in $M_{d_1 d_2}(k)$ the tensor product of a primitive idempotent in $M_{d_1}(k)$ and a primitive idempotent in $M_{d_2}(k)$?

The answer to this question is no. An idempotent $e \in M_d(k)$ is primitive iff it has rank $1$, meaning that its image $\text{im}(e) \subseteq k^d$ is a line (a $1$-dimensional subspace). An idempotent $e \in M_{d_1 d_2}(k)$ which is the tensor product of an idempotent $e_1 \in M_{d_1}(k)$ and an idempotent $e_2 \in M_{d_2}(k)$ has the property that its image $\text{im}(e)$ is a tensor product

$$\text{im}(e_1) \otimes \text{im}(e_2) \subseteq k^{d_1} \otimes k^{d_2} \cong k^{d_1 d_2}$$

of the images $\text{im}(e_1)$ and $\text{im}(e_2)$. If $d_1, d_2 \ge 2$ then there are lines in $k^{d_1 d_2}$ which are not tensor products of lines in $k^{d_1}$ and $k^{d_2}$ (in fact "almost all" lines, or the "generic" line, have this property), so a primitive idempotent whose image is such a line cannot be a tensor product. We can be quite explicit about this for $d_1 = d_2 = 2$: a tensor product line in $k^4$ is spanned by a vector of the form

$$(x_1, y_1) \otimes (x_2, y_2) = (x_1 x_2, y_1 x_2, x_1 y_2, y_1 y_2)$$

and the four components $(a, b, c, d)$ of such a vector must satisfy $ad - bc = 0$. So any vector $(a, b, c, d) \in k^4$ such that $ad - bc \neq 0$ spans a line which is not a tensor product. There's a more general version of this criterion involving vanishing of the $2 \times 2$ minors of an element of $k^{d_1 d_2}$ regarded as a matrix in $M_{d_1 \times d_2}(k)$.

Since $S_n$ is nonabelian for $n \ge 3$, $k[S_n]$ contains a simple factor $M_d(k)$ where $d \ge 2$ and so this argument applies.

However, the answer is yes if instead of primitive idempotents we consider centrally primitive idempotents, since $M_{d_1 d_2}(k)$ has a unique centrally primitive idempotent, namely the identity, which is the tensor product of the identities of $M_{d_1}(k)$ and $M_{d_2}(k)$. This corresponds to the statement that every irreducible representation of $S_n \times S_m$ is the tensor product of an irreducible representation of $S_n$ and an irreducible representation of $S_m$.


问:
Inverse of Trefoil Knot Paramaterization
I'm working on a coding project at the moment where I have to integrate a function around a trefoil knot. I'm using the following parametrization to do so.

$x = \sin(t) + 2\sin(2t)$

$y = \cos(t) - 2 \cos(2t)$

$z = -\sin(3t)$

which is copied from wikipedia. When I actually integrate my function in code I want to make sure that I don't go ahead and pass a point on the knot to the integration routine. If I do that it blows up. I'm trying to integrate the electric potential around a charged knot. It has the following general form

$$\varphi=\int_0^{2\pi} \frac{|r'(t)|}{|x-r(t)|}\,dt$$
where the denominator of the fraction is the distance from a point to the knot itself. You can see that if the point is a position on the knot itself you get $0$ in the denominator and the function explodes.

To test if a point is on the knot I thought I could go ahead and find the inverses of my three functions. Then I'd just do a bit of linear algebra to see if there was a single solution for my point with $x,y$ and $z$ and I could reject the point if there was. However, while finding the inverse for $z$ is pretty trivial, for $x$ and $y$ I'm a little stumped. After noodling with it for awhile I gave up and used Wolfram, and in particular got a super gnarly response for $x$. Am I making some basic mistake in my trig identities that would simplify these? Or is there no simple way to deal with this?
The function itself is the following:

$$\phi=\int_0^{2\pi} \frac{\sqrt{8cos(3t)+4.5cos(6t)+21.5}}{\sqrt{(x-(sin(t) + 2sin(2t)))^2+(y-(cos(t) - 2 cos(2t)))^2+(z-(-sin(3t)))^2)}}dt$$

答:
I guess I don't understand your problem.  You plot this function and it looks like the knot:

Then you take your function (which you haven't specified) and integrate along this knot.


问:
All possible tournament pairing such that you get no pair from the same group.
I thought about this problem for a while, but I have no idea how to approach it.

You have 8 groups, with 4 of the groups having 6 people and rest of the 4 groups having 3 people. So you have 36 people in total.

Now we want to pick 18 pairs from 36 people to form a tournament.

I believe there are $\frac{36!}{18! 2^{18}}$(I don't really understand how to get this number though) as can be seen here: Number of ways you can form pairs with a group of people when certain people cannot be paired with each other.

Now, I want pairings to be such that no people from the same group play against each other. How many possible pairings exist under this constraint?

This is a very similar question: UEFA Champions League quarterfinals 2018 draw - pairing of same country teams

However, I don't think the approach there would work.

Thanks!

EDIT: The most general form of this question would be to let the number of groups and number of people in each group vary, and to find the formula for this. I am now wondering if such a formula exists. So for example, what if you have 11 groups, and 4 of them have 5 people, 5 of them have 4 people, and 2 of them have 12 people.

EDIT:

I ran some simulation, I keep getting about 0.11 instead of Henry's 0.245. Here is my code.

```<code>team_list = c(rep(1:6, 4), rep(1:3,4))

for (i in 1:6){
  team_list[i] = paste("A", team_list[i], sep = "")
}

for (i in 7:12){
  team_list[i] = paste("B", team_list[i], sep = "")
}

for (i in 13:18){
  team_list[i] = paste("C", team_list[i], sep = "")
}

for (i in 19:24){
  team_list[i] = paste("D", team_list[i], sep = "")
}

for (i in 25:27){
  team_list[i] = paste("E", team_list[i], sep = "")
}

for (i in 28:30){
  team_list[i] = paste("F", team_list[i], sep = "")
}

for (i in 31:33){
  team_list[i] = paste("G", team_list[i], sep = "")
}

for (i in 34:36){
  team_list[i] = paste("H", team_list[i], sep = "")
}

check_pair = function(x){
  for (i in seq(from = 1, to = length(x), by = 2)){
    if (substr(x[i],1,1) == substr(x[i+1],1,1)){
      return (TRUE)
    }
  }
  return (FALSE)
}

count = 0

for (i in 1:10000){
  x = sample(team_list, size = 36)
  if (!check_pair(x)){
    count = count+1
  }
}

count/10000

team_list = c("A1", "A2", "B1", "B2", "C1", "C2")

pair_combn <- function(x) {
  Filter(function(e) all(unique(x) %in% unlist(e)),
         combn(as.data.frame(combn(x, 2)),
               length(x)/2, simplify = FALSE))
}

pair_combn(team_list)

check_pair = function(x){
  for (i in seq(from = 1, to = length(x), by = 2)){
    if (substr(x[i],1,1) == substr(x[i+1],1,1)){
      return (TRUE)
    }
  }
  return (FALSE)
}

count = 0

for (i in 1:10000){
  x = sample(team_list, size = 6)
  if (!check_pair(x)){
    count = count+1
  }
}

count/10000

team_list = c("A1", "A2", "B1", "B2", "C1", "D1")

pair_combn <- function(x) {
  Filter(function(e) all(unique(x) %in% unlist(e)),
         combn(as.data.frame(combn(x, 2)),
               length(x)/2, simplify = FALSE))
}

pair_combn(team_list)

check_pair = function(x){
  for (i in seq(from = 1, to = length(x), by = 2)){
    if (substr(x[i],1,1) == substr(x[i+1],1,1)){
      return (TRUE)
    }
  }
  return (FALSE)
}

count = 0

for (i in 1:10000){
  x = sample(team_list, size = 6)
  if (!check_pair(x)){
    count = count+1
  }
}

count/10000

z = pair_combn(team_list)

team_list = c("A1", "A2", "B1", "B2", "C1", "D1", "E1", "E2")

pair_combn <- function(x) {
  Filter(function(e) all(unique(x) %in% unlist(e)),
         combn(as.data.frame(combn(x, 2)),
               length(x)/2, simplify = FALSE))
}

combination = pair_combn(team_list)

check_pair = function(x){
  for (i in seq(from = 1, to = length(x), by = 2)){
    if (substr(x[i],1,1) == substr(x[i+1],1,1)){
      return (TRUE)
    }
  }
  return (FALSE)
}

count = 0
for (i in 1:105){
  to_check = as.vector(unlist(combination[[i]]))
  if (!check_pair(to_check)){
    count = count+1
  }
}

print (count)

count = 0

for (i in 1:10000){
  x = sample(team_list, size = 8)
  if (!check_pair(x)){
    count = count+1
  }
}

count/10000

team_list = c("A1", "A2", "A3", "A4", "B1", "B2", "C1", "C2")

pair_combn <- function(x) {
  Filter(function(e) all(unique(x) %in% unlist(e)),
         combn(as.data.frame(combn(x, 2)),
               length(x)/2, simplify = FALSE))
}

combination = pair_combn(team_list)

check_pair = function(x){
  for (i in seq(from = 1, to = length(x), by = 2)){
    if (substr(x[i],1,1) == substr(x[i+1],1,1)){
      return (TRUE)
    }
  }
  return (FALSE)
}

count = 0
for (i in 1:105){
  to_check = as.vector(unlist(combination[[i]]))
  if (!check_pair(to_check)){
    count = count+1
  }
}

print (count)

count = 0

for (i in 1:10000){
  x = sample(team_list, size = 8)
  if (!check_pair(x)){
    count = count+1
  }
}

count/10000

team_list = c("A1", "A2", "A3", "B1", "B2", "B3", "C1", "C2")

pair_combn <- function(x) {
  Filter(function(e) all(unique(x) %in% unlist(e)),
         combn(as.data.frame(combn(x, 2)),
               length(x)/2, simplify = FALSE))
}

combination = pair_combn(team_list)

check_pair = function(x){
  for (i in seq(from = 1, to = length(x), by = 2)){
    if (substr(x[i],1,1) == substr(x[i+1],1,1)){
      return (TRUE)
    }
  }
  return (FALSE)
}

count = 0
for (i in 1:105){
  to_check = as.vector(unlist(combination[[i]]))
  if (!check_pair(to_check)){
    count = count+1
  }
}

print (count)

count = 0

for (i in 1:10000){
  x = sample(team_list, size = 8)
  if (!check_pair(x)){
    count = count+1
  }
}

count/10000

</code>```

And some results I get:

For 3 group of 4 people, 2 people, and 2 people, I get 24 out of 105

For 3 group of 3 people, 3 people and 2 people, I get 36 out of 105

For 5 group of 2 people, 2 people, 2 people, 1 person and 1 person, I get 68 out of 105.

答:
The number is 24855678464505984000.

Suppose we have $k$ different groups, of sizes $N_1, N_2 ... N_k$. Define $F(N_1, N_2, ... N_k)$ to be the number of possible tournaments. So the answer to your particular problem is $F(3, 3, 3, 3, 6, 6, 6, 6)$.

How to compute $F$? We can come up with a recurrence relation, and hopefully a computer should be compute it. Here's the recurrence relation:

$$
F(N_1...N_k) = \frac{2}{\sum_l N_l}\sum_i\sum_{j < i} N_j \times N_i \times F(N_1, N_2\dots N_j-1 \dots N_i-1 \dots N_k)
$$

The idea is that we choose a pair (from different groups), then figure out the subproblem with that pair removed. The factor $2 / \sum_l N_l$ comes from the fact that we can choose any of the pairs to be the first one, which would lead to over-counting without dividing by the number of pairs.

For the base cases, we have $F(0, 0, \dots 0) = 1$, and $F=0$ if any of its arguments are 0.

I used the following code, which takes about a minute to run.

```<code>from functools import lru_cache

@lru_cache(maxsize = 1000000)
def F(M, ntup, k):
    if M < 0: return 0
    for n in ntup:
        if n < 0: return 0
    if M == 0:
        return 1
    ans = 0
    for i in range(1, k):
        for j in range(0, i):
            ans += ntup[i] * ntup[j] * F(M-2, ntup[:j] + (ntup[j]-1,) + ntup[j+1:i] + (ntup[i]-1,) + (ntup[i+1:] if i+1 < k else ()), k)
    return (2 * ans) // M

print(F(36, (3, 3, 3, 3, 6, 6, 6, 6), 8))
</code>```

This prints 24855678464505984000. That means the probability of finding a successful tournament (meaning no pairs from the same group) by randomly sampling from all possible pairings is about 0.11, as expected.

