问:
Image measure space
Is it possible to come up with a measure space  $(X, \mathcal{M}, \mu)$ such that
$\{ \mu(E) : E \in \mathcal{M}\}=[0,+\infty]\setminus \Bbb Q^+$ , where $\Bbb Q^+$ denotes positive rationals

I suspect yes, although I have no idea how to construct such a measure space like this

答:
There is no measure space $(X, \mathcal{M}, \mu)$ such that $\{ \mu(E) : E \in \mathcal{M}\}=[0,+\infty]\setminus \Bbb Q^+$ (where $\Bbb Q^+$ denotes the positive rational numbers).

Proof:

Let $(X, \mathcal{M}, \mu)$ be a measure space. Suppose that $\{ \mu(E) : E \in \mathcal{M}\}=[0,+\infty]\setminus \Bbb Q^+$.

By a result from  Sierpinski, we know that the image of an non-atomic measure is an interval (see for instance, this article in Wikipedia). In fact, from Sierpinski's result, we can easy prove that: if the image of a measure $\mu$ does not contains any interval, then for every $E \in \mathcal{M}$ such that $\mu(E)>0$, $E$ contains (at least) one atom.

Since $\{ \mu(E) : E \in \mathcal{M}\}=[0,+\infty]\setminus \Bbb Q^+$, we can take a $E \in \mathcal{M}$ such that $0<\mu(E)<+\infty$. Let $A \subseteq E$ be an atom. We have that $0<\mu(A)  <+\infty$.

Now take $r\in \Bbb Q$ such that $\mu(A)< r < 2 \mu(A)$. So, we have that
$$0< r-\mu(A) <  \mu(A)$$
Since $\mu(A) \notin \Bbb Q$ and $r\in \Bbb Q$, we have that $ r-\mu(A) \in (0, +\infty) \setminus \Bbb Q^+$. So there is $B \in \mathcal{M}$, such that $\mu(B) = r-\mu(A)$.

Since $A$ is an atom and $\mu(B) < \mu(A)$, we have that $\mu(A\cap B)=0$. So
$$\mu(A \cup B) = \mu(A) + \mu(B)=  \mu(A) + r-\mu(A) = r \in \Bbb Q$$
Contradiction to the assumption that $\{ \mu(E) : E \in \mathcal{M}\}=[0,+\infty]\setminus \Bbb Q^+$.

Title: Image measure space


问:
If $X$ and $Y$ are ordinals and $Y\subset X$, then $Y$ is a section of $X$
I'm studying ordinals via Cameron's Sets, Logic and Categories. On p. 42, Cameron works to prove that

If $X$ and $Y$ are ordinals and $Y\subset X$, then $Y$ is a section of $X$.

Proof: Take $a$ to be the least element of $X\setminus Y$. Then $X_a\subseteq Y$. [The proof follows by showing that $Y\subseteq X_a$ by applying the trichotomy rule to $a$.]

I have two questions about this proof.

How does $X_a\subseteq Y$ follow immediately from $a$ being the least element of $X\setminus Y$? It isn't intuitive to me.

Wouldn't the proof proceed more simply as follows? Let $a$ be the least element of $X\setminus Y$. So $\forall y\in Y$, $y<a$ since $Y$ is an ordinal. Now $\forall x\in X$, if $x>a$ or $x=a$, $x\notin Y$. So $Y=\{x\in X:x<a\}=X_a$

答:
By our choice of $a$ as the least element of $X\setminus Y,$ we have for any $b\in X_a$ that $b\in X=(X\cap Y)\cup(X\setminus Y)$ and $b<a,$ so $b\notin X\setminus Y,$ and so $b\in X\cap Y=Y.$ Thus $X_a\subseteq Y.$

Your approach starts off fine, applying the trichotomy rule to $a,$ and concluding that $Y\subseteq X_a.$ However, it doesn't show that $X_a\subseteq Y,$ only that $Y\subseteq X_a$ in another way (by contrapositive).

ContentLicense: CC BY-SA 4.0


问:
Complexity class of comparison of power towers
Consider the following decision problem: given two lists of positive integers $a_1, a_2, \dots, a_n$ and $b_1, b_2, \dots, b_m$ the task is to decide if $a_1^{a_2^{\cdot^{\cdot^{\cdot^{a_n}}}}} < b_1^{b_2^{\cdot^{\cdot^{\cdot^{b_m}}}}}$.

Is this problem in the class $P$?

If yes, then what is the algorithm solving it in polynomial time?

Otherwise, what is the fastest algorithm that can solve it?

Update:

I mean polynomial type with respect to the size of the input, i.e. total number of digits in all $a_i, b_i$.

$p^{q^{r^s}}=p^{(q^{(r^s)})}$, not $((p^q)^r)^s$.

答:
For readability I'll write $[a_1,a_2,\ldots,a_n]$ for the tower $a_1^{a_2^{a_3^\cdots}}$.

Let all of the $a_i,b_i$ be in the interval $[2,N]$ where $N=2^K$ (if any $a_i$ or $b_i$ is 1 we can truncate the tower at the previous level, and the inputs must be bounded to talk about complexity).

Then consider two towers of the same height
$$
T=[N,N,\ldots,N,x] \quad \mathrm{and} \quad S=[2,2,\dots,2,y]
$$
i.e. T is the largest tower in our input range with $x$ at the top, and S is the smallest with $y$ at the top.

With $N, x\ge 2$ and $y>2Kx$ then
$$
\begin{aligned}
2^y & > 2^{2Kx} \\
    & = N^{2x} \\
    & > 2log(N)N^x &\text{ as $x \gt 1 \ge \frac{1+log(log(N))}{log(N)}$} \\
    & = 2KN^x
\end{aligned}
$$
Now write $x'=N^x$ and $y'=2^y>2Kx'$ then
$$
[N,N,x]=[N,x']<[2,y']=[2,2,y]
$$
Hence by induction $T<S$ when $y>2Kx$.

So we only need to calculate the exponents down from the top until one exceeds the other by a factor of $2K$, then that tower is greater no matter what values fill in the lower ranks.

If the towers have different heights, wlog assume $n>m$, then first we reduce
$$
[a_1,a_2,\ldots,a_n] = [a_1,a_2,\ldots,a_{m-1},A]
$$
where $A=[a_m,a_{m+1},\ldots,a_n]$. If we can determine that $A>2Kb_m$ then the $a$ tower is larger.

If the towers match on several of the highest exponents, then we can reduce the need for large computations with a shortcut. Assume $n=m$, that $a_j> b_j$ for some $j<m$ and $a_i=b_i$ for $j2K$. So we don't need to compute $X$ fully if we can determine that it exceeds $\log(2K)/\log(a_j/b_j)$.

These checks need to be combined with a numeric method like the one @ThomasAhle gave. They can solve the problem that method has with deep trees that match at the top, but can't handle $[4,35,15],[20,57,13]$ which are too big to compute but don't allow for one of these shortcuts.

ContentLicense: CC BY-SA 3.0

Body:
For readability I'll write $[a_1,a_2,\ldots,a_n]$ for the tower $a_1^{a_2^{a_3^\cdots}}$.

Let all of the $a_i,b_i$ be in the interval $[2,N]$ where $N=2^K$ (if any $a_i$ or $b_i$ is 1 we can truncate the tower at the previous level, and the inputs must be bounded to talk about complexity).

Then consider two towers of the same height
$$
T=[N,N,\ldots,N,x] \quad \mathrm{and} \quad S=[2,2,\dots,2,y]
$$
i.e. T is the largest tower in our input range with $x$ at the top, and S is the smallest with $y$ at the top.

With $N, x\ge 2$ and $y>2Kx$ then
$$
\begin{aligned}
2^y & > 2^{2Kx} \\
    & = N^{2x} \\
    & > 2log(N)N^x &\text{ as $x \gt 1 \ge \frac{1+log(log(N))}{log(N)}$} \\
    & = 2KN^x
\end{aligned}
$$
Now write $x'=N^x$ and $y'=2^y>2Kx'$ then
$$
[N,N,x]=[N,x']<[2,y']=[2,2,y]
$$
Hence by induction $T<S$ when $y>2Kx$.

So we only need to calculate the exponents down from the top until one exceeds the other by a factor of $2K$, then that tower is greater no matter what values fill in the lower ranks.

If the towers have different heights, wlog assume $n>m$, then first we reduce
$$
[a_1,a_2,\ldots,a_n] = [a_1,a_2,\ldots,a_{m-1},A]
$$
where $A=[a_m,a_{m+1},\ldots,a_n]$. If we can determine that $A>2Kb_m$ then the $a$ tower is larger.

If the towers match on several of the highest exponents, then we can reduce the need for large computations with a shortcut. Assume $n=m$, that $a_j> b_j$ for some $j<m$ and $a_i=b_i$ for $j<i\le m$.
Then
$$
[a_1,a_2,\ldots,a_m] = [a_1,a_2,\ldots,a_j,X] \\
[b_1,b_2,\ldots,b_m] = [b_1,b_2,\ldots,b_j,X]
$$
and the $a$ tower is larger if $(a_j/b_j)^X>2K$. So we don't need to compute $X$ fully if we can determine that it exceeds $\log(2K)/\log(a_j/b_j)$.

These checks need to be combined with a numeric method like the one @ThomasAhle gave. They can solve the problem that method has with deep trees that match at the top, but can't handle $[4,35,15],[20,57,13]$ which are too big to compute but don't allow for one of these shortcuts.


问:
Does the standard normal distribution have a heavy right tail?
I read about heavy right tail and I saw  that a distribution is said to have a heavy right tail if its tail probabilities vanish slower than any exponential
$$\forall >0: \lim_{x\to\infty}e^{tx}(>)=\infty.$$

Does the standard normal distribution have a heavy right tail?

答:
The definition of a heavy right tailed distribution is that the moment generating function $M_X(t)$ is infinite for all $t > 0$ (see here). This is not the case for the standard normal distribution, where we have $$M_X(t)=\exp\left(\frac{t^2}{2}\right).$$
What you are writing about the limit $\lim_{x\to\infty}e^{tx}P(X>x)=\infty$ is only a necessary condition (an implication of the definition). But since it was part of your question: To calculate $\lim_{x\to\infty}e^{tx}P(X>x)$ you can use $P(X>x)=1-F_X(x)$ and then apply L'Hôpital's rule:
$$\lim_{x\to\infty}e^{tx}P(X>x)=\lim_{x\to\infty}\frac{-\exp(-x^2/2)/\sqrt{2\pi}}{-t\cdot e^{-tx}}=0.
$$
So by calculating this you also see that the standard normal distribution is not a heavy right tailed distribution.


问:
Complex Analysis: If f is an odd function, find an expression for f.
Complex Analysis

Suppose that $f$ is a holomorphic function in $\mathbb{C}\setminus\{0\}$ and satisfies $$\vert f(z) \vert \leq \vert z \vert ^{2} +\frac{1}{\vert z \vert ^{2}}$$

If f is an odd function, find an expression for f.

I thought about calculating the Laurent Serie of f, but I can't calculate Laurent's series explicitly (there is no way to calculate the coefficients), Dai thought about estimating, using the invariance of the circle to know which one cancels out. I cannot develop this question

答:
The inequality implies that for $|z|<1$
$$
|z^2f(z)|\le |z|^4+1\le 2\ .
$$

It follows from Riemann's theorem that $z=0$ is a removable singularity for $g(z):=z^2f(z)$. Thus, $g(z)=\sum_{n=0}^\infty a_nz^n$ on the punctured complex plane.

The polynomial growth
$$
|g(z)|\le |z|^4+1
$$
implies that $g$ is a polynomial (or order at most $4$) by Liouville’s theorem

So you can write
$$
z^2f(z)=a_0+a_1z+a_2z^2+a_3z^3+a_4z^4.
$$

The parity of $f$ then implies that $a_0=a_2=a_4=0$ by the fundamental theorem of algebra: $g(z)+g(-z)=0$ implies that
$$
a_0+a_2z^2+a_4z^4=0
$$
for all $z$. (But a nonzero polynomial has only finitely many roots.)


问:
How to find the volume of this region?
I want to find the volume of the following subset of $\mathbb R^3$:
\begin{equation}
(x^2+y^2)^2\leq x,\quad 0\leq z\leq 2x-\sqrt{x^2+y^2}\text{.}
\end{equation}

I tried to draw a picture of the given region in my head, but I couldn't. So I changed it into a cylindrical coordinate system, and the region became
\begin{equation}
r^4\leq r\cos\theta,\quad 0\leq z\leq 2r\cos\theta-r,\quad r>0,\quad 0<\theta<2\pi\text{.}
\end{equation}

But I still don't know what to do.

答:
To define correctly the region you only need to see for what values of $\theta$ the given conditions holds. As $0<r\leqslant (\cos \theta)^{1/3}$ from here you know that valid values of $\theta $ are just possible in the region $[-\pi/2,\pi/2]$ because otherwise $\cos \theta <0$.

Now from the condition $0\leqslant z\leqslant r(2\cos \theta -1)$, as $r>0$, then the valid values of $\theta $ for this condition are these such that $2\cos \theta -1\geqslant 0\Leftrightarrow \cos \theta \geqslant 1/2 \Leftrightarrow \theta \in [-\pi/3,\pi/3]$.

Then, all together, gives that the valid range for $\theta $ for the two conditions is $[-\pi/3,\pi/3]$, so the integral of the volume take the form

$$
\int_{-\pi/3}^{\pi/3}\int_0^{(\cos \theta )^{1/3}}\int_{0}^{r(2\cos \theta -1)}r\mathop{}\!d z\mathop{}\!d r\mathop{}\!d \theta
$$

Title: How to find the volume of this region?


问:
Proof of binomial coefficient formula.
How can we prove that the number of  ways choosing $k$ elements among $n$ is    $\frac{n!}{k!(n-k)!} = \binom{n}{k}$    with $k\leq n$?
This is an accepted fact in every book but i couldn't find a proof.</p>

答:
If you have 5 people (A, B, C, D & E) but wants to build a team of three people, the most number of ways you can do is 5x4x3= 60 ways. But realize that team ABC and CBA is basically same team.

