问:
Finding unknown variance in order to maximize probability given a Gaussian distribution
Given a Gaussian (Normal) Random Variable that has a known mean, μ, and unknown variance $\sigma^2,$ how would we determine the variance in order to maximize the probability, $P[X_1<X<X_2]?$

The way I think about this is in order to maximize the probability, I set the derivative with respect to sigma of, $${\int\displaystyle {\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma  }}\right)^{2}}}$$ equal to 0. However, I don't know where to go after that.

答:
Letting $Z \sim N(0,1)$ and noting that $Z$ and $\frac{X-\mu}{\sigma}$ are equal in distribution, then

\begin{align*}
P(x_1 \le X \le X_2)
&= P \left (\frac{x_1-\mu}{\sigma} \le \frac{X-\mu}{\sigma}
 \le \frac{x_2-\mu}{\sigma}\right)\\
&= P \left (\frac{x_1-\mu}{\sigma} \le Z
 \le \frac{x_2-\mu}{\sigma}\right)\\
&= \Phi \left ( \frac{x_2-\mu}{\sigma}\right) - \Phi \left ( \frac{x_1-\mu}{\sigma}\right)
\end{align*}

where $\Phi$ denotes the cdf of the standard normal, i.e. $\Phi(z) = P(Z \le z)$. Now, differentiate this expression with respect to $\sigma$ and set equal to zero yields:

$$
0 = -\frac{(x_2-\mu)}{\sigma^2} \phi \left ( \frac{x_2-\mu}{\sigma}\right)
+\frac{(x_1-\mu)}{\sigma^2} \phi \left ( \frac{x_1-\mu}{\sigma}\right)
$$
where $\phi(z) = \frac{1}{\sqrt{2 \pi}} e^{-z^2/2}$ denotes the pdf of the standard normal, and $\frac{d}{dz} \Phi(z) = \phi(z)$. Rearranging the above yields
\begin{align*}
&(x_2-\mu) \phi \left ( \frac{x_2-\mu}{\sigma}\right)
=(x_1-\mu) \phi \left ( \frac{x_1-\mu}{\sigma}\right)\\
\implies &
(x_2-\mu) \frac{1}{\sqrt{2 \pi}} \exp \left \{ -\frac{1}{2 \sigma^2} (x_2-\mu)^2 \right \}
=(x_1-\mu)  \frac{1}{\sqrt{2 \pi}} \exp \left \{ -\frac{1}{2 \sigma^2} (x_1-\mu)^2 \right \}\\
\implies &
\frac{x_2-\mu}{x_1 - \mu}
=   \frac{\exp \left \{ -\frac{1}{2 \sigma^2} (x_1-\mu)^2 \right \}}{
\exp \left \{ -\frac{1}{2 \sigma^2} (x_2-\mu)^2 \right \}
}\\
\implies &
\frac{x_2-\mu}{x_1 - \mu}
=   \exp \left \{ -\frac{1}{2 \sigma^2}[ (x_1-\mu)^2 - (x_2-\mu)^2] \right \}\\
\implies &
\log \left ( \frac{x_2-\mu}{x_1 - \mu}  \right)
=   -\frac{1}{2 \sigma^2}[ (x_1-\mu)^2 - (x_2-\mu)^2]\\
\implies & \sigma^2 = \frac{- [ (x_1-\mu)^2 - (x_2-\mu)^2]}{ \log \left ( \frac{x_2-\mu}{x_1 - \mu}  \right)}
\end{align*}

and as noted in the comment, we must have $\frac{x_2-\mu}{x_1-\mu}>0$ for the final expression to be defined


问:
Proof of the identity $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$
I just found this identity but without any proof, could you just give me an hint how I could prove it?
$$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$

I know that $$2^n = \sum\limits_{k=0}^n \binom{n}{k}$$ but that didn't help me

ContentLicense: CC BY-SA 3.0

Body:
$$\sum\limits_{k=0}^{n}\binom{k+m}{m}z^k=\frac{1}{(1-z)^{m+1}}-\frac{z^{n+1}}{(1-z)^{m+1}}\sum\limits_{k=0}^{m}\binom{n+k}{k}(1-z)^k$$
$$n=m, z=\frac{1}{2}$$
$$\sum\limits_{k=0}^{n}\binom{k+n}{n}\frac{1}{2^k}=\frac{1}{\frac{1}{2^{n+1}}}-\frac{\frac{1}{2^{n+1}}}{\frac{1}{2^{n+1}}}\sum\limits_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}$$
$$S=2^{n+1}-S, 2S = 2^{n+1}, S=2^n$$

ContentLicense: CC BY-SA 4.0


问:
Way where $K$ elements (numbered $1$ to $K$) can be aligned in $N$ places where only one element is allowed to repeat?
I was asked this question while I was giving an interview for a software engineer role but simply couldn't find the correct answer. I am pretty sure this can be solved using permutation and combination but still always arrived at the wrong answer. Any help would be appreciated.

What are the number of ways in which, $K$ elements which are numbered from $1$ to $K$ like $1,2,....,K$ can be aligned in $N$ places such that no $2$ same number are adjacent except the one specified as variable $X$ ?

For the context of the question, please consider the following example:

$$N = 3 K = 2 X = 1$$

Explanation: Arrange $K = 2$ so $1,2$ in $3$ places such that $2$ cannot be placed next to each other but $1$ can like $1,1$ but not $2,2$.

Possible ways: $\{1,1,1\} , \{1,1,2\} , \{1,2,1\} , \{2,1,1\} , \{2,1,2\}$.

Another example can be:

$$N = 5 K = 4 X = 3$$ And there are $469$ possible ways in total.

So, how do I generate a common formula for greater numbers?

答:
To solve this question, we start by building two sequences, $a_n,b_n$ such that $a_n$ is the number of allowed sequences of length $n$ and $b_n$ is the number of allowed sequences of length $n$ assuming that you can't start with one of the regular numbers (i.e. can't start with some specific number that's not $X$).

We are therefore interested in a formula for $a_n$.

Then $a_n = a_{n-1} + (k-1)b_{n-1}$ as we can either choose $X$ as the first number (and then there are $a_{n-1}$ options to continue the sequence) or one of the other $k-1$ options (and then there are $b_{n-1}$ options to continue the sequence for each of the choices).

Quite similarly, we get $b_n = a_{n-1}+(k-2)b_{n-1}$ as we can either choose $X$ or any of the other $k-2$ elements to start the sequence.

We can therefore obtain a matrix equality given by
$$\left[ \array{1\ k-1\newline1\ k-2} \right]\cdot\left[ \array{a_{n-1}\newline b_{n-1}} \right]=\left[ \array{a_{n}\newline b_{n}} \right]$$

After finding $a_1=k,b_1=k-1$, we could calculate the $n$'th element of $a$ by diagonalising the matrix above (it has 2 eigenvalues for every real, positive value of $k$, thus is diagonalisable for such values), and, denoting by $P$ the change of basis matrix and $D$ the diagonal matrix, taking this form to the $n$'th power, then calculating $PD^{n-1}P^{-1}\left[ \array{a_{1}\newline b_{1}} \right]$, or if you rather, since $\left[ \array{1\ k-1\newline1\ k-2} \right]\left[ \array{1\newline 1} \right]=\left[ \array{a_{1}\newline b_{1}} \right]$, one could calculate $$PD^{n}P^{-1}\left[ \array{1\newline 1} \right]$$ and return the first coordinate.

There might be an easier way to do this, but unfortunately I couldn't find one :)


问:
Proving $2\cos\left(2\pi ft + \phi\right)\cos(2\pi ft) = \cos(4\pi ft + \phi) + \cos(\phi)$
I need to prove this identity:

$$2\cos\left(2\pi ft + \phi\right)\cos(2\pi ft) = \cos(4\pi ft + \phi) + \cos(\phi)$$

I know I have to use some identity or property but I can't find any to do it.

答:
We will use: $\cos(a+b) = \cos a. \cos b - \sin a. \sin b$

and also $\cos^2 \frac{a}{2} =\frac{1+\cos a}{2}$ and $ \sin (2a) = 2 \sin a \cos a$

$2 \cos(2 \pi ft + \phi) \cos(2 \pi ft) = 2 cos(2 \pi ft)(cos(2 \pi ft) \cos\phi - \sin(2 \pi ft) \sin \phi) $

$= 2 cos²(2 \pi ft) \cos\phi - 2 \cos(2 \pi ft) \sin(2 \pi ft) \sin \phi$

$= 2 cos²(2 \pi ft) \cos\phi - \sin(4 \pi ft) \sin \phi$

$= 2. \frac{1+ \cos(4 \pi ft)}{2} \cos \phi - \sin(4 \pi ft) \sin \phi$

$= \cos \phi + \cos (4 \pi ft + \phi)$


问:
Linear equation | Finding the unknown values
I have a set of equations as follow,

$a_1\cdot x_1 + a_2\cdot x_2 + a_3\cdot x_3 + a_4\cdot x_4 + \dots + a_{11} \cdot x_{12} + a_{12}\cdot x_{12}  = 6000$

$x_1 + x_2 + x_3 + x_4 + \dots x_{11} + x_{12} = 100 $

$\{a_1,a_2,a_3, \dots , a_{11}, a_{12}\} \leq 60$

How can I find the values for all $a_i, \text{ and }  x_i$,not all values of $a_i$ can be equal to $60$ also if possible integer, else the general solution.

How the question get affected if I change the constrain on $a_i$ to some $X$, say $0 < a_i \leq X$,

Also if the constrain on $a_i$ is , say $a_i>0$.

If solvable in Matlab please show the code also .

If not solvable please mention why.

What I tried is
$$
\begin{bmatrix}
a_1&a_2 & a_3 &\dots & a_{11} & a_{12} \\
1 & 1 & 1& \dots & 1 & 1
\end{bmatrix}\begin{bmatrix}
x_1 \\ x_2 \\ x_3  \\ \vdots\\ x_{11} \\ x_{12}
\end{bmatrix} = \begin{bmatrix}
6000 \\ 100
\end{bmatrix}
$$

The rank will be almost $2$ so atmost I can get $10$ free variables. but since the values for $a_i$'s are also unknown I don't know how to proceed.

答:
So we have
$$
\left\{ \matrix{
  0 \le a_{\,k}  \le 60\quad \left| {\,1 \le k \le 12} \right. \hfill \cr
  0 \le x_{\,k}  \le 100\quad \left| {\,1 \le k \le 12} \right. \hfill \cr
  x_{\,1}  + x_{\,2}  +  \cdots  + x_{\,12}  = 100 \hfill \cr
  {\bf a} \cdot {\bf x} = 6000 \hfill \cr}  \right.
$$

If we put $b_k = 60 -a_k$ then
$$
\left\{ \matrix{
  b_{\,k}  = 60 - a_{\,k} \quad \left| {\,1 \le k \le 12} \right. \hfill \cr
  u_{\,k}  = 1\quad \left| {\,1 \le k \le 12} \right. \hfill \cr
  0 \le x_{\,k}  \le 100 \hfill \cr
  0 \le b_{\,k}  \le 60\quad \left| {\,1 \le k \le 12} \right. \hfill \cr
  {\bf u} \cdot {\bf x} = 100 \hfill \cr
  6000 = {\bf a} \cdot {\bf x} = \left( {60\,{\bf u} - {\bf b}} \right) \cdot {\bf x} =  \hfill \cr
   = 6000 - {\bf b} \cdot {\bf x}\quad  \Rightarrow \quad {\bf b} \cdot {\bf x} = 0 \hfill \cr}  \right.
$$

and clearly, with ${\bf b}, \, {\bf x}$ having non-negative components we must have
$$
\left\{ \matrix{
  {\bf b} = {\bf 0}\; \Rightarrow \;{\bf a} = 60\,{\bf u} \hfill \cr
  0 \le x_{\,k}  \le 100 \hfill \cr
  {\bf u} \cdot {\bf x} = 100 \hfill \cr}  \right.
$$

That is, $\bf x$ are the vectors lying on the diagonal plane of a 12-D hypercube of side  $[0,100]$.

There are  $101^{12}  \approx  10^{24}$ points inside that cube, and those on the diagonal plane
are exactly the number of weak compositions
of $100$ into $12$ parts , i.e. $\binom {100+12-1}{11} \approx 2.8 \cdot 10^{14}$.

So your first care should be how to manage such a huge amount of data, if you want to list all the solutions.

You can somewhat reduce the size by taking advantage of the permutation invariance.

If $(x_1,x_2, \cdots)$ is a solution, then any permutation of the twelve components will be.

So you can limit to list the solutions in order, e.g.  non-decreasing.

The number in this case will be reduced to that of the partitions of $100$ into up to twelve parts
(because the components may be null), which is $ \approx 1.7 \, \cdot 10^7$.


问:
Modular number theory problem
Let $k\in\mathbb{N}, 2<p\in\mathbb{P}, p^{\alpha}\mid\mid n,\alpha\ge1$. Then prove that $$1^k+2^k+\ldots+n^k\equiv\begin{cases}\hfill 0\pmod{p^{\alpha}},& p-1\nmid k\\-\frac{n}{p}\pmod{p^{\alpha}},& p-1\mid k\end{cases}$$

Firstly, I proved for $n=p$. If $p-1\mid k$, then it can easily proved by Fermat's little theorem. Let $p-1\nmid k$. Since $p$ is a prime, it has a primitive root $g$ and it satisfies $\{1,g,g^2,\ldots,g^{p-2}\}=\{1,2,3,\ldots,p-1\}$. So it's sufficient to prove that $1+g^k+g^{2k}+\ldots+g^{(p-2)k}\equiv 0\pmod p\iff p\mid\frac{g^{(p-1)k}-1}{g^k-1}\iff p\cdot (g^k-1)\mid g^{(p-1)k}-1$. Since $(p, g^k-1)\neq1\iff p-1\nmid k$ and $p\mid g^{(p-1)k}-1,g^k-1\mid g^{(p-1)k}-1$, it's true.