Body:
The number is 24855678464505984000.

Suppose we have $k$ different groups, of sizes $N_1, N_2 ... N_k$. Define $F(N_1, N_2, ... N_k)$ to be the number of possible tournaments. So the answer to your particular problem is $F(3, 3, 3, 3, 6, 6, 6, 6)$.

How to compute $F$? We can come up with a recurrence relation, and hopefully a computer should be compute it. Here's the recurrence relation:

$$
F(N_1...N_k) = \frac{2}{\sum_l N_l}\sum_i\sum_{j < i} N_j \times N_i \times F(N_1, N_2\dots N_j-1 \dots N_i-1 \dots N_k)
$$

The idea is that we choose a pair (from different groups), then figure out the subproblem with that pair removed. The factor $2 / \sum_l N_l$ comes from the fact that we can choose any of the pairs to be the first one, which would lead to over-counting without dividing by the number of pairs.

For the base cases, we have $F(0, 0, \dots 0) = 1$, and $F=0$ if any of its arguments are 0.

I used the following code, which takes about a minute to run.

from functools import lru_cache

@lru_cache(maxsize = 1000000)
def F(M, ntup, k):
    if M < 0: return 0
    for n in ntup:
        if n < 0: return 0
    if M == 0:
        return 1
    ans = 0
    for i in range(1, k):
        for j in range(0, i):
            ans += ntup[i] * ntup[j] * F(M-2, ntup[:j] + (ntup[j]-1,) + ntup[j+1:i] + (ntup[i]-1,) + (ntup[i+1:] if i+1 < k else ()), k)
    return (2 * ans) // M

print(F(36, (3, 3, 3, 3, 6, 6, 6, 6), 8))

This prints 24855678464505984000. That means the probability of finding a successful tournament (meaning no pairs from the same group) by randomly sampling from all possible pairings is about 0.11, as expected.


问:
Proving an equality involving harmonic numbers
I am working through the book (Almost) Impossible Integrals, Sums, and Series, and I am on the following problem:

Let $n$ be a positive integer. Prove that the following equality holds.
$$K_n = \int_0^1 x^{n-1} \log^3 (1-x) \, dx = -\frac{3}{n} \sum_{k=1}^n \frac{H_k^2 + H_k^{(2)}}{k} = - \frac{H_n^3 + 3H_n H_n^{(2)} + 2H_n^{(3)}}{n},$$ where $H_n^{(m)} = 1 + \frac{1}{2^m} + \cdots + \frac{1}{n^m}, \, m \geq 1,$ is the $n$th generalized harmonic number of order $m$.

I was able to evaluate the integral, but I was unable to evaluate the resulting series. Here is what I have so far:

Using Abel’s summation, we get that

$$\begin{align} \sum_{k=1}^{n} \frac{H_k^2 + H_k^{(2)}}{k} &= H_n(H_{n+1}^2 + H_{n+1}^{(2)}) + \sum_{k=1}^{n}H_k \left(H_k^2 + H_k^{(2)} - H_{k+1}^2 - H_{k+1}^{(2)} \right) \tag{1} \\ &= H_n(H_{n+1}^2 + H_{n+1}^{(2)}) + \sum_{k=1}^{n}H_k \left(H_k^2  - H_{k+1}^2 - \frac{1}{(k+1)^2} \right) \tag{2} \end{align} $$

I got stuck here so I went and looked at the author’s solution. In the author’s solution, he skips right over (1) and (2), and writes

$$\sum_{k=1}^{n} \frac{H_k^2 + H_k^{(2)}}{k} = H_n(H_{n+1}^2 + H_{n+1}^{(2)}) - 2 \sum_{k=1}^n \left( \frac{H_k^2}{k+1} + \frac{H_k}{(k+1)^2} \right) \tag{3},$$

but I do not see how does this follows. How does one get from (2) to (3)?

答: Their simplification amounts to $$H_k^2-H_{k+1}^2-\frac{1}{(k+1)^2}=-\frac{2H_k}{k+1}-\frac{2}{(k+1)^2}.$$ Now recall that $H_{k+1}=H_k+\dfrac{1}{k+1}$.


问:
Is there an encyclopedia of mathematical notation?
For example - for set complements, it would include all common notations such as 'A - B' or 'A \ B'</p>

答:
https://en.wikipedia.org/wiki/List_of_mathematical_symbols
You can most probably get all mathematical symbols here.</p>

ContentLicense: CC BY-SA 3.0


问:
Let $P$ and $Q$ be prime ideals of a ring $R$ such that every element of $R\setminus (P\cup Q)$ is a unit. Prove that either $P$ or $Q$ is maximal.

Let $R$ be a commutative ring with a unit. Let $P$ and $Q$ be prime ideals of $R$ such that every element of $R\setminus( P\cup Q)$ is a unit. We want to show that either $P$ or $Q$ is maximal.

I am not really sure what to try here. I can see that if $a$ is an element of a maximal ideal which contains $P$, but $a\notin P$, then $a\in Q$, and vice versa, but I am not sure how that helps me. I have mostly tried to come up with contradictions by supposing that neither $P$ nor $Q$ is maximal, then using things like $R/P$ and $R/Q$ are both integral domains with noninvertible elements, but I haven't had any luck. Any hints on how to proceed would be appreciated (for reference, I haven't taken a full commutative algebra course yet).

答:
By Krull's theorem, there exists $M$ a maximal ideal of $R$. Since $M$ is proper and hence no element of $M$ is a unit, we must have $M\subseteq P\cup Q$. Now, suppose for contradiction that $M\nsubseteq P$ and $M\nsubseteq Q$. Then there is some $m_p\in M\setminus P$ and $m_q\in M\setminus Q$, and we have $m:=m_p+m_q\in M$. Since $M\subseteq P\cup Q$, this means $m\in P\cup Q$, and so either $m\in P$ or $m\in Q$; without loss of generality suppose $m\in P$. Then, since $m_q\in M\subseteq P\cup Q$, and $m_q\notin Q$, we must have $m_q\in P$. But this means that $$m_p=m-m_q\in P,$$ a contradiction. Thus we must have $M\subseteq P$ or $M\subseteq Q$, and so – by maximality of $M$, and since $P$ and $Q$ are prime and hence proper – either $M=P$ or $M=Q$, as desired.

You do not need $P$ and $Q$ to be prime here, just proper, and the argument above shows a more general result: if $R$ is any ring, and $I$, $J_1$, and $J_2$ are any ideals of $R$ with $I\subseteq J_1\cup J_2$, then either $I\subseteq J_1$ or $I\subseteq J_2$. We can extend this result even further by reintroducing the primality hypothesis; if $P_1,\dots,P_n$ are prime ideals of $R$, and $I\subseteq P_1\cup\dots\cup P_n$, then there is some $i$ such that $I\subseteq P_i$. Exercise: try to prove this by induction! (Note, however, that the primality hypothesis really is necessary. For instance, $$(\overline{x},\overline{y})=(\overline{x})\cup (\overline{x+y})\cup (\overline{y})$$ in $\mathbb{F}_2[x,y]\big/(x^2,xy,y^2)$, but $(\overline{x},\overline{y})$ is not contained in any of the ideals on the right.)


答: Firsly I supposed that the ideals are nontrivial. Let $P' \supset P$ and $Q' \supset Q$ be ideals. Obviously, if $P'$ or $Q'$ contains an element in $R \setminus P \cup Q$, then one of them must be $R$. Assume the other case, which is $P' \subset P \cup Q$ and $Q' \subset P \cup Q$. $P'$ contains $q \in Q \setminus P$ and $Q'$ contains $p \in P \setminus Q$. Note that $P'$ and $Q'$ contain both of $p$ and $q$, so their all combinations. Look at the element $p-q$. This is not element of $P$ and $Q$, by using the fact that ideals are additive subgroups. So $p-q \notin P \cup Q$, implies that $p-q$ is unit. Both of $P'$ and $Q'$ become $R$. This shows $P$ and $Q$ are maximals. I wonder that I did not use being prime.

Title: Let $P$ and $Q$ be prime ideals of a ring $R$ such that every element of $R\setminus (P\cup Q)$ is a unit. Prove that either $P$ or $Q$ is maximal.

Body:
Let $R$ be a commutative ring with a unit. Let $P$ and $Q$ be prime ideals of $R$ such that every element of $R\setminus( P\cup Q)$ is a unit. We want to show that either $P$ or $Q$ is maximal.

I am not really sure what to try here. I can see that if $a$ is an element of a maximal ideal which contains $P$, but $a\notin P$, then $a\in Q$, and vice versa, but I am not sure how that helps me. I have mostly tried to come up with contradictions by supposing that neither $P$ nor $Q$ is maximal, then using things like $R/P$ and $R/Q$ are both integral domains with noninvertible elements, but I haven't had any luck. Any hints on how to proceed would be appreciated (for reference, I haven't taken a full commutative algebra course yet).


问:
Alternative to Heron's Formula , through Proof by Pythagorean Theorem.
(This works for all triangles)

The area ($A$) of a right triangle, can be  given by the inner radius ($r$) multiplied by the semi perimeter ($s$), where $r = \frac12(a + b- c)$ and $s = \frac12(a + b + c)$.

$$A = \tfrac14\left(a+ b - c\right)\left(a + b + c\right)$$

Proof through Pythagorean Theorem:

$$\begin{align}
\tfrac12ab &= \tfrac14\left(a+ b - c)(a + b + c\right) \\[8pt]
2ab &= (a+ b - c)(a + b + c)\\[4pt]
&= \left(a+ b)^2 - c^2\right)\\[4pt]
&= \left(a^2 + 2ab + b^2 - c^2\right)\\[4pt]
0 &= a^2 + b^2 - c^2\\[4pt]
c^2 &= a^2 + b^2
\end{align}$$

答:
If we are dealing with Pythagorean triples it pays to look at Euclid' formula, shown here as

$$A=m^2-k^2\qquad B=2mk\qquad C=m^2+k^2$$

Heron's formula is great for finding area of an arbitrary triangle, but there is no need for it if we are dealing with Pythagorean triples. Instead, we can straight to the area/perimeter ratio. These occur in multiples of 1/2 starting with triangle $(3,4,5)$. To find triples with any such ration (here R-1), we may begin by solving equations for $k$ and testing a define range of m-values to see which, if any, yield integers.

$$\text{Using $F(m,k)$: }\qquad R=\frac{area}{perimeter}=\frac{AB/2P}=\frac{2mk(m^2-k^2)}{2(2m^2+2mk)}=\frac{mk-k^2}{2}
\qquad\qquad\qquad\qquad\qquad $$

\begin{equation}
R=\frac{mk-k^2}{2}\quad\implies k=\frac{m\pm\sqrt{m^2-8R}}{2}\\
\text{for}\quad \big\lceil\sqrt{8R}\big\rceil\le m \le (2R+1)
\end{equation}
The lower limit insures that $k\in\mathbb{N}$. The upper limit ensures that $m> k$.
$$R=1\implies  \lceil\sqrt{8}\rceil=3\le m \le (2+1)=3 \qquad\land\qquad   m\in\{ 3\}\implies k\in\{ 2,1\}$$
$$F(3,2)=(5,12,13)\quad\land\quad \frac{30}{30}=1\qquad\qquad\qquad F(3,1)=(8,6,10)\quad\land\quad \frac{24}{24}=1$$

It appears that $R=1$ for perfect triangles but any ratio can be found this way. for example, $R=1.5$ yields
\begin{align*}
R=1.5 &\rightarrow \space (7,24,25)\quad\space\space (15,8,17)\\
R=2.0 &\rightarrow  \space (9,40,41)\quad\space (12,16,20)\quad (24,10,26)\\
R=2.5 &\rightarrow  (11,60,61)\quad (35,12,37)\\
R=3.0 &\rightarrow    (13,84,85)\quad (16,30,34)\quad (21,20,29)\quad (48,14,50)
\end{align*}


问:
Evaluating a triple summation ${\sum \sum \sum}_{1 \le i < j \le k \le 20} a_ia_ja_k$

If $$a_i= \begin{cases} 1 && 1 \le i \le 10 \\ 2 && 11 \le i \le 20 \end{cases}  $$

Find the value of $$\displaystyle {\sum \sum \sum}_{1 \le i < j \le k \le 20} a_ia_ja_k $$

I could only evaluate the summation when $1 \le i < j \le k \le 10$, by writing it as $\displaystyle {\sum _{i=1}^9\sum_{j=i+1}^{10} \sum_{k=j}^{10}} 1$ which came out to be $\displaystyle \binom{11}{3}$.