So for three people, you can have 3x2x1=6 different combinations.

Meaning that, making a team with 5 poeple with 3 positions, you can have total (5x4x3) / (1x2x3) = 10 different combinations. This number 10 is the 3rd number on 5th row of Pascal's triangle. This basically tells us the number of combination we can have.

And (5x4x3) / (1x2x3) = 5! / (3! x 2!) = C (5, 3)

Title: Proof of binomial coefficient formula.


问:
Find the modular residue of this product..
Please help me solve this and please tell me how to do it..

$12345234 \times 23123345 \pmod {31} = $?

edit: please show me how to do it on a calculator not a computer thanks:)

答:
We want to replace these big numbers by much smaller ones that have the same remainder on division by $31$.

Take your first big number $12345234$. Divide by $31$ with the calculator. My display reads $398233.3548$. Subtract $398233$. My calculator shows $0.354838$. Multiply by $31$. The calculator gives $10.999978$. If it were perfect, it would say $11$.

Do the same with the other big number. My calculator again says $10.99978$, and if perfect would say $11$.

Multiply $11$ by $11$, and find the remainder when $121$ is divided by $31$. Again, we could use the calculator. But it can be done in one's head. The remainder when we divide by $31$ is $28$.

Remark: It can be fairly important not to write down an intermediate calculator result, and rekey. The reason is that the calculator inernally keeps guard digits, that is, it is calculating to greater precision than it displays. If you rekey, typing in only digits that you see, you will lose some of the built-in accuracy that you paid for. For similar reasons, it is useful to learn to make use of the memory features of your calculator.

Let's ee why the procedure we used works. When we divide $12345234$ by $31$, the stuff before the decimal point is the quotient. The remainder is unfortunately not given directly, but what's "left over" is (approximately) $0.354838$. This is a decimal representation of $\frac{r}{31}$, where $r$ is the (integer) remainder. To recover $r$, we multiply $0.354838$ by $31$. Because of internal roundoff, we usually don't get an integer, but most of the time, if the divisor (here $31$) is not too large, we get an answer that is very close to an integer, so we can quickly decide what $r$ must be.


问:
If an algebra is free on a set, that set is linearly independant
Let $R$ be a ring (commutative with unit), and $E$ a set.
We denote $A$ the free $R$-algebra on $E$.

Note. $A$ is a non-commutative algebra.

I wanted to show that if we consider $A$ as a left $A$-module, then the set $E$ is $A$-linearly independant. That is, if $\sum_{e \in E} a_e e_e = 0$, then $a_e = 0$ for all $e \in E$ (Note. $a_e \in A$).

Since $A$ is free on $E$, we already know that the set of all words in E and the empty word form an $R$-basis of $A$. I think with that I have all I need for the proof, but I miss something on how to use this fact.

Any hint or help would be great.

Best regards.

K. Y.

答:
Write $E = \{e_1, \ldots, e_n\}$ and assume $\sum_{i=1}^n a_ie_i = 0$ with $a_i \in A$. Every $a_i$ can be uniquely written as
\begin{equation}
a_i = \sum_{k=0}^\infty \sum_{i_1, \ldots, i_k \in \{1, \ldots, n\}} a^{(i)}_{i_1,\ldots, i_k} e_{i_1}\ldots e_{i_k}\tag{1}\end{equation}
for some $a^{(i)}_{i_1,\ldots, i_k} \in R$ with at most finitely many of them nonzero. Plugging this in we get
\begin{align}
0 &= \sum_{i=1}^n a_ie_i \\
&= \sum_{i=1}^n\left(\sum_{k=0}^\infty \sum_{i_1, \ldots i_k \in \{1, \ldots, n\}} a^{(i)}_{i_1,\ldots, i_k} e_{i_1}\ldots e_{i_k}\right)e_i\\
&= \sum_{i=1}^n\sum_{k=0}^\infty \sum_{i_1, \ldots i_k \in \{1, \ldots, n\}} a^{(i)}_{i_1,\ldots, i_k} e_{i_1}\ldots e_{i_k}e_i\\
&= \sum_{k=0}^\infty \sum_{i_1, \ldots, i_k,i \in  \{1, \ldots, n\}} a^{(i)}_{i_1,\ldots, i_k} e_{i_1}\ldots e_{i_k}e_i.
\end{align}
Now the monomials $e_{i_1}\ldots e_{i_k}e_i$ are all different since they are parameterized by $(i_1, \ldots, i_k,i) \in \{1,\ldots, n\}^{k+1}$ so by definition it follows that all scalars are $0$:
$$a^{(i)}_{i_1,\ldots, i_k}=0, \quad \forall i_1, \ldots, i_k,i \in  \{1, \ldots, n\}.$$

Therefore $(1)$ implies  $a_i = 0$ for all $1 \le i \le n$ so $E$ is $A$-linearly independent.


问:
Evaluate the integral using the given data

Let $f:[0,1]\to [0,1]$ be a continuous function such that $x^2 +(f(x))^2\le 1$ for all $x\in [0,1]$ and $\int_0^1 f(x).dx=\frac{\pi}{4}$, then find $\int_{-1/2}^{1/\sqrt 2} \frac{f(x)}{1-x^2}.dx$

The part that’s throwing me off the inequality function. I have no idea how to solve such that problems. Can I get a hint?

答:
Hint :

$$f(x) \leq \sqrt{1-x^2}$$

It means graph of $f(x)$ always lies on or below the unit circle in the first quadrant.

Now what can you conclude from following? $$\int_{0}^{1} f(x) \, dx = \frac{\pi}{4}$$

Think area!


问:
Number of roots of the equation $f(x)= \int_0^x (t-1)(t-2)(t-3)(t-4)dt =0$
I have the following question before me:
Find the number of roots of the equation $f(x)= \int_0^x (t-1)(t-2)(t-3)(t-4)dt =0$ in the interval $[0,5]$.
$0$ is clearly one of the roots. But how can I find other roots, if any? I tried evaluating the integral and came up with a fifth degree polynomial in $x$ having no constant term. The equation seemed quite daunting to me. How can I get the roots quicker? Please suggest.

答:
$f(x)=\dfrac{x^5}{5}-\dfrac{5x^4}{2} +\dfrac{35x^3}{3}-25x^2+24x$ so $x=0$ is a root in $[0,5]$

On the other hand $$\dfrac{df}{dx}=\dfrac{d}{dx}\int_0^x((t-1)(t-2)(t-3)(t-4))dt=(x-1)(x-2)(x-3)(x-4)$$ which gives two maximums $(1,\dfrac{251}{30})$ and $(3,\dfrac{81}{10})$ and two minimums $(2,\dfrac{116}{15})$ and $(4,\dfrac{112}{15})$ so all of the four extremums being positive there are not real solutions in $[1,5]$. We can not say that a non-real solution is in $[0,5]$ because it would implies an order in $\mathbb C$ where there is not any structure of order.

Consequently the only root is $x=0$-


问:
Calculate the following complex integral
I try to calculate the following integral:

$$\oint_{\{\vert z\vert=3\}}\frac{e^z}{(z-1)(z-2i)}$$

The integral encloses 2 poles in $1$ at $2i$. Using the homotopy we see that expect the two circle of radius $\epsilon$ around both pole, the rest is cancelled out. Now my idea is to try to split the integral into something like

$$\oint_{\{\vert z\vert=3\}}f(z)dz+\oint_{\{\vert z\vert=3\}}f(z)dz$$
and use the cauchy integral formular but I go stucked...

EDIT: If I write now $$
\frac{1}{(z-1)(z-2i)}=\frac{A}{z-1}+\frac{B}{z-2i}\ .
$$
as in the answer and I solve it. Then we get$A=-B$ and $A=-\frac{1}{2i-1}$. Okay now we get
$$A\oint_{\{\vert z\vert=3\}}\frac{e^z}{z-1}-A\oint_{\{\vert z\vert=3\}}\frac{e^z}{z-2i}$$
By the Cauchy Formular this is equal to:
$$2\pi iA(e^1+e^{-2i})$$
Is this right?

答:
You can use the Cauchy integral formula by writing
$$
\frac{1}{(z-1)(z-2i)}=\frac{A}{z-1}+\frac{B}{z-2i}\ .
$$

Alternatively, you can use the residue theorem. The answer should be
$$
2\pi i (\textrm{Res}(f;1)+\textrm{Res}(f;2i))
$$
where
$$
\textrm{Res}(f;1)=\frac{e}{1-2i},\quad
\textrm{Res}(f;2i)=\frac{e^{2i}}{2i-1}
$$


问:
Proof verification: Atiyah-Macdonald Exercise 3.7(i)
A multiplicatively closed subset $S$ of a ring $A$ is said to be saturated if:
$$ xy\in S \Leftrightarrow x\in S \text{ and } y\in S $$

Exercise 3.7(i): $S$ is saturated if and only if $A-S$ is a union of prime ideals.

$(\Leftarrow)$: Let $A-S$ be a union of prime ideals $\{ \mathfrak{p}_i \}_{i\in I}$.
$$ S = A-(A-S) = A-\bigcup_{i\in I} \mathfrak{p}_i = \bigcap_{i\in I} A\setminus \mathfrak{p}_i $$
Let $xy\in S$ then $xy\in A\setminus \mathfrak{p}_i$ for all $i\in I$. Let $\mathfrak{m}_i$ be the maximal ideal in the local ring $A_{\mathfrak{p}_i}$ and consider the maps:
$$ A \overset{\mu}{\to} A_{\mathfrak{p}_i} \overset{\pi}{\to} A_{\mathfrak{p}_i} / \mathfrak{m}_i $$
Thus, $\pi(\mu(xy)) \neq 0$ since $\mu(xy)$ is a unit. Therefore, $\pi(\mu(x)), \pi(\mu(y)) \neq 0$. Suppose that $x\in \mathfrak{p}_i$ (any $i$) for a contradiction, then $(x)\subset \mathfrak{p}_i$ and hence $(x)_{\mathfrak{p}_i} \subset \mathfrak{p}_i^e = \mathfrak{m}_i$; therefore, $\pi(\mu(x)) = 0$. Contradiction. Therefore, $x\in S$; similarly for $y$.

($\Rightarrow$):  Let $x \in A$ be a unit, then $x\cdot 1/x = 1\in S$. Therefore, $x\in S$. Therefore, $A\setminus S$ contains only non-units. Consider:
$$ U := \bigcup_{\mathfrak{p} \text{ prime} \\ \mathfrak{p} \subset A\setminus S} \mathfrak{p} $$
I claim that $A\setminus S \subset U$. Let $x\in A\setminus S$, then $x\in \mathfrak{m}$ for some maximal ideal $\mathfrak{m}$ (since $x$ is a non-unit). Therefore, $(x)\subset \mathfrak{m}$ and $S^{-1}(x) \subset S^{-1}\mathfrak{m}$ where $S^{-1}\mathfrak{m}$ is a prime ideal in $S^{-1}A$ by the correspondence theorem. The contraction of $S^{-1}\mathfrak{m}$ back into $A$ is a prime ideal which contains $x$ and is contained in $A\setminus S$. Therefore, $x\in U$.

I hope I used the correspondence theorem correctly. Please let me know if my reasoning is sound. Thank you.

EDIT: I think I have not guaranteed that $\mathfrak{m}$ does not meet $S$, so this proof might not work as stated. I will try to fix it tomorrow morning. Please let me know your thoughts anyway.

答:
Let $A-S$ be a union of prime ideals. Suppose $xy\in S$, but $x\notin S$. Then $x\in A-S$, which is a union of prime ideals, and hence $x\in\mathfrak{p}\subseteq A-S$. So $xy\in\mathfrak{p}$ too. So $xy\in A-S$. Contradiction.

Let $S$ be saturated. Let $x\in A-S$. You showed $x$ is not a unit. So $x$ remains a non-unit in $S^{-1}A$. Then $x$ is in a maximal (hence prime) ideal in $S^{-1}A$. So $x$ is in a prime ideal in $A-S$ by the correspondence.


问:
Integration of $\displaystyle \int\frac{1}{1+x^8}\,dx$

Compute the indefinite integral
  $$
\int\frac{1}{1+x^8}\,dx
$$

My Attempt:

First we will factor $1+x^8$

$$
\begin{align}
1+x^8 &= 1^2+(x^4)^2+2x^4-2x^4\\
&= (1+x^4)^2-(\sqrt{2}x^2)^2\\
&= (x^4+\sqrt{2}x^2+1)(x^4-\sqrt{2}x^2+1)
\end{align}
$$

Then we can rewrite the integral as

$$
\int\frac{1}{1+x^8}\,dx = \int \frac{1}{(x^4+\sqrt{2}x^2+1)(x^4-\sqrt{2}x^2+1)}\,dx$$

To use partial fractions let $t = x^2$ to get

$$
\frac {1}{(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)} = \frac{At+B}{t^2+\sqrt{2}t+1}+\frac{Ct+D}{t^2-\sqrt{2}t+1}
$$

This method of solving the problem becomes very complex. Is there a less complex approach to the problem?

答:
Why not splitting up in fractions until you have first degree polynomials in the nominators?

$$\frac{1}{1+x^8}=\frac{A}{x-e^{i\pi/8}}+\frac{B}{x-e^{-i\pi/8}}+\frac{C}{x-e^{i3\pi/8}}+\frac{D}{x-e^{-i3\pi/8}}+\frac{E}{x-e^{i5\pi/8}}+\frac{F}{x-e^{-i5\pi/8}}+\frac{G}{x-e^{i7\pi/8}}+\frac{H}{x-e^{-i7\pi/8}}$$

or if you prefer without the complex numbers

$$\frac{1}{1+x^8}=\frac{ax+b}{x^2-2\cos(\pi/8)x+1}+\frac{cx+d}{x^2-2\cos(3\pi/8)x+1}+\frac{ex+f}{x^2-2\cos(5\pi/8)x+1}+\frac{gx+h}{x^2-2\cos(7\pi/8)x+1} \; .$$

With the complex formula, you can find the coefficients easily as follows

$$A=\lim_{x \to e^{i\pi/8}}\frac{x-e^{i\pi/8}}{1+x^8}\overset{\text{H}}{=}\lim_{x \to e^{i\pi/8}}\frac{1}{8x^7}=\frac{e^{-i7\pi/8}}{8}$$

where I used de l'Hôpital's rule.