And I can't continue that for $n=p^{\alpha}$ (for $p-1\nmid k$, it's similar to $n=p$. But I can't prove for $p-1\mid k$), $n=p^{\alpha}\cdot n_1$, where $n_1>1, (n_1,p)=1$ and so on. Can anyone help me?

答:
Viewing everything as elements of $\Bbb{Z}/p^a\Bbb{Z}$,

If $p-1\nmid k$  then take $p\nmid s, s^k \ne 1\bmod p$ $$\sum_{m\bmod p^a} m^k =\sum_{m\bmod p^a} (sm)^k=s^k\sum_{m\bmod p^a} m^k$$ which means that $\sum_{m\bmod p^a} m^k=0$.

If $p-1|k$ then take the largest $b$ such that $(p-1)p^b | k$, since $m\to m^k$ , restricted to the units, is surjective to the subgroup of elements of the form $1+p^{b+1}l$ we get that $$\sum_{m\bmod p^a} m^k=\sum_{m\bmod p^a,p\ \nmid\ m} m^k+\sum_{m\bmod p^{a-1}} (pm)^k\\=(p-1)\sum_{l\bmod p^{a-1}} (1+p^{b+1} l)+p^k \sum_{m\bmod p^{a-1}} m^k$$
$$ = -p^{a-1}+p^k \sum_{m\bmod p^{a-1}} m^k$$
$$ = -p^{a-1}-p^k p^{a-2}-p^{2k} p^{a-3}-\ldots- p^{(a-1)k} p^{a-a}$$
$$ = -p^{a-1}$$

Title: Modular number theory problem


问:
Find the inverse Laplace transform of $f(t) = \int_t^\infty \frac{e^{ - u}}{u}du$
Find the inverse Laplace transform of the integral:$$f(t) = \int_t^\infty  {\frac{{{e^{ - u}}}}{u}du} $$
If the integral: $$f(t) = \int_0^\infty  {\frac{{{e^{ - u}}}}{u}du} $$
I had done. However the integral from t (not ZERO) to infinity???</p>

Body: $\mathcal{L}^{-1}_{t\to x}\left\{\int_t^\infty\dfrac{e^{-u}}{u}du\right\}=\dfrac{\mathcal{L}^{-1}_{t\to x}\left\{\dfrac{e^{-u}}{u}\right\}}{x}=\dfrac{H(x-1)}{x}$


问:
Singular Points of $\sin(x)y''+y=0$. Review my work
I have to classify the singular points of the second order
$$\sin(x)y´´+y=0$$

Where, in its standar form:
$$y´´+\dfrac{1}{\sin(x)}y=0$$

With

$$P(x)=0,  Q(x)=\dfrac{1}{\sin(x)}$$

As far as I can tell, the singular points are
$$x_0=n\pi, \forall n\in \mathbb{Z}$$

So, there are infinite singular points? Which all of them are irregular since there is at most 1 power of $(x-x_0)$ in $P(x)$ (there is non) and there is not a power of 2 in $Q(x)$.

I other words, both $(x-x_0) P(x)$ remains finite as $x\to n\pi$ but $(x-x_0)^2 Q(x)$ doesn´t.

$$\lim_{x\to n\pi}{(x-n\pi)P(x)}=\lim_{x\to n\pi}{0}=0$$
and
$$\lim_{x\to n\pi}{(x-n\pi)^2Q(x)}=\lim_{x\to n\pi}{(x-n\pi)^2\dfrac{1}{\sin(x)}}$$

Does not exist.

答:
$$L=\lim_{x\to n\pi}{(x-n\pi)^2Q(x)}=\lim_{x\to n\pi}{(x-n\pi)^2\dfrac{1}{\sin(x)}}$$
Note that:
$$L=\lim_{z\to 0}{\dfrac{z^2}{\sin(z+n\pi)}}$$
And we also have:
$$\sin (z+n \pi)=(-1)^n\sin z$$
$$\lim_{z\to 0}{\dfrac{z}{\sin(z)}}=1$$
So that we have:
$$L=\lim_{z\to 0}{\dfrac{(-1)^nz^2}{\sin(z)}}=\lim_{z\to 0}{{(-1)^nz}} \dfrac z{\sin(z)}$$
Looks like the limit tends to zero as $z \to 0$


问:
Please give feed back on my proof that $S$ must be a basis for $V$.
Let $S$ be a maximal linearly independent subset of a finite vector
space $V$. That is, $S$ has the property that if a vector not in $S$
is adjoined to $S$, then the new set will no longer be linearly
independent. Prove that $S$ must be a basis for $V$.

Proof:

We have $S$={$v_1...v_n$}$⊆V$, and because $S$ is linearly independent, there scalars $c_1...c_n$ must all zero if $c_1v_1+...+c_nv_n=0$

Let $v_{n+1}∈V$. by the problem statement, the set {$v_1...v_n, v_{n+1}$} is linearly dependent, so we can write $c_1v_1+...+c_nv_n+c_{n+1}v_{n+1}=0$ such that $c_1...c_n, c_{n+1}$ are not all $0$.

If $c_{n+1}=0$ then we have $c_1v_1+...+c_nv_n=0$ and because {${v_1,\ldots,v_n}$} is linearly independent, $c_1...c_n$ must all be $0$, which contradicts the fact that $c_1...c_n, c_{n+1}$ are not all $0$, so $c_{n+1}≠0$.

So we can write $v_{n+1}=(c_1/-c_{n+1})v_1+...+(c_n/-c_{n+1})v_n$.

Now, $v_{n+1}$ is an arbitrary element of $V$, so $SPAN${$v_1...v_n$}=$V$

So, $S$ must be a basis for $V$.

答: It is mostly correct, but you made a mistake when you wrote that, if $c_{n+1}=0$, then we have $c_1v_1+\cdots+c_nv_n=0$, where $c_1,\ldots,c_n$ must all be $0$, which contradicts the fact that $\{v_1,\ldots,v_n,v_{n+1}\}$ is linearly dependent. There is no contradiction between $c_1v_1+\cdots+c_nv_n=0$ and the fact that $\{v_1,\ldots,v_n,v_{n+1}\}$ is linearly dependent. The reason why we cannot have $c_{n+1}=0$ is that because then $c_1v_1+\cdots+c_nv_n=0$ which would imply (since $\{v_1,\ldots,v_n\}$ is linearly independent) that $c_1=\cdots=c_n=0$. But $c_{n+1}=0$ too and we are assuming that not all $c_k$'s are equal to $0$.


问:
How do you define “independence” in combinatorics?
I feel like most definitions of “independence” are circular. Consider how we count the number of cards in a standard deck of cards: $|S \times R| = |S||R|$, where $S$ is the set of suits, and $R$ is the set of ranks. That is, $$S = \{\text{Hearts, Diamonds, Spades, Clubs}\}$$ and $$R = \{ 2,3,\dots,\text{King},\text{Ace} \}.$$ We know that $|S| = 4$ and $|R| = 13$, so $|S \times R| = |S||R| = 4 \cdot 13 = 52.$ In such an example, we define that they are independent, because they are disjoint subsets. But do we know that? In this example, it is “obvious.” What about examples where it is not obvious?

答:
To answer the question in the title: you don't. Independence is a concept in probability theory, not in combinatorics. In your card example, there is no concept of the sets $S$ and $R$ being independent. What you can say is that, if all $52$ cards are equally likely to be chosen in some experiment, then the probability of a particular rank, say ace, is independent of a particular suit, say clubs. But if one of the cards is sticky, and hence more likely to be chosen than the others, independence is lost.

You can look up the standard definitions of independence: for events $E$ and $F$, the events are independent if $\Pr(E\cap F)=\Pr(E)\Pr(F)$, or equivalently, if $\Pr(E\vert F)=\Pr(E)$.

Also, be careful to distinguish the concepts of disjointness and independence. They are not related.

Added: It looks like your question is really about sets that have the structure of a Cartesian product. If $S=A\times B$ and all elements of $S$ are equally likely, then the events $X=$"first element of tuple is $x$" and $Y=$"second element of tuple is $y$" are independent in the usual probability sense. These events are the sets $X=\{(x,b)\vert b\in B\}$ and $Y=\{(a,y)\vert a\in A\}$. Now $\Pr(X\cap Y)=\Pr(\{(x,y)\})=\frac{1}{\lvert S\rvert}=\frac{1}{\lvert A\rvert\lvert B\rvert}$, while $\Pr(X)\Pr(Y)=\frac{\lvert B\rvert}{\lvert S\rvert}\frac{\lvert A\rvert}{\lvert S\rvert}=\frac{\lvert B\rvert\lvert A\rvert}{(\lvert A\rvert\lvert B\rvert)^2}=\frac{1}{\lvert A\rvert\lvert B\rvert}$, so that the first definition of independence is satrisfied. This independence holds regardless of whether $A$ and $B$ are disjoint or not. (They might even be the same set.)


问:
Let $A = \begin{pmatrix} a & b & c \\ 0 & a & b \\ 0 & 0 & a \\ \end{pmatrix} $. Find $A^n$
I am new here so please let me know if I must resentence the exercice. I considered it too short not to include it in the title too.

Let $ A=     \begin{pmatrix}     a & b & c \\     0 & a & b \\     0 & 0 & a \\     \end{pmatrix}$

Find $A^n$

I tried finding the matrix's first 4 or 5 powers and it looks like a pettern can be noticed but I can't seem to find a form for the element on line 3 row 3. I haven't tried using a binomial expansion though, writing the matrix as $$     \begin{pmatrix}     0 & b & c \\     0 & 0 & b \\     0 & 0 & 0 \\     \end{pmatrix}+ a    \begin{pmatrix}     1 & 0 & 0 \\     0 & 1 & 0 \\     0 & 0 & 1 \\     \end{pmatrix} $$

the second one being the identity matrix times $a.$

答:
Hint

If $$N=\begin{pmatrix}0&b&c\\0&0&b\\0&0&0\end{pmatrix},$$
then $$A=aI+N$$

(where $I$ is the identity matrix), and $N^3=0$. Applying Binomial theorem yields $$A^n=(aI+N)^n=\sum_{k=0}^n\binom{n}{k} a^{n-k}N^k=\cdots$$


问:
Finite-dimensionality of Ext¹ of simple modules
For simplicity let $\Bbbk$ be a field of characteristic $0$ and let $A$ be a finitely generated unital associative $\Bbbk$-algebra.

Is it true that for any two simple $A$-modules $S_1, S_2$, we have that $\operatorname{Ext}^1_A (S_1, S_2)$ is finite-dimensional?

If not, what would be a simple counterexample, and what sort of conditions do we need to ensure this?

(As far as I understand, the Weyl algebra $\Bbbk \langle x, y \rangle / (xy - yx - 1)$ has only infinite-dimensional simple modules, but their first extension groups are still finite-dimensional.)

答:
It's true if $A$ is commutative. We can compute $\text{Ext}^1(S_1, S_2)$ using the first three terms of a free resolution of $S_1$; since $S_1$ is simple it's cyclic so the first term of this resolution is $A$ and the second term of this resolution is any finite set of generators of the kernel of $A \to S_1$, which is finitely generated because $A$ is Noetherian; Noetherianness also implies that we can take the third term (and in fact every term) to be finite free. So we get a levelwise finite free resolution

$$\cdots \to A^n \to A^m \to A \to S_1 \to 0$$

which computes $\text{Ext}^1(S_1, S_2)$ as a subquotient of $\text{Hom}(A^m, S_2) \cong S_2^m$; by the Nullstellensatz $\dim_k S_2$ is finite so $\dim_k \text{Ext}^1(S_1, S_2)$ must also be finite. More generally $\dim_k \text{Ext}^n(S_1, S_2)$ is finite.

If $A$ is noncommutative $\dim_k S_2$ may be infinite and the kernel of $A \to S_1$ may be infinitely generated (if $A$ is not Noetherian) so this argument fails in two places.


问:
Project Euler #731_part2
In this link : Project Euler problem #731 , I got a nice solution for my question but i'm missing a crucial part .

I want to get an explanation of this part of the answer :

My question : Why did we calculate the numerator $mod$ $3^k$ ? I would like to get a demonstration , because i'm not familiar with modular arithmetic

答:
The author of the answer states: "We only need the fractional part of $S_2$": Any number $n$ can be written as $n=m+r$ where $m$ is divisible by $3^k$ and $r=n\mod 3^k$. But $\frac{n}{3^k} = \frac{m}{3^k} + \frac{r}{3^k}$.
Since $m$ is divisible by $3^k$ the first summand on the rhs is an integer and thus does not contribute to the fractional part of $\frac{n}{3^k}$. So it suffices to compute the $n\mod 3^k$. In this particular instance that seems to be more efficient computationally. EDIT: For a demonstration check Wikipedia or google

Title: Project Euler #731_part2


问:
Are all of the irreducible representations of (any) symmetric group over $\mathbb{C}$ also irreducible over a finite splitting field.
This is probably an incredibly stupid question, but I'm a novice to representation theory and finite field theory, as I've just been introduced to these concepts, so all I really need is confirmation of the following:

Let $S_n$ be the symmetric group of $n$ symbols and let $F=\mathbb{F}_q$ be a finite field of $q$ elements and characteristic $p>n$. Then $F$ is a splitting field for $S_n$ (that is to say every irreducible $FS_n$-module is absolutely irreducible). Furthermore, every irreducible representation of $S_n$ over $F$ is also irreducible over $\mathbb{C}$.

This seems almost tautological to me, but it would be nice to have confirmation just to make sure I haven't misunderstood thinks spectacularly.

As a further (potentially stupid) question, are these irreducible representations over $F$ the 'same' as those over $\mathbb{C}$, albeit with reduction modulo $p$. Do they have the same degree for instance? Again from the definitions, this would appear to me to be an emphatic yes, but again it would be nice to have confirmation.