Should I make different cases like $i \le 10, 11 \le j \le k \le 20$ , etc.?

How do I evaluate it for the other cases? Is there any easier way to evaluate it without evaluating each sum one by one? Could there possibly be a combinatorial approach? I'd be delighted to see different approaches to this problem.

Body:
Let

$ A = \sum_{i < j < k } a_ia_ja_k$

$ B = \sum_{i = j < k } a_ia_ja_k$

$ C = \sum_{i < j = k } a_ia_ja_k$

$ D = \sum_{i = j = k } a_ia_ja_k$

We're after $ A  + C$.

Clearly,

$ 30^3 = ( \sum a_i)^3  = D + 3C + 3B + 6A$.

$ D = 10 ( 1^3 + 2^3 ) = 90$

$C = 2^2 ( 10 \times 10\times 1 + \frac{10 \times 9 }{ 2} \times 2 ) + 1^2 (\frac{10 \times 9 } { 2} \times 1) = 805$

$ B = 1^2 (10\times 10 \times 2 + \frac{10 \times 9} { 2} \times 1) + 2^2 (\frac{10 \times 9 }{ 2} \times 2)  = 605$

So, $A = \frac{30^3 - 90 - 3 \times 805 - 3 \times 605 }{ 6} = 3780$.

Hence, $ A + C = 4585$.


问:
What do cones have to do with quadratics? Why is $2$ special?
I've always been nagged about the two extremely non-obviously related definitions of conic sections (i.e. it seems so mysterious/magical that somehow slices of a cone are related to degree 2 equations in 2 variables). Recently I came across the following pages/videos:

This 3B1B video about ellipses, which reignited my desire to understand conics

Why are quadratic equations the same as right circular conic sections?, which offers a very computational approach to trying to resolve this question

Another 3B1B video on visualizing Pythagorean triples (i.e. finding the rational points of a circle)

and Manjul Bhargava's lecture on the Birch-Swinnerton-Dyer Conjecture, where minutes ~10-15 discuss the complete solution to the problems of rational points on conics.

While 3B1B's video makes a lot of sense and is very beautiful from a geometric standpoint, it does not talk about any of the other conics, or discuss the relationship with "degree 2". Moreover, the 2nd 3B1B video I linked and then Bhargava's lecture highlights "degree 2" as something we understand well, compared to higher degrees (reminds me a little bit of Fermat's last theorem and the non-existence of solutions for $n>2$).

So, I suppose my questions are as follows:

Why, from an intuitive standpoint, should we expect cones to be deeply related to zero-sets of degree 2 algebraic equations?

and more generally:

Is there some deep reason why "$2$" is so special? I've often heard the quip that "mathematics is about turning confusing things into linear algebra" because linear algebra is "the only subject mathematicians completely understand"; but it seems we also understand a lot of nice things about quadratics as well  -- we have the aforementioned relationship with cones, a complete understanding of rational points, and the Pythagorean theorem (oh! and I just thought of quadratic reciprocity). 2 is also special in all sorts of algebraic contexts, as well as being the only possible finite degree extension of $\mathbb R$, leading to in particular $\mathbb C$ being 2-dimensional.

Also interesting to note that many equations in physics are related to $2$ (the second derivative, or inverse square laws), though that may be a stretch. I appreciate any ideas you share!

$$\rule{5cm}{0.4pt}$$

EDIT 3/12/21: was just thinking about variances, and least squares regression. "$2$" is extremely special in these areas: Why square the difference instead of taking the absolute value in standard deviation?, Why is it so cool to square numbers (in terms of finding the standard deviation)?, and the absolutely mindblowing animation of the physical realization of PCA with Hooke's law: Making sense of principal component analysis, eigenvectors & eigenvalues.

In these links I just listed, seems like the most popular (but still not very satisfying to me) answer is that it's convenient (smooth, easy to minimize, variances sum for independent r.v.'s, etc), a fact that may be a symptom of a deeper connection with the Hilbert-space-iness of $L^2$. Also maybe something about how dealing with squares, Pythagoras gives us that minimizing reconstruction error is the same as maximizing projection variance in PCA. Honorable mentions to Qiaochu Yuan's answer about rotation invariance, and Aaron Meyerowitz's answer about the arithmetic mean being the unique minimizer of sum of squared distances from a given point. As for the incredible alignment with our intuition in the form of the animation with springs and Hooke's law that I linked, I suppose I'll chalk that one up to coincidence, or some sort of SF ;)

$$\rule{5cm}{0.4pt}$$

EDIT 2/11/22:
I was thinking about Hilbert spaces, and then wondering again why they behave so nice, namely they have the closest point lemma (leading to orthogonal decomposition $\mathcal H = \mathcal M \oplus \mathcal M^\perp$ for closed subspaces $\cal M$), or orthonormal bases (leading to Parseval's identity, convergence of a series of orthogonal elements if and only if the sum of the squared lengths converge), and I came to the conclusion that the key result each time seemed to be the Pythagorean theorem (e.g. the parallelogram law is an easy corollary of Pythag). So that begs the questions, why is the Pythagorean theorem so special? The linked article in the accepted answer of this question:  What does the Pythagorean Theorem really prove? tells us essentially the Pythagorean theorem boils down to the fact that right triangles can be subdivided into two triangles both similar to the original.

The fact that this subdivision is reached by projecting the vertex onto the hypotenuse (projection deeply related to inner products) is likely also significant... ahh, indeed by the "commutativity of projection", projecting a leg onto the hypotenuse is the same as projecting the hypotenuse onto the leg, but by orthogonality of the legs, the projection of the hypotenuse onto the leg is simply the leg itself! The square comes from the fact that projection scales proportionally to the scaling of each vector, and there are two vectors involved in the operation of projection.

I suppose this sort of "algebraic understanding" of the projection explains the importance of "2" more than the geometry, since just knowing about the "self-similarity of the subdivisions" of the right triangle, one then has to wonder why say tetrahedrons or other shapes in other dimensions don't have this "self-similarity of the subdivisions" property. However it is still not clear to me why projection seems to be so fundamentally "2-dimensional". Perhaps 1-dimensionally, there is the "objective" perception of the vector, and 2-dimensionally there is the "subjective" perception of one vector in the eyes of another, and there's just no good 3-dimensional perception for 3 vectors?

There might also be some connection between the importance of projection and the importance of the Riesz representation theorem (all linear "projections" onto a 1-dimensional subspace, i.e. linear functionals, are actually literal projections against a vector in the space).

$$\rule{5cm}{0.4pt}$$

EDIT 2/18/22: again touching on the degree 2 Diophantine equations I mentioned above, a classical example is the number of ways to write $k$ as the sum of $n$ squares $r_n(k)$. There are a number of nice results for this, the most famous being Fermat's 2-square theorem, and Jacobi's 4-square theorem. A key part of this proof was the use of the Poisson summation formula for the Euler/Jacobi theta function $\theta(\tau) := \sum_{n=-\infty}^\infty e^{i \pi n^2 \tau}$, which depends on/is heavily related to the fact that Gaussians are stable under the Fourier transform. I still don't understand intuitively why this is the case (see Intuitively, why is the Gaussian the Fourier transform of itself?), but there seems to be some relation to Holder conjugates and $L^p$ spaces (or in the Gaussian case, connections to $L^2$), since those show up in generalizations to the Hardy uncertainty principle (“completing the square”, again an algebraic nicety of squares, was used in the proof of Hardy, and the Holder conjugates may have to do with the inequality $-x^p + xu \leq u^q$ -— Problem 4.1 in Stein and Shakarchi’s Complex analysis, where the LHS basically comes from computing the Fourier transform of $e^{-x^p}$) Of course why the Gaussian itself appears everywhere is another question altogether: https://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics.

This (squares leading to decent theory of $r_n(k)$, and squares leading to nice properties of the Gaussian) is probably also connected to the fact that $\int_{\mathbb R} e^{-x^2} d x$ has a nice explicit value, namely $\sqrt \pi$. I tried seeing if there was a connection between this value of $\pi$ and the value of $\pi$ one gets from calculating the area of a circle "shell-by-shell" $\frac 1{N^2} \sum_{k=0}^N r_2(k) \to \pi$, but I couldn't find anything: Gaussian integral using Euler/Jacobi theta function and $r_2(k)$ (number of representations as sum of 2 squares).

$$\rule{5cm}{0.4pt}$$

EDIT 10/13/22: I was recently learning about Riemannian geometry, and indeed the metric tensor is a bilinear form (cf. above discussion on inner products), and the Riemann curvature tensor (or curvature in general) is all about the second (covariant) derivative. Taking traces we arrive at the Ricci curvature tensor (used in no less important things than Einstein's general relativity, whose "classical approximation" Newtonian gravity follows an inverse square law; and Perelman's proof of the Poincare conjecture!) or scalar curvature, which can be interpreted geometrically as the second-order change in volume of balls/base of cones (see https://math.stackexchange.com/a/469005/405572 or Equation (10) of https://arxiv.org/pdf/2201.04923v1.pdf).

And of course like Ricci flow we also have the heat/diffusion differential equations (and the Schrodinger equation, quoting James Gleck, is "diffusion through imaginary time"), and endless equations involving the Laplacian, all second-order differential equations (how AMAZING that the two great pillars of 20th century physics, general relativity and quantum mechanics both have at their hearts second-order differential equations! And the fact that observables in quantum mechanics are modeled as operators on Hilbert spaces). Relating to the Laplacian, we have the important concept of harmonicity, with beautiful manifestations/consequences in complex analysis, or PDEs, in the form of elliptic regularity. Besides tensors of 2nd derivatives, we have the ever-present Hessian matrix of second derivatives. I'll end with some quotes from a MO answer I linked here in a comment several months ago (https://mathoverflow.net/a/171743/112504) which tries to argue that "Polynomials are useful because quadratic polynomials are useful.":

So the question becomes: why are quadratic polynomials useful? There seem to be two different but interacting reasons. The first is that quadratic functions of a real variable are always either convex or concave and therefore have a unique maximum or minimum. The second is that quadratic functions are intimately related to bilinear forms and therefore can be accessed using linear algebra. The combination of these two reasons seems to explain the success of quadratic algebra in analysis and geometry (e.g. Hilbert spaces, Riemannian manifolds)

答:
A cone itself is a quadratic! Just in three variables rather than two. More precisely, conical surfaces are "degenerate hyperboloids," such as

$$x^2 + y^2 - z^2 = 0.$$

Taking conic sections corresponds to intersecting a cone with a plane $ax + by + cz = d$, which amounts to replacing one of the three variables with a linear combination of the other two plus a constant, which produces a quadratic in two variables. The easiest one to see is that if $z$ is replaced by a constant $r$ then we get a circle $x^2 + y^2 = r^2$ (which is how you can come up with the above equation; a cone is a shape whose slice at $z = \pm r$ is a circle of radius $r$). Similarly if $x$ or $y$ is replaced by a constant we get a hyperbola.

I don't know that I have a complete picture to present about why quadratics are so much easier to understand than cubics and so forth. Maybe the simplest thing to say is that quadratic forms are closely related to square (symmetric) matrices $M$, since they can be written $q(x) = x^T M x$. And we have lots of tools for understanding square matrices, all of which can then be brought to bear to understand quadratic forms, e.g. the spectral theorem. The corresponding objects for cubic forms is a degree $3$ tensor which is harder to analyze.

Maybe a quite silly way to say it is that $2$ is special because it's the smallest positive integer which isn't equal to $1$. So quadratics are the simplest things that aren't linear and so forth.


问:
Why is expectation undefined?
My textbook says

Consider a random variable $X$ that takes the values $2^k$ and $-2^k$ with probability $2^{-k}$, for $k = 2,3, \ldots$ . The expectation is undefined, even though the PMF is symmetric around zero and one might be tempted to say that $E[X]$ is zero.