问:
Limit $t\to \infty$ of a convolution
I would like to expect that, if $h(t)=f(t)\ast g(t)$, ie,

$$h(t)=\int_0^t f(t-t')g(t')dt',$$

and $\lim_{t\to \infty} f(t)=\lim_{t\to \infty} g(t)=0,$ so $\lim_{t\to \infty} h(t)=0$ (or maybe I have at least some other information).

However, I have uncertanties for proceed.

Thank you so much.

答:
As written in the comments it is not true. A counterexample would be with $f(s)=1/\sqrt{s}=g(s)$. Then we get
$$ h(t)=\int_0^1 \frac{1}{\sqrt{1-x}\sqrt{x}} dx = \pi.$$
Thus, $\lim_{t\rightarrow\infty} h(t)=\pi \neq 0$. With a different power, $f(s)=s^{-1/4}=g(s)$ we get that $\lim_{t\rightarrow\infty} h(t)=\infty$. Indeed, we have in this case
$$ h(t) = \int_0^t \frac{1}{(t-s)^{1/4}} \frac{1}{s^{1/4}} ds
= \int_0^t \frac{1}{t^{1/4}}\frac{1}{(1-s/t)^{1/4}} \frac{1}{s^{1/4}} ds
\stackrel{x=s/t, ds=tdx}{=} t^{3/4} \int_0^1 \frac{1}{(1-x)^{1/4} x^{1/4}} dx.   $$

Title: Limit $t\to \infty$ of a convolution


问:
Prove that every natural number is either even or odd using induction
It is a pretty basic question at first sight. It seems to be intuitively correct but I can't figure out a way to prove it. I think we have to use strong induction but even after sitting on this question for a long time, I still haven't got any leads.</p>

答:
Define an even number as a multiple of 2, and an odd number as a multiple of 2 plus 1.

Induction. Your inductive hypothesis is $H_n$: the positive integer $n$ is even or odd. $H_1$ is true, because 1 is a multiple of 2 plus 1. Suppose $H_n$ is true. Then either (i) $n$ is a multiple of 2 or (ii) $n$ is a multiple of 2 plus 1. In case (i), $n+1$ is a multiple of 2 plus 1; in case (ii) $n+1$ is a multiple of 2. So in both cases $H_{n+1}$ is true. So $H_n$ is true for all positive integers $n$.


问:
Question about row equivalent matrices
I am training for my exam, and I don't understand very well what does it means that two matrices are row equivalent. I understand that row equivalence mean that they have the same solution. But how this option I write are linked to the row equivalence. If there any of the statement that are true? If you can explain how they are related to row equivalence.

Thanks

If $A$ is invertible so $B$ invertible

if $Ax = 0$ has one solution $\implies$ $Bx = 0$ has one solution

$\text{det}A = \text{det}B$

$\text{rank A} = \text{rank} B$

答:
Let me explain why is enough to prove statement number $3$. (At least when we are talking about square matrices, otherwise $det(A)$ has no meaning)

$3\implies 1$: A matrix is invertible if and only if its determinant is non-zero. If $A$ is invertible, we have $det(A)\neq 0$, using $3$, we get $det(B)\neq 0$ and hence $B$ is invertible.

$3\implies 2$: A matrix $A$ is invertible if and only if the only solution to the equation $Ax=0$ is $x=0$ . That is $1\iff 2$.

$3\implies 4$:  In general a $m×n$ matrix has rank $r$ if and only if some $r×r$ minor does not vanish and every $(r+1)×(r+1)$ minor does vanish. We already know that vanishing (having determinant equal to 0) is not chaged by row operations (we are assuming $3$) hence the rank of a matrix is also invariant.

Try to prove now that if two square matrices are row equivalent then they have the same determinant. If you can't prove it, tell me and I will add it to the answer later.

Title: Question about row equivalent matrices


问:
A matrix with all entries being positive, each row has the same sum.
A matrix $A$ with all entries being positive, each row has the same sum, that is, $\sum_j a_{ij}=c>0$. Show that up to constant multiplier, the eigenvector $(1,\cdots,1)'$ is the unique eigenvector with all entries being positive.

Let $x=(x_1,\cdots,x_n)'$ being the eigenvector of $A$, with eigenvalue $\mu>0$, $x_i$ being not identically the same, How to derive a contradiction?

答:
We can start from here. Let's take an eigenvector x with all positive entries and let $m=min_i x_i$ and $M=max_i x_i$ the minimum and maximum of the components of the eigenvector. Than:

$\lambda x_i=\sum_j a_{i,j}x_j$ [1]

$Cm \le \lambda x_i \le CM$ [2]

for every $i$. Since this is valid for every $i$ it must be valid for $x_i=m$ and $x_i=M$, so that we have $C\le \lambda$ from the left inequality and $C \ge \lambda$ from the right. This shows that $\lambda=C$.

Now we can look back at the system [1] where $x_i=m$ with $\lambda=\sum_j a_{i,j}$:

$m \sum_j a_{i,j}=\sum_j a_{i,j}x_j$

, that is:

$\sum_j a_{i,j} (x_j-m)=0$

,which shows that we must have $x_j=m$ for each $j$ since all entries $a_{i,j}$ are positive.


问:
Does $f, f' \in L^1([0, \infty))$ imply that $\lim_{x \to \infty} xf(x) = 0$?
Does $\int_0^\infty |f(x)| \, dx$ and $\int_0^\infty |f'(x)| \, dx$ being finite imply that $\lim_{x \to \infty} xf(x) = 0$? </p>

<p>(Context: I am working through an analytic number theory textbook. In a lemma, the book assumes the conditions on $f$ and $f'$. The term $xf(x)$ appears from an integration by parts, but is tossed out, which must mean that $\lim_{x\to\infty} xf(x) = 0$. However, I've tried to prove this and was not successful. I also cannot think of any counterexamples.)</p>

答:
One way to think about problems of such type is in terms of linear interpolation.
Define your function $f(n)=a_n>0$ let $f$ to be linear between $[n,n+1].$ Then, $f\in L_1$ is equivalent to $\sum_{n=1}a_n<\infty$ whereas  $f'\in L_1$ is equivalent to $\sum_{n=1}|a_{n+1}-a_n|<\infty.$ Now it is clear that the standard example of series mentioned by Nick satisfies both conditions. The only thing left is to "smoothen" $f$ at the corners to make sure that $f'$ exists.</p>


问:
Definition of limits in $\mathbb R^3$
Let$$ f: \ D_f \longrightarrow \mathbb{R},$$
where
$$ D_f \subset \mathbb{R}^n, \ n\ge 2.$$
In some materials, it is said that for  function f to have a limit at point $ x_0 \in \mathbb{R}^n$,
for all positive real numbers d the following has to be true: $$ D_f \cap  \left\{  x\in \mathbb{R}^n \mid 0< \left \Vert x-x_0\right \Vert <d \right\} \ne \varnothing, \ \ x\in \mathbb{R} $$

I wonder if this condition is enough. For example, let
$$ D_f=\left\{(x, y) \mid x\ge 0, y\ge 0 \right\} $$

and $ x_0=(0,0 )$.

Now the condition
$$ \left[0, \infty \right) \times \left[0, \infty \right) \cap  \left\{  x\in \mathbb{R}^n \mid 0< \left \Vert (x,y)-(0,0)\right \Vert <d \right\} \ne \varnothing$$
holds for all $d>0$, but the limit does not exist for sure, since one can not approach the point $(0,0)$ from all directions.

Could someone please clarify?

答:
Yes, the limit may well exist if $D_f=[0,\infty)\times[0,\infty)$. For instance, if $f\colon[0,\infty)\times[0,\infty)\longrightarrow\Bbb R$ is defined by $f(x,y)=x$, then $\lim_{(x,y)\to(0,0)}f(x,y)=0$. What this means is that$$(\forall\varepsilon>0)(\exists\delta>0):\|(x,y)\|<\delta\implies\bigl|f(x,y)\bigr|<\varepsilon,$$which is true: for each $\varepsilon>0$, just take $\delta=\varepsilon$.

The condition$$(\forall\delta>0):D_f\cap\{x\in\Bbb R^n\mid\|x-x_0\|<\delta\}\ne\emptyset$$is stated only to make sure that the limit, when it exists, is unique. And the existence of a limit at $x_0$ only has to do with what happens when $x$ approaches $x_0$ from some direction assuming that $x\in D_f$. So, if $D_f$ is such you cannot approach $x_0$ from some directions, that is not a problem.

Title: Definition of limits in $\mathbb R^3$


问:
Prove by induction that $9^{n+1}+2^{6n+1}$ is divisible by 11
I need to prove that $9^{n+1}+2^{6n+1} $ is divisible by $11$ $\forall n\in N$

Steps I did:

basis step
$$\\ P(0)\ 11\mid(9+2 ) \ (True)$$

inductive step
$$\\ P(n)\implies P(n+1) \ where: \\  P(n): 9^{n+1}+2^{6n+1} = 11k\\P(n+1): 9^{n+2}+2^{6n+7} = 11k$$
so $$9^{n+2}+2^{6n+7}=11k\\9^n\times9^2+2^{6n}\times2^7=11k$$
Then I don't know how to proceed

答:
Hypothesis gives that if $9^{n+1} \equiv r \pmod{11}$ then $2^{6n+1} \equiv -r \pmod {11}$

The induction step follows:
$$\begin{align}P(n+1) &=9^{n+2}+2^{6n+7} \\
&= 9\cdot 9^{n+1}+2^6\cdot2^{6n+1} \\
&\equiv 9 r + 2^6(-r) \\
&= - 55r \\
&\equiv 0 \pmod{11}\end{align}$$


Title: Prove by induction that $9^{n+1}+2^{6n+1}$ is divisible by 11

Body:
I need to prove that $9^{n+1}+2^{6n+1} $ is divisible by $11$ $\forall n\in N$

Steps I did:

basis step
$$\\ P(0)\ 11\mid(9+2 ) \ (True)$$

inductive step
$$\\ P(n)\implies P(n+1) \ where: \\  P(n): 9^{n+1}+2^{6n+1} = 11k\\P(n+1): 9^{n+2}+2^{6n+7} = 11k$$
so $$9^{n+2}+2^{6n+7}=11k\\9^n\times9^2+2^{6n}\times2^7=11k$$
Then I don't know how to proceed

Body:
Hint:

$$9^{n+2}+2^{6n+7}=9\times9^{n+1}+2^6\times2^{6n+1}=9\times(9^{n+1}+2^{6n+1})+55\times2^{6n+1}$$


问:
Normal subgroup notation
Let $G$ be a group and $H$ a subgroup, then it is standard to say that $H$ is normal in $G$ if and only if for all $x \in G$, $xHx^{-1} \subseteq H$. My question here is, if $G$ and $H$ are additive groups, it is okay notation to write $x + H + x^{-1}$? Or is this something that is not done.

答:
For any binary operation $\ast$ that defines a group $G$ with a subgroup (or even a subset) $H$, it is acceptable to write

$$x\ast H\ast x^{-1}=\{ x\ast h\ast x^{-1}\mid h\in H\}$$

for any $x\in G$. This is due to the associativity of $\ast$. Note that $x^{-1}$ is with respect to $\ast$, so if $\ast=+$, then use $-x$.

Title: Normal subgroup notation


问:
Eigenvalues of $B = u v^t$
I want to find all the eigenvalues of $B= u v^t$ where $u,v$ are column vectors in $\mathbb{R}^m$.

With some trick $\det(I + B) = 1 + \operatorname{tr}(B)$ [This is due to rank $B$ is $1$], I can do
\begin{align}
\det(\lambda I -B) = \lambda^m \det(I - \lambda^{-1} B) = \lambda^m \left( 1  + \operatorname{tr}(-\lambda^{-1}B)\right) = \lambda^m \left(1 - \lambda^{-1} \operatorname{tr}(B) \right) = \lambda^m \left(1-\lambda^{-1} \lambda \right) =0?
\end{align}
It seems this does not that useful.

答:
Notice that $v^tu \in \Bbb{R}$ so
$$B^2 - (v^t u)B = (uv^t)(uv^t)  - (v^tu)(uv^t) = u(v^tu)v^t  - (v^tu)(uv^t) = (v^tu)(uv^t) - (v^tu)(uv^t) = 0.$$
Hence the polynomial $x^2 - (v^tu)x = x(x-v^tu)$ annihilates $B$ and hence the eingenvalues are contained in its zeroes: $0$ and $v^tu$.

Both are indeed eigenvalues. $B$ has rank $1$ so it clearly noninvertible and hence $0$ is an eigenvalue. On the other hand
$$(uv^t)u = u(v^tu) = (v^tu)u$$
so $v^tu$ is also an eigenvalue. Note that $\operatorname{Tr} B = v^tu$ since
$$\operatorname{Tr} B = \operatorname{Tr}(uv^t) = \operatorname{Tr}(v^tu) = v^tu.$$

Title: Eigenvalues of $B = u v^t$


问:
For what values of $l$ and $j$ (simultaneously) is the following congruence solvable?
Here is the congruence I want to know for what values of $l$ and $j$ (simultaneously) is it solvable?

$$ l (xp^s)^2 \equiv xp^{s}j p^{a-b} + yp^t\pmod{p^b} $$ where $x,y$ are not divisible by $p.$ and $s,a, b$ are any integers and $a \geq b.$

We have delt here what values of $j$ makes this congruence solvable? with the case when $l=0,$ but I need now to know for what values of $l$ and $j$ (simultaneously) is it solvable?

Could anyone help me in studying this?

答:
A way forward

A mechanical way of proceeding would be to analyse separately all the different possibilities such as the following one.
$$2s= s+a-b>t= b.$$

Fortunately you can group large numbers of these possibilities together. I will first do this below for all  possibilities that have $b$ as the right-hand number.

If $b$ is the least of the four numbers

Then $p^b$ divides all the terms and so the equations are satisfied for all pairs $(l,j)$.