Thank you.

答:
One can show that for every $\def\bb{\mathbb}\bb CS_n$-module $V$ there is a $\bb ZS_n$-module $V'$ which is free as an abelian group and of rank equal to the dimension of $V$ as a vectr space, such that $V'\otimes_{\bb Z}\bb C\cong V$ as $\bb CS_n$-modules. This is a rather  special property of $S_n$.

Now, if we start with $V$, construct $V'$ as above, then we can also construct $V''=V'\otimes_{\bb Z}\bb F_p$, and this is a representation of $\bb F_pS_n$. The passage from $V$ to $V''$ involves a choice, that of $V'$. When one understands what it entails, one can wonder if the mapping $V \leadsto V''$ gives a bijection between irreps of $\bb CS_n$ an irreps of $\bb F_pS_n$. It does, as long as $p>n$.

Other finite groups which have this property that its complex irreps are defined over $\bb Z$ are the Weyl groups (of which $S_n$ is an example) and surely others.
For a general group $G$, a complex representation $V$ is always defined over the ring $\mathcal O$ of integers of some splitting field of $G$ (one can show that there is a splitting field which is cyclotomic) and then one can "reduce" $V$ modulo prime ideals of $\mathcal O$. The theory is extremely rich, complicated and deep. You can find this explain in the classic book by Curtis and Reiner, for example. The litle book of Serre has lots of information on the problem of rationality of representations, that is, of trying to restrict the field over which a representation is defined, but here we want considerably more; we want to restrict to a ring whose field of quotients is a field of definition of the representation. Already for cyclic groups this is a non-obvious problem.


问:
Proof of the matrix form of the Lie algebra of $SU(2)$
The Lie algebra elements fulfill

$$\mathfrak su(2) =\{X: \text {Tr} (X) = 0 \;, X^\dagger = - X \}.\tag 1$$

$$\mathfrak su(2) = \left\{\begin{bmatrix} ix & y + iz \\ -y+iz & -ix \end{bmatrix}\;x,y,z\in \mathbb R \right\},$$

On the other hand the Lie group is

$$SU(2)=\{A: \det(A)=1 \;, A^\dagger A=1 \}.$$

This implies matrices of the form

$$SU(2) = \left\{\begin{bmatrix}a & b \\ -\bar b & \bar a \end{bmatrix}\;a,b\in \mathbb C,\vert a \vert^2+ \vert b \vert^2=1 \right\},$$

since $A^\dagger A=1 \implies A^\dagger A^{-1},$ and the determinant is $1.$

I guess that to prove $(1)$ either the log or the exponential mapping would have to be used with some Taylor expansion at least to the linear term, but I don't know where to start.

答: You have$$\mathfrak{su}(2)=\left\{X\in GL(2,\Bbb C)\,\middle|\,(\forall t\in\Bbb R):\exp(tX)\in SU(2)\right\}.$$But, if $X\in GL(2,\Bbb C)$ and $t\in\Bbb R$,\begin{align}\exp(tX)\in SU(2)&\iff\exp(tX)^\dagger\exp(tX)=\operatorname{Id}\text{ and }\det\bigl(\exp(tX)\bigr)=1\\&\iff\exp\left(tX^\dagger\right)=\exp(tX)^{-1}\text{ and }\exp\bigl(\operatorname{tr}(tX)\bigr)=1\\&\iff\exp\left(tX^\dagger\right)=\exp(-tX)\text{ and }\exp\bigl( t(\operatorname{tr}(X)\bigr)=1\end{align}and, since these equalities occur for each $t\in\Bbb R$,$$\left.\frac{\mathrm d}{\mathrm dt}\exp\left(tX^\dagger\right)\right|_{t=0}=\left.\frac{\mathrm d}{\mathrm dt}\exp(-tX)\right|_{t=0}\quad\text{and}\quad\left.\frac{\mathrm d}{\mathrm dt}\exp\bigl( t(\operatorname{tr}(X)\bigr)\right|_{t=0}=0,$$which is the same thing as asserting that $X^\dagger=-X$ and that $\operatorname{tr}(X)=0$.


问:
Darboux Sums and Mesh P
I am currently learning about the analysis basics of Riemann Integration and integration as a whole. For one of my practice exercises, I was able to prove that if f is uniformly continuous over $[a,b]$, we can in fact show that the the difference between the Darboux Upper Sum and the Darboux Lower Sum is less than or equal to $\epsilon (b-a)$.

Knowing that $U(f,P) - L(f,P) \leq \epsilon(b-a)$, how can I go about using that to prove that the limit of the difference between the Darboux Upper Sum and Darbou Lower sum, as the mesh(P) $\to 0$, in fact equals $0$.

That is, how can I conclude lim $(U(f,P) - L(f,P))$  as mesh(P) $\to 0$ actually equals $0$?

Thank you!

答:
hint

You want to prove that

$$\forall \epsilon>0\;\exists \delta>0\;:$$
$$\;||P||<\delta \implies U(f,P)-L(f,P)<\epsilon$$

You can take $ \delta $ as the $ \eta$ of the uniform continuity :

$$\forall \epsilon\;\exists \eta>0\;:$$
$$\;|x-y|<\eta\implies|f(x)-f(y)|<\frac{\epsilon}{b-a}$$

Title: Darboux Sums and Mesh P


问:
Parametrization of a circle, why is it $(\sin t, \cos t)$ instead of $(\sin^2t,\cos^2t)$?

The equation of a circle is $x^2 + y^2 = r^2$.

Then why is the parameterization of a circle $(r\cos(t), r\sin(t))$ and not $(r\cos^2(t), r\sin^2(t))$?

I thought what I'm supposed to do here is to eliminate $x$ and $y$ and write them in a way where I can represent them both with $t$ and then put it back into the equation? If I do that, I get

$$x = r\cos(t)$$
$$y = r\sin(t)$$

then if I put it back into the equation of the circle I get them both squared. Shouldn't the result then be $(\sin^2t,\cos^2t)$?

What's wrong here?

答:
When we use the notation
$(x,y)$ with regards to Cartesian co-ordnates we mean that the $x$-coordinate is $x$ and the $y$-coordiante is $y$ (unsurprisingly). If we have $x$ and $y$ expressed parametrically then we can replace each variable by the equation in the given parameter it is equivalent to; eg in your case we'd replace $x$ by $r\cos{t}$.


问:
What is an integral number?
This has nothing to do with integration :)

I've just seen the words 'integral number' used a couple of times and it seems that it means the same thing as an integer; is that correct?

Thanks for the verification.

EDIT
For example, this is a question I saw:

How many differently shaped triangles exist in which no two sides are the
same length, each side is of integral unit length and the perimeter of the triangle is
less than 13 units?

答: Your guess was right, but the quoted example is especially idiosyncratic because it says "unit". In normal use, integral length would be equal to some integer, while unit length would be of length $1$ (see "unit number" here). Presumably the author meant, "in the unit (with a different meaning!) we use to measure lengths, these lengths are integer-valued".

Title: What is an integral number?


问:
Problem with right angle triangle, circumscribed circle, tangent and the half of its height
(An interesting problem inspired by this one but still different. And, no, I'm not looking for your help to solve a detail here in order to provide a full solution elsewhere. I'll stop here).

A right-angled triangle ABM is given ($\angle M = 90^\circ)$. A tangent $t$ is drawn at point $M$ to the circumscribed circle with center $O$. Perpendicular bisector $OE$ of side $AM$ meets tangent $t$ at point $D$. Prove that line $DB$ cuts triangle height $MH$ in half ($CM=CH$)

Ok, one way to solve this is to use analytical geometry. It's doable but tedious. More ellegant way is to solve the problem by using complex geometry and I will demonstrate that solution here:

We'll take circumscribed circle to be unit circle. Let's assign the following complex numbers to various points: $z_M=z,z_0=0,z_A=-1,z_B=1$. It's a well know formula that:

$$z_D=\frac{2z_Mz_A}{z_M+z_A}=\frac{-2z}{z-1}$$

Also notice that:

$$Re(C)=Re(M)=\frac{z+\bar{z}}{2}$$

$$Im(C)=\frac12 Im(M)=\frac12\frac1{2i}(z-\bar{z})$$

$$z_C=Re(C)+i\cdot Im(C)=\frac{z+\bar{z}}{2}+\frac14(z-\bar{z})=\frac 14(3z+\bar z)$$

The last thing we have to prove is that points $D,C,B$ are colinear which is true if:

$$z_D-z_B=k(z_C-z_B)$$

...for some real $k$. In other words:

$$\frac{-2z}{z-1}-1=k(\frac 14(3z+\bar z)-1)\iff$$

$$\frac{-2z-(z-1)}{z-1}=\frac{k}{4}(3z+\bar z-4)\iff$$

$$\frac{-3z+1}{z-1}\frac{\bar z-1}{\bar z-1}=\frac{k}{4}(3z+\bar z-4)\iff$$

$$\frac{-3z+1}{z-1}\frac{\bar z-1}{\bar z-1}=\frac{k}{4}(3z+\bar z-4)\iff$$

$$\frac{3z+\bar z-4}{l}=\frac{k}{4}(3z+\bar z-4)\iff$$

$$k=\frac4l, \quad l=(z-1)(\bar z - 1)\in R$$

Question: I think we can prove this with Euclid too, without complex or any other numbers. Can someone come up with one such solution?

答:
Let $AM=b$ and $BM=a$.

Thus, $$CH=HB\tan\measuredangle CBH=\frac{a^2}{\sqrt{a^2+b^2}}\cdot\tan\measuredangle DBA=\frac{a^2}{\sqrt{a^2+b^2}}\cdot\frac{DA}{AB}=$$
$$=\frac{a^2}{a^2+b^2}\cdot\frac{b}{2\cos\measuredangle DAE}=\frac{a^2b}{2(a^2+b^2)\cos\measuredangle ABM}=\frac{a^2b}{2(a^2+b^2)\cdot\frac{a}{\sqrt{a^2+b^2}}}=$$
$$=\frac{ab}{2\sqrt{a^2+b^2}}=\frac{1}{2}MH.$$


问:
What is the name for the function that returns the number of times one number is a factor of another
I'm looking for the function that for a number n returns the number of times a number k is a factor of n. So for example for k=3, the following values are returned for a n of 1 through 27:

$$\begin{array}{c:c}
n & \text{# of times $k=3$ is a factor of $n$}\\
\hline
27 & 3 \\
9, 18 & 2 \\
3, 6, 12, 15, 21, 24 & 1 \\
\text{the rest} & 0
\end{array}$$

What is the name of that function and how is it written? I hope this is an easy question, my google-fu has failed me.

Edit: after I refined my title a bit a relevant question came up in the suggestions. (Notation for the number of times one element divides another.) However I couldn't read the answer so go easy on me. I'm just using real, positive integers today.

答:
From Ireland and Rosen, "A classical introduction to modern number theory," (page 3):

Let $n\in\mathbb{Z}$ and let $p$ be prime. Then if $n$ is not zero, there is a nonnegative integer $a$ such that $p^a\mid n$ but $p^{a+1}\not\mid n$. [...] The number $a$ is called the order of $n$ at $p$ and is denoted by $\operatorname{ord}_p n$. Roughly speaking, $\operatorname{ord}_pn$ is the number of times $p$ divides $n$. If $n=0$, we set $\operatorname{ord}_pn=\infty$. Notice that $\operatorname{ord}_pn=0$ if and only if (iff) $p\not\mid n$.

The omitted part was about how this is well defined for positive and negative $n$.

The language "order of $n$ at $p$" evokes a modern idea that $n$ is a function that can be evaluated at different primes (by modulo), and $\operatorname{ord}_pn$ is the order of the zero at $p$.  It's like how the polynomial $(x-1)^2$ has an order-$2$ zero at $1$.


问:
Minimize $x^2y^2z^2$ using classical inequalities.

Minimize $x^2y^2z^2$ if $x^2+y^2+z^2=4$

Well I want to minimize it using classical inequalities; because I have easily maximized it with $AM\ge GM$ as follows:
$$x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2 z^2}$$
$$\iff 4 \ge3\sqrt[3]{x^2y^2 z^2} \iff \frac{4}{3}\ge \sqrt[3]{x^2 y^2z^2}$$
$$\iff \frac{64}{27}\ge x^2y^2z^2$$
but I don't know how to find the minimum using classical inequalities. Is there a way?

Any help would be appreciated!

答:
Let $z=0$ and $y=z=\sqrt2$.

Thus, $$x^2y^2z^2=0$$ and since $$x^2y^2z^2\geq0$$ for any $x$, $y$ and $z$ such that $x^2+y^2+z^2=4$, we got a minimal value.

Id est, $$\min_{x^2+y^2+z^2=4}x^2y^2z^2=0.$$


答:
Let $X=aI,Y=bJ$ and $Z=cJ^2$, where $J$ denotes the $3\times3$ nilpotent Jordan block. Then $X,Y,Z$ commute. Since $J^3=0$, among all degree-$n$ monomials of them, only $X^n,\,X^{n-1}Y,\,X^{n-1}Z$ and $X^{n-2}Y^2$ are nonzero. It follows that
\begin{aligned}
A^n
&=(X+Y+Z)^n\\
&=X^n+nX^{n-1}Y+nX^{n-1}Z+\binom{n}{2}X^{n-2}Y^2\\
&=\pmatrix{a^n&na^{n-1}b&na^{n-1}c+\binom{n}{2}a^{n-2}b^2\\ 0&a^n&na^{n-1}b\\ 0&0&a^n}.
\end{aligned}

Title: Let $A = \begin{pmatrix} a & b & c \\ 0 & a & b \\ 0 & 0 & a \\ \end{pmatrix} $. Find $A^n$

Body:
I am new here so please let me know if I must resentence the exercice. I considered it too short not to include it in the title too.