My question: why is the expectation undefined?

My thoughts: pmf is as below, which has a center of mass at zero. So expectation should be zero.

Calculation of $E[X]:$ $$E[X]=\left(2^2\cdot \frac{1}{2^2}+2^3\cdot \frac{1}{2^3}+\cdots\right)-\left(2^2\cdot \frac{1}{2^2}+2^3\cdot \frac{1}{2^3}+\cdots\right)$$ $=0$
if we add the first term of the figure within the left side bracket with the first term in right side bracket, then second term etc..

Is this undefined because $\infty -\infty $ is undefined?

答: Infinite sums are defined only as the limit of the finite sums leading up to it.  As you say, each of your parenthesized terms is infinite.  If the infinite sum is absolutely convergent, you can commute the terms however you want and the sum will remain the same.  Absolutely convergent demands that the sum converge when all the terms are made positive.  Your sum is not absolutely convergent because if we change the central minus sign to plus the sum diverges.  If the sum is not absolutely convergent the result depends on the order we sum the terms.  You chose an order, which seems to you to be the natural one, but there is nothing special about it.  That order cancels the positive terms with the negative terms one by one, giving a result of $0$.  Another order takes two terms from the first group for every term in the second.  This sum diverges to $+\infty$.  As we cannot get a unique consistent answer, we say the sum is undefined.

Title: Why is expectation undefined?


问:
Calculus - Continuous Function Proof Help
Regarding this question:

Let $f$ be function that is continuous on $[0,1]$.

Suppose that $f(x)>0$ for every $x \in [0,1]$.

Then there exists an $\varepsilon > 0$ such that $f(x)>\varepsilon$ for every $x \in [0,1]$.

--

So I want to say: because $f$ is continuous on $[0,1]$, and $f(x)>0$ for every $x \in [0,1]$, so $f(0)>0$ and $f(1)>0$. So $f(0)f(1)>0$.

From the intermediate value theorem, there is NO such a point $c \in R$ that $0<c<1$ such that $f(c)=0$.

From here I need some hint to continue.

Thanks a lot!

答: This is immediate using the extreme value theorem.</p>


问:
How to find $ \tan \left(\frac{x}{2}\right) $ knowing that $\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $
Good evening to everyone. I don't know how to find $ \tan \left(\frac{x}{2}\right) $ knowing that $$\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $$ and x$\in (0,\frac{\pi}{3})$ Here's what I've tried:
$$\tan \left(\frac{x}{2}\right) = \frac{1-\cos \left(x\right)}{\frac{7}{5}- \cos \left(x\right)}$$ But I don't know what to do from here. Can someone explain to me how to solve this? Thanks for any answers.</p>

答:
\begin{align}
   \cos x +\sin x &= \frac 75 \\
   \cos^2 x + 2 \cos x \sin x + \sin^2 x &= \frac{49}{25} \\
   2 \cos x \sin x &= \frac{24}{25} \\
\hline
   \cos^2 x - 2 \cos x \sin x + \sin^2 x &= 1 - 2 \cos x \sin x \\
   (\cos x - \sin x)^2 &= \dfrac{1}{25} \\
   \cos x - \sin x &= \pm \frac 15 \\
\hline
   (\cos x, \sin x) &\in \left\{ \left(\frac 45, \frac 35\right),
                         \left(\frac 35, \frac 45\right) \right\} \\
\hline
   \tan \frac x2 &= \frac{\sin x}{1 + \cos x} \\
   \tan \frac x2 &\in \left\{\frac 39, \frac 48\right\} \\
   \tan \frac x2 &\in \left\{\frac 13, \frac 12\right\}
\end{align}


问:
Cotangent map is not surjective
This is Vakil 19.1 B, self-study.

We have to show that the normalization of the cusp at the tacnode:

$$\operatorname{Spec} k[t] \to \operatorname{Spec} k[x, y]/(y^2 - x^4)$$

given by $x \mapsto t^2, y \mapsto t^4$ does not have an injective tangent map at closed points.

Here is what I have so far: this is equivalent to not having a surjective contangent map at closed points. If we choose the closed point $(x, y)$ in the image, this is mapped to by $(t)$. The residue fields are both $k$, so the cotangent map makes sense.

I believe the cotangent map is then a map

$$(x, y)/(x, y)^2 \to (t)/(t)^2$$

but the problem is I do not know what this map actually does, and to be honest, I am not sure I ever did. How do I proceed from here?

答:
I think it's more intuitive to first think about the tangent map. At a point $t_0 \in \mathbb{A}^1$ it is literally given by the derivative of the polynomial function $t \mapsto (t^2, t^4)$, which means it is $(2t_0, 4t_0^3)$, and in particular it vanishes at the origin (but not at any other point).

The cotangent map is the infinitesimal version of pullback of regular functions so at the origin it sends $f(x, y) \in (x, y)/(x, y)^2$ to $f(t^2, t^4) \in t/t^2$. If $f(x, y) = ax + by + \text{higher terms}$ then $f(t^2, t^4) = at^2 + \text{higher terms}$ vanishes in $t/t^2$.

Title: Cotangent map is not surjective


问:
Why are we having factorial sign when there is already the $\Pi$ notation in use?
https://mathmaine.com/2010/04/01/sigma-and-pi-notation/
According to the above website the Pi notation is doing exactly what the factorial symbol does then what is the significance of Having two symbols which have the same function ? Please see that I am a 8th Grader</p>

答:
The short answer is that it makes everything shorter, easier to read, easier to write, and easier to think about. That's what good notation should do. We would much rather write

$${n \choose m} = \frac{n!}{m! (n-m)!}$$

than

$${n \choose m} = \frac{\left( \prod_{i=1}^n i \right)}{\left( \prod_{i=1}^m i \right) \left( \prod_{i=1}^{n-m} i \right)}$$

and similarly we would much rather write

$$e^x = \sum_{n \ge 0} \frac{x^n}{n!}$$

than

$$e^x = \sum_{n \ge 0} \frac{x^n}{\prod_{i=1}^n i}$$

or, God forbid,

$$e^x = \sum_{n \ge 0} \frac{\prod_{i=1}^n x}{\prod_{i=1}^n i}.$$

There is actually some interesting mathematics to talk about here. When assigning meanings to different words in a language you inevitably only have a limited supply of short words; how should their meanings be assigned? It's actually not hard to see that if there are some meanings you want to use very frequently, then to make your messages as short as possible, the shortest words should be assigned the meanings you want to use the most frequently, and you can prove a precise statement to this effect, related to Huffman coding. It's worth meditating on why the shortest words in English (or whatever other languages you may speak) have the meanings that they do: "eat," "fire," "dog," etc.

$n!$ is very short notation and that's appropriate because factorials are a concept we constantly invoke all the time in mathematics and we want to be able to do it as quickly as possible.


问:
Can the nonexistence of a constructive proof be proven when an existential proof exists?
Proofs are usually constructive or existential.  For example, we know there are an infinite number of primes, and therefore the centillionth prime exists.  We don't know what it is, though we can make bounds on its location.

Are there any existential proofs where it has been proven that there is NOT a constructive proof of the same fact?

答:
Certainly, there are many such situations.  For example, given a continuous function $f$ on $[0,1]$, it is known that a maximum of $f$ "exists".  However, the proof relies essentially on an argument by contradiction, whose key ingredient is the law of excluded middle.  It can be shown that such a maximum could never be constructed without reliance on the latter.  See e.g., Constructivism in Mathematics.  By A.S. Troelstra, D. van Dalen (http://www.google.co.il/books?id=-tc2qp0-2bsC&source=gbs_navlinks_s) Thus, in an intuitionistic framework, there exists a real $a$ such that $\neg ((a\leq 0)\vee (a\geq 0))$.  Now consider the function $f(x)=ax$.  If we could find a maximum of $f$, we could also pin down the sign of $a$, which is impossible by definition.

This type of construction is known as a Brouwerian counterexample in the literature.  Of course, the first theorem "refuted" in this fashion is... Brouwer's fixed point theorem :-)

The counterexample above is discussed in Troelstra-van Dalen on page 295 which unfortunately is missing from the online version.


答:
What is a cone?

It is a solid so that every cross section perpendicular to its center axis is a circle, and the radii of the these cross section circles a proportional to the the distance from the cone's vertex.

And that's it.  the surface of the cone are the points $(x,y,z)$ where $z = h= $ the height of the the cross-section $= r = $ the radius of the cross section.  And $(x,y)$ are the points of the circle with radius $r = h = z$.

As the equation of a circle is $\sqrt{x^2 +y^2} = r$ or $x^2 + y^2 = r^2$ the equation of a cone is $x^2 + y^2 = z^2$.

Every conic section is a matter intersecting the cone with a plane.  A plane is a restriction of the three variable to be related by restraint $ax +by + cz= k$ and that is a matter of expressing any third variable as a linear combination of the other two.

So the cross section of a plane and cone will be a derivation of the 2 degree equation $x^2 = y^2 = z^2$ where one of the variables will be linear combination of the other two.  In other words a second degree equation with two variables.

And that's all there is to it.

Of course the real question is why is the equation of a circle $x^2 + y^2 =r^2$? and why is that such an important representation of a second degree equation?

And that is entirely because of the Pythagorean theorem.   If we take any point $(x,y)$ on a plane and consider the three points  $(x,y), (x,0)$ and $(0,0)$ they for the three vertices of a right triangle.  The legs of this triangle are of lengths $x$ and $y$ and therefore by the Pythagorean theorem the hypotenuse will have length $\sqrt{x^2 + y^2}=h$ and that is the distance of $(x,y)$ to $(0,0)$.

Now a circle is the collection of points where the distance from $(x,y)$ to $(0,0)$ is the constant value $r = h$.  And so it will be all the points $(x,y)$ where $\sqrt{x^2 + y^2} =r$.

And that's it. That's why: distances are related to right triangles, right triangles are related to 2nd degree equations, circles are related to distances, cones are related to circles and all of them are related to 2nd degree equations.

That's it.

Title: What do cones have to do with quadratics? Why is $2$ special?

Body:
I've always been nagged about the two extremely non-obviously related definitions of conic sections (i.e. it seems so mysterious/magical that somehow slices of a cone are related to degree 2 equations in 2 variables). Recently I came across the following pages/videos:

This 3B1B video about ellipses, which reignited my desire to understand conics

Why are quadratic equations the same as right circular conic sections?, which offers a very computational approach to trying to resolve this question

Another 3B1B video on visualizing Pythagorean triples (i.e. finding the rational points of a circle)

and Manjul Bhargava's lecture on the Birch-Swinnerton-Dyer Conjecture, where minutes ~10-15 discuss the complete solution to the problems of rational points on conics.

While 3B1B's video makes a lot of sense and is very beautiful from a geometric standpoint, it does not talk about any of the other conics, or discuss the relationship with "degree 2". Moreover, the 2nd 3B1B video I linked and then Bhargava's lecture highlights "degree 2" as something we understand well, compared to higher degrees (reminds me a little bit of Fermat's last theorem and the non-existence of solutions for $n>2$).

So, I suppose my questions are as follows:

Why, from an intuitive standpoint, should we expect cones to be deeply related to zero-sets of degree 2 algebraic equations?

and more generally:

Is there some deep reason why "$2$" is so special? I've often heard the quip that "mathematics is about turning confusing things into linear algebra" because linear algebra is "the only subject mathematicians completely understand"; but it seems we also understand a lot of nice things about quadratics as well  -- we have the aforementioned relationship with cones, a complete understanding of rational points, and the Pythagorean theorem (oh! and I just thought of quadratic reciprocity). 2 is also special in all sorts of algebraic contexts, as well as being the only possible finite degree extension of $\mathbb R$, leading to in particular $\mathbb C$ being 2-dimensional.

Also interesting to note that many equations in physics are related to $2$ (the second derivative, or inverse square laws), though that may be a stretch. I appreciate any ideas you share!

$$\rule{5cm}{0.4pt}$$

EDIT 3/12/21: was just thinking about variances, and least squares regression. "$2$" is extremely special in these areas: Why square the difference instead of taking the absolute value in standard deviation?, Why is it so cool to square numbers (in terms of finding the standard deviation)?, and the absolutely mindblowing animation of the physical realization of PCA with Hooke's law: Making sense of principal component analysis, eigenvectors & eigenvalues.