Let's now look at a more difficult case.

If $t\ge b>2s=s+a-b$

Then the equation simplifies to
$$lx^2 \equiv xj \pmod{p^{b-2s}}.$$

So the equation is solvable if and only if either $l \equiv j \equiv 0\pmod{p^{b-2s}}$ or $l$ and $j$ are divisible by the same power (possibly zero) of $p$.

Apart from applying some ingenuity in lumping cases together there is little you can do to avoid some hard work which has to be carried out very carefully.


问:
expectation of logarithm under generalised inverse gaussian
I want to follow the following integral:
$$\frac{1}{C}\int_0^\infty \log(z)\,z^{p-1}\exp\left(-\frac{az+b/z}{2}\right)\,dz$$
where C is the normalising constant.

The following might be useful (expectation of logarithm under gamma distribution): http://en.wikipedia.org/wiki/Gamma_distribution#Logarithmic_expectation

Thanks in advance,
Sachin

答:
This is what I managed so far, and the basic idea is to use sufficient statistics and also take note of the constant term as well.

Let the integral of GIG, $$I=\int_0^\infty \frac{(a/b)^{p/2}}{2K_p(\sqrt{ab})} z^{p-1}\exp\left(-\frac{1}{2}(az+b/z) \right)dz=1$$ (http://en.wikipedia.org/wiki/Generalized_inverse_Gaussian_distribution) where, $K_p$ is the modified bessel function of the second kind.

Let us differentiate the above w.r.t. "p". Note that $z^{p-1}=\exp((p-1)\ln z)$ and $(a/b)^{p/2}=\exp(\frac{p}{2}\ln(a/b))$.
$$\frac{1}{2}\ln(a/b)\,I-\frac{\frac{dK_p(\sqrt{ab})}{dp}}{(K_p(\sqrt{ab}))^2}\int_0^\infty \frac{(a/b)^{p/2}}{2} \dots dz+\int_0^\infty \ln(z)\frac{(a/b)^{p/2}}{2K_p(\sqrt{ab})} \dots dz =0 \\
\therefore \frac{1}{2}\ln(a/b)-\frac{\frac{dK_p(\sqrt{ab})}{dp}}{K_p(\sqrt{ab})} +E(\ln(z))=0 \\
E(\ln(z))=\frac{\frac{dK_p(\sqrt{ab})}{dp}}{K_p(\sqrt{ab})}-\frac{1}{2}\ln(a/b)$$

Note that there is no analytic solution or function, that could be found at the time of writing with regards to the differential of the Modified Bessel function (w.r.t. the order p). Numeric differentiation would be suggested. This is the closest solution I could find to this differentiation: http://dlmf.nist.gov/10.38


问:
Can I say that $\int f_n\ d\mu \to \int f\ d\mu$?
Let $(X,\mathcal F, \mu)$ be a finite measure space. Let $\{f_n\}_{n \geq 1}$ be a sequence of $\mathcal F$-measurable functions converges to a $\mathcal F$-measurable function $f.$ Let $1 \lt p \lt \infty$ and $\int |f_n - f|^p\ d\mu \to 0.$ Can we say that $\int f_n\ d\mu \to \int f\ d\mu\ $?

What I tried is as follows $:$ $$\left \lvert \int f_n\ d\mu - \int f\ d\mu \right \rvert = \left \lvert \int (f_n - f)\ d\mu \right \rvert \leq \int |f_n - f|\ d\mu.$$

Now can I able to write $$\int |f_n - f|\ d\mu \leq M \int |f_n - f|^p\ d\mu\ $$ for some $M \gt 0.$ Any suggestion regarding this will be highly appreciated.

Thanks in advance.

答:
Hint: Use Holder's inequality and the fact that your space is of finite measure.

Notes 1.

Holder's inequality implies that for
$
\frac{1}{q}=\frac1p+\frac1r
$, you have
$$
\|g\|_q\le (\mu(X))^{1/r}\|g\|_p
$$

Here you want $q=1$.

Notes 2.
In general, see $L^p$ and $L^q$ space inclusion


答:
Just use Holder's inequality. If $\frac{1}{p}+\frac{1}{q}=1$ then:

$|\int\limits_X (f_n-f)|\leq(\int\limits_X |f_n-f|^p)^{\frac{1}{p}}(\int\limits_X 1)^{\frac{1}{q}}=(\int\limits_X |f_n-f|^p)^{\frac{1}{p}}(\mu(X))^{\frac{1}{q}}\to 0$

Title: Can I say that $\int f_n\ d\mu \to \int f\ d\mu$?

Body:
Let $(X,\mathcal F, \mu)$ be a finite measure space. Let $\{f_n\}_{n \geq 1}$ be a sequence of $\mathcal F$-measurable functions converges to a $\mathcal F$-measurable function $f.$ Let $1 \lt p \lt \infty$ and $\int |f_n - f|^p\ d\mu \to 0.$ Can we say that $\int f_n\ d\mu \to \int f\ d\mu\ $?

What I tried is as follows $:$ $$\left \lvert \int f_n\ d\mu - \int f\ d\mu \right \rvert = \left \lvert \int (f_n - f)\ d\mu \right \rvert \leq \int |f_n - f|\ d\mu.$$

Now can I able to write $$\int |f_n - f|\ d\mu \leq M \int |f_n - f|^p\ d\mu\ $$ for some $M \gt 0.$ Any suggestion regarding this will be highly appreciated.

Thanks in advance.


问:
An interesting problem on progression and series
In my text book I found a problem which asked to sum of the following series $$1\times 2\times 3+2\times 3 \times 4+ 3\times 4\times 5+\cdots + n(n+1)(n+2)$$ which I found to be $\frac{n(n+1)(n+2)(n+3)}{4}$ which is indeed true.

Now as a general thought it came to my mind what if the problem asks to find the closed form of this$-$ $$\sum_{k=1}^{n}\underbrace{k\times(k+1)\times(k+2)\times(k+3)\times(k+4)\cdots\times(k+m-1)}_{m \text { terms}}$$ where $m$ is any integer for instance in the aforementioned problem it was $4$. Now, following the pattern our intuition suggest that the answer should be $$\boxed{\frac{n(n+1)(n+2)\cdots(n+m)}{m+1}}$$ Now, in order to check one can easily say that for $m=1$ it's obviously true. It's also true for $m=2$. So now my natural question is that Is the closed sum true for all $m \in\mathbb{N}$ ? I have tried with induction but at the end I mess it up and the idea to proof in that way seems to be went into vain and I also have no further idea to proceed, any help? Thanks for your attention.

答:
The result that $$\frac{n(n+1)(n+2)\cdots(n+m)}{m+1}-\frac{(n-1)n(n+1)\cdots(n+m-1)}{m+1}$$ $$=n(n+1)\cdots(n+m-1)$$
can be seen easily by taking out (mentally) the common factor $n(n+1)\cdots(n+m-1)$ of all terms.

You can then prove the result either by induction or by the method of differences.


问:
How to find the volume of this region?
I want to find the volume of the following subset of $\mathbb R^3$:
\begin{equation}
(x^2+y^2)^2\leq x,\quad 0\leq z\leq 2x-\sqrt{x^2+y^2}\text{.}
\end{equation}

I tried to draw a picture of the given region in my head, but I couldn't. So I changed it into a cylindrical coordinate system, and the region became
\begin{equation}
r^4\leq r\cos\theta,\quad 0\leq z\leq 2r\cos\theta-r,\quad r>0,\quad 0<\theta<2\pi\text{.}
\end{equation}

But I still don't know what to do.

答:
You have written the equations correctly in cylindrical coordinates. Now please note that the cylinder $r^3 = \cos \theta$ is between $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

But $(2r\cos\theta-r)$ is not always $\geq 0$ in that range of $\theta$. As the lower bound of $z$ is $0$,

we must have $2r\cos\theta-r \geq 0 \implies \cos\theta \geq \frac{1}{2} $ or $|\theta| \leq \frac{\pi}{3}$

So your integral to find the volume is -

$\displaystyle \int_{-\pi/3}^{\pi/3} \int_0^{(\cos\theta)^{1/3}} \int_0^{2rcos\theta-r} r \ dz \ dr \ d\theta$

Title: How to find the volume of this region?


问:
What is wrong with my solution to the Monty Hall problem?
I'm trying to develop an intuitive sense of why the suggestion of the Monty Hall problem is that you should switch doors when an informed host opens one of the two dummy doors.

So, I'm trying to think of this without using any formulas or rigorous proofs by breaking the problem down into three steps:

the first choice of door

the door opened

the final choice of door

And just tabulating how many times a switch results in a win, and how many times a nonswitch results in a win. This is the result:

Car
Goat
Goat

S

O

S

Car
Goat
Goat

S

O

S

Car
Goat
Goat

S

O

S

Car
Goat
Goat

S

O

S

Car
Goat
Goat

S

O

S

Car
Goat
Goat

S

O

S

Car
Goat
Goat

S

O

S

Car
Goat
Goat

S

O

S

This is all eight outcomes that satisfy the condition:

the participant can choose any of the three doors originally at random

the host can only subsequently open a door that the participant did not choose

the host can only subsequently open a door that hides a goat

the participant is free to choose either of the remaining doors

From counting the results we see that there are four scenarios in which switching the door will result in a win, and four scenarios in which not switching the door will result in a win.

I am not debating what the correct answer to this problem is. I know that the correct answer has been both proven and demonstrated by simulation, and that the solution states that you double your chances of winning by choosing to switch doors. I simply seek to understand what is wrong about my assumptions that's leading me astray. Thanks!

答:
The first four scenarios each have probability $\frac13\times \frac12 \times \frac12 =\frac1{12}$ if the decisions are random within the constraints

The last four scenarios each have probability $\frac13\times 1 \times \frac12 =\frac1{6}$ if the decisions are random within the constraints

So they are not equally likely, though combined they give a probability of winning the car of $\frac12$ for somebody who switches or not at random.

For a determined switcher, the first and third scenarios  each have probability $\frac13\times \frac12 \times 1 =\frac1{6}$, making the probability of not winning the car $\frac13$.  For the same determined switcher, the fifth and seventh scenarios  each have probability $\frac13\times 1 \times 1 =\frac1{3}$, making the probability of winning the car $\frac23$.


问:
cardinality of a set of finite sequences of natural numbers
I recently got into set theory and i was wondering what is the cardinality of a set of all finite sequences of natural numbers?
I know that it is N for natural numbers and 2^N is for real numbers but how can i prove it?</p>

答:
Theorem: The set of all finite-length sequences of natural numbers is
countable.

More is true: that follows from

Theorem: (Assuming the axiom of countable choice) The union of
countably many countable sets is countable.

You can find   proofs on this wikipedia page:
https://en.wikipedia.org/wiki/Countable_set


答: Every finite sequence can be thought of as the digits in a terminating decimal, a subset of the rational numbers, so the set of all finite sequences of natural numbers is countably infinite.</p>

Title: cardinality of a set of finite sequences of natural numbers

Body:
I recently got into set theory and i was wondering what is the cardinality of a set of all finite sequences of natural numbers?
I know that it is N for natural numbers and 2^N is for real numbers but how can i prove it?</p>


问:
Prove that $\text{Aut}(\mathbb{Z}_{p^{n}} )\cong \mathbb{Z}_{\phi(p^{n})}$ where p is a prime and $\phi(.)$ denotes the euler totient function.
Can anyone tell me how to prove this?...I was asked to find the number of homomorphims from $Aut(\mathbb{Z}_{121} )$ to $Aut(\mathbb{Z}_{25} )$ .
In the solution they used the result that I asked in the title and then calculated the number of homomorphims  from $\mathbb{Z}_{110}$ to $\mathbb{Z}_{20}$ which is easy enough to do as it is just the gcd(20,110)=10 But how so I prove the above theorem?.

答:
An automorphism of $\Bbb Z_n$ is determined by sending a generator, say $1$, to a  generator, which is to say to $k$ such that $(k,n)=1$.  We pretty easily get the nice result that $\rm{Aut}(\Bbb Z_n)\cong\Bbb Z_n^\times$.

(Note that, in particular, when $n=1,2,4,p^\alpha,2p^\alpha$ for $p\gt2$ prime, we get a cyclic automorphism group.  This part is nontrivial.  See for instance I.M. Vinogradov's Elements of Number Theory.)


答:
Take a finite sequence of natural numbers such as $764\ 32\ 87\ 12\ 922$. Write them in base $10$ and replace the space between the numbers with the letter 'a', giving $764a32a87a12a922$. Interprete this as a number written in base $11$. This correspondance is an injective map from the set of finite sequences of natural numbers into ${\mathbb N}$. Hence the set of such sequences is countable.

You can even make this correspondance a bijection if you replace the potential sequence $0a$ at the beginning of a number by $a$.


答:
Suppose we want to evaluate the sum which you have already found out.

We evaluate the sum $$\displaystyle \sum_{k=1}^n k(k+1)(k+2)$$
Now if we multiply up and down by $4$, and write $4$ as $(k+3)-(k-1)$ (why? To make it a telescopic sequence!)

We get the sum as $$\dfrac{1}4 \left[\sum_{k=1}^n k(k+1)(k+2)(k+3)-\sum_{k=1}^n (k-1)(k)(k+1)(k+2)\right]$$

Now it is clear that it telescopes...so this method can be extended for higher terms as well.

Title: An interesting problem on progression and series

Body:
In my text book I found a problem which asked to sum of the following series $$1\times 2\times 3+2\times 3 \times 4+ 3\times 4\times 5+\cdots + n(n+1)(n+2)$$ which I found to be $\frac{n(n+1)(n+2)(n+3)}{4}$ which is indeed true.