Let $ A=     \begin{pmatrix}     a & b & c \\     0 & a & b \\     0 & 0 & a \\     \end{pmatrix}$

Find $A^n$

I tried finding the matrix's first 4 or 5 powers and it looks like a pettern can be noticed but I can't seem to find a form for the element on line 3 row 3. I haven't tried using a binomial expansion though, writing the matrix as $$     \begin{pmatrix}     0 & b & c \\     0 & 0 & b \\     0 & 0 & 0 \\     \end{pmatrix}+ a    \begin{pmatrix}     1 & 0 & 0 \\     0 & 1 & 0 \\     0 & 0 & 1 \\     \end{pmatrix} $$

the second one being the identity matrix times $a.$


问:
Show that the contracted multiplication U = ${{U_k}^j}_m$ = $S^{uj}T_{kum}$ is a (1,2)-tensor.
Let $S^{ij}$ be the coordinates of a contravariant tensor S of the second order and $T_{klm}$ is the coordinates for a covariant tensor of third order. Show that the contracted multiplication U = ${{U_k}^j}_m$ = $S^{uj}T_{kum}$ is a (1,2)-tensor.

I want to show this by writing out the full tensor including the basisvectors.
I do understand that it is possible to show it by showing that the contracted tensor product transforms coordinates as a (1,2)-tensor, but I want to do it in a slightly other way, just by inspection of the expression of the tensor.

The coordinates S^{ij} are the product of two vectors and the coordinates $T_{klm}$ are the product of three covectors.

Contracting over the indices i and l , i = l = u we find :

$$ U = S^{uj}T_{kum}e_u \oplus e_j \oplus f^k \oplus f^u \oplus f^m $$

what I want my result to be is the following :

= ${{U_k}^j}_m f^k \oplus e_j \oplus f^m$

I do not understand the argument for how to get here. I understand that we sum over the indices u, but I dont understand why the two basisvectors $e_u$ and $f^u$ dissapear from the expression when we sum.

答:
You should be using $\otimes$, not $\oplus$, in the tensor product formulas.

Do you know how to think about vectors and covectors as being in a vector space $V$ and its dual space $V^*$? The tensor $S$ belongs to $V^{\otimes 2} = V \otimes V$ and the tensor $T$ belongs to $(V^*)^{\otimes 3} = V^* \otimes V^* \otimes V^*$. The idea of contraction is to just evaluate elements of $V^*$ on elements of $V$ in order to product scalars in a nice (bilinear) way.  The evaluation function $V \times V^* \to \mathbf R$ where $(v,\varphi) \mapsto \varphi(v)$ is a bilinear function of $v$ and $\varphi$ (linear in $v$ when $\varphi$ is fixed and linear in $\varphi$ when $v$ is fixed).  Therefore you get a linear function $V \otimes V^* \to \mathbf R$ that on elementary tensors $v \otimes \varphi$ has the value $\varphi(v)$.  Do you understand how bilinear or multilinear functions become linear functions when using tensor products?  This is the key point about tensor products of vector spaces: they linearize multilinear functions.

In any case, you are asking about how a contraction mapping
$$
V^{\otimes 2} \otimes (V^*)^{\otimes 3} \to V \otimes (V^*)^{\otimes 2}
$$
really works.  Specifically, you want to contract with the "first" piece of $V^{\otimes 2}$ and the "second" piece of $(V^*)^{\otimes 3}$ and you are confused about the basis vectors in those pieces disappear.  They do not really disappear, but rather you are evaluating the second piece of $(V^*)^{\otimes 3}$ on the first piece of $V^{\otimes 2}$ to get a scalar.  When you choose to use for your bases a basis of $V$ and its dual basis for the basis of $V^*$ then things appear to disappear because the very definition of a dual basis $e^1, \ldots, e^n$ in $V^*$ (you write $f^1, \ldots, f^n$) to a basis $e_1, \ldots, e_n$ of $V$ is the rule $e^i(e_j) = \delta_{ij}$, so $e^i(e_i) = 1$.

Here is how I would describe the contraction you ask about.  First we define the multilinear mapping
$$
V \times V \times V^* \times V^* \times V^* \to V  \otimes V^* \otimes V^*
$$
by combining the first coordinate from $V \times V$ and the second coordinate from $V^* \times V^* \times V^*$ using evaluation:
$$
(v_1,v_2,\varphi_1,\varphi_2,\varphi_3) \mapsto \varphi_2(v_1)v_2 \otimes \varphi_1 \otimes \varphi_3.
$$
The values of this function are multilinear in all 5 coordinates of $V \times V \times V^* \times V^* \times V^*$, so we get automatically a linear mapping
$$
V \otimes V \otimes V^* \otimes V^* \otimes V^* \to V  \otimes V^* \otimes V^*
$$
that has the following effect on elementary tensors:
$$
v_1 \otimes v_2 \otimes \varphi_1 \otimes \varphi_2 \otimes \varphi_3 \mapsto \varphi_2(v_1)v_2 \otimes \varphi_1 \otimes \varphi_3.
$$
This is what the contraction you write about really is.  Nothing is disappearing anywhere.  Suppose we pick a basis $e_1, \ldots, e_n$ of $V$ and use as a basis of $V^*$ the dual basis $e^1, \ldots, e^n$.  Then the 5-fold tensor product $V^{\otimes 2} \otimes (V^*)^{\otimes 3}$ has as a basis the 5-fold elementary tensors $e_i \otimes e_j \otimes e^k \otimes e^{\ell} \otimes e^m$.  The effect on these particular elementary tensors of your chosen contraction (evaluate 2nd dual tensorand in $(V^*)^{\otimes 3}$ on 1st tensorand in $V^{\otimes 2}$) is
$$
e_i \otimes e_j \otimes e^k \otimes e^{\ell} \otimes e^m \mapsto e^{\ell}(e_i) e_j \otimes e^k \otimes e^m = \delta_{i\ell} e_j \otimes e^k \otimes e^m.
$$
So if we have general tensors $S = \sum S^{ij} e_i \otimes e_j$ in $V^{\otimes 2}$ and $T = \sum T_{k\ell m} e^k \otimes e^\ell \otimes e^m$ in $(V^*)^{\otimes 3}$, then the above contraction $V^{\otimes 2} \otimes (V^*)^{\otimes 3} \to V \otimes (V^*)^{\otimes 2}$
has the following effect on $S \otimes T$:
\begin{eqnarray*}
S \otimes T = \sum_{ijk\ell m} S^{ij}T_{k\ell m} e_i \otimes e_j \otimes e^k \otimes e^\ell \otimes e^m  & \mapsto & \sum_{ijk\ell m} S^{ij}T_{k\ell m} \, e^\ell(e_i) \, e_j \otimes e^k \otimes e^\ell \otimes e^m  \\
& = & \sum_{ijk\ell m} S^{ij}T_{k\ell m}\delta_{i\ell} \, e_j \otimes e^k \otimes e^m  \\
& = & \sum_{ijk m} S^{ij}T_{k i m} e_j \otimes e^k \otimes e^m
\end{eqnarray*}
where the index $\ell$ goes away in the last calculation because if $\ell \not= i$ then the $(ijk\ell m)$-term is zero on account of $\delta_{i\ell} = 0$.
The coordinates of this final tensor are the numbers $S^{ij}T_{kim}$ in your post, but you chose to create a new index letter $u$ in place of $i$. Otherwise it is exactly the same thing.

When written in terms of a basis and dual basis, this contraction appears to be making things "disappear", but what is really going on is that you're just combining one tensorand (element of $V$) from $V^{\otimes 2}$ and one dual tensorand (in $V^*$) from $(V^*)^{\otimes 3}$, via evaluation, to produce a scalar so everything is depending multilinearly on the initial components.  If you did not use a dual basis in $V^*$ for calculations then the disappearing phenomenon would disappear: what's fundamental is evaluation of $V^*$ on $V$, a concept that does not depend on a choice of basis of the vector space or its dual space.

You say the tensor coordinates $S^{ij}$ are "products of two vectors" and $T_{k\ell m}$ are "products of three covectors". This is false: the coordinates are just numbers. They are not products of anything. The tensors $S$ and $T$ are in $V^{\otimes 2}$ and $(V^*)^{\otimes 3}$, but they are usually not elementary tensors $v \otimes v'$ or $\varphi \otimes \varphi' \otimes \varphi''$, so in no way are $S$ or $T$ a tensor product of two vectors or a tensor product of three covectors.
A simple analogy: polynomials in two variables are sums of monomials $c_{ij}x^iy^j$, but most polynomials in two variables are not products $g(x)h(y)$ of two single-variable polynomials.

Think about the dot product $\mathbf v \cdot \mathbf w$ on $\mathbf R^n$.  It is not making anything disappear: it's just a dot product.  But when you write it out in terms of the standard basis $\mathbf e_1, \ldots, \mathbf e_n$ then it looks like things disappear because $\mathbf e_i \cdot \mathbf e_j = \delta_{ij}$.

You are asking how to show the contracted tensor is a tensor. I did that by actually defining the contraction operation so it takes values in a tensor product space $V \otimes (V^*)^{\otimes 2}$, and thus the whole issue of "checking" the formula has values that are tensors goes away, just as it is tautological that the polyomial $1 - 2x + 7x^3$ is real-valued when $x$ is real.  What would you think if someone asked how to show the function $f : \mathbf R \to \mathbf R$ where $f(x) = 1 - 2x + 7x^3$ is real-valued? It's just automatic because the function is defined that way from the start.


答:
Here:

$BEF \sim BAD$ so $${a\over x} = {t\over z+t} = {y\over x+y}\implies a = {xy\over x+y}$$

$BGD\sim FCD$ so $${b\over y} = {x\over x+y}\implies b = {yx\over x+y}$$

Title: Problem with right angle triangle, circumscribed circle, tangent and the half of its height

Body:
(An interesting problem inspired by this one but still different. And, no, I'm not looking for your help to solve a detail here in order to provide a full solution elsewhere. I'll stop here).

A right-angled triangle ABM is given ($\angle M = 90^\circ)$. A tangent $t$ is drawn at point $M$ to the circumscribed circle with center $O$. Perpendicular bisector $OE$ of side $AM$ meets tangent $t$ at point $D$. Prove that line $DB$ cuts triangle height $MH$ in half ($CM=CH$)

Ok, one way to solve this is to use analytical geometry. It's doable but tedious. More ellegant way is to solve the problem by using complex geometry and I will demonstrate that solution here:

We'll take circumscribed circle to be unit circle. Let's assign the following complex numbers to various points: $z_M=z,z_0=0,z_A=-1,z_B=1$. It's a well know formula that:

$$z_D=\frac{2z_Mz_A}{z_M+z_A}=\frac{-2z}{z-1}$$

Also notice that:

$$Re(C)=Re(M)=\frac{z+\bar{z}}{2}$$

$$Im(C)=\frac12 Im(M)=\frac12\frac1{2i}(z-\bar{z})$$

$$z_C=Re(C)+i\cdot Im(C)=\frac{z+\bar{z}}{2}+\frac14(z-\bar{z})=\frac 14(3z+\bar z)$$

The last thing we have to prove is that points $D,C,B$ are colinear which is true if:

$$z_D-z_B=k(z_C-z_B)$$

...for some real $k$. In other words:

$$\frac{-2z}{z-1}-1=k(\frac 14(3z+\bar z)-1)\iff$$

$$\frac{-2z-(z-1)}{z-1}=\frac{k}{4}(3z+\bar z-4)\iff$$

$$\frac{-3z+1}{z-1}\frac{\bar z-1}{\bar z-1}=\frac{k}{4}(3z+\bar z-4)\iff$$

$$\frac{-3z+1}{z-1}\frac{\bar z-1}{\bar z-1}=\frac{k}{4}(3z+\bar z-4)\iff$$

$$\frac{3z+\bar z-4}{l}=\frac{k}{4}(3z+\bar z-4)\iff$$

$$k=\frac4l, \quad l=(z-1)(\bar z - 1)\in R$$

Question: I think we can prove this with Euclid too, without complex or any other numbers. Can someone come up with one such solution?


问:
Modular number theory problem
Let $k\in\mathbb{N}, 2<p\in\mathbb{P}, p^{\alpha}\mid\mid n,\alpha\ge1$. Then prove that $$1^k+2^k+\ldots+n^k\equiv\begin{cases}\hfill 0\pmod{p^{\alpha}},& p-1\nmid k\\-\frac{n}{p}\pmod{p^{\alpha}},& p-1\mid k\end{cases}$$

Firstly, I proved for $n=p$. If $p-1\mid k$, then it can easily proved by Fermat's little theorem. Let $p-1\nmid k$. Since $p$ is a prime, it has a primitive root $g$ and it satisfies $\{1,g,g^2,\ldots,g^{p-2}\}=\{1,2,3,\ldots,p-1\}$. So it's sufficient to prove that $1+g^k+g^{2k}+\ldots+g^{(p-2)k}\equiv 0\pmod p\iff p\mid\frac{g^{(p-1)k}-1}{g^k-1}\iff p\cdot (g^k-1)\mid g^{(p-1)k}-1$. Since $(p, g^k-1)\neq1\iff p-1\nmid k$ and $p\mid g^{(p-1)k}-1,g^k-1\mid g^{(p-1)k}-1$, it's true.

And I can't continue that for $n=p^{\alpha}$ (for $p-1\nmid k$, it's similar to $n=p$. But I can't prove for $p-1\mid k$), $n=p^{\alpha}\cdot n_1$, where $n_1>1, (n_1,p)=1$ and so on. Can anyone help me?

答:
It's not difficult to check that it's enough to show it for $n=p^{\alpha}$.

We'll show it by induction over $\alpha$.

If $\alpha=1$ then it works. Assume the result holds for $\alpha-1$.