In these links I just listed, seems like the most popular (but still not very satisfying to me) answer is that it's convenient (smooth, easy to minimize, variances sum for independent r.v.'s, etc), a fact that may be a symptom of a deeper connection with the Hilbert-space-iness of $L^2$. Also maybe something about how dealing with squares, Pythagoras gives us that minimizing reconstruction error is the same as maximizing projection variance in PCA. Honorable mentions to Qiaochu Yuan's answer about rotation invariance, and Aaron Meyerowitz's answer about the arithmetic mean being the unique minimizer of sum of squared distances from a given point. As for the incredible alignment with our intuition in the form of the animation with springs and Hooke's law that I linked, I suppose I'll chalk that one up to coincidence, or some sort of SF ;)

$$\rule{5cm}{0.4pt}$$

EDIT 2/11/22:
I was thinking about Hilbert spaces, and then wondering again why they behave so nice, namely they have the closest point lemma (leading to orthogonal decomposition $\mathcal H = \mathcal M \oplus \mathcal M^\perp$ for closed subspaces $\cal M$), or orthonormal bases (leading to Parseval's identity, convergence of a series of orthogonal elements if and only if the sum of the squared lengths converge), and I came to the conclusion that the key result each time seemed to be the Pythagorean theorem (e.g. the parallelogram law is an easy corollary of Pythag). So that begs the questions, why is the Pythagorean theorem so special? The linked article in the accepted answer of this question:  What does the Pythagorean Theorem really prove? tells us essentially the Pythagorean theorem boils down to the fact that right triangles can be subdivided into two triangles both similar to the original.

The fact that this subdivision is reached by projecting the vertex onto the hypotenuse (projection deeply related to inner products) is likely also significant... ahh, indeed by the "commutativity of projection", projecting a leg onto the hypotenuse is the same as projecting the hypotenuse onto the leg, but by orthogonality of the legs, the projection of the hypotenuse onto the leg is simply the leg itself! The square comes from the fact that projection scales proportionally to the scaling of each vector, and there are two vectors involved in the operation of projection.

I suppose this sort of "algebraic understanding" of the projection explains the importance of "2" more than the geometry, since just knowing about the "self-similarity of the subdivisions" of the right triangle, one then has to wonder why say tetrahedrons or other shapes in other dimensions don't have this "self-similarity of the subdivisions" property. However it is still not clear to me why projection seems to be so fundamentally "2-dimensional". Perhaps 1-dimensionally, there is the "objective" perception of the vector, and 2-dimensionally there is the "subjective" perception of one vector in the eyes of another, and there's just no good 3-dimensional perception for 3 vectors?

There might also be some connection between the importance of projection and the importance of the Riesz representation theorem (all linear "projections" onto a 1-dimensional subspace, i.e. linear functionals, are actually literal projections against a vector in the space).

$$\rule{5cm}{0.4pt}$$

EDIT 2/18/22: again touching on the degree 2 Diophantine equations I mentioned above, a classical example is the number of ways to write $k$ as the sum of $n$ squares $r_n(k)$. There are a number of nice results for this, the most famous being Fermat's 2-square theorem, and Jacobi's 4-square theorem. A key part of this proof was the use of the Poisson summation formula for the Euler/Jacobi theta function $\theta(\tau) := \sum_{n=-\infty}^\infty e^{i \pi n^2 \tau}$, which depends on/is heavily related to the fact that Gaussians are stable under the Fourier transform. I still don't understand intuitively why this is the case (see Intuitively, why is the Gaussian the Fourier transform of itself?), but there seems to be some relation to Holder conjugates and $L^p$ spaces (or in the Gaussian case, connections to $L^2$), since those show up in generalizations to the Hardy uncertainty principle (“completing the square”, again an algebraic nicety of squares, was used in the proof of Hardy, and the Holder conjugates may have to do with the inequality $-x^p + xu \leq u^q$ -— Problem 4.1 in Stein and Shakarchi’s Complex analysis, where the LHS basically comes from computing the Fourier transform of $e^{-x^p}$) Of course why the Gaussian itself appears everywhere is another question altogether: https://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics.

This (squares leading to decent theory of $r_n(k)$, and squares leading to nice properties of the Gaussian) is probably also connected to the fact that $\int_{\mathbb R} e^{-x^2} d x$ has a nice explicit value, namely $\sqrt \pi$. I tried seeing if there was a connection between this value of $\pi$ and the value of $\pi$ one gets from calculating the area of a circle "shell-by-shell" $\frac 1{N^2} \sum_{k=0}^N r_2(k) \to \pi$, but I couldn't find anything: Gaussian integral using Euler/Jacobi theta function and $r_2(k)$ (number of representations as sum of 2 squares).

$$\rule{5cm}{0.4pt}$$

EDIT 10/13/22: I was recently learning about Riemannian geometry, and indeed the metric tensor is a bilinear form (cf. above discussion on inner products), and the Riemann curvature tensor (or curvature in general) is all about the second (covariant) derivative. Taking traces we arrive at the Ricci curvature tensor (used in no less important things than Einstein's general relativity, whose "classical approximation" Newtonian gravity follows an inverse square law; and Perelman's proof of the Poincare conjecture!) or scalar curvature, which can be interpreted geometrically as the second-order change in volume of balls/base of cones (see https://math.stackexchange.com/a/469005/405572 or Equation (10) of https://arxiv.org/pdf/2201.04923v1.pdf).

And of course like Ricci flow we also have the heat/diffusion differential equations (and the Schrodinger equation, quoting James Gleck, is "diffusion through imaginary time"), and endless equations involving the Laplacian, all second-order differential equations (how AMAZING that the two great pillars of 20th century physics, general relativity and quantum mechanics both have at their hearts second-order differential equations! And the fact that observables in quantum mechanics are modeled as operators on Hilbert spaces). Relating to the Laplacian, we have the important concept of harmonicity, with beautiful manifestations/consequences in complex analysis, or PDEs, in the form of elliptic regularity. Besides tensors of 2nd derivatives, we have the ever-present Hessian matrix of second derivatives. I'll end with some quotes from a MO answer I linked here in a comment several months ago (https://mathoverflow.net/a/171743/112504) which tries to argue that "Polynomials are useful because quadratic polynomials are useful.":

So the question becomes: why are quadratic polynomials useful? There seem to be two different but interacting reasons. The first is that quadratic functions of a real variable are always either convex or concave and therefore have a unique maximum or minimum. The second is that quadratic functions are intimately related to bilinear forms and therefore can be accessed using linear algebra. The combination of these two reasons seems to explain the success of quadratic algebra in analysis and geometry (e.g. Hilbert spaces, Riemannian manifolds)


问:
Diferences between $V^3$, $E^3$ and $\mathbb{R}^3$
I am studying linear algebra (link) and at a certain point it is stated that $V^3$, $E^3$ (the space of the points) and $\mathbb{R}^3$ are diferents, but $E^3$ was never explicitly defined and I couldn't find anything about it on the internet. Can someone help me?

答: When we talk about vector spaces, we use dimension and base terms to say something about structure of the space, later we may define a transformation on that space. However, if we talk about affine spaces, we do not care the structure. $E^3$ means just the set of all points in 3 dimensional space, this is an affine space. Here we do not say more.


问:
Evaluate this complex integration
Let u and v be the real and imaginary parts of the function f(z)=1/(z^2-6z+8). C is |z|=3 counterclockwise. Evaluate   $\int $ udy+vdx over C.

My attempt: I was trying to apply greens theorem by writing f(z) in terms of x and y. But i think it will not work as for greens theorem u and v must have first order partial derivative exists and continuous. U and v doesn't exist at (2,0).

Another attempt is$ \int $f(z)dz =$\int $ (u+iv) d(x+iy)
=    $\int $ (u+iv) dx+idv
=  $\int $ udx -vdy +i     $\int $ udy +vdx (after multiplication)

Now applying cauchys integral formula   $\int $ g(z)/(z-2) dz =g(2)2$ \pi $ * i= $ \pi $* i
I cant proceed further.  Where g(z) =1/(z-4) .Also please elaborate how can i apply greens theorem.

Please let me know what i have to apply in this problem?

答:
Your second attempt is the way to go and you are almost done. So, there is no need to find explicit expressions for $u$ and $v$.

Denoting the imaginary part of a complex number $\zeta$ by $\Im(\zeta)$, you found already that

$$\int_{x^2+y^2=9}u\,dy + v\,dx = \Im\left(\int_{|z|=3}\frac 1{z^2-6z+8}dz\right)$$

Now, as you already did, applying the Cauchy integral formula you get

\begin{eqnarray*} \int_{|z|=3}\frac 1{z^2-6z+8}dz
& = & \int_{|z|=3}\frac 1{(z-2)(z-4)}dz \\
& = & \int_{|z|=3}\frac{\frac 1{z-4}}{z-2}dz \\
& \stackrel{Cauchy}{=} & 2\pi i \frac 1{2-4} \\
& = &-\pi i
\end{eqnarray*}

Hence,

$$\int_{x^2+y^2=9}u\,dy + v\,dx = \Im(-\pi i) = -\pi$$

Title: Evaluate this complex integration


问:
Multiplicities of Polynomials Multiplied
This is a follow-up to one of my past questions, Multiplicities of Polynomials. I thought it was more appropriate to ask in a new question. This question is self-contained, I have written the relevant details here.

I'm trying to comprehend the explanation:

Let $P$ be a non-zero polynomial in $\mathbb{C}\left[x_{0}, \ldots, x_{n}\right]$ and let $a=\left(a_{0}, \ldots, a_{n}\right)$ be a point in $\mathbb{C}^{n+1}$. The multiplicity $\operatorname{mult}_{a} P$ of $P$ at $a$ is the smallest non-negative integer $m$ such that there exists $\alpha \in \mathbb{N}^{n+1}$ with $|\alpha|=m$ and $\partial^{\alpha} P(a) \neq 0$.
Such a non-negative integer $m$ exists, because, up to some non-zero factors, the partial derivatives $\partial^{\alpha} P(a)$ are the coefficients in the Taylor expansion of $P$ around the point $a .$ Thus if $P$ is non-zero, then at least one of these partial derivatives is different from zero. This argument also shows that mult $_{a} P \leq \operatorname{deg}(P),$ and that $\operatorname{mult}_{a}(P Q)=\operatorname{mult}_{a} P+\operatorname{mult}_{a} Q$ for every non-zero polynomial $Q$ in $\mathbb{C}\left[x_{0}, \ldots, x_{n}\right],$ since the Taylor expansion of $P Q$ around $a$ is the product of the Taylor expansions of $P$ and $Q$.

We have the identity $$ \frac{\partial^\gamma (PQ)(a)}{\gamma!} = \sum_{\alpha + \beta = \gamma} \frac{\partial^\alpha P(a)}{\alpha!} \frac{\partial^\beta Q(a)}{\beta!}. \tag{1} $$
But why does this imply that $\operatorname{mult}_{a}(P Q)=\operatorname{mult}_{a} P+\operatorname{mult}_{a} Q$? The RHS of the equation $(1)$ is a multiple sum, therefore non-zero summands may cancel out to give $0$. In other words, suppose that $\operatorname{mult}_a PQ = r$, $\operatorname{mult}_a P = s$, $\operatorname{mult}_a Q = t$. I get that $r \ge st$, but am not able to conclude that $r = st$.

答:
Thinking in terms of derivatives is making this harder than it needs to be; we are just talking about multiplying polynomials here. Translate the polynomial so that $a = 0$ WLOG. The multiplicity $\text{mult}_0(P)$ is the smallest $m$ for which $P$ has a nonzero term of degree $m$ (meaning multidegree $\alpha$ such that $|\alpha| = m$). Among all the nonzero terms of degree $m$ there is a unique one which is largest in lex order, and the product of this term in $P$ with the corresponding term in $Q$ is the unique nonzero term in $PQ$ of minimal degree which is largest in lex order; an explicit example to make the reasoning clear here is

$$P(x, y) = x^2 + y^2 + \text{higher terms}$$
$$Q(x, y) = x^2 y - y^3 + \text{higher terms}$$
$$P(x, y) Q(x, y) = x^4 y - y^5 + \text{higher terms}$$

where, if we pick the lex order in which $x \ge y$, the lex biggest term in $P$ of minimal degree is $x^2$, the lex biggest term in $Q$ of minimal degree is $x^2 y$, and the lex biggest term in $PQ$ of minimal degree is their product $x^4 y$.