Now as a general thought it came to my mind what if the problem asks to find the closed form of this$-$ $$\sum_{k=1}^{n}\underbrace{k\times(k+1)\times(k+2)\times(k+3)\times(k+4)\cdots\times(k+m-1)}_{m \text { terms}}$$ where $m$ is any integer for instance in the aforementioned problem it was $4$. Now, following the pattern our intuition suggest that the answer should be $$\boxed{\frac{n(n+1)(n+2)\cdots(n+m)}{m+1}}$$ Now, in order to check one can easily say that for $m=1$ it's obviously true. It's also true for $m=2$. So now my natural question is that Is the closed sum true for all $m \in\mathbb{N}$ ? I have tried with induction but at the end I mess it up and the idea to proof in that way seems to be went into vain and I also have no further idea to proceed, any help? Thanks for your attention.


问:
Prove that $\overline{A \times B}=\overline{A} \times \overline{B}$
How I can prove that $\overline{A \times B}=\overline{A} \times \overline{B}$?

I start proving that $\overline{A} \times \overline{B} \subseteq \overline{A \times B}$.

Taking $(x,y) \in \overline{A} \times \overline{B},\ \ \ \forall U \times V \in \mathscr{N}_{(x,y)}$,

$x \in \overline{A} \Rightarrow U\cap A \not= \emptyset \ $  and $\ y \in \overline{B} \Rightarrow V\cap  B\not= \emptyset \ $ so

$U \times V \cap A \times B \not= \emptyset \Rightarrow (x,y) \in \overline{A \times B}$

Is this part correct? And how can I prove the other part?

答:
That part is correct, but you could use more words and fewer symbols: let $(x,y) \in \overline{A} \times \overline{B}$. To see that $(x,y) \in \overline{A \times B}$, we take a basic open open neighbourhood of $(x,y)$, so one of the form $U \times V$ where $U$ is an open neighbourhood of $x$ and $V$ an open neighbourhood of $y$ and show it intersects $A \times B$: as $U$ is an open neighbourhood of $x$ and $x \in \overline{A}$ we know that there is some $a \in A \cap U$ and similarly, as $y \in \overline{B}$ we have some $b \in B \cap V$, but then $(a,b) \in (A \times B) \cap (U \times V) \neq \emptyset$ and we're done.

For the reverse, we take $(x,y) \in \overline{A \times B}$ and we will show that $x \in \overline{A}$ and $y \in \overline{B}$, so that $(x,y) \in \overline{A} \times \overline{B}$, showing the inclusion.

To see that $x \in \overline{A}$: let $U$ be an open neighbourhood of $x$. Then $U \times Y$ is an open neighbourhood of $(x,y)$ so the assumption that $(x,y) \in \overline{A \times B}$ gives us $(a,b) \in (U \times Y) \cap (A \times B)$. In particular we have that $a \in U \cap A \neq \emptyset$. As $U$ was arbitrary, we have that $x \in \overline{A}$ as claimed. To show that $y \in \overline{B}$ is similar and left to the reader, using sets of the form $X \times V$ instead.

As an alternative to that last proof, note that $$\overline{A} \times \overline{B} = \pi_X^{-1}[\overline{A}] \cap \pi_Y^{-1}[\overline{B}]$$
where $\pi_X$ and $\pi_Y$ are the continuous projections so that $\overline{A} \times \overline{B}$ is closed in $X \times Y$. It obviously contains $A \times B$ so also $\overline{A \times B}$, by minimality of the closure.


问:
Module Isomorphism
Let X, Y , and Z be $R$- Modules, for $R$ a commutative ring. Furthermore, let $X$ and $Y$ be submodules of $Z$. I want to show that if $X + Y = Z$ and $X \cap Y = \{0 \}$, then $Z \cong X \oplus Y$.

To start, I am defining a map $\phi$: $X \oplus Y \rightarrow Z$ by $(x,y) \mapsto x + y$.

My question is, will this map do it for me? In otherwords, it is possible to prove that $\phi$ is an isomorphism of $R$ modules, or am I going down the wrong path with this mapping. Thank you.

答:
Just use your proposed map. To see surjectivity, consider any $z \in Z$. Since $Z = X + Y$, there exists $x \in X$ and $y \in Y$ such that $z = x + y$. Now $\phi(x, y) = z$, proving surjectivity.

To see injectivity, suppose $\phi(x_1, y_1) = \phi(x_2, y_2)$. Then $x_1 + y_1 = x_2 + y_2$. This implies $x_1 - x_2 = y_2 - y_1$. Since both $X, Y$ are submodules, we see that the left-hand side is in $X$ and the right-hand side is in $Y$. Thus they are both in the intersection $X \cap Y = \{0\}$. So $x_1 - x_2 = y_2 - y_1 = 0$, i.e. $(x_1, x_2) = (y_1, y_2)$, proving injectivity.

It is also easy to see that this is a $R$-linear map.
$$\phi(rx, ry) = rx+ry = r(x+y) = r\phi(x, y).$$

So your map is indeed an $R$-module isomorphism.

Title: Module Isomorphism


答:
Let your sum be denoted $S_{n,m}$. If we increment $m$, the last factor in every term is $k+m=(k-1)+(m+1)$. This gives you two new sums, namely $S_{n-1,m+1}$ (by shifting the index, the first terms being $0$) and $(m+1)S_{n,m}$. Hence the recurrence

$$S_{n,m+1}=S_{n-1,m+1}+(m+1)S_{n,m}$$

from which

$$S_{n,m}=\frac{S_{n,m+1}-S_{n-1,m+1}}{m+1}.$$

The numerator is the last term of $S_{n,m+1}$.


问:
Prove that every natural number is either even or odd using induction
It is a pretty basic question at first sight. It seems to be intuitively correct but I can't figure out a way to prove it. I think we have to use strong induction but even after sitting on this question for a long time, I still haven't got any leads.</p>

答:
Alternative approach:

For positive integer $n$,

$n$ is odd $\iff 2|(n+1)~$ and

$n$ is even $\iff 2|n.$

$1$ is odd and $2$ is even.

Inductively assume that $(n+1)$ is odd and $(n+2)$ is even.

Then, $2|([n+1] + 1) \implies 2|([n+3] + 1) \implies (n+3)$ is odd.

Thus, $(n+1)$ is odd $\implies (n+3)$ is odd.

By virtually identical analysis, $(n+2)$ even $\implies (n+4)$ even.

Therefore, you have the base cases of $1$ odd, $2$ even and, by inductive analysis, whenever $(n+1)$ is odd, so is $(n+3)$ and whenever $(n+2)$ is even, so is $(n+4).$


问:
Prove that $( X,τ)$ is homeomorphic to a subspace of $\Bbb{R}$ if and only if $X$ is countable
"Let $(X,τ)$ be a discrete topological space. Prove that $( X,τ)$ is homeomorphic to a subspace of $\Bbb R$ if and only if $X$ is countable."

How do I get started on this one? I know that R has both countable and uncountable subspace, and you can only have a homeomorphism if the cardinality is the same. Now sure where to go from there.

答: $\Bbb R$ has a countable base, so all subspaces of $\Bbb R$ have a countable base too. So if $D$ is a subspace of $\Bbb R$ that has the discrete topology, any base for $D$ (in the subspace topology) must contain all sets of the form $\{d\}, d \in D$, and so $D$ is at most countable.


问:
Is a curve homologous to zero according to Ahlfors actually homologous to zero?
The presentation of the homology version of Cauchy's theorem in Ahlfors is slick, but sweeps a lot of the topology under the rug using clever arguments. This question is an attempt to reconcile Ahlfors' analytic notion of a curve being homologous to zero (presented in his book Complex Analysis and originally due to E. Artin, I believe) with the standard definition in homology as found in Hatcher.

We work in the complex plane and fix $a\in \mathbb C$. Let $\gamma$ be a continuous map $[0,1]\rightarrow C\backslash \{ a\}$. Following Munkres in his book Topology, we define the winding number of $\gamma$ with respect to the point $a$ by considering

$$g(t)=\frac{\gamma(t)-a}{|\gamma(t)-a|}.$$

This is clearly a loop in $S^1$ and corresponds to some multiple of of the generator of the fundamental group of $S^1$. If the generator is $\tau$ and $g(t)$ corresponds to $m\tau$, $m\in\mathbb Z$, we define the winding number $n(\gamma, a)$ to be $m$. This is the definition of winding number I will use in this question, but in case $\gamma$ is piecewise differentiable, it corresponds to the analytic definition by integration given in Ahlfors.

Ahlfors calls a curve contained in an open region $\Omega$ homologous to zero if $n(\gamma,a)=0$ for all $a\in \Omega^c$.

In homology theory, as I understand it, we would call a curve homologous to zero if it represents the zero element in $H_1(\Omega, \mathbb Z)$. That is, $\gamma$ is the boundary of some singular $2$-chain.

My question: Why do all of these notions agree? Why is Ahlfors' definition of being homologous to zero (using Munkres' definition of winding number) agree with the usual homological one?

I would like to use Munkres' definition because it works for continuous curves, not just piecewise differentiable ones, and it seems to me the equivalence should hold in this generality.

Edit: This result appears as proposition 1.9.13 in Berenstein and Gay's book on complex analysis.

答: This result appears as proposition 1.9.13 in Berenstein and Gay's book on complex analysis.</p>


问:
Number of multisets with at most twice the same element
I know that the number of multisets of cardinality $K$ with elements taken from the set $\{1,...,N\}$ is $\binom{N+K-1}{K}$.

I only want to consider the multisets such that an element appears at most twice. How do I count them?

Is there a general formula for multisets such that an element appears at most $M$ times?

答:
The generating function for number $f(K,N,M)$ of multisets of cardinality $K$ from $\{1,\ldots,N\}$ with each element appearing at most $M$ times is

$$ g_{N,M}(z) = \sum_{K=0}^\infty f(K,N,M) z^K = (1+z+\ldots+z^M)^N = \left(\frac{1-z^{M+1}}{1-z}\right)^N$$

Now $$(1-z^{M+1})^N = \sum_{n=0}^N (-1)^n {N \choose n} z^{(M+1)n}$$
while $$(1-z)^{-N} = \sum_{k=0}^\infty {N +k-1 \choose N-1} z^k$$
so we get
$$ f(K,N,M) = \sum_{n=0}^{\min(N,\lfloor K/(M+1) \rfloor)} (-1)^n {N \choose n} {N+K-(M+1)n-1 \choose N-1}$$


答:
If $a \equiv b \left(\bmod m \right)$, then $a^n \equiv b^n \left(\bmod m \right)$ where $n \in \mathbb{N}$ and $a,b,m \in \mathbb{Z}$.

The reason being $a^n - b^n$ factors as $(a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + b^{n-1})$ and hence if $m | (a-b)$, $m | (a^n-b^n)$

If you want to prove by induction, first note that

if $a \equiv b \left(\bmod m \right)$ and $c \equiv d \left(\bmod m \right)$, then $a \times c  \equiv b \times d \left(\bmod m \right)$.

This is true since $ac-bd = a(c-d) + d(a-b)$

Hence at the inductive step, if you have that $10^k \equiv (-1)^k \left(\bmod m \right)$, and we have that $10 \equiv (-1) \left(\bmod m \right)$ and from the above result we have
$$10^{k+1} \equiv (-1)^{k+1} \left(\bmod m \right)$$

Title: Complete induction of $10^n \equiv (-1)^n \pmod{11}$

Body:
To prove $10^n \equiv (-1)^n\pmod{11}$, $n\geq 0$, I started an induction.

It's $$11|((-1)^n - 10^n) \Longrightarrow (-1)^n -10^n = k*11,\quad k \in \mathbb{Z}. $$

For $n = 0$:
$$ (-1)^0 - (10)^0 = 0*11 $$

$n\Rightarrow n+1$
$$\begin{align*}
 (-1) ^{n+1} - (10) ^{n+1} &= k*11\\
(-1)*(-1)^n - 10*(10)^n &= k*11
\end{align*}$$
But I don't get the next step.


问:
Right/Left inverse mapping
Can someone clearly explain the difference between right and left inverse mapping, if $f: X \to Y$ and $g: Y \to X$ in set theory?

I've read the definitions, but I still have no awareness of what is it, their difference, and how if differs from just inverse mapping.

答:
Let $X$ be the set of all integers and $Y$ the set $\{\text{even},\text{odd}\}$ with just two elements.

Suppose $f: X \to Y$ is defined in the obvious way and let
$g: Y \to X$ be
$$
g(\text{even}) = 0 \text{ and  } g(\text{odd}) = 1.
$$

the the composition $f \circ g$ is the identity on $Y$ so $f$ is a left inverse for $g$ and $g$ is a right inverse for $f$.

Title: Right/Left inverse mapping


问:
Convergence of $\sum_{n=1}^{\infty} \frac{1}{n^3}$
I know the alternating series $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}$$ converges by doing the alternating series test, but I can't seem to find a way to prove $$\sum_{n=1}^{\infty} \frac{1}{n^3}$$ converges. I tried it with  $\lim_{n \to +\infty}\sqrt[n] \frac{1}{n^3} $ and $\lim_{n \to +\infty}\frac{a_{n+1}}{a_n}$, but both of these result in 1, which means you can't conclude anything from them.

Any help would be appreciated!

答: $\displaystyle \sum_{n=1}^{\infty}{\dfrac{1}{n^3}}$ converges by the P-Series Test.


答:
You might notice that your set is $\bigcup_{n\in\mathbb{N}}\mathbb{N}^n$. I will assume you know that for every $n\in\mathbb{N}$, $|\mathbb{N}^n|=|\mathbb{N}|$. Then, for each $n\in\mathbb{N}$ there exists a bijection $h_n:\mathbb{N}^n\longrightarrow\mathbb{N}$. The function:
$$\Psi:(x_1,\ldots,x_k)\in\bigcup_{n\in\mathbb{N}}\mathbb{N}^n\longrightarrow\Psi(x_1,\ldots,x_k):=(k,h_k(x_1,\ldots,x_k))\in\mathbb{N}\times\mathbb{N}$$
is bijective. To check it's injective, if $\Psi(x_1,\ldots,x_k)=\Psi(y_1,\ldots,y_p)$ then $(k,h_k(x_1,\ldots,x_k))=(p,h_p(y_1,\ldots,y_p))$, so $k=p$ and $h_k=h_p$. As $h_k$ is a bijection $h_k(x_1,\ldots,x_k)=h_k(y_1,\ldots,y_k)\implies(x_1,\ldots,x_k)=(y_1,\ldots,y_k)$. To check it's surjective, given $(m,n)\in\mathbb{N}\times\mathbb{N}$, $\Psi(h_m^{-1}(n))=(m,h_m(h_m^{-1}(n))=(m,n)$.
Then, $|\bigcup_{n\in\mathbb{N}}\mathbb{N}^n|=|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|$.