Then, mod $p^{\alpha}$, $\sum_{m=0}^{p^{\alpha}-1}{m^k} = \sum_{m=0}^{p^{\alpha-1}-1}{\sum_{l=0}^{p-1}{(m+p^{\alpha-1}l)^k}} = \sum_{m=0}^{p^{\alpha-1}-1}{\sum_{l=0}^{p-1}{m^k+kp^{\alpha-1}l}}=p\sum_{k=0}^{p^{\alpha-1}-1}{m^k}+kp^{\alpha-1}\sum_{l=0}^{p-1}{l} = p\sum_{k=0}^{p^{\alpha-1}-1}{m^k} = -p \cdot p^{\alpha-1}/p = -p^{\alpha}/p$.

Title: Modular number theory problem


问:
Computing Fundamental Solution
How can I compute the fundamental solution for $$\frac{1}{4}\partial_x^2\partial_y^2=0$$ Can anyone assist me in computing this.

So far I have: $(\frac{1}{4}\partial_x^2\partial_y^2)u=\delta$ $\implies$ $(\frac{1}{4}|x|^2|y|^2)\hat{u}=1$ $\implies$ $\hat{u}=\frac{4}{|x|^2|y|^2}$, how can I get the inverse FT?

答:
Solving $\partial_x^2\partial_y^2u=\delta$ by removing one differentiation at a time:

$$
\partial_x^2\partial_y^2u(x,y)=\delta(x)\delta(y) \\
\partial_x\partial_y^2u(x,y)=\frac12\operatorname{sign}(x)\delta(y)+A''(y) \\
\partial_y^2u(x,y)=\frac12|x|\delta(y)+xA''(y)+B''(y) \\
\partial_yu(x,y)= \frac12|x|\frac12\operatorname{sign}(y)+xA'(y)+B'(y)+C(x)\\
u(x,y)=\frac12|x|\frac12|y|+xA(y)+B(y)+C(x)y+D(x)
$$
where $A,B,C,D$ are arbitrary distributions.

Note: To reduce the amount of introduced names, first introducing one name for a distribution, then another for its primitive distribution, and yet another for the second primitive distribution, I "backtraced" the names and use derivatives early on.

Title: Computing Fundamental Solution


问:
Dense Subsets Of Topological Subspaces
Suppose that $(X, T)$ be a topological space and $A$ be an arbitrary non empty subset of $X$, then $A$ is dense in the subspace $(\bar A, T')$, where $T'$ is the relative topology.
$$$$I want to verify my proof. Please tell me about any mistakes or any corrections (minor or major) is required??Thank You
$$$$To prove that above statement we just have to prove that $$\bar{A'}=\bar A$$, where $\bar{A'}$ is the new closure of $A$ under the topology $T'$. It is easy to see that $$\bar{A'} \subset \bar A$$, so we just have to prove that $$\bar A \subset \bar{A'}$$. Suppose this is not true then there is a $x \in \bar A$ such that $x$ does not lie in $\bar{A'}$. So there exists a set $D$ such that $$A \subset D \subset \bar A$$ and $D^c \in T'$ and $x$ does not lie in $D$. Now as $D^c \in T'$, so $$D^c = \bar A \cap P$$ for some $P \in T$, by the definition of relative topology. Now as $x$ is not in $D$, so $x \in D^c$. So $x \in P$, also as $P \in T$ so $P^c$ is closed under the topology $T$. Also it is easy to see that $D^c \subset P$ and hence $P^c \subset D$. Now as for any $y \in A$ we have $$y \in D \subset \bar A$$, and hence $y$ does not lie in $D^c$, and hence $y$ does not lie in $P$ and hence $y \in P^c$. So we have $A \subset P^c$ and hence we have $$A \subset P^c \subset D \subset \bar A$$. Now as $x \in \bar A$, so by the definition $x$ lies in every superset of $A$ which is closed under $T$ but as $x$ is not in $P^c$ which is a superset of $A$ and closed under $T$, so a contradiction. And hence $$\bar A \subset \bar{A'}$$. And hence $A$ is a dense subset of the topological space $(\bar A, T')$.

答: $\newcommand{\cl}{\operatorname{cl}}$Your proof is correct, but it’s more complicated than necessary. Let $x\in\cl_TA$, and suppose that $x\in U\in T'$. By the definition of $T'$ there is a $V\in T$ such that $U=V\cap\cl_TA$, so $x\in V$. $V$ is an open nbhd of $x$ in $X$, and $x\in\cl_TA$, so $V\cap A\ne\varnothing$, and therefore $U\cap A=V\cap(\cl_TA)\cap A=V\cap A\ne\varnothing$. Thus, every $T'$-open nbhd of $x$ intersects $A$, so $x\in\cl_{T'}A$. This shows that $\cl_TA\subseteq\cl_{T'}A$, and as you say, it’s easy to see that $\cl_{T'}A\subseteq\cl_TA$.

Title: Dense Subsets Of Topological Subspaces


问:
How can I find the value of the coefficients of an inverse exponential graph?
If I have a set of discrete data points and I want to find the line of best fit of the form $$y=Ae^{-Bx}$$

then I can do the following:
$$\ln(y)=-Bx+\ln(A)$$

Therefore, if I plot $(x,\ln(y)$), the gradient is the $-B$ coefficient of my line of best fit and my $y$-intercept is $\ln(A)$ of my line of best fit.

Note: The above method, finds the values of the coefficient of my line of best fit using all data points, and not just choosing two and solving a set of equations.

Now, if I have a set of data points and I want to fit a function of the following form:
$$y=A(e^{-Bx}-1)+C$$

how can I find the value of the coefficients? The method above would not work for this scenario. Instead how could I identify the value of the coefficients $A, B$ and $C$?

Note: ${A,B,C \in \mathbb R }$ and $x$ and $y$ can be either positive or negative.

答: Performing a linear best fit to the log of the data works if you assume that the measurements' noise are Gaussian distributed after taking the log. While there may be some trick to reduce the second case to a linear fit, it is probably better to assume that the noise is Gaussian in the original form and perform a non linear regression, based on a maximum likelihood estimation.


问:
Proving the identity $\sum_{k=1}^n {k^3} = \big(\sum_{k=1}^n k\big)^2$ without induction
I recently proved that

$$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$$

using mathematical induction.  I'm interested if there's an intuitive explanation, or even a combinatorial interpretation of this property. I would also like to see any other proofs.

答:
You've gotta see it to believe it.

In formulas:

$$(\sum k)^2=\sum j^2 + 2 \sum_{i<j} ij =\sum_j (j^2+2 \sum_{i<j} ij)=\sum_j j(j+j(j-1))=\sum_j j(j^2)=\sum_j j^3$$

In words: we can assemble the square with side $(1+...+k)$ from
$k$ smaller squares and pairs of rectangular blocks.  If we look only at the rectangular blocks with the largest side fixed to some size $j$, they all pair up into $j-1$ "complementary" pairs (based on the size of the smaller side, pairing $1$ with $j-1$, $2$ with $j-2$,  untill $j-1$ with $j$; this is Gauss' trick), each pair of rectangles forming a $j$ by $j$ square together. Thus, together with the $j^2$ piece they assemble into a cube of side $j$.


问:
Can this problem have solutions for $N>3$:Find three $(N=3)$ positive integers a, b and c such that: $a+b+c=k^2$, $a+b=t^2$, $b+c=m^2$ and $a+c=n^2$
Can this question have solutions for N>3:

Find three ($N =3$) positive integers a, b and c such that: $a+b+c=k^2$, $a+b=x^2$, $b+c=m^2$ and $a+c=n^2$; $x, m, n∈\mathbb   N$

Solution:

Let $a+b+c=(x+1)^2=x^2+2x+1$

If $a+b=x^2$ then third number $c=2x+1$.

If $b+c=(x-1)^2=x^2-2x+1=m^2$ then :

$a=4x$, so $b=x^2-4x$

Now due to statement we must have $a+c=6x+1=n^2$

$n^2=6x+1$ can have infinite solutions such as:

$(x, n^2)=(20, 121=11^2), (60, 361=19^2), (140, 841=29^2)\cdot\cdot\cdot$

Which give:

$(a, b, c, k)=(80, 320, 41, 21), (240, 3360, 121, 61), (560, 19040, 281, 141),\cdot\cdot\cdot$

I tried to solve this problem for $N=4$, four numbers but no success. Now I have two questions:

1-Does this problem have solutions for N>3?

2-Any idea for better method?

答:
The solution given by "@ player3236" has $k=325$.

The integer $325$ can be written as sum of two squares

in three way's & is shown below:

$k= x_1^2+x_2^2=x_3^2+x_4^2=x_5^2+x_6^2$

($x_1,x_2,x_3,x_4,x_5,x_6$)=$(15,10,17,6,18,1)$

$325= 15^2+10^2=17^2+6^2=18^2+1^2$

Above can help in arriving at a parametric solution.

Also mathematician "Seiji Tomita has given parametric

solution in which $(a,b,c,d)$ taken two at a time is a

square & the link is given below.

http://www.maroon.dti.ne.jp/fermat/dioph115e.html

And click on (dioph115e)


问:
Intuition of critical values in t-distribution below $N=50$ or below sufficiently large $N$

I understand how the z-scores are defined in the standard normal
distribution and I also understand how when $N$ extends to infinity
why the t-distribution approaches and equals the standard normal
distribution.

My question: How are the values in the statistical table for the t-distribution calculated/defined for low values of $N$, because they seem arbitrary to me?

答:
Suppose $n = 10.$ One can compute many t statistics from normal samples of size $n$ where the null hypothesis $H_0: \mu-0$ is true in order
to get a good idea of the shape of Student's t distribution
with $\nu = n-1 = 9$ degrees of freedom. By finding quantile $0.975$
of the results one can find the upper critical value for a two-sided test at the 5% level. [The R code below generates a million normal samples and finds their corresponding t statistics.
With a million samples one can expect about 2-place accuracy.]

```<code>set.seed(1031)
t = replicate(10^6, t.test(rnorm(10))$stat)
quantile(t, .975)
   97.5%
2.256101       # aprx 2.262
qt(.975, 9)
[1] 2.262157   # exact value
</code>```

You can find the exact value of quantile 0.975 of $\mathsf{T}(\nu=9);$ look on line 9 of a printed t table. By symmetry,
the lower critical value for a two-sided test at level 5% is
$-2.262.$

```<code>hdr = "Simulated Dist'n of T(9) with PDF"
hist(t, prob=T, br=50, ylim=c(0,.5), col="skyblue2", main=hdr)
curve(dt(x,9), add=T, col="orange", lwd=2)
abline(v = 2.262, col="blue", lty="dotted")
</code>```

Below is a plot of the density functions of $\mathsf{T}(\nu=0)$ and
$\mathsf{Norm}(0,1).$ They are very nearly the same.
However, they are sufficiently different that $2.262$ cuts
2.5% of the probability from the upper tail of the t distribution, while $1.96$ cuts 2.5% from the upper tail of the standard normal distribution. [Respective standard deviations are $\sqrt{9/7} = 1.1339$ and
$1.]$