You are correct that there can be cancellation for some $\gamma$, which happens in the example above: $x^2 y (y^2)$ cancels with $- x^2 y^3$. But cancellation doesn't occur for the biggest (or smallest) term in any monomial order because the RHS of your product formula has a single term in this case.


问:
Deductions for the sets $A = \left\{(-1)^n + \frac{2}{n}:n=1,2,3,\ldots\right\}$ and $B = \{x \in \mathbf{Q}:0<x<1\}$
I would like to ask, if my proof and deductions for the below sets with respect to openness, closure checks out and is technically correct.

Exercise 3.2.2 from Understanding Analysis by Stephen Abbot.

Let
\begin{align*}
 A = \left\{(-1)^n + \frac{2}{n}:n=1,2,3,\ldots\right\}
\end{align*}
and
\begin{align*}
 B = \{x:\mathbf{Q}:0<x<1\}
\end{align*}

Answer the following questions for each set.
(a) What are the limit points?
(b) Is the set open? Closed?
(c) Does the set contain any isolated points?
(d) Find the closure of the set.

Proof.

(a) Writing out a few elements of the set $A$, we have:
\begin{align*}
 A = \left\{1,2,-\frac{1}{3},\frac{5}{4},-\frac{3}{5},\frac{4}{3},\ldots\right\}
\end{align*}

The points $-1$ and $1$ are the limit points of $A$. There exists a subsequence $a_{2n-1} = -1 + \frac{2}{2n-1}$ in $A$ such that $\lim_{n\to\infty} a_{2n-1} = -1$ and $a_{2n-1} \ne 1$ for all $n \in \mathbf{N}$. Similarly, look at the subsequence $(a_{2n})$. $\lim_{n\to\infty} a_{2n} = 1$.

Every real number in $[0,1]$ is the limit of a sequence of rational numbers in $B$. So, the closed interval $[0,1]$ is the set of all limit points of $B$.

(b) The set $A$ is open. For all $a \in A$, there exists an $\epsilon$-neighbourhood of $a$, $V_\epsilon(a)$ that is contained in $A$, $V_\epsilon(a) \subseteq A$.

The limit points of $A$ are not elements of the set $A$. $A$ is not closed.

The set $B$ is open. $B$ has no largest or smallest element. For every rational number $x \in B$, there exists an $\epsilon$-neighbourhood of $B$, that intersects $B$ and contains points other than $x$.

The set $B$ is not closed, because $B$ does not contain its limit points.

(c) Pick any point $a = (-1)^n + 2/n$. Let $\epsilon = 2(\frac{1}{n} - \frac{1}{n+1})$. Then, $V_\epsilon(a) \cap A = \{a\}\ \subseteq A$. So, every point in $A$ is an isolated point.

The set $B$ does not contain any isolated points.

(d) \begin{align*}
 \bar{A} &= A \cup \{-1,1\}\\
 \bar{B} &= [0,1]
\end{align*}

答:
Overall, if this were a test and I was grading your results, you'd get 6 out of 10 points.

For (a), you are correct.

For (b), not so much.

$A$ is not open. Take $a=\frac54\in A$. Then, clearly, $a+\frac\epsilon2$ is not a member of $A$, meaning that no neighborhood of $a$ can ever be fully included in $A$.

You claim that the limit points of $A$ are not elements of the set $A$. This is partially true. One of the limit points is an element of $A$. For the other, you should show that it is not (the proof is easy, but it should be present).

Similarly to 1, $B$ is also not open. For any rational number $q$ and any $\epsilon > 0$, there exists an irrational number $x$ such that $|x-q|<\epsilon$.

For (c), you are again sloppy. Not every point of $A$ is an isolated point. One of them is not.

For (d), you are correct.


答: Hint. Only one of the poles of the integrand is inside the circle of interest. Thus you only need to find the residue of the integrand at the pole of interest, and then the integral is just $2πi\times \text{residue of interest}.$

Title: Evaluate this complex integration

Body:
Let u and v be the real and imaginary parts of the function f(z)=1/(z^2-6z+8). C is |z|=3 counterclockwise. Evaluate   $\int $ udy+vdx over C.

My attempt: I was trying to apply greens theorem by writing f(z) in terms of x and y. But i think it will not work as for greens theorem u and v must have first order partial derivative exists and continuous. U and v doesn't exist at (2,0).

Another attempt is$ \int $f(z)dz =$\int $ (u+iv) d(x+iy)
=    $\int $ (u+iv) dx+idv
=  $\int $ udx -vdy +i     $\int $ udy +vdx (after multiplication)

Now applying cauchys integral formula   $\int $ g(z)/(z-2) dz =g(2)2$ \pi $ * i= $ \pi $* i
I cant proceed further.  Where g(z) =1/(z-4) .Also please elaborate how can i apply greens theorem.

Please let me know what i have to apply in this problem?


问:
Tilted rectangles in a rectangle (problem from a Swedish 12th grade ‘Student Exam’ from 1934)
The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1934.

$\mathit{ABCD}$ is a rectangle with the sides $\mathit{AB}=15$ cm and $\mathit{BC}=12$ cm. On $\mathit{AB}$, between $A$ and $B$, is a point $E$ placed such that $\mathit{AE}=2$ cm. There are two rectangles, inscribed in the given rectangle, which has a corner at $E$ and corners on each of the three other sides of the given rectangle. Calculate the area of the two rectangles.

Below I present two solutions which have the correct answer in two different ‘formats’. The latter is the one given as answer in a book. The first solution, that gives the more ‘convoluted’ format, also includes an equation that is, perhaps, not as easy to solve unless one observes a particular (imaginary) solution first. Also, the calculations are pretty cumbersome to do by hand, as they were in 1934, so I lean towards thinking there is a simpler way to arrive at the answer, in its simplest form. I look forward to hear your thoughts on how to solve this problem.

As basis for both solutions we use the following figure;

where $l_1$ and $l_2$ are straight lines representing the longer sides and $n_1$ and $n_2$ the perpendicular lines (called ‘normal’ in Swedish) representing the shorter sides.

Solution

Method 1

Let $l_1$ and $l_2$ have the slope $k$.

We have the following equations for the straight lines;
\begin{align*}
\text{\(l_1\):} & \quad y=k(x-2)\\
\text{\(n_1\):} & \quad y=-\tfrac{1}{k}(x-2)=-\tfrac{1}{k}x+\tfrac{2}{k}\\
\text{\(l_2\):} & \quad y=kx+\tfrac{2}{k}
\end{align*}

The $y$-coordinate of $F$ is
$$ y_F=k(x-2)\big|_{x=15}=13k $$
and the $x$-coordinate of $G$ is given by the equation
$$kx_G+\frac{2}{k}=12 \quad\Leftrightarrow\quad x_G=\frac{12-2/k}{k}.$$

The slope for $n_2$ is given by
$$ \frac{y_F-y_G}{x_F-x_G}=\frac{13k-12}{15-\tfrac{12-2/k}{k}}.$$

Since $n_1$ is parallell to $n_2$ we get
$$
\frac{13k-12}{15-\tfrac{12-2/k}{k}}=-\frac{1}{k}
\quad\Leftrightarrow\quad
13k^4-12k^3+15k^2-12k+2=0.
$$

This equation is not that easy to solve explicitly at ’first sight’, but we observe that $k=\pm i$ solves the equation and thus we can, after polynomial division, write
$$13k^4-12k^3+15k^2-12k+2=(13k^2-12k+2)(k^2+1)$$
and we find the remaining solutions to be
$$k=\tfrac{1}{13}(6\pm\sqrt{10}\,).$$

For the two $k$ we get the following figure;

The dimensions of the two rectangles are given by the ‘distance formula’ and we get
\begin{align*}
|\mathit{EF}|^2&
=(2-15)^2+(0-13k)^2\big|_{k=\frac{1}{13}(6-\sqrt{10}\,)}=215-12\sqrt{10}\\
|\mathit{EH}|^2&
=(2-0)^2+(0-\tfrac{2}{k})^2\big|_{k=\frac{1}{13}(6-\sqrt{10}\,)}=50+12\sqrt{10}\\
|\mathit{EF'}|^2&
=(2-15)^2+(0-13k)^2\big|_{k=\frac{1}{13}(6+\sqrt{10}\,)}=215+12\sqrt{10}\\
|\mathit{EH'}|^2&
=(2-0)^2+(0-\tfrac{2}{k})^2\big|_{k=\frac{1}{13}(6+\sqrt{10}\,)}=50-12\sqrt{10}
\end{align*}

The areas can now be calculated to be (in $\text{cm}^2$)
\begin{align*}
A_1&=|\mathit{EF}|\cdot|\mathit{EH}|=\sqrt{9310+1980\sqrt{10}}\\
A_2&=|\mathit{E'F'}|\cdot|\mathit{E'H'}|=\sqrt{9310-1980\sqrt{10}}
\end{align*}

This is the more ‘convoluted’ answer. It can be simplified further, but that is not obvious(?) at ‘first sight’. The simpler format is given as result in Method 2 presented below.

Method 2

We proceed as above but we calculate the areas of the 4 triangles;
\begin{align*}
A_1&=\frac{1}{2}\cdot(15-2)\cdot y_F=\frac{169}{2}k\\
A_2&=\frac{1}{2}\cdot\frac{2}{k}\cdot2=\frac{2}{k}\\
A_3&=\frac{1}{2}\cdot x_G\cdot(12-y_H)=\frac{2(1-6k)^2}{k^3}\\
A_4&=\frac{1}{2}\cdot(12-13k)\cdot\Bigl(15-\frac{12-2/k}{k}\Bigr)=\frac{(12-13k)(15k^2-12k+2)}{2k^2}
\end{align*}
and add up the total triangle area to be
$$ A=A_1+A_2+A_3+A_4=168-13k-\frac{11}{k}-\frac{12}{k^2}+\frac{2}{k^3}, $$
hence the area of the rectangle is
$$ R=12\cdot15-A=12+13k+\frac{11}{k}+\frac{12}{k^2}-\frac{2}{k^3}.$$

For the values $k=\frac{1}{13}(6\pm\sqrt{10}\,)$ we get
$$ R=90\pm11\sqrt{10}~\text{cm}^2. $$

This is the answer given in the book.

The original exam

答:
The triangles $A_1,A_2,A_3$, and $A_4$ are easily seen to be similar. Let $x$ be your $y_F$. Then

$$\frac{x}{13}=\frac2{12-x}\,,$$

so $x^2-12x+26=0$, and

$$x=\frac{12\pm\sqrt{40}}2=6\pm\sqrt{10}\,.$$

The total area of the four triangles is therefore

$$13\left(6\pm\sqrt{10}\right)+2\left(6\mp\sqrt{10}\right)=90\pm11\sqrt{10}\,,$$

and the areas of the rectangles are

$$15\cdot12-\left(90\pm11\sqrt{10}\right)=90\mp11\sqrt{10}\,.$$


答:
The point to note is that in any parallelogram in the rectangle ABCD with vertices on each sides and with point $E$ being one of the vertices, $AE = CH$. So $E$ and $H$ are fixed. Now two rectangles are special cases where points $F$ and $G$ are chosen such that $\angle EFH$, $\angle EGH$ and opposite angles are $90^0$.

We then have $\angle CFH = \angle BEF$ and hence $\triangle BEF \sim \triangle CFH$. You also see two pairs of congruent triangles.

So area of $4$ triangles $= 2 \times \frac{1}{2} \times 13 \times x + 2 \times \frac{1}{2} \times 2 \times (12-x) = 11x + 24$

Finding $x$ from similar triangles is easy,

$\frac{2}{12-x} = \frac{x}{13} \implies x = 6 \pm \sqrt{10}$

Title: Tilted rectangles in a rectangle (problem from a Swedish 12th grade ‘Student Exam’ from 1934)

Body:
The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1934.

$\mathit{ABCD}$ is a rectangle with the sides $\mathit{AB}=15$ cm and $\mathit{BC}=12$ cm. On $\mathit{AB}$, between $A$ and $B$, is a point $E$ placed such that $\mathit{AE}=2$ cm. There are two rectangles, inscribed in the given rectangle, which has a corner at $E$ and corners on each of the three other sides of the given rectangle. Calculate the area of the two rectangles.