答: Since $\displaystyle\int_1^\infty\frac1{x^3}\,\mathrm dx$ converges, your series converges too, by the integral test.

Title: Convergence of $\sum_{n=1}^{\infty} \frac{1}{n^3}$

Body:
I know the alternating series $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3}$$ converges by doing the alternating series test, but I can't seem to find a way to prove $$\sum_{n=1}^{\infty} \frac{1}{n^3}$$ converges. I tried it with  $\lim_{n \to +\infty}\sqrt[n] \frac{1}{n^3} $ and $\lim_{n \to +\infty}\frac{a_{n+1}}{a_n}$, but both of these result in 1, which means you can't conclude anything from them.

Any help would be appreciated!


问:
Normal subgroup notation
Let $G$ be a group and $H$ a subgroup, then it is standard to say that $H$ is normal in $G$ if and only if for all $x \in G$, $xHx^{-1} \subseteq H$. My question here is, if $G$ and $H$ are additive groups, it is okay notation to write $x + H + x^{-1}$? Or is this something that is not done.

答: As Mark points out in the comments, it would be more appropriate to write $x+H+(-x)$ in such a case, but we typically use the additive notation for abelian groups, in which case every subgroup is normal, so we wouldn't bother to write it.

Title: Normal subgroup notation


问:
Functions & Points
Suppose there are three functions f(x), g(x) & h(x) and there is a point A. Is there any method by which I may know that how many of the given functions pass via point A, and which are those functions ?

答:
Take a function $f(x)=x^3$ and consider two points $P_1=(2,4) and P_2=(-2,-4$. Now, check if the function $f$ passing through the points $P_1$ and $P_2$. For $P_1$,
$f(2)=4$ so f passes through the point $P_1$. For $P_2$, $f(-2)=4=\neq -4$ so $f$ does not pass through the point $P_2$. I hope that helps

Title: Functions & Points


问:
Discrete maths and circles
Finitely many Real numbers are written along a circle. If some four successive numbers $a ,b ,c$ and $d$ satisfy the inequality $(a-d)(b-c)>0$,then it is permitted to transpose(exchange positions of $b$ and $c$) the numbers b and c. Prove that it is impossible to perform infinitely many such operations. I tried to see that to have infinite operations we must return to the original permutations of numbers, then tried to prove by contradiction,but couldn't. Also,I observed that $a>b>c....$ works for most part except for last part giving contradiction $a>b>c....>a$

答:
$(a-d)(b-c)>0$

implies $ab+cd>ac+bd$

implies $ab+cd+bc+ad>ac+bd+bc+ad$

implies $(a+c)(b+d)>(a+b)(c+d)$

implies $(a+c)(c+b)(b+d)>(a+b)(b+c)(c+d)$

This means when we perform a transpose action, the consecutive product of pairwise sums strictly increases.

Let the numbers be $x_1,x_2,x_3,...,x_n$. We can deduce that the product $(x_1+x_2)(x_2+x_3)(x_3+x_4)...(x_{n-1}+x_n)(x_n+x_1)$ always increases after a transpose action. Since there are only a finite number of possible products the result is proven.

Edit: the above proves the case where all numbers are positive. However since the result does not change if we add a constant to all numbers so we can add "the smallest number $+ 1$" to all numbers to make them all positive without affecting the result.

Title: Discrete maths and circles


问:
Why doesn't integrating the area of the square give the volume of the cube?
I had a calculus course this semester in which I was taught that the integration of the area gives the size (volume):

$$V = \int\limits_a^b {A(x)dx}$$

But this doesn't seem to work with the square. Since the size of the area of the square is $x^2$ then $A(x) = {x^2}$, then:

$$V = \int\limits_{ - r}^r {{x^2}dx}  = \left[ {\frac{{{x^3}}}{3}} \right]_{ - r}^r = \frac{{{r^3}}}{3} - \frac{{ - {r^3}}}{3} = \frac{2}{3}{r^3}$$

It's clear that this is not the volume of the cube. Why is this the case? Am I misunderstanding something?

答:
Can you tell what your proposed solution represents? I think it's far more interesting (and important) for you to be able to look at the integral, and be able tell what it means.

If you think of $x^2$ in terms of the area of a square of sides $x$, then your proposed integral calculates the volume of a solid that consists of two identical square-base pyramids (like an Egyptian pyramid) placed tip-against-tip, fashioning an hourglass-looking solid (but with flat sides). The tips of both pyramids meet at the origin and each pyramid is $r$-units high. Do you see why that is the case?

The volume also happens to be the same as that of an irregular octahedron of edge $r$ that measures $2r$ vertex-to-vertex. It's trivial to go from square hourglass to the octahedron by grabbing one of the pyramids and flipping it over, so they meet at the base instead of the vertexes. Note that this is a "pointy" octahedron, not a regular (platonic) one because the faces of the pyramids don't come out to be regular triangles (here's how I convinced myself of that: the section of the pyramids is a triangle of base $r$ and height $r$. That's "taller" than a regular triangle; the faces of the pyramids are like taking one of these sections and stretching it outward by pulling the base, so they are necessarily even "taller" than a regular triangle).

But $x^2$ doesn't have to be the area of a square (in fact, it doesn't have to be an area at all; we just like geometric interpretations because they are fun).

Among other things, $x^2$ can be the area of a circle of radius $x\over\sqrt{\pi}$ (do you see why?). In that interpretation, your integral calculates the volume of a cone (again, in the shape of an hourglass). The volume of a cone is indeed $1\over 3$ of the volume of its enclosing cylinder, which in this case would have a base of $r^2$ (the largest of the circles) and height $2r$ (one $r$ for each half of the "hourglass"). That is, ${1\over3} \times (r^2 \times 2r) = {2\over 3} r^3$.

What's so special about the circle and square interpretation of the formula? Absolutely nothing! Even after deciding that we are looking only at geometric interpretations of the integral, the integral makes no statement about the shape of the figure whose area the integrand calculates. As long as the area equals to the square of its height from the horizontal plane, the figure can be shaped like a circle, a square, or a flattened pretzel, and the volume of the extruded hourglass-looking volume will always be the same.

Playing with these geometric interpretations is a great way to become comfortable with integrals, and should teach you over time a sense for when "something looks wrong".


问:
Is it possible to prove Infinite Primes using Picks Theorem?
Lemma: Triangles containing exactly the same number of grid points (excluding boundary), then $n$ is prime. (see Pick's Theorem and Primes.)

Claim: There exists infinite such triangles, so infinite primes.

Is that claim provable?

Note: I only know basic high-school level maths. Thanks.

答:
Your question is completely unclear. But presumably it refers to your earlier question. The answer there proved that each of the $n-2$ triangles contains the same number of lattice points in its interior if and only if $n=4$ or $n$ is prime.

If you had some independent way of proving that there were infinitely many $n$ for which the $n-2$ triangles each contain the same number of points in their interiors, then you would be able to deduce that there were infinitely many primes. But until you can find such a way, you will have to be content with the existing one-line proof that there are infinitely many primes.


答: Suppose that $A=(x_0,y_0).$ To find out which functions pass through the point, if any, simply calculate $f(x_0),$ $g(x_0),$ and $h(x_0)$ to see if they are equal to $y_0.$

Title: Functions & Points

Body: Suppose there are three functions f(x), g(x) & h(x) and there is a point A. Is there any method by which I may know that how many of the given functions pass via point A, and which are those functions ?


答:
If we have a function $f: X \to Y$ a function $g: Y \to X$ is called a left inverse of $f$ iff $g \circ f = 1_X$, where $1_A: A \to A$ is the identity map on a set $A$ (so $1_A(a)=a$ for all $a$) and it's called a right inverse of $f$ iff $f \circ g = 1_Y$.

The type (left/right) of inverse can be seen by whether you compose with $f$ on the left or the right in the condition giving the identity map.

The map $f: \Bbb N \to \Bbb N$ defined by $f(n)=n+1$ has a left inverse (what is it?) but not a right inverse.

$g$ is an inverse if both a left and a right inverse.

Title: Right/Left inverse mapping

Body:
Can someone clearly explain the difference between right and left inverse mapping, if $f: X \to Y$ and $g: Y \to X$ in set theory?

I've read the definitions, but I still have no awareness of what is it, their difference, and how if differs from just inverse mapping.


答:
If you've read the definitions then you already know how they're different.  I guess what you're really looking for is a demonstration.

Define two functions $f,g$ on $\mathbb N$ by $f(n)=2n$ and $g(n)=\lfloor n/2\rfloor$ (that's the floor function of $n/2$, in case you haven't seen it.)  Clearly $g\circ f(n)=n$ for all $n\in\mathbb N$, and that means $g\circ f$ is the identity function on $\mathbb N$. By definition $g$ is a left inverse of $f$, and $f$ is a right inverse of $g$.

But $f\circ g(3)= 2$, so $f\circ g$ is not the identity.  $f$ isn't a left inverse to $g$, and $g$ isn't a right inverse to $f$.

In general when you have two functions $f,g$ such that $f\circ g$ and $g\circ f$ are the identity functions on their respective domains, that's when you say "$g$ and $f$ are inverses of each other" which means that they are both right and left inverses of each other.  The definition of an inverse function is just that it has a counterpart which is both a left and a right inverse.

Note: I really advise not using the phrasing "what is the difference between $X$ and $Y$" when you already do indeed know the definitions are completely different. One could rightly respond to that phrasing by "why do you think they are the same?!"

In a case like this, just ask for examples demonstrating the difference between left inverses and inverse functions. That makes sense because there is a difference to be made.

Title: Right/Left inverse mapping


问:
calculation of divergence of a covariant vector

$(x,y)$ are rectangular Cartesian coordinates in a plane and
$(\bar{x},\bar{y})$ are coordinates defined by
$$x=\frac{\bar{x}^2+\bar{y}^2}{2} \ \ , \ \ y=\bar{x}\bar{y}$$ If
$A_1,A_2$ are coordinates of a covariant vector in the
$\bar{x},\bar{y}$ coordinates defined by
$A_1=A_2=(\bar{x}^2-\bar{y}^2)^2$ , show that its divergence is
$$\frac{2(\bar{x}-\bar{y})^2}{\bar{x}+\bar{y}}$$

Clearly the metric tensor in the new coordinate is $$g_{11}=g_{22}=\bar{x}^2+\bar{y}^2$$ Hence the conjugate metric tensors are $$g^{11}=g^{22}=\frac{1}{\bar{x}^2+\bar{y}^2}$$ Hence $$\text{div}(A_i)=g^{ij}A_{i \ , \ j}=g^{11}A_{1,1}+g^{22}A_{2,2}$$ "$,$" is the covariant differentiation operator . So $$\text{div}(A_i)=g^{11}\bigg(\frac{\partial A_1}{\partial\bar{x}}-\Gamma^1_{11}A_1-\Gamma^2_{11}A_2\bigg)+g^{22}\bigg(\frac{\partial A_2}{\partial\bar{y}}-\Gamma^1_{22}A_1-\Gamma^2_{22}A_2\bigg)$$ But upon calculation of the Christoffel symbols and substituting in the expression gives $$\text{div}(A_i)=\frac{4(\bar{x}^2-\bar{y}^2)(\bar{x}-\bar{y})}{\bar{x}^2+\bar{y}^2}$$ I don't get what is going possibly wrong there , any help will be appreciated .

答:
You have already correctly calculated $g_{11}$ and $g_{22}$ of metric tensor $g_{ik}$. The non-diagonal component $g_{12}=g_{21}=2xy$ (to save time I don't write barr above x and y). $G=det|g_{ik}|=(x^2-y^2)^2$, $A^1=A^2=x^2+y^2-2xy$ $(A^1=g^{11}A_1+g^{12}A_2)$. Then you can calculate Christoffel symbols (from learning point this is a good approach), but let me go directly to the goal: in case of divergence of a vector we may use the formula for $A_{i;j}g^{ij}$ (; means covariant differentiation): $A_{i;j}g^{ij}=\frac{1}{\sqrt{G}}\frac{d}{dx^i}(\sqrt{G}A^i)$(summation over repeated indexes i)$=\frac{1}{\sqrt{G}}\frac{d}{dx}(\sqrt{G}A^1)+\frac{1}{\sqrt{G}}\frac{d}{dy}(\sqrt{G}A^2)=\frac{1}{x^2-y^2}\frac{d}{dx}\left((x^2-y^2)(x-y)^2\right)+\frac{1}{x^2-y^2}\frac{d}{dy}\left((x^2-y^2)(x-y)^2\right)=2\frac{(x-y)^3}{x+y}$

Hopefully this will be helpful.


问:
Finding invariant factors of finitely generated Abelian group
There is this question that I wasn't sure how to do but somehow got the answers partially correct (maybe).

Suppose that the abelian group $M$ is generated by three elements $x,y,z$ subject to the relations $4x+y+2z=0, 5x+2y+z=0,6y-6z=0$. Determine the invariant factors of $M$.

My attempt:

I wasn't sure what theorem or reasoning to use, but I have a feeling to write the equations in matrix form and find the equivalent matrix. My question is normally if we write in matrix form, we should be priorly given a map or homomorphism, but this question did not, or whether somehow we can interpret the equations as map? if so, what is the map?

Let me try to make one. Let $M$ be the $\mathbb{Z}$-module $F/N$ where $F=\mathbb{Z}^3$ and $N=\langle(4,5,0),(1,2,6),(2,1,-6)\rangle\le\mathbb{Z}^3$. Consider the homomorphism $\varphi:\mathbb{Z}^3\to\mathbb{Z}^3$ given by $(x,y,z)\to x(4,5,0)+y(1,2,6)+z(2,1,-6)$. Am I correct? Please correct me if I am wrong.