R code for figure just above:

```<code>hdr="PDFs of T(9) [orange] and NORM(0,1) [dashed]"
 curve(dt(x,9), -5, 5, col="orange", lwd=2, main=hdr)
 curve(dnorm(x), add=T, lty="dashed")
 abline(h=0, col="green2"); abline(v=0, col="green2")
</code>```

Body:
Suppose $n = 10.$ One can compute many t statistics from normal samples of size $n$ where the null hypothesis $H_0: \mu-0$ is true in order
to get a good idea of the shape of Student's t distribution
with $\nu = n-1 = 9$ degrees of freedom. By finding quantile $0.975$
of the results one can find the upper critical value for a two-sided test at the 5% level. [The R code below generates a million normal samples and finds their corresponding t statistics.
With a million samples one can expect about 2-place accuracy.]

set.seed(1031)
t = replicate(10^6, t.test(rnorm(10))$stat)
quantile(t, .975)
   97.5%
2.256101       # aprx 2.262
qt(.975, 9)
[1] 2.262157   # exact value

You can find the exact value of quantile 0.975 of $\mathsf{T}(\nu=9);$ look on line 9 of a printed t table. By symmetry,
the lower critical value for a two-sided test at level 5% is
$-2.262.$

hdr = "Simulated Dist'n of T(9) with PDF"
hist(t, prob=T, br=50, ylim=c(0,.5), col="skyblue2", main=hdr)
curve(dt(x,9), add=T, col="orange", lwd=2)
abline(v = 2.262, col="blue", lty="dotted")

Below is a plot of the density functions of $\mathsf{T}(\nu=0)$ and
$\mathsf{Norm}(0,1).$ They are very nearly the same.
However, they are sufficiently different that $2.262$ cuts
2.5% of the probability from the upper tail of the t distribution, while $1.96$ cuts 2.5% from the upper tail of the standard normal distribution. [Respective standard deviations are $\sqrt{9/7} = 1.1339$ and
$1.]$

R code for figure just above:

hdr="PDFs of T(9) [orange] and NORM(0,1) [dashed]"
 curve(dt(x,9), -5, 5, col="orange", lwd=2, main=hdr)
 curve(dnorm(x), add=T, lty="dashed")
 abline(h=0, col="green2"); abline(v=0, col="green2")


问:
Evaluate the integral $\int_0^{\infty} \frac{b\ln{(1+ax)}-a\ln{(1+bx)}}{x^2} \,dx$
Evaluate the following integral

$$I=\int_0^{\infty} \frac{b\ln(1+ax)-a\ln(1+bx)}{x^2} \,dx$$ with $\ a,b\in\mathbb{R},\ 0<a<b$.

My first attempt was to write $b\ln(1+ax)-a\ln(1+bx)$ as another integral, so I could substitute in the initial integral and then, by reversing the order of integration, try to calculate $I$.

I've tried writing $$b\ln(1+ax)-a\ln(1+bx)=\frac{ab}{t}\cdot\ln(1+tx)\Biggr|_{t=b}^{t=a}=ab\int_b^a\left(\frac{x}{t(1+tx)}-\frac{\ln(1+tx)}{t^2}\right)\,dt$$ but it doesn't seem to help me evaluating $I$.

What else could I try?

答:
hint

With the substitution, $$t=\frac 1x$$

$$I=\int_0^{+\infty}(b\ln(t+a)-a\ln(t+b)+(a-b)\ln(t))dt$$

and
$$\int \ln(X+c)dX=$$
$$(X+c)\ln(X+c)-X$$


问:
Show that a sequence in a normed space is bounded
Show that $(x_{k})$ a sequence in a normed space is bounded if and only if there exists a number $c$ such that $||x_{k}||<c$  for all $k \in \mathbb{N}$

The definitions to use are the following:

A sequence in $\mathbb{R^n}$ is a function $X: \mathbb{N} \rightarrow \mathbb{R^n}$

It is said that $(x_{k})$ is bounded when there is a ball that contains all the terms of $(x_{k})$

答:
"It is said that $(x_k)$ is bounded when there is a ball that contains all the terms of $(x_k)$"

That just means, that all values are in a ball $B_r(y)$, where $r$ is the radius and $y$ is the middle point. If you take $c=r+\|y\|$, then you get what you wanted, since
$$ \|x_k\|=\|x_k-y+y\| \leq \|x_k-y\| + \|y\| \leq r + \|y\| $$
holds. The other direction is trivial, since you can take the ball $B_c(0)$.


问:
Show that there exists a set $E \in \mathbb{R} \times \mathbb{R}$ such that the cross sections are borel sets, but $E \notin B\otimes B$
From Axler's MIRA

Let $B$ denote the $\sigma$-algebra of Borel subsets of $\mathbb{R}$. Show that there exists a set $E \subset \mathbb{R} \times \mathbb{R}$ such that $[E]_a \in B$ and $[E]^a \in B$ for every $a \in \mathbb{R}$, but $E \notin B \otimes B$.

Here $[E]_a$ is the cross section with a fixed second coordinate and $[E]^a$ is the cross section with a fixed first coordinate. Currently my approach to this problem is to use a non-measurable set such as the vitali set $V$ so that $V \times V \notin B \oplus B$ and then having the projection onto an axis make it trivially a Borel subset, however everytime I try to write this up it ends up being wrong so I am wondering if I am even going in the right direction with this problem now.

edit: Here is what I wrote up so far btw

答:
Following Sheldon Axler's hints, we construct $E$ without invoking the Continuum Hypothesis.
Define an equivalence relation $\sim$ on $[0,1]$ by $x\sim y$ iff
$x-y\in\mathbb{Q}$. Denote the quotient set $[0,1]/\sim$ by $\{A_{i}\mid i\in I\}$.
By the Axiom of Choice, there exists a map $\theta:I\rightarrow\cup_{i\in I}A_{i}=[0,1]$
such that $\theta(i)\in A_{i}$ for each $i$. Denote $V=\theta(I)$.
We go to show that $V$ is not Lebesgue measurable by contradiction.

Suppose the contrary that $V$ is Lebesgue measurable. Let $\mathbb{Q}_{1}=\mathbb{Q}\cap[-1,1]$.
We assert that $[0,1]\subseteq\cup_{r\in\mathbb{Q}_{1}}\left(V+r\right)\subseteq[-1,2]$
and the union is disjoint. Denote $x_{i}=\theta(i)$. Let $x\in[0,1]$,
then $x\sim x_{i}$ for some $i$. Clearly $r_{0}:=x-x_{i}\in\mathbb{Q}_{1}$.
Therefore, $x=x_{i}+r_{0}\in V+r_{0}\subseteq\cup_{r\in\mathbb{Q}_{1}}\left(V+r\right)$.
Let $x\in\cup_{r\in\mathbb{Q}_{1}}\left(V+r\right)$, then $x=x_{i}+r$
for some $i\in I$ and $r\in\mathbb{Q}_{1}$. Note that $x_{i}\in[0,1]$
and $r\in[-1,1]$, so $x\in[-1,2]$. Finally, we go to show that the
union $\cup_{r\in\mathbb{Q}_{1}}\left(V+r\right)$ is disjoint. Suppose
that there exist $r_{1}\neq r_{2}$ such that $(V+r_{1})\cap(V+r_{2})\ne\emptyset$.
Choose $\xi\in(V+r_{1})\cap(V+r_{2})$. $\xi\in(V+r_{1})$ implies
that $\xi=x_{i}+r_{1}$ for some $i$. $\xi\in(V+r_{2})$ implies
that $\xi=x_{j}+r_{2}$ for some $j$. Hence, $x_{i}+r_{1}=x_{j}+r_{2}\Rightarrow x_{i}-x_{j}=r_{2}-r_{1}\in\mathbb{Q}\Rightarrow x_{i}\sim x_{j}\Rightarrow x_{i}=x_{j}$.
It follows that $r_{1}=r_{2}$, which is a contradiction.

Let $l=m(V)$. Since Lebesgue measure is translation invariant, we
have that
\begin{eqnarray*}
 &  & m\left(\cup_{r\in\mathbb{Q}_{1}}\left(V+r\right)\right)\\
 & = & \sum_{r\in\mathbb{Q}_{1}}m(V+r)\\
 & = & \sum_{r\in\mathbb{Q}_{1}}m(V)\\
 & = & \begin{cases}
0, & \mbox{ if }l=0\\
\infty, & \mbox{ if }l>0
\end{cases},
\end{eqnarray*}
which is impossible because $1=m\left([0,1]\right)\leq m\left(\cup_{r\in\mathbb{Q}_{1}}\left(V+r\right)\right)\leq m([-1,2])=3$.

Finally, we define $E=\{(x,x)\mid x\in V\}$. For each $t\in\mathbb{R}$,
\begin{eqnarray*}
E_{t} & = & \{x\mid(x,t)\in E\}\\
 & = & \begin{cases}
\{t\}, & \mbox{ if }t\in V\\
\emptyset, & \mbox{ if }t\notin V
\end{cases}.
\end{eqnarray*}
Therefore $E_{t}$ is Borel. Similarly, we can prove that $E^{t}=\{y\mid(t,y)\in V\}$
is Borel. Next, we go to show that $E\notin\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$.
Let $\pi_{X}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the canonical
projection onto the first coordinate, i.e., $\pi_{X}(x,y)=x$. Let
$\pi_{Y}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the canonical projection
onto the second coordinate. We recall the universal property of joint
measurability: For any map $f:\mathbb{R}\rightarrow\mathbb{R}^{2}$,
$f$ is $\mathcal{B}/\mathcal{B}\otimes\mathcal{B}$-measurable iff
$\pi_{X}\circ f$ and $\pi_{Y}\circ f$ are measurable. Define $f:\mathbb{R}\rightarrow\mathbb{R}^{2}$
by $f(t)=(t,t)$. Clearly $\pi_{X}\circ f=\pi_{Y}\circ f=\mbox{id}_{\mathbb{R}}$,
where $\mbox{id}_{\mathbb{R}}:\mathbb{R}\rightarrow\mathbb{R}$, $\mbox{id}_{\mathbb{R}}(t)=t$,
which is clearly a Borel function. Therefore, $t\mapsto(t,t)$ is
$\mathcal{B}/\mathcal{B}\otimes\mathcal{B}$-measurable. Finally,
$V=f^{-1}(E)$. Since $V\notin\mathcal{B}$, we must have $E\notin\mathcal{B}\otimes\mathcal{B}$.


问:
Prove that there is no 333 consecutive prime numbers such that sum of their squares is also a square.

Prove that there is no 333 consecutive prime numbers such that sum of their squares is also a square.

Such square is of course not divisble by $2$.

Let $p$ be prime number not equal to $2$ than:
$$333 \times p^2 \equiv 333\equiv 1\pmod{4}$$
Not proven.

Let $p$ be prime number not equal to $3$ than:
$$333 \times p^2 \equiv 333\equiv 0\pmod 3$$
That means that this square would be divisible by $9$

I did not know how to use fact that those primes are consecutive.

答:
You have the right idea regarding checking modulo values, but using $8$ works better for the general case. If the first prime number is not $2$, then each of the consecutive prime numbers is odd. Thus, each of their squares would be congruent to $1$ modulo $8$. Also, the sum of these squares must be odd, so if that sum is a perfect square, then it must also be congruent to $1$ modulo $8$.

However, the sum of the $333$ squares would be congruent to $333 \equiv 5 \pmod 8$, so it can't be a perfect square.

If, instead, the first prime is $2$, then there would be $332$ primes which are not a multiple of $3$, so each of their squares would be congruent to $1$ modulo $3$. Thus, the sum of all the $333$ prime squares would be congruent to $332 \equiv 2 \pmod{3}$, which is not congruent to any perfect square.


问:
Singular Points of $\sin(x)y''+y=0$. Review my work
I have to classify the singular points of the second order
$$\sin(x)y´´+y=0$$

Where, in its standar form:
$$y´´+\dfrac{1}{\sin(x)}y=0$$

With

$$P(x)=0,  Q(x)=\dfrac{1}{\sin(x)}$$

As far as I can tell, the singular points are
$$x_0=n\pi, \forall n\in \mathbb{Z}$$

So, there are infinite singular points? Which all of them are irregular since there is at most 1 power of $(x-x_0)$ in $P(x)$ (there is non) and there is not a power of 2 in $Q(x)$.

I other words, both $(x-x_0) P(x)$ remains finite as $x\to n\pi$ but $(x-x_0)^2 Q(x)$ doesn´t.

$$\lim_{x\to n\pi}{(x-n\pi)P(x)}=\lim_{x\to n\pi}{0}=0$$
and
$$\lim_{x\to n\pi}{(x-n\pi)^2Q(x)}=\lim_{x\to n\pi}{(x-n\pi)^2\dfrac{1}{\sin(x)}}$$

Does not exist.

答:
$$\underset{z\to k \pi }{\text{lim}}\frac{z-k \pi }{\sin (z)}=\underset{z\to k \pi }{\text{lim}}\frac{1 }{\cos (z)}=(-1)^k$$
As the limit is nonzero for any $k\in\mathbb{Z}$, $k\pi$ are simple  poles. Furthermore
$$\underset{z\to k \pi }{\text{lim}}\frac{(z-k \pi )^2}{\sin (z)}=0$$
thus $z=k\pi$ are  regular singular points.


问:
Determine the support of an infinite divisible probability measure on $[0,\infty)$
Let $\mu$ be a probability measure on $\mathbb R$ and $$\mathcal L_\mu(t):=\int e^{-tx}\:\mu({\rm d}x)\;\;\;\text{for }t\in\mathbb R$$ denote the Laplace transform of $\mu$. Assume $$\mu((-\infty,0))=0\tag1$$ and $$-\ln\mathcal L_\mu(t)=\alpha t+\int 1-e^{-tx}\:\nu({\rm d}x)\;\;\;\text{for all }t\ge0\tag2$$ for some $\alpha\ge0$ and some $\sigma$-finite measure $\nu$ on $[0,\infty)$ with $$\nu(\{0\})=0.\tag3$$

I want to show that $$\alpha=\sup\left\{x\ge0:\mu([0,x))=0\right\}\tag4.$$

Intuitively, the claim shouldn't be hard to prove. And I think it's related to the support $\operatorname{supp}\mu$ of $\mu$. In the present context, this notion should reduce to $$\operatorname{supp}\mu=\left\{x\ge0:\mu((x-\varepsilon,x+\varepsilon))>0\text{ for all }\varepsilon>0\right\}\tag5.$$ Maybe we can show that the right-hand side of $(4)$ is equal to $\inf\operatorname{supp}$. However, even if it's possible to show that, I still wouldn't know how we can conclude $(4)$ from this result.

答:
Define

$$\beta := \sup\{x \geq 0\::\: \mu([0,x))=0\},$$

then

$$\mathcal{L}_{\mu}(t) = \int_{[\beta,\infty)} e^{-tx} \, \mu(dx) = e^{-\beta t} \underbrace{\int_{[0,\infty)} e^{-tx} \, \tilde{\mu}(dx)}_{\mathcal{L}_{\tilde{\mu}}(t)} \tag{1}$$

for the shifted measure $\tilde{\mu}$ defined by $\tilde{\mu}(B) := \int 1_{B}(x-\beta)\, \mu(dx)=\mu(B+\beta)$. Note that, by construction, $\tilde{\mu}([0,x])>0$ for any $x>0$. Since $\tilde{\mu}$ is also a probability measure with $\tilde{\mu}(-\infty,0)=0$, its Laplace transform satisfies

$$- \ln \mathcal{L}_{\tilde{\mu}}(t) = \tilde{\alpha} t + \int (1-e^{-tx}) \tilde{\nu}(dx) \tag{2}$$ for some $\tilde{\alpha} \geq 0$ and some $\sigma$-finite measure $\tilde{\nu}$ on $[0,\infty)$. By $(1)$, we are done if we can show that $\tilde{\alpha}=0$. As $1-e^{-tx} \geq 0$, it follows from $(2)$ that

$$\mathcal{L}_{\tilde{\mu}}(t) \leq e^{-\tilde{\alpha} t}.$$

On the other hand, the monotonicity of the exponential function yields

$$\mathcal{L}_{\tilde{\mu}}(t) \geq \int_{[0,r]} \underbrace{e^{-tx}}_{\geq e^{-rt}} \, \tilde{\mu}(dx) \geq e^{-rt} \tilde{\mu}[0,r],$$

implying

$$\tilde{\mu}[0,r] e^{-rt} \leq e^{-\tilde{\alpha} t}$$ for all $t \geq 0$. If $\tilde{\mu}([0,r])>0$, then this gives $\tilde{\alpha} \leq r$. Recalling that $\tilde{\mu}([0,r])>0$ holds, by construction, for any $r>0$, we conclude that $\tilde{\alpha}=0$.


问:
Find the maximum of $2xy^2 - 2x^2$
I got the constraint $g(x,y)=x^2+y^2$ (since the domain is $x^2+y^2\ge 1$ and $x\ge 0$) (it is a half circle domain) and the function $f(x,y)=2xy^2-2x^2$. I got the critical point to be $(0,0)$. The Hesse gave me $0$ which is of course no conclusion. I have a problem expressing $x$ and $y$ in terms of $\lambda$ but here is where my problem occurs. I get $\lambda=y^2-2$ and $\lambda=-y^2-2$. What am I doing wrong, is it the start?

答:
Before we start using Lagrange multipliers, lets look at the objective function.

$f(x,y) = x^2y - 2x^2$

If we set $y$ equal to zero, we can make $x$ large and get a very negative value for $f(x).$

If we set $x$ equal to something moderately small, say 1, and let $y$ become large, we get a very large positive value for $f(x).$  The function is unbounded in both directions within the limits of the constraint.

There is no absolute minimum or maximum.

We can use the method of Lagrange multipliers, to find critical points on the boundary of the constraint.

The constraints $x^2 + y^2 \ge 1$ and $x\ge 0$ give a half-plane, excluding a half-disk.

I am inclined to ignore the constraint $x\ge 0$ for the moment.

$F(x,y,\lambda) = 2xy^2 - 2x^2 + \lambda (x^2 + y^2 - 1)$

$\frac {\partial f}{\partial x} = 2y^2 - 4x + 2\lambda x  = 0\\
\frac {\partial f}{\partial y} = 4xy + 2\lambda y = 0\\
\frac {\partial f}{\partial \lambda} = x^2 + y^2 - 1 = 0$

from the second line:

$2\lambda y = -4xy$

Either $y = 0$ or $\lambda = -2x$

$y=0$ gives us $(1,0)$ as a critical point.

Setting $\lambda = -2x$ in the first line:

$2y^2 - 4x - 4x^2 = 0\\
y^2 = 2x^2 + 2x$

Substituting into the equation of the circle:

$3x^2 + 2x - 1 = 0\\
(3x - 1)(x+1)\\
x = \frac 13, -1$

$x = \frac 13$

$(\frac 13 , \pm \frac {\sqrt 8}{3})$ is a second set of critical points to check.

Title: Find the maximum of $2xy^2 - 2x^2$


问:
Probability of picking at least 3 green balls and atmost 1 red and 1 orange ball from 4 different balls in 6 tries
Few days ago I asked almost same question about 3 or more and atmost 1 red balls and thanks to very helpful answers I can calculate any kind of combinations using only two different items(balls).

Now I am facing problem with generalizing this approach. E.I. how this formula $$p=\sum_{j=3}^6 P(\text{green$=j$ and red$=0$})+\sum_{j=3}^5 P(\text{green$=j$ and red$=1$}).$$

transform for the case where ${green}\ge3, \text{red}\le1,\text{orange}\le1$ is required.

Calculating "static" values (E.I ${green}=2, red=1, orange=1$) is breeze with multinomial distribution but these cumulative values really trips me off

答:
In my opinion, the key to this problem is to forgo elegance and count, focusing directly on the # of green balls selected.

Probability of picking at least 3 green balls and at most 1 red and 1 orange ball from 4 different balls in 6 tries

My answer will be

$$ \frac{N\text{(umerator)}}{D\text{(enominator)}}.$$

Clearly,

$$D = 4^6.$$

I will compute various numerators $N_1, N_2, N_3, N_4$. Then

$$N = N_1 + N_2 + N_3 + N_4.$$

Assume that the colors are represented by G,R,O, and B (for black).

On each line below, the number of ways that a distribution can occur will be counted.

$\underline{\text{3 Greens} : N_1}$

GGGBBB : $~ \binom{6}{3}.$

GGGBBR : $~ \binom{6}{3} \times 3.$

GGGBBO : $~ \binom{6}{3} \times 3.$

GGGBRO : $~ \binom{6}{3} \times 3!.$

$$N_1 = \binom{6}{3} \times 13.$$

$\underline{\text{4 Greens} : N_2}$

GGGGBB : $~ \binom{6}{4}.$

GGGGBR : $~ \binom{6}{4} \times 2.$

GGGGBO : $~ \binom{6}{4} \times 2.$

GGGGRO : $~ \binom{6}{4} \times 2.$

$$N_2 = \binom{6}{4} \times 7.$$

$\underline{\text{5 Greens} : N_3}$

GGGGGB : $~ \binom{6}{5}.$

GGGGGR : $~ \binom{6}{5}.$

GGGGGO : $~ \binom{6}{5}.$

$$N_3 = \binom{6}{5} \times 3.$$

$\underline{\text{6 Greens} : N_4}$

GGGGGG : $~ \binom{6}{6}.$

$$N_4 = \binom{6}{6} \times 1.$$

final answer:

$$\frac{
\left[\binom{6}{3} \times 13\right] ~+~
\left[\binom{6}{4} \times 7\right] ~+~
\left[\binom{6}{5} \times 3\right] ~+~
\left[\binom{6}{6} \times 1\right]
}
{4^6}.
$$


问:
How to get the jacobian of this change of variables?
I am reading an article where we have
$$\iint_{S^2\times\mathbb{R}^3}f(w,v)\,dw\,dv$$
where $w\in S^2$ and $v\in\mathbb{R}^3$ are variable vectors. A change of variable was done $(w,v)\mapsto(p,q)$ where
$$p=(v\cdot w)w$$
and
$$q=v-p=v-(v\cdot w)w.$$

That is, $p$ is the projection of $v$ on $w$, and $q$ is the projection of $p$ on $w^\perp$, where $w^\perp$ is the rotation of $w$ in the plane $\{w,v\}$ by $\pi/2$. Then $p\in\mathbb{R}^3$ and $q\in\{p\}^\perp$.

They computed the Jacobian to be $$\,dw\,dv=\frac{2}{p^2\sin(p,p+q)}\,dp\,dq.$$
Does anyone have an idea how to compute this Jacobian explicitly?

The article can be found here

The part I am asking about is in page 12 proposition 5.

It has to be related to spherical coordinate, but since $\omega$ is a vector I don't understand how they compute the Jacobian.

答:
I'll use the paper's notation. First, rearrange the claim as $dpdq=\tfrac12p^2\sin(\omega,\,V_\ast)d\omega dV_\ast$. We actually don't need a matrix of partial derivatives to prove this, any more than you need them to prove the more famous $dxdy=rdrd\theta$; in both cases, we can make a geometric argument, by comparing two coordinate systems' infinitesimal area elements.

In our example, the left-hand side is such an element. The right-hand side is half the area of an infinitesimal parallelogram with angle $(\omega,\,V_\ast)$ between sides of length $|p|d\omega,\,|p|dV_\ast$. Don't let this factor of $\tfrac12$ surprise you; it's a consequence of the degeneracy in $p=\pm|p|\omega$. The paper itself makes essentialy this argument, between Eqs. (36) & (37).


问:
Compute least hypothenuse of more than n distinct Pythagorean triples
I'm interested in a variation of https://oeis.org/A006339 (Least hypotenuse of n distinct Pythagorean triangles).

Basically, if A006339 is $u(n)$, I want $v(n) := \min_{n \leq i} u(i)$

In other words, $v(n)$ is increasing and for any $k \leq n$, there are less than $v(n)$ triples with hypothenuse $k$.

Are there tables for that sequence? How can I compute it?

A possible way would be to have lower bounds for A006339 and compute it until the lower bound is big enough. I'm not even sure about how I can compute A006339. There is an easy enumeration algorithm for Pythagorean triples due to a result by Hall (1970) and Roberts (1977) (https://stackoverflow.com/a/8263898/5133167), and it allows us to stop the search as any multiplication by $U$, $A$ or $D$ will increase the value of $c$. Is there a more performant way?

答:
Worked it out. Taking the consecutive (1 mod 4) primes $$ 5, 13, 17, 29, 37, 41, 53,...$$
we calculate the best exponent for each such prime $p$ in terms of the real positive exponent $\delta.$ As I said, this is Ramanujan's approach, used in his Superior Highly Composite Numbers. I learned it from Nicolas

Given real $\delta > 0,$ we demand the exponent of $p$ to be
$$ k=\left\lfloor \frac{1}{p^\delta - 1}  \right\rfloor  $$
Meanwhile, with prime $p$ and desired exponent $k,$ we use
$$ \delta = \frac{\log (k+1)  - \log k}{\log p}  $$
This defines a sequence of numbers that rapidly increase  $r(n) .$  Indeed, $n$ is guaranteed to have more representations as the sum of two squares than any smaller number has.

As was Ramanujan's case, there may be new champions  of $r(n)$ that lie between two of the numbers defined above. In that case, there is a procedure due to Guy Robin, I think in his dissertation (supervised by J.-L. Nicolas).  I don't know how complicated his "benefit method" is, but I programmed such things (operations research) as an undergraduate, and recall that a fair amount of work was involved. The constraints include: the prime exponents are non-increasing and always non-negative.

Alright, Nicolas gives complete details of the "benefit" method in ACTA 1988 . The approach of Robin is to combine numerous tools. This is the article where he displays the smallest number with more than $10^{1000}$ divisors.

I can write a program to show the first few such numbers...