Below I present two solutions which have the correct answer in two different ‘formats’. The latter is the one given as answer in a book. The first solution, that gives the more ‘convoluted’ format, also includes an equation that is, perhaps, not as easy to solve unless one observes a particular (imaginary) solution first. Also, the calculations are pretty cumbersome to do by hand, as they were in 1934, so I lean towards thinking there is a simpler way to arrive at the answer, in its simplest form. I look forward to hear your thoughts on how to solve this problem.

As basis for both solutions we use the following figure;

where $l_1$ and $l_2$ are straight lines representing the longer sides and $n_1$ and $n_2$ the perpendicular lines (called ‘normal’ in Swedish) representing the shorter sides.

Solution

Method 1

Let $l_1$ and $l_2$ have the slope $k$.

We have the following equations for the straight lines;
\begin{align*}
\text{\(l_1\):} & \quad y=k(x-2)\\
\text{\(n_1\):} & \quad y=-\tfrac{1}{k}(x-2)=-\tfrac{1}{k}x+\tfrac{2}{k}\\
\text{\(l_2\):} & \quad y=kx+\tfrac{2}{k}
\end{align*}

The $y$-coordinate of $F$ is
$$ y_F=k(x-2)\big|_{x=15}=13k $$
and the $x$-coordinate of $G$ is given by the equation
$$kx_G+\frac{2}{k}=12 \quad\Leftrightarrow\quad x_G=\frac{12-2/k}{k}.$$

The slope for $n_2$ is given by
$$ \frac{y_F-y_G}{x_F-x_G}=\frac{13k-12}{15-\tfrac{12-2/k}{k}}.$$

Since $n_1$ is parallell to $n_2$ we get
$$
\frac{13k-12}{15-\tfrac{12-2/k}{k}}=-\frac{1}{k}
\quad\Leftrightarrow\quad
13k^4-12k^3+15k^2-12k+2=0.
$$

This equation is not that easy to solve explicitly at ’first sight’, but we observe that $k=\pm i$ solves the equation and thus we can, after polynomial division, write
$$13k^4-12k^3+15k^2-12k+2=(13k^2-12k+2)(k^2+1)$$
and we find the remaining solutions to be
$$k=\tfrac{1}{13}(6\pm\sqrt{10}\,).$$

For the two $k$ we get the following figure;

The dimensions of the two rectangles are given by the ‘distance formula’ and we get
\begin{align*}
|\mathit{EF}|^2&
=(2-15)^2+(0-13k)^2\big|_{k=\frac{1}{13}(6-\sqrt{10}\,)}=215-12\sqrt{10}\\
|\mathit{EH}|^2&
=(2-0)^2+(0-\tfrac{2}{k})^2\big|_{k=\frac{1}{13}(6-\sqrt{10}\,)}=50+12\sqrt{10}\\
|\mathit{EF'}|^2&
=(2-15)^2+(0-13k)^2\big|_{k=\frac{1}{13}(6+\sqrt{10}\,)}=215+12\sqrt{10}\\
|\mathit{EH'}|^2&
=(2-0)^2+(0-\tfrac{2}{k})^2\big|_{k=\frac{1}{13}(6+\sqrt{10}\,)}=50-12\sqrt{10}
\end{align*}

The areas can now be calculated to be (in $\text{cm}^2$)
\begin{align*}
A_1&=|\mathit{EF}|\cdot|\mathit{EH}|=\sqrt{9310+1980\sqrt{10}}\\
A_2&=|\mathit{E'F'}|\cdot|\mathit{E'H'}|=\sqrt{9310-1980\sqrt{10}}
\end{align*}

This is the more ‘convoluted’ answer. It can be simplified further, but that is not obvious(?) at ‘first sight’. The simpler format is given as result in Method 2 presented below.

Method 2

We proceed as above but we calculate the areas of the 4 triangles;
\begin{align*}
A_1&=\frac{1}{2}\cdot(15-2)\cdot y_F=\frac{169}{2}k\\
A_2&=\frac{1}{2}\cdot\frac{2}{k}\cdot2=\frac{2}{k}\\
A_3&=\frac{1}{2}\cdot x_G\cdot(12-y_H)=\frac{2(1-6k)^2}{k^3}\\
A_4&=\frac{1}{2}\cdot(12-13k)\cdot\Bigl(15-\frac{12-2/k}{k}\Bigr)=\frac{(12-13k)(15k^2-12k+2)}{2k^2}
\end{align*}
and add up the total triangle area to be
$$ A=A_1+A_2+A_3+A_4=168-13k-\frac{11}{k}-\frac{12}{k^2}+\frac{2}{k^3}, $$
hence the area of the rectangle is
$$ R=12\cdot15-A=12+13k+\frac{11}{k}+\frac{12}{k^2}-\frac{2}{k^3}.$$

For the values $k=\frac{1}{13}(6\pm\sqrt{10}\,)$ we get
$$ R=90\pm11\sqrt{10}~\text{cm}^2. $$

This is the answer given in the book.

The original exam


问:
What are examples of isomorphisms of Lie algebras?
I'm looking for an example of an isomorph Lie Algebra. 2 algebras are isomorph, if there exists an bijective linear function $g_1 \rightarrow g_2$ which maps all $X,Y \in g_1$ like $\phi([X,Y]) = [\phi(X),\phi(Y)]$.

So 2 Lie algebras I could think of would be the cross product in ${\rm I\!R}^3$ and the Commutator algebra of a left invariant Vectorfield but I can't think of a function that maps them like I stated before.

答:
Examples, roughly ordered from easy to hard:

Let $\mathfrak g$ be any Lie algebra. The identity map $x \mapsto x$ is an isomorphism from $\mathfrak g$ to itself.

Let $V$, $W$ be vector spaces over a field $k$, and define Lie brackets on them as $[v_1, v_2] = 0$ and $[w_1,w_2]=0$ for all $v_1,v_2 \in V$, $w_1,w_2 \in W$. Show that the Lie algebras $V$ and $W$ (with these brackets) are isomorphic if and only if $V$ and $W$ have the same dimension. (This should be just a check you understand isomorphisms of vector spaces, the absolute basis of linear algebra.)

Let $k$ be any field and $\mathfrak{gl}_n(k)$ the Lie algebra given by all $n \times n$-matrices over $k$, with the Lie bracket given by the matrix commutator $[A,B]:= A\cdot B-B\cdot A$ (where $\cdot$ is the usual matrix multiplication). Let $g$ be any invertible $n\times n$-matrix over $k$, i.e. an element of $\mathrm{GL}_n(k)$. Show that the map
$$ A \mapsto g\cdot A \cdot g^{-1}$$
is an isomorphism from $\mathfrak{gl}_n(k)$ to itself, i.e. an automorphism of $\mathfrak{gl}_n(k)$.

Let $\mathfrak{gl}_n(k)$ be as in the previous example. The map which sends each matrix to its negative transpose,
$$ A \mapsto -A^T$$
is an isomorphism from $\mathfrak{gl}_n(k)$ to itself, i.e. an automorphism of $\mathfrak{gl}_n(k)$.

Let $k$ be any field, $c \in k^\times$, $\mathfrak g_1$ a two-dimensional $k$-vector space with basis $v_1, v_2$ and Lie bracket $[v_1, v_2] = v_2$. Let $\mathfrak g_2$ be another two-dimensional $k$-vector space with basis $w_1,w_2$ and $[w_1,w_2]= c\cdot w_2$. Find an isomorphism of the Lie algebras $\mathfrak g_1$ and $\mathfrak g_2$.

Let $\mathfrak g_1$ and $\mathfrak g_2$ be as in the previous example, except that now the Lie bracket on $\mathfrak g_2$ is given by $[w_1,w_2] = a w_1 + c w_2$ where $c \in k^\times$ and $a \in k$. Again find an isomorphism $\mathfrak g_1 \simeq \mathfrak g_2$. (For this and the previous example, cf. Classsifying 1- and 2- dimensional Algebras, up to Isomorphism, How to get an explicit isomorphism (explicitly defined) between any two nonabelian Lie algebras of dimension $2$, What are the possible two-dimensional Lie algebras?, Two dimensional Lie Algebra - what do we know without knowing the Bracket? .)

Let $k$ be any field of characteristic $\neq 2$, $\mathfrak{sl}_2(k) := \{ A \in \mathfrak{gl}_2(k): Tr(A)=0\}$ the Lie algebra of traceless $2 \times 2$-matrices (with Lie bracket given as in example 3). Let $\mathfrak{so}_3(k) := \{ \pmatrix{a&0&-f\\0&-a&-e\\e&f&0} : a,e,f \in k \}$ (the "split form of $\mathfrak{so}_3$") also with Lie bracket given by matrix commutator. Find an isomorphism between these two Lie algebras. (Compare Is there an intelligent way to find the Lie algebra isomorphism $\mathfrak{sl}_2(\mathbb{C})\simeq\mathfrak{o}_3(\mathbb{C})$?, What is a direct proof of the isomorphism $\mathfrak{so}(3)_{\mathbb C}\simeq\mathfrak{sl}(2,\mathbb C)$?, An Explicit Isomorphism Between the Three Dimensional Orthogonal Lie Algebra and the Special Linear Lie Algebra of Dimension $3$ and links therein.)

Let $\mathfrak{su}_2 := \{\pmatrix{ai&b+ci\\-b+ci&-ai} : a,b,c \in \mathbb R \}$ (a three-dimensional real subspace of the $2 \times 2$ complex matrices); convince yourself that again with the Lie bracket given by the matrix commutator (as in example 3), this is a Lie algebra. Show it is isomorphic to $(\mathbb R^3, \times)$ i.e. the three-dimensional real Lie algebra with Lie bracket given by the cross product. (Compare Why is there a factor of $2$ in the isomorphism $\operatorname{Lie}(S^3)\cong\mathbb{R}^3$?. This seems to be what you allude to in the question.)

Find an isomorphism between $\mathfrak{sl}_2(\mathbb C) \oplus \mathfrak{sl}_2(\mathbb C)$ and the skew-symmetric $4\times 4$ matrices over $\mathbb C$. (Cf. Explicit isomorphism between the four dimensional orthogonal Lie algebra and the direct sum of special linear Lie algebras of dimension 3. )

Find an isomorphism between: a) the direct sum of skew-symmetric $3 \times 3$ real matrices with itself; and b) the $4 \times 4$ real skew-symmetric matrices. (Cf. Isomorphism between $\mathfrak o(4,\mathbb R)$ and $\mathfrak o (3,\mathbb R) \oplus\mathfrak o (3,\mathbb R) $ )

10.5. Find an isomorphism between: a) the skew-symmetric complex $6 \times 6$ matrices b) the traceless complex $4 \times$ matrices. (Cf. How should I show that the Lie algebra $\mathfrak{sl}(4,\mathbb{C})$ is isomorphic to the Lie algebra $\mathfrak{so}(6,\mathbb{C})$? and after going through the next one, come back and compare this to How should I show that the Lie algebra so(6) of SO(6) is isomorphic to the Lie algebra su(4) of SU(4)? .)

For $\mathfrak g$ a real Lie algebra, the scalar extension / complexification $\mathbb C \otimes \mathfrak g$ is a complex Lie algebra with Lie bracket given by bilinear extension of $[a \otimes x, b \otimes y]:=ab\otimes [x,y]$. Easy: Show that the complexification of $\mathfrak{sl}_2(\mathbb R)$ is isomorphic to $\mathfrak{sl}_2(\mathbb C)$. Harder: For $\mathfrak{su}_2$ as defined in example 8, show that the complexification $\mathbb C \otimes \mathfrak{su}_2$ is also isomorphic to $\mathfrak{sl}_2(\mathbb C)$. Bonus: Show that in spite of that, the real Lie algebras $\mathfrak{sl}_2(\mathbb R)$ and $\mathfrak{su}_2$ are not isomorphic to each other. (Compare Precise connection between complexification of $\mathfrak{su}(2)$, $\mathfrak{so}(1,3)$ and $\mathfrak{sl}(2, \mathbb{C})$, Are Lie algebra complexifications $\mathfrak g_{\mathbb C}$ equivalent to Lie algebra structures on $\mathfrak g\oplus \mathfrak g$?, and probably many more.)