Then I find the equivalent matrix:

$\begin{bmatrix}4 & 1 & 2\\5 & 2 & 1\\0 & 6 & -6\end{bmatrix}$ $4R_2-5R_1$ $\begin{bmatrix}4 & 1 & 2\\0 & 3 & -6\\0 & 6 & -6\end{bmatrix}$ $R_3-2R_2$ $\begin{bmatrix}4 & 1 & 2\\0 & 3 & -6\\0 & 0 & 6\end{bmatrix}$ $3R_1+R_2$ $\begin{bmatrix}12 & 6 & 0\\0 & 3 & -6\\0 & 0 & 6\end{bmatrix}$ $R_2+R_3$ $\begin{bmatrix}12 & 6 & 0\\0 & 3 & 0\\0 & 0 & 6\end{bmatrix}$ $R_1-2R_2$ $\begin{bmatrix}12 & 0 & 0\\0 & 3 & 0\\0 & 0 & 6\end{bmatrix}$. Then by performing column and row swaps, the matrix becomes $\begin{bmatrix}3 & 0 & 0\\0 & 6 & 0\\0 & 0 & 12\end{bmatrix}$. Since 3,6,12 are non-units in $\mathbb{Z}$, then the invariant factors of M are 3, 6, 12.

However, the solution says that the invariant factors are 1,3,6.

Was my method correct? Where did I make the mistake?

Thanks!

答:
When you do the operation $4 R_2 - 5 R_1$, you are altering the determinant by a factor of $4$, since you are multiplying your matrix on the left by
$$
\begin{bmatrix}1 & 0 & 0\\-5&4&0\\0&0&1\end{bmatrix}.
$$
Ditto for $3 R_1 + R_2$, which alters the determinant by a factor of $3$. These two operations have introduced the spurious factor $12$.

I don't have the time to post the correct calculations (which are a lengthier alternative to the use of the determinant in this case), perhaps later.

So here's how you do it:
$$\begin{align}
&\begin{bmatrix}4 & 1 & 2\\5 & 2 & 1\\0 & 6 & -6\end{bmatrix} C_1 \leftrightarrow C_2 \begin{bmatrix}1 & 4 & 2\\2 & 5 & 1\\6 & 0 & -6\end{bmatrix} R_2 - 2 R_1 \begin{bmatrix}1 & 4 & 2\\0 & -3 & -3\\6 & 0 & -6\end{bmatrix} R_3 - 6 R_1\\& \begin{bmatrix}1 & 4 & 2\\0 & -3 & -3\\0 & -24 & -18\end{bmatrix} C_2 - 4 C_1, C_3 - 2 C_1 \begin{bmatrix}1 & 0 & 0\\0 & -3 & -3\\0 & -24 & -18\end{bmatrix} R_3 - 8 R_2\\& \begin{bmatrix}1 & 0 & 0\\0 & -3 & -3\\0 & 0 & 6\end{bmatrix}  C_3 - C_2 \begin{bmatrix}1 & 0 & 0\\0 & -3 & 0\\0 & 0 & 6\end{bmatrix} {-R_2} \begin{bmatrix}1 & 0 & 0\\0 & 3 & 0\\0 & 0 & 6\end{bmatrix}
\end{align}$$


答:
Any sine curve can be written as a  vertically shifted sinusoidal function, $f(t) = B\sin(\omega t+\phi) + C,\ $ where $B, C, \omega$ and $\phi$ are real constants.

The amplitude of $f(t)$ is $B$ because the function $\sin$ takes all values between $-1$ and $1$. Therefore the range of the $f(t)$ is $[C-B,C+B]$.

So if the y-coordinate of $A \in [C-B,C+B],$ then $f(t)$ passes through $A$. Else it doesn't.

Title: Functions & Points


问:
Find all singular points of f(z), classify them and inspect the behaviour at infinity
Find all singular points of f(z), classify them and inspect the behaviour at infinity

$f(z)=sin(1/z)+1/z^2$

Here i'm stuck to find singularities since z=0'd be a singularity however we have $+sin(1/z)$ so i guess z has to be a value to make whole function singular. Then i realized i may be rewrite this function as a series expansion but i failed. So i need some suggestions or guidance.

答:
You've got the right idea.

Note that since you need to inspect the behavior at infinity, you'll want to consider the function $$g(z):=f\left(\frac1z\right)=\sin(z)+z^2.$$

I would start by finding the series expansion for $g$ about $0,$ then noting that $f(z)=g(1/z).$ That will help you show that $z=0$ is the only singularity of $f,$ as well as to classify it.


问:
Dirac delta in polar coordinates
Given

$$x=r\,\cos\theta\\y=r\,\sin\theta$$ and
$$x'=r'\,\cos\theta'\\y'=r'\,\sin\theta'$$

how can I express

$$\delta(x'-x)\delta(y'-y)$$ in terms of the polar coordinates?

And the more general case:

$$\delta(x'-x-a)\delta(y'-y-b)$$

答:
If the transformation between coordinates ${\bf x}$ and ${\boldsymbol \xi}$ is not singular then
$$\delta({\bf x}-{\bf x_0})
= \frac{1}{|J|}\delta({\boldsymbol \xi}-{\boldsymbol \xi}_{0}),$$
where $J$ is the Jacobian of the transformation.
This is analogous to
$\delta(f(x)) = \delta(x-x_0)/|f'(x_0)|$,
for $x$ near an isolated zero $x_0$ of $f$.

The Jacobian is $r$ so, assuming $r'\ne 0$,
$$\delta(x-x')\delta(y-y') = \frac{1}{r}\delta(r-r')\delta(\theta-\theta').$$
(We take $\theta'\in[0,2\pi)$.)
Notice that
$$\int_0^\infty r dr\int_0^{2\pi}d\theta \
\frac{1}{r}\delta(r-r')\delta(\theta-\theta') = 1$$
as required.

If $r'=0$ we must integrate out the ignorable coordinate $\theta$,
$J\to \int_0^{2\pi}d\theta \ J = 2\pi r$.
Thus
$$\delta(x)\delta(y) = \frac{1}{2\pi r}\delta(r).$$
Again, notice
$$\int_0^\infty r dr\int_0^{2\pi}d\theta \
\frac{1}{2\pi r}\delta(r) = 1.$$

Title: Dirac delta in polar coordinates


问:
How do I find the Laurent series expansion?
I want to find the order of the pole of $\sqrt{1+\frac{1}{z}}$ at 0. But how can I expand this function at 0? Can I take the Taylor expansion of $\sqrt{1+z}$  and plug in $\frac{1}{z}$ in the place of z? And how can one find the Laurent series expansion of a function in general? I mean since the function diverges toward infinity at poles, how can the Laurent series be "defined"?

答:
You're quite right that to find the Laurent series expansion for $f(z)=\sqrt{1+\frac1z}$ about $0,$ we need only start with the Taylor series expansion of $g(z)=\sqrt{1+z}$ about $0,$ then note that $f(z)=g(1/z).$ In fact, this will show you that it isn't a pole, but rather an essential singularity.

As for how the Laurent series can be thought of as being defined, recall that a power series is defined wherever it converges. In this case, it converges at all values of $z$ such that $|z|>1.$


问:
Integrate $\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx$.

Integrate the following integral:
$$\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx$$

I did this question a while ago and the answer is correct. However, when I started revising this topic again, I can't seem to figure out how I went from the first step $\left( \text{for this integral}\int \log _{3x}\left(x^2\right)dx\right)$ to the second step. Here's my attempt:

\begin{align}
\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx  &= {\int{\frac{2}{\sin{\left(2 x \right)}} d x}} \\
&= {\int{\frac{1}{\sin{\left(x \right)} \cos{\left(x \right)}} d x}} \\
&= {\int{\frac{\sec^{2}{\left(x \right)}}{\tan{\left(x \right)}} d x}} \\
&\stackrel{{u = \tan{\left(x\right)}}}{=} {\int{\frac{1}{u} d u}} \\
&= {\ln{\left(u \right)}} + c \\
&= \ln{\left(\left|{\tan{\left(x \right)}}\right| \right)}+
\end{align}

\begin{align}
\int \log _{3x}\left(x^2\right)dx &= x\log _{3x}\left(x^2\right)-\int \frac{2\ln \left(3\right)}{\ln ^2\left(3x\right)}dx \\
&= x\log _{3x}\left(x^2\right)-\left(-\frac{2\ln \left(3\right)x}{\ln \left(3x\right)}+\frac{2}{3}\ln \left(3\right)\text{Ei}\left(\ln \left(3x\right)\right)\right) \\
&= 2x\log _{3x}\left(x\right)+\frac{2\ln \left(3\right)x}{\ln \left(3x\right)}-\frac{2}{3}\ln \left(3\right)\text{Ei}\left(\ln \left(3x\right)\right) + c
\end{align}

Therefore:

$$\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx = \left(\ln \left|\tan \left(x\right)\right|+c\right)\left(2x\log _{3x}\left(x\right)+\frac{2\ln \left(3\right)x}{\ln \left(3x\right)}-\frac{2}{3}\ln \left(3\right)\text{Ei}\left(\ln \left(3x\right)\right)+c\right)$$

答:
I think you are asking to verify that
$$\frac{\tan \left(x\right)}{\sin ^2\left(x\right)}.$$ RIGHT?
if so, $$\frac{\sin(x)/\cos(x)}{\sin ^2\left(x\right)}=\frac{1}{\sin(x)\cos(x)}=\frac{2}{2 \sin(x)\cos(x)}=\frac{2}{\sin(2x)}$$


答: $ \sin{(2x)} = 2 \cos{(x)} \sin{(x)}$ if that's what you are asking for...

Title: Integrate $\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx$.

Body:
Integrate the following integral:
$$\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx$$

I did this question a while ago and the answer is correct. However, when I started revising this topic again, I can't seem to figure out how I went from the first step $\left( \text{for this integral}\int \log _{3x}\left(x^2\right)dx\right)$ to the second step. Here's my attempt:

\begin{align}
\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx  &= {\int{\frac{2}{\sin{\left(2 x \right)}} d x}} \\
&= {\int{\frac{1}{\sin{\left(x \right)} \cos{\left(x \right)}} d x}} \\
&= {\int{\frac{\sec^{2}{\left(x \right)}}{\tan{\left(x \right)}} d x}} \\
&\stackrel{{u = \tan{\left(x\right)}}}{=} {\int{\frac{1}{u} d u}} \\
&= {\ln{\left(u \right)}} + c \\
&= \ln{\left(\left|{\tan{\left(x \right)}}\right| \right)}+
\end{align}

\begin{align}
\int \log _{3x}\left(x^2\right)dx &= x\log _{3x}\left(x^2\right)-\int \frac{2\ln \left(3\right)}{\ln ^2\left(3x\right)}dx \\
&= x\log _{3x}\left(x^2\right)-\left(-\frac{2\ln \left(3\right)x}{\ln \left(3x\right)}+\frac{2}{3}\ln \left(3\right)\text{Ei}\left(\ln \left(3x\right)\right)\right) \\
&= 2x\log _{3x}\left(x\right)+\frac{2\ln \left(3\right)x}{\ln \left(3x\right)}-\frac{2}{3}\ln \left(3\right)\text{Ei}\left(\ln \left(3x\right)\right) + c
\end{align}

Therefore:

$$\int \frac{\tan \left(x\right)}{\sin ^2\left(x\right)}dx\cdot \int \log _{3x}\left(x^2\right)dx = \left(\ln \left|\tan \left(x\right)\right|+c\right)\left(2x\log _{3x}\left(x\right)+\frac{2\ln \left(3\right)x}{\ln \left(3x\right)}-\frac{2}{3}\ln \left(3\right)\text{Ei}\left(\ln \left(3x\right)\right)+c\right)$$


问:
Proof of binomial coefficient formula.
How can we prove that the number of  ways choosing $k$ elements among $n$ is    $\frac{n!}{k!(n-k)!} = \binom{n}{k}$    with $k\leq n$?
This is an accepted fact in every book but i couldn't find a proof.</p>

答:
Perhaps it is worth mentioning a method of solution that might appear a bit far-fetched, but which is an excellent introductory example to the theory of generating functions, for which the book generatingfunctionology by Herbert S. Wilf is highly recommended.

Write $\dbinom{n}{k}$ for the number of choosing $k$ objects out of $n$. Consider the polynomial
$$
(1 + x_{1}) (1 + x_{2}) \cdots (1 + x_{n})
$$
in the $n$ indeterminates $x_{1}, \dots, x_{n}$. It is easy to convince oneself that this polynomial has exactly $\dbinom{n}{k}$ monomials of degree $k$. If one now sets $x_{1} = \dots = x_{n} = x$, where $x$ is another indeterminate, each such monomial reduces to $x^{k}$, so that one obtains that
$$
(1 + x)^{n} = \sum_{k = 0}^{n} \dbinom{n}{k} x^{k}.
$$
Now use Taylor's theorem for polynomials to obtain the required formula.

Let me stress that this might look totally overkill in this relatively simple setting. But the method has tremendous potential.

Title: Proof of binomial coefficient formula.


问:
Proving a property on Hilbert-Schmidt operator 's norm.
Let $E\subseteq \Bbb R^n$ be measurable and with $|E|>0$ (positive measure), we know $L^2(E)$ is a Hilbert space; so, given $K\in L^2(E\times E)$, we can define the Hilbert-Schmidt operator
$$T_K:L^2(E) \to L^2(E)$$
such that, $\forall f\in L^2(E)$,
$$(T_Kf)(x) := \int_E K(x,y) f(y) dy.$$

Since $L^2(E)$ is a Hilbert space, you can use Schwarts' inequality to determine that $\Vert T_k \Vert_{\mathcal{L}(L^2(E))} \le \Vert K\Vert_{L^2(E\times E)}$, in fact
$$\Vert T_Kf \Vert_{L^2(E)}^2 =\int_E \Bigg( \int_E k(x,y) f(y) dy \Bigg)^2 dx \le \int_E \Bigg( \int_E \big(K(x,y)\big)^2dy \cdot \int_E \big(f(y)\big)^2 dy \Bigg) dx =$$
$$= \int_{E\times E} \big(K(x,y)\big)^2 dxdy \cdot \Vert f \Vert_2^2 = \Vert K \Vert_{L^2(E\times E)}^2 \cdot \Vert f \Vert_{L^2(E)}^2,$$
so that, by definition of norm of an operator, we get that inequality.