```<code>delta      r(n)           n = factored
0.3 ;;;; 4 times 2 ;;;; 5  =   5
0.27 ;;;; 4 times 4 ;;;; 65  =   5 13
0.25 ;;;; 4 times 6 ;;;; 325  =   5^2 13
0.24 ;;;; 4 times 12 ;;;; 5525  =   5^2 13 17
0.2 ;;;; 4 times 24 ;;;; 160225  =   5^2 13 17 29
0.19 ;;;; 4 times 48 ;;;; 5928325  =   5^2 13 17 29 37
0.18 ;;;; 4 times 96 ;;;; 243061325  =   5^2 13 17 29 37 41
0.175 ;;;; 4 times 128 ;;;; 1215306625  =   5^3 13 17 29 37 41
0.17 ;;;; 4 times 256 ;;;; 64411251125  =   5^3 13 17 29 37 41 53
0.168 ;;;; 4 times 512 ;;;; 3929086318625  =   5^3 13 17 29 37 41 53 61
0.161 ;;;; 4 times 1024 ;;;; 286823301259625  =   5^3 13 17 29 37 41 53 61 73
0.158 ;;;; 4 times 1536 ;;;; 3728702916375125  =   5^3 13^2 17 29 37 41 53 61 73
0.154 ;;;; 4 times 3072 ;;;; 331854559557386125  =   5^3 13^2 17 29 37 41 53 61 73 89
0.151 ;;;; 4 times 6144 ;;;; 32189892277066454125  =   5^3 13^2 17 29 37 41 53 61 73 89 97
0.15 ;;;; 4 times 12288 ;;;; 3251179119983711866625  =   5^3 13^2 17 29 37 41 53 61 73 89 97 101
0.147 ;;;; 4 times 24576 ;;;; 354378524078224593462125  =   5^3 13^2 17 29 37 41 53 61 73 89 97 101 109
0.146 ;;;; 4 times 49152 ;;;; 40044773220839379061220125  =   5^3 13^2 17 29 37 41 53 61 73 89 97 101 109 113
delta         r(n)                 n           =        factored
</code>```

$$ \bigcirc  \bigcirc \bigcirc  \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc   \bigcirc  \bigcirc \bigcirc  \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc   $$

We get an ordered list of consecutive useful $\delta's$ by solving for each prime and each exponent up to $5.$  To make sure it works properly, let the $\delta$ used be in between two consecutive reals indicated in the list below. An alternative is to read the sorted file, ignore the actual value of $\delta$ on that line, and multiply the number $n$ by the prime on that line. If done properly, factoring the resulting $n$ at each step should show the increased exponent for that prime $p \; . \; \; $
$$ \bigcirc  \bigcirc \bigcirc  \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc   \bigcirc  \bigcirc \bigcirc  \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc   $$

```<code> 0.4306765580733931  5 ^ 1
  0.2702381544273197  13 ^ 1
  0.2519296364125923  5 ^ 2
   0.244650542118226  17 ^ 1
  0.2058468324604344  29 ^ 1
  0.1919587200065601  37 ^ 1
  0.1866524112389434  41 ^ 1
  0.1787469216608008  5 ^ 3
  0.1745834300480449  53 ^ 1
  0.1686130986895011  61 ^ 1
  0.1615554674429964  73 ^ 1
   0.158079186604075  13 ^ 2
  0.1544226628011101  89 ^ 1
  0.1515171524096389  97 ^ 1
   0.150190483223688  101 ^ 1
  0.1477501131786861  109 ^ 1
  0.1466237184553111  113 ^ 1
  0.1431113929202641  17 ^ 2
  0.1408841194731412  137 ^ 1
  0.1386468838532138  5 ^ 4
  0.1385201075671774  149 ^ 1
  0.1370873187823978  157 ^ 1
  0.1345057169479844  173 ^ 1
  0.1333360731748201  181 ^ 1
  0.1317096685686114  193 ^ 1
  0.1311982683517524  197 ^ 1
  0.1275639755045533  229 ^ 1
  0.1271587398372755  233 ^ 1
  0.1263760881150453  241 ^ 1
  0.1249121781636255  257 ^ 1
  0.1238932866290727  269 ^ 1
  0.1232476925609442  277 ^ 1
  0.1229342993142804  281 ^ 1
  0.1220292459879827  293 ^ 1
  0.1206269875212659  313 ^ 1
  0.1204126778815846  29 ^ 2
  0.1203610007213705  317 ^ 1
  0.1190957566850307  337 ^ 1
  0.1183840591148901  349 ^ 1
  0.1181540875758708  353 ^ 1
  0.1170544627527939  373 ^ 1
  0.1162300592682939  389 ^ 1
  0.1158346526511383  397 ^ 1
   0.115640914438019  401 ^ 1
  0.1152610595238196  409 ^ 1
  0.1147094651682719  421 ^ 1
  0.1141784098763921  433 ^ 1
  0.1135000135583575  449 ^ 1
  0.1132827525593783  5 ^ 5
  0.1131727366659431  457 ^ 1
  0.1130119349340791  461 ^ 1
  0.1122886528902697  37 ^ 2
  0.1121589678232448  13 ^ 3
  0.1112158783706281  509 ^ 1
  0.1108016106675626  521 ^ 1
  0.1101384080494621  541 ^ 1
  0.1096306885348829  557 ^ 1
  0.1092623336390804  569 ^ 1
   0.109184661243966  41 ^ 2
   0.109022393369416  577 ^ 1
  0.1085553762730775  593 ^ 1
  0.1083280294582822  601 ^ 1
  0.1079943548673958  613 ^ 1
  0.1078850289369187  617 ^ 1
  0.1072480305374002  641 ^ 1
  0.1069411297181192  653 ^ 1
  0.1067405999140516  661 ^ 1
  0.1064456832186591  673 ^ 1
  0.1063489015533347  677 ^ 1
  0.1057834942757523  701 ^ 1
  0.1056006150284899  709 ^ 1
  0.1050677386679089  733 ^ 1
  0.1045571277453435  757 ^ 1
  0.1044740746281937  761 ^ 1
  0.1043096601296809  769 ^ 1
  0.1042282850833419  773 ^ 1
   0.103751273435192  797 ^ 1
   0.103519712755121  809 ^ 1
  0.1032925703785417  821 ^ 1
  0.1031435227654369  829 ^ 1
  0.1027073457992096  853 ^ 1
  0.1026361963957853  857 ^ 1
  0.1022867948546604  877 ^ 1
  0.1022181522110647  881 ^ 1
  0.1021247598253814  53 ^ 2
   0.101539149197962  17 ^ 3
  0.1014246637053439  929 ^ 1
  0.1012975688001153  937 ^ 1
  0.1012345460491029  941 ^ 1
  0.1010475355097117  953 ^ 1
  0.1006824786789692  977 ^ 1
  0.1003869949333126  997 ^ 1
  0.1002133496204834  1009 ^ 1
  0.1001560585753007  1013 ^ 1
  0.1000423464158209  1021 ^ 1
 0.09987391378937605  1033 ^ 1
 0.09965321714259505  1049 ^ 1
 0.09949051958609365  1061 ^ 1
 0.09938336460724531  1069 ^ 1
 0.09906799126575533  1093 ^ 1
 0.09901629479181956  1097 ^ 1
 0.09886264801081832  1109 ^ 1
 0.09876139901991253  1117 ^ 1
 0.09863233986375369  61 ^ 2
 0.09861125988744147  1129 ^ 1
</code>```

Body:
Worked it out. Taking the consecutive (1 mod 4) primes $$ 5, 13, 17, 29, 37, 41, 53,...$$
we calculate the best exponent for each such prime $p$ in terms of the real positive exponent $\delta.$ As I said, this is Ramanujan's approach, used in his Superior Highly Composite Numbers. I learned it from Nicolas

Given real $\delta > 0,$ we demand the exponent of $p$ to be
$$ k=\left\lfloor \frac{1}{p^\delta - 1}  \right\rfloor  $$
Meanwhile, with prime $p$ and desired exponent $k,$ we use
$$ \delta = \frac{\log (k+1)  - \log k}{\log p}  $$
This defines a sequence of numbers that rapidly increase  $r(n) .$  Indeed, $n$ is guaranteed to have more representations as the sum of two squares than any smaller number has.

As was Ramanujan's case, there may be new champions  of $r(n)$ that lie between two of the numbers defined above. In that case, there is a procedure due to Guy Robin, I think in his dissertation (supervised by J.-L. Nicolas).  I don't know how complicated his "benefit method" is, but I programmed such things (operations research) as an undergraduate, and recall that a fair amount of work was involved. The constraints include: the prime exponents are non-increasing and always non-negative.

Alright, Nicolas gives complete details of the "benefit" method in ACTA 1988 . The approach of Robin is to combine numerous tools. This is the article where he displays the smallest number with more than $10^{1000}$ divisors.

I can write a program to show the first few such numbers...

delta      r(n)           n = factored
0.3 ;;;; 4 times 2 ;;;; 5  =   5
0.27 ;;;; 4 times 4 ;;;; 65  =   5 13
0.25 ;;;; 4 times 6 ;;;; 325  =   5^2 13
0.24 ;;;; 4 times 12 ;;;; 5525  =   5^2 13 17
0.2 ;;;; 4 times 24 ;;;; 160225  =   5^2 13 17 29
0.19 ;;;; 4 times 48 ;;;; 5928325  =   5^2 13 17 29 37
0.18 ;;;; 4 times 96 ;;;; 243061325  =   5^2 13 17 29 37 41
0.175 ;;;; 4 times 128 ;;;; 1215306625  =   5^3 13 17 29 37 41
0.17 ;;;; 4 times 256 ;;;; 64411251125  =   5^3 13 17 29 37 41 53
0.168 ;;;; 4 times 512 ;;;; 3929086318625  =   5^3 13 17 29 37 41 53 61
0.161 ;;;; 4 times 1024 ;;;; 286823301259625  =   5^3 13 17 29 37 41 53 61 73
0.158 ;;;; 4 times 1536 ;;;; 3728702916375125  =   5^3 13^2 17 29 37 41 53 61 73
0.154 ;;;; 4 times 3072 ;;;; 331854559557386125  =   5^3 13^2 17 29 37 41 53 61 73 89
0.151 ;;;; 4 times 6144 ;;;; 32189892277066454125  =   5^3 13^2 17 29 37 41 53 61 73 89 97
0.15 ;;;; 4 times 12288 ;;;; 3251179119983711866625  =   5^3 13^2 17 29 37 41 53 61 73 89 97 101
0.147 ;;;; 4 times 24576 ;;;; 354378524078224593462125  =   5^3 13^2 17 29 37 41 53 61 73 89 97 101 109
0.146 ;;;; 4 times 49152 ;;;; 40044773220839379061220125  =   5^3 13^2 17 29 37 41 53 61 73 89 97 101 109 113
delta         r(n)                 n           =        factored

$$ \bigcirc  \bigcirc \bigcirc  \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc   \bigcirc  \bigcirc \bigcirc  \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc   $$

We get an ordered list of consecutive useful $\delta's$ by solving for each prime and each exponent up to $5.$  To make sure it works properly, let the $\delta$ used be in between two consecutive reals indicated in the list below. An alternative is to read the sorted file, ignore the actual value of $\delta$ on that line, and multiply the number $n$ by the prime on that line. If done properly, factoring the resulting $n$ at each step should show the increased exponent for that prime $p \; . \; \; $
$$ \bigcirc  \bigcirc \bigcirc  \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc   \bigcirc  \bigcirc \bigcirc  \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc\bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc   $$

 0.4306765580733931  5 ^ 1
  0.2702381544273197  13 ^ 1
  0.2519296364125923  5 ^ 2
   0.244650542118226  17 ^ 1
  0.2058468324604344  29 ^ 1
  0.1919587200065601  37 ^ 1
  0.1866524112389434  41 ^ 1
  0.1787469216608008  5 ^ 3
  0.1745834300480449  53 ^ 1
  0.1686130986895011  61 ^ 1
  0.1615554674429964  73 ^ 1
   0.158079186604075  13 ^ 2
  0.1544226628011101  89 ^ 1
  0.1515171524096389  97 ^ 1
   0.150190483223688  101 ^ 1
  0.1477501131786861  109 ^ 1
  0.1466237184553111  113 ^ 1
  0.1431113929202641  17 ^ 2
  0.1408841194731412  137 ^ 1
  0.1386468838532138  5 ^ 4
  0.1385201075671774  149 ^ 1
  0.1370873187823978  157 ^ 1
  0.1345057169479844  173 ^ 1
  0.1333360731748201  181 ^ 1
  0.1317096685686114  193 ^ 1
  0.1311982683517524  197 ^ 1
  0.1275639755045533  229 ^ 1
  0.1271587398372755  233 ^ 1
  0.1263760881150453  241 ^ 1
  0.1249121781636255  257 ^ 1
  0.1238932866290727  269 ^ 1
  0.1232476925609442  277 ^ 1
  0.1229342993142804  281 ^ 1
  0.1220292459879827  293 ^ 1
  0.1206269875212659  313 ^ 1
  0.1204126778815846  29 ^ 2
  0.1203610007213705  317 ^ 1
  0.1190957566850307  337 ^ 1
  0.1183840591148901  349 ^ 1
  0.1181540875758708  353 ^ 1
  0.1170544627527939  373 ^ 1
  0.1162300592682939  389 ^ 1
  0.1158346526511383  397 ^ 1
   0.115640914438019  401 ^ 1
  0.1152610595238196  409 ^ 1
  0.1147094651682719  421 ^ 1
  0.1141784098763921  433 ^ 1
  0.1135000135583575  449 ^ 1
  0.1132827525593783  5 ^ 5
  0.1131727366659431  457 ^ 1
  0.1130119349340791  461 ^ 1
  0.1122886528902697  37 ^ 2
  0.1121589678232448  13 ^ 3
  0.1112158783706281  509 ^ 1
  0.1108016106675626  521 ^ 1
  0.1101384080494621  541 ^ 1
  0.1096306885348829  557 ^ 1
  0.1092623336390804  569 ^ 1
   0.109184661243966  41 ^ 2
   0.109022393369416  577 ^ 1
  0.1085553762730775  593 ^ 1
  0.1083280294582822  601 ^ 1
  0.1079943548673958  613 ^ 1
  0.1078850289369187  617 ^ 1
  0.1072480305374002  641 ^ 1
  0.1069411297181192  653 ^ 1
  0.1067405999140516  661 ^ 1
  0.1064456832186591  673 ^ 1
  0.1063489015533347  677 ^ 1
  0.1057834942757523  701 ^ 1
  0.1056006150284899  709 ^ 1
  0.1050677386679089  733 ^ 1
  0.1045571277453435  757 ^ 1
  0.1044740746281937  761 ^ 1
  0.1043096601296809  769 ^ 1
  0.1042282850833419  773 ^ 1
   0.103751273435192  797 ^ 1
   0.103519712755121  809 ^ 1
  0.1032925703785417  821 ^ 1
  0.1031435227654369  829 ^ 1
  0.1027073457992096  853 ^ 1
  0.1026361963957853  857 ^ 1
  0.1022867948546604  877 ^ 1
  0.1022181522110647  881 ^ 1
  0.1021247598253814  53 ^ 2
   0.101539149197962  17 ^ 3
  0.1014246637053439  929 ^ 1
  0.1012975688001153  937 ^ 1
  0.1012345460491029  941 ^ 1
  0.1010475355097117  953 ^ 1
  0.1006824786789692  977 ^ 1
  0.1003869949333126  997 ^ 1
  0.1002133496204834  1009 ^ 1
  0.1001560585753007  1013 ^ 1
  0.1000423464158209  1021 ^ 1
 0.09987391378937605  1033 ^ 1
 0.09965321714259505  1049 ^ 1
 0.09949051958609365  1061 ^ 1
 0.09938336460724531  1069 ^ 1
 0.09906799126575533  1093 ^ 1
 0.09901629479181956  1097 ^ 1
 0.09886264801081832  1109 ^ 1
 0.09876139901991253  1117 ^ 1
 0.09863233986375369  61 ^ 2
 0.09861125988744147  1129 ^ 1


问:
Do the premises logically imply the conclusion?
$$b\rightarrow a,\lnot c\rightarrow\lnot a\models\lnot(b\land \lnot c)$$

I have generated an 8 row truth table, separating it into $b\rightarrow a$, $\lnot c\rightarrow\lnot a$ and $\lnot (b\land\lnot c)$. I know that if it was

$$\lnot c\rightarrow\lnot a\models\lnot(a \land \lnot c)$$

I would only need to check the right side with  every value that makes the left side true to make sure the overall statement is true. How do I deal with more than one premise?

答:
So you can show this a couple of ways. One of them is to use truth table semantics- your hypotheses are connected by an "and". Another equivalent, but faster way is to use the method of analytic tableaux, which essentially works by trying to find ways to falsify the statement (if they all work out to contradictions, then it must be true).

A third method you can use is called sequent calculus. This works in Intuitionistic Propositional Calculus, which does not use the law of excluded middle. The cool thing about IPC is that using sequent calculus can actually produce functions that verify the formula.

$b \Rightarrow a, \lnot c \Rightarrow \lnot a \vdash \lnot (b \land \lnot c) \;\;\; \mathrm{notR}$

$\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, (b \land \lnot c) \vdash false \;\;\; \mathrm{andL}$

$\;\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, b,\lnot c \vdash false \;\;\; \mathrm{impL}$

$\;\;\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, b,\lnot c \vdash b \;\;\; \mathrm{axiom}$

$\;\;\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, b, \lnot c, a \vdash false \;\;\; \mathrm{impL}$

$\;\;\;\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, b, \lnot c, a \vdash \lnot c \;\;\; \mathrm{axiom}$

$\;\;\;\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, b, \lnot c, a, \lnot a \vdash false \;\;\; \mathrm{notL}$

$\;\;\;\;\;b \Rightarrow a, \lnot c \Rightarrow \lnot a, b, \lnot c, a, \lnot a \vdash a \;\;\; \mathrm{axiom}$

The evidence that this produces is the following function (unless I made a mistake in forming the evidence term): $\lambda f. \lambda x. (f_1(x_1),f_2(x_2))$


问:
Why can't $y=xe^x$ be solved for $x$?
I apologize for my mathematical ignorance regarding this, but could someone help me understand why it isn't possible to (symbolically) find an inverse function for $f(x)=xe^x$?

The most obvious (but presumably the most trivial) is that $f$ does not pass the "horizontal line test". However, if we restrict