Also, try Finding Lie algebra isomorphisms, and Question regarding isomorphisms in low rank Lie algebras .


问:
for any polynomial $f(t)$,we have that $f(A)$ is the matrix representation of $f(T)$.
Let $A$ be the matrix representation of a linear transformation $T$ from $V$ to $V$. Prove that, for any polynomial $f(t)$,we have that $f(A)$ is the matrix representation of $f(T)$.

How can I start proving such a statement? Can I prove it without using induction.

Thanks.

答:
If $A$ is the matrix representation of $T$ then there is some basis $v_k$ such that
$\sum_k [Ae_i]_k v_k = T v_i$.

If we apply $T$ to both sides we see that
\begin{eqnarray}
T(\sum_k [Ae_i]_k v_k) &=& \sum_k [Ae_i]_k Tv_k \\
&=&  \sum_k [Ae_i]_k \sum_j [Ae_k]_j v_j \\
&=& \sum_k \sum_j [Ae_i]_k [Ae_k]_j v_j \\
&=& \sum_k [A^2e_i]_k v_k \\
&=& T^2 v_i
\end{eqnarray}
It is not hard to see that if we repeat this process we get
$\sum_k [A^ne_i]_k v_k = T^n v_i$, and from there that
$f(T)v_i = \sum_{n=0}^N a_n T^nv_i =  \sum_{n=0}^N a_n  \sum_k [A^n e_i]_k v_k = \sum_k [\sum_{n=0}^N a_n A^n e_i]_k v_k = \sum_k [f(A) e_i]_k v_k$


问:
Is the derivative of $|x|$ defined for every x in the neighborhood?

Is the derivative of $f(x)= |x|$ defined for every $x$ in the neighborhood?

I know the derivative of $x=0^+$ and $x=0^-$ is not the same. But is it correct to say that the derivative is defined for every $x$ in the neighborhood of $0$?

答: No, it is not correct, since $0$ belongs to every neighborhood of $0$, and $f$ is not differentiable at $0$.


问:
Boundaries in heat equation
I have this heat equation:

$u_t = 9u_{xx} - 7u + f(x,t),$

$f(x,t) = 1; 0 < x < l; 0 < t < T $

$u(x,0) = 6x^2 - 5x +2$

$u(0,t) = 3t + 2$

$u(l,t) = t + 3$

$l = 1$

My problem is that I know how to solve when IC and BC are zeros. Maybe someone knows (or can point me to) a detailed method to transform(?) my equation and what to do then?

Thank you very much.

答: Find any function $h(x,t)$ such that $h(0,t)=u(0,t)$ and $h(l,t)=u(l,t)$. Then make the change of variables $u=v+h$. This will give you an equation for $v$ with homogeneous boundary conditions (but a different initial value $v(x,0)$).

Title: Boundaries in heat equation


问:
Function spaces and transitive group actions
Note: this question is really a subquestion of this one, but I decided to ask it separately since it seems it might be attacked first.

Let $B$ be a topological space and $G$ a topological group acting on a space $F$. Then the space $G^B$ of continuous maps from $B$ to $G$ has a group structure given by pointwise multiplication. Moreover, for nice enough spaces the compact-open topology gives $G^B$ the structure of a topological group. This group acts on the space $F^B$ (again endowed with the compact open-topology) in the obvious pointwise way and the action is continuous.

My question is the following: assume $G$ acts transitively; can we infer the same holds for $G^B$? If not, under what assumptions is transitivity preserved?

Notice that freeness and faithfulness are (separately) preserved (it is enough to think pointwise).

答:
No. Consider the case $G=(\Bbb R,+),~B=F=S^1$, and $G$ acts on $F$ via
$$t\cdot z=\exp(it)\cdot z$$
Then there is no map $g:B\to G$ such that $g\cdot 1=\mathrm{id}$, where $1$ is the constant map $B\to F, z\mapsto 1$. Indeed any map $g: S^1\to\Bbb R$ is homotopic to a constant map because $\Bbb R$ is contractible, however $1$ and $\mathrm{id}$ are not homotopic as maps $S^1\to S^1$.

More precisely, in this example we have that two maps $B\to F$ are in the same $G^B$ orbit iff they are homotopic as maps $S^1\to S^1$.


问:
Is dependent probability the same as conditional probability?
I can't find any solid definition online. A few websites however say "dependent probability is defined when a previous event will affect the next event. This gives rise to conditional probability" which is really vague. What's the difference? Are they the same?

I am also confused with a few definitions. If independent probability is defined as $P(A) = P(A\mid B) = P(A\mid B')$, how can you show that in a Venn/Euler diagram? It seems in that case, the probability theorems cannot be translated into a diagram!

答:
Independence and dependence have nothing to do with Venn diagram. Since $A,B$ are independent does not mean $A$ and $B$ are separated.

Here is an example, Let $\Omega = \{a,b,c\}$ and $\mathcal{A}$ be a sigma algebra with $\mathcal{A} = \{\varnothing,\Omega, \{a\},\{b,c\}\}$. Clearly, $\mathcal{A}$ is the sigma algebra generated by $\{a\}$. We let the probability of events in $\mathcal{A}$ be equally distributed. Then for the events $A,B\in\mathcal{A}$, $P(A=\{a\}) =  0.25$, $P(B = \Omega)=0.25$. Clearly, the event $A=\{a\}$ and $B=\{\Omega\}$ is independent, but $A\cap B = \{a\}\neq\varnothing$.


问:
limits of marginal probability of polar coordinate probability
I'm self-studying probability and came across a problem of finding the marginal probability $f_\theta$ of a pdf in polar coordinate $f_{r,\theta}$. To find the $f_r$, I know we do $\int^{2\pi}_{0} f_{r,\theta} d\theta$, but how about $f_\theta$? Is it from 0 to 1? Or 0 to infinity?

EDIT: the original problem is
find $f_\theta = \int_{0}^{?} \frac{r}{2\pi\sigma^2}exp(\frac{-r^2}{2\sigma^2})dr$

答:
Clearly the domain of $\Theta$ is $[0;2\pi]$ and the domain of $R$ is $[0;+\infty)$

The problem arises from a vector $(X,Y)$ where $X\perp \!\!\!\!\!\! \perp Y$, are identically Centered Gaussian distributed

$$X \sim N(0;\sigma^2)$$

$$Y \sim N(0;\sigma^2)$$

Thus the joint density is

$$f_{XY}(x,y)=f_X(x)\cdot f_Y(y)=\frac{1}{2\pi\sigma^2}e^{-(x^2+y^2)/(2\sigma^2)}$$

Starting from this vector, and passing in polar coordinates you get

$$ \bbox[5px,border:2px solid black]
{
f_{R;\Theta}(r;\theta)=\frac{1}{2\pi}\cdot \frac{r}{\sigma^2}e^{-r^2/(2\sigma^2)}
\qquad (1)
}
$$

Analyzing this joint density it is evident that it can be factorized in the following way

$$ \bbox[5px,border:2px solid black]
{
f_{R;\Theta}(r;\theta)=\underbrace{\frac{1}{2\pi}\cdot\mathbb{1}_{[0;2\pi]}(\theta)}_{\text{uniform density}}\times\underbrace{\frac{r}{\sigma^2}e^{-r^2/(2\sigma^2)}\cdot\mathbb{1}_{[0;+\infty)}(r)}_{\text{rayleigh density}}=f_{\Theta}(\theta)\times f_R(r)
\qquad (2)
}
$$

Now observe that:

The uniform density is the density of the angle, uniform over all the circle (the circumference of a circle centered in the Origin)

$$\Theta\sim U[0;2\pi]$$

The Rayleigh density is the density of the radius, with support $[0;+\infty)$

$$R\sim Rayl(\sigma)$$

Angle and Radius are independent rv's

The following drawing is useful to understand the problem

As you can see, the radius that is $R=\sqrt{X^2+Y^2}$ can takes any real value from zero to $\infty$ while the angle can takes any real value in the interval $[0;2\pi]$

Your original problem starts from the joint density Angle - Radius, independents rv's. Thus to find $f_{\Theta}(\theta)$ you can integrate the joint density in

$$f_{\Theta}(\theta)=\int_0^{\infty}f_{R\Theta}(r,\theta)dr$$

Or simply factorize the joint density (1) as I showed in (2).

Note for the O.P.: when writing a density

$$f_R(r)$$

the subscrtipt letter indicates the rv thus it must be a capital letter. The letter in the brackets indicates the value that the random variable assumes thus it is a small letter.


问:
A sequence where the nth term is the cumulative sum over n-1
Consider the following sequence:

$u_{n}=-\frac{u_{1}}{n}-\frac{\beta\gamma}{n}(u_{1}+...+u_{n-1})$, where $u_1$, $\beta$ and $\gamma$ are some none zero constant.

In this sequence, can I obtain the expression for $u_{n}$ as a function of $u_{1}$? If I can, what would it be? At first, it seems pretty straightforward but I couldn't get a nice closed form.

答:
Here is an approach via generating functions (though it mostly proves that this method is not always the best one!). We define $U(x):=\sum_{n=1}^\infty u_n x^n$, and for convenience I'll introduce $r:=\beta\gamma$ as well. To obtain a relation on this GF, we first rearrange the recurrence as $$u_n+u_1+r (u_1+\cdots+u_{n-1})=0.$$
We now multiply both sides of the recurrence by $x^{n-1}$ and sum from $n=2$ to infinity to obtain

$$\sum_{n=2}^\infty n u_n x^{n-1} + u_1\sum_{n=2}^\infty x^{n-1}+\sum_{n=2}^\infty r(u_1+\cdots +u_{n-1})x^{n-1}=0.$$
We deal with these sums one-by-one. For the first, we use the power rule for derivatives to write

$$\sum_{n=2}^\infty n u_n x^{n-1}=\frac{d}{dx}\sum_{n=2}^\infty n u_n x^n=\frac{d}{dx}[U(x)-u_1 x]=U'(x)-u_1.$$
For the second, we have the geometric sum

$$u_1\sum_{n=2}^\infty x^{n-1}=\frac{u_1 x}{1-x}.$$

For the third, we may rewrite the series as

\begin{align}
\sum_{n=2}^\infty r(u_1+\cdots +u_{n-1})x^{n-1}
&=\sum_{n=1}^\infty r(u_1+\cdots +u_n)x^{n}\\
&=\sum_{n=1}^\infty \sum_{k=1}^n r u_k x^n \\
&=\sum_{1\leq k \leq n\leq \infty} r u_k x^n \\
&=\sum_{k=1}^\infty \sum_{n=k}^\infty r u_k x^n \\
&=\sum_{k=1}^\infty r u_k \frac{x^k}{1-x} \\
&= \frac{r}{1-x}U(x)
\end{align}

Therefore we have the differential equation

$$U'(x)-u_1+\frac{u_1 x}{1-x}+\frac{r}{1-x}U(x)=0.$$

This may be solved by seeking an appropriate integrating factor. Indeed, we have
\begin{align}
(1-x)^{-r}\frac{d}{dx}[(1-x)^r U(x)]
&=U'(x)+\frac{r}{1-x}U(x)\\
&=u_1-\frac{u_1 x}{1-x}\\
&=u_1\frac{1-2x}{1-x}
\end{align}
We move $(1-x)^{-r}$ over, integrate both sides with base point $x=0$, and move $(1-x)^r$ over. (Note that $U(0)=0$.) We obtain

$$U(x) = (1-x)^{-r}u_1 \int_0 (1-2x)(1-x)^{r-1}\,dx=u_1\frac{(1+r)-2rx -(1+r)(1-x)^r}{r(r-1)}.$$

If we fearlessly expand this using the binomial expansion, we obtain

\begin{align}
U(x)
&=\frac{u_1}{r(r-1)}\left[(1+r)-2rx-(1+r)\sum_{k=0}^\infty \binom{r}{k}(-x)^k\right]\\
&=u_1\left[x+\sum_{k=2}^\infty \binom{r}{k}(-x)^k\right]
\end{align}
Hence we finally conclude that, for $n\geq 2$, we have $$u_n=u_1 (-1)^n \binom{r}{n}=u_1 (-1)^n \frac{r(r-1)\cdots(r-n+1)}{n!}=-u_1 \frac{r(1-r)\cdots(n-1-r)}{n!}$$
which agrees with the (much shorter) deduction by Bumblebee.