My request: does somebody know how to prove that actually equality holds; in other words that, with the generality of the Hilbert-Schmidt operators, we always have
$$\Vert T_k \Vert_{\mathcal{L}(L^2(E))} = \Vert K\Vert_{L^2(E\times E)} .$$

I am not sure that this is true, but in particular cases I seem to be able to show that equality holds and, if not, I can never show that it doesn't.

答:
Conceptually, the $L^2$-norm of $K$ is the Hilbert-Schmidt norm of $T_K$. You are looking for the operator norm of $T_K$, which is usually smaller. In matrices, if you have
$$
K=\begin{bmatrix} 1&0\\0&2\end{bmatrix},
$$
then $\|K\|_2=\sqrt5$, while $\|K\|=2$. Below is an adaptation of this idea to your context.

Write $E=E_1\cup E_2$ with $E_1\cap E_2=\emptyset$ and $|E_1|=|E_2|=|E|/2$.  Let
$$
K(x,y)=1^\vphantom{E}_{E_1\times E_1}+2\,1^\vphantom{E}_{E_2\times E_2}.
$$
So
$$
T_Kf(x)=\Big(\int_{E_1}f\Big)\,1^\vphantom{E}_{E_1}+2\,\Big(\int_{E_2}f\Big)\,1^\vphantom E_{E_2}.
$$
We have
\begin{align}
\|T_kf\|_2^2&=\Big|\int_{E_1}f\Big|^2\,|E_1|+4\,\Big|\int_{E_2}f\Big|^2\,|E_2|
\leq |E_1|^2\,\int_{E_1}|f|^2+4|E_2|^2\,\int_{E_2}|f|^2\\[0.3cm]
&=\frac{|E|^2}4\,\bigg[\int_{E_1}|f|^2+4\int_{E_2}|f|^2\bigg]
=\frac{|E|^2}4\,\bigg[4\|f\|^2_2-3\int_{E_1}|f|^2\bigg]\\[0.3cm]
&\leq |E|^2\,\|f\|_2^2.
\end{align}
The equality is achieved when $f=\frac{\sqrt2}{|E|^{1/2}}\,\,1_{E_2}$. Thus $\|T_K\|=|E|$.

Meanwhile,
$$
\|K\|_2^2=\int_E\int_E |K(x,y)|^2\,dy\,dx
=\int_{E\times E}1_{E_1\times E_1}+4\int_{E\times E}1_{E_2\times E_2}
=|E_1|^2+4|E_2|^2=\frac{5|E|^2}4.
$$
That is,
$$
\|K\|_2^2=\frac{\sqrt5\,|E|}2>|E|=\|T_K\|.
$$


问:
An interesting problem on progression and series
In my text book I found a problem which asked to sum of the following series $$1\times 2\times 3+2\times 3 \times 4+ 3\times 4\times 5+\cdots + n(n+1)(n+2)$$ which I found to be $\frac{n(n+1)(n+2)(n+3)}{4}$ which is indeed true.

Now as a general thought it came to my mind what if the problem asks to find the closed form of this$-$ $$\sum_{k=1}^{n}\underbrace{k\times(k+1)\times(k+2)\times(k+3)\times(k+4)\cdots\times(k+m-1)}_{m \text { terms}}$$ where $m$ is any integer for instance in the aforementioned problem it was $4$. Now, following the pattern our intuition suggest that the answer should be $$\boxed{\frac{n(n+1)(n+2)\cdots(n+m)}{m+1}}$$ Now, in order to check one can easily say that for $m=1$ it's obviously true. It's also true for $m=2$. So now my natural question is that Is the closed sum true for all $m \in\mathbb{N}$ ? I have tried with induction but at the end I mess it up and the idea to proof in that way seems to be went into vain and I also have no further idea to proceed, any help? Thanks for your attention.

答:
The product
$$
\prod\limits_{k = 0}^{m - 1} {\left( {z + k} \right)}  = z^{\,\overline {\,m\,} }
  = {{\Gamma \left( {z + m} \right)} \over {\Gamma \left( z \right)}}
 = \left( {z + m - 1} \right)^{\,\underline {\,m\,} }
$$

is known as the Rising factorial
and is in general defined for $z,m \in \mathbb C$.

Here we will consider for simplicity the case $m \in \mathbb Z$,
and $z^{\,\underline {\,m\,} } ,\quad z^{\,\overline {\,m\,} } $ will represent respectively the Falling and Rising Factorial.

One of its properties is that the Finite Difference (unitary step) is
$$
\Delta _{\,z} \;z^{\,\overline {\,m\,} }
  = \left( {z + 1} \right)^{\,\overline {\,m\,} }  - z^{\,\overline {\,m\,} }
  = m\left( {z + 1} \right)^{\,\overline {\,\,m - 1\,\,} }
$$
or
$$
\Delta _{\,z} \;\left( {z - 1} \right)^{\,\overline {\,m + 1\,} }
  = z^{\,\overline {\,m + 1\,} }  - \left( {z - 1} \right)^{\,\overline {\,m + 1\,} }
  = \left( {m + 1} \right)z^{\,\overline {\,\,m\,\,} }
$$

Then the sum has a clean and straight formulation
$$
\eqalign{
  & \sum\limits_{k = 1}^n {\left( {z + k} \right)^{\,\overline {\,m\,} } }
  = {1 \over {m + 1}}\sum\limits_{k = 1}^n
 {\Delta _{\,z} \;\left( {z + k - 1} \right)^{\,\overline {\,m + 1\,} } }  =   \cr
  &  = {1 \over {m + 1}}\sum\limits_{k = 0}^{n - 1}
 {\Delta _{\,z} \;\left( {z + k} \right)^{\,\overline {\,m + 1\,} } }
  = {1 \over {m + 1}}\sum\limits_{k = 0}^{n - 1} {\left( {\left( {z + 1 + k} \right)^{\,\overline {\,m + 1\,} }
  - \left( {z + k} \right)^{\,\overline {\,m + 1\,} } } \right)}  =   \cr
  &  = {1 \over {m + 1}}\left( {\left( {z + n} \right)^{\,\overline {\,m + 1\,} }
  - z^{\,\overline {\,m + 1\,} } } \right) \cr}
$$

That is, the Falling/Rising factorials have difference / sum results that are the discrete
analog of the derivative /integral results for $z^m$.


问:
Prove or give a counterexample: $ null\ T^m=null\ T^{m+1}\;\;\Leftrightarrow\;\;range\ T^m=range\ T^{m+1} $
V denotes an infinite-dimensional vector space over F (R or C). Suppose $ T\in L(V)$ (i.e. a linear map from V to itself ) and $m$ is a nonnegative integer. Prove or give a counterexample:

$ null\ T^m=null\ T^{m+1}\;\;\Leftrightarrow\;\;range\ T^m=range\ T^{m+1} $ .

BTW: It is true when V is finite-dimensional. This can be proved using the Fundamental Theorem of Linear Maps.

I would be really grateful if someone could help.

答:
It is false in general. Consider the left shift $L : \ell^2 \to \ell^2$ given by
$$L(x_1,x_2,x_3, \ldots) = (x_2,x_3, \ldots).$$

Then $\operatorname{null} L^n = \{(x_1, \ldots, x_n, 0, 0, \ldots): x_1, \ldots, x_n\in\Bbb{F}\}$ but $\operatorname{range} L^n = \ell^2$.


问:
What does the comma notation mean?
I am confused by the notation comma.

I know that the comma means 'AND' in Set theory as gate ($a \land b=a$ AND $b$),

But we write solution of equation as $x=1$, $2$ (the equation: $x^2-3x+2=0$)

The question is whether $x=1,2$ is WRONG?

$x=1,2$ $\iff$ $x=1$ AND $x=2$, so rewritten as $x=1$ or $x=2$.

But in so many books, it is written as $x=1$, $2$.

I am confused.....

(PS. I think the comma of $x=1$, $2$ is only notation of classification...? Is it ok?)

答:
saying x = 1 and x = 2 can be misleading as x can't be both at the same time. So, IMO, don't use the comma as it might be interpreted as AND.

x^2-3x+2 = 0

x=1 or x=2

i.e one or the other but not both at the same time

It's not a lot to ask given that it immediately avoids any miscommunication

Title: What does the comma notation mean?


问:
If $A, B$ are Morita equivalent, then $\underline{\text{Nat}}(I_A, I_A)\cong\underline{\text{Nat}}(I_B, I_B)$.
Let $A, B$ two rings and $I_A: {}_A\text{Mod} \to{}_A\text{Mod}$ the identity functor. I am trying to show that if $A, B$ are Morita equivalent, then $\underline{\text{Nat}}(I_A, I_A)\cong\underline{\text{Nat}}(I_B, I_B)$ (where $\text{Nat}(I_A, I_A)$ is the set of all natural transformations from $I_A$ to $I_A$).

My idea is the following: If the $A, B$ are Morita equivalent, then ${}_A \text{Mod}$ is equivalent to ${}_B \text{Mod}$ and the equivalence is given by the adjunction
$$(P \otimes_A - , Q\otimes_B -)$$
where $P \in {}_B \text{Mod}_A$ and $Q \in {}_A \text{Mod}_B$. Consequently, $\lambda \in \underline{\text{Nat}}(I_A, I_A)$ can be send to $\overline{\lambda} \in \underline{\text{Nat}}(I_B, I_B)$ such that for each $N \in{}_B\text{Mod}$, there exists a unique (up to isomorphism) $M \in {}_A \text{Mod}$ with
$$\varphi: N \cong P\otimes_A M,$$
and we define
$$\overline{\lambda}_N: N \overset{\varphi}{\longrightarrow} P\otimes_A M \overset{\text{id}_P \otimes \lambda_M}{\longrightarrow} P\otimes_A M  \overset{\varphi^{-1}}{\longrightarrow}N.$$
With the same idea, we can send $\gamma \in \underline{\text{Nat}}(I_B, I_B)$ to $\overline{\gamma} \in \underline{\text{Nat}}(I_A, I_A)$. But here I am not able to show that these applications are mutual inverse (I even have some difficulties to show that $\overline{\lambda}$ is indeed in $\underline{\text{Nat}}(I_B, I_B)$). Is it the right way to show the statement ? Can someone help me?

答:
Note that the given data should already ensure such a canonical isomorphism $\varphi:N\to P\otimes_A M$, namely using the specific $M=Q\otimes_B N$ and the natural isomorphism $P\otimes_A(Q\otimes_B N)\cong N$. If you use that, everything should work as intended.

Nevertheless, as this ought to be the case for general equivalent categories as well, a more abstract approach could be more beneficial:

Suppose $F:\mathcal A\to\mathcal B$ and $G:\mathcal B\to\mathcal A$ are an equivalence pair of functors with adjoint natural isomorphisms $\eta:1_{\mathcal A}\to GF,\ \varepsilon:FG\to 1_{\mathcal B}$.

Then for a natural transformation $\lambda:1_{\mathcal A}\to 1_{\mathcal A}$ we can define $\overline\lambda$ by
$$\overline\lambda_B:=\varepsilon_B\circ F(\lambda_{GB})\circ \varepsilon^{-1}_B\,,$$
and similarly you can define the other map ${\rm Nat}(1_{\mathcal B},1_{\mathcal B})\to{\rm Nat}(1_{\mathcal A},1_{\mathcal A})$ using $\eta$, and verify that they are inverses of each other.


问:
Mathematical Systems Question Help
Alright, this is another question for my math for teachers course. This question is not actually in the homework, but there are problems similar to it. I'd really like to learn how to do problems like this one, so it would be much appreciated if I was given a solid explanation/demonstration. I don't really have a clue where to start, so I've defined the terms as best as I can under each question... The question should be answered using the data from the table.

A mathematical system is defined by the following table:

Is there an identity element in this system? If so, what is it?
Identity is: aXe=a and eXa=a
Is closure satisfied by this system? Explain.
Closure is: aXb
Is this system commutative? Explain.
Commutative is: aXb=bXa
Does every element have an inverse? Explain.
Inverse is: aXx=e and xXa=e
Is this system a group? Explain.
Not sure what they mean by this...
Are they referring to another property?.

答:
As we can see through the table below:

According to the table if we want to have an identity element, we should see one of the following cases:

```<code>        a    b    c
     ---------------
   a .  a    b    c
   b .  b
   c .  c
</code>```

Or:

```<code>        a    b    c
     ---------------
   a .       a
   b .  a    b    c
   c .       c
</code>```

Or:

```<code>        a    b    c
     ---------------
   a .            a
   b .            b
   c .  a    b    c
</code>```

so we see the second form above is coincide the table. It is not difficult that $b$ plays a role of an identity here. Just check the definition you had written for elements $a, b$ and $c$.

If $id=b$ as it happened above, so search for an element $x$ among $a,b,c$ such that $$a*x=b,~~x*a=b$$ It is really $x=a$ and $x=c$. This shows that the structure is not a Group since the inverse is unique in any Group and for any element in it. Now do the same way for $b$ and $c$ to find additional information.

Title: Mathematical Systems Question Help

Body:
As we can see through the table below:

According to the table if we want to have an identity element, we should see one of the following cases:

        a    b    c
     ---------------
   a .  a    b    c
   b .  b
   c .  c

Or:

        a    b    c
     ---------------
   a .       a
   b .  a    b    c
   c .       c

Or:

        a    b    c
     ---------------
   a .            a
   b .            b
   c .  a    b    c

so we see the second form above is coincide the table. It is not difficult that $b$ plays a role of an identity here. Just check the definition you had written for elements $a, b$ and $c$.

If $id=b$ as it happened above, so search for an element $x$ among $a,b,c$ such that $$a*x=b,~~x*a=b$$ It is really $x=a$ and $x=c$. This shows that the structure is not a Group since the inverse is unique in any Group and for any element in it. Now do the same way for $b$ and $c$ to find additional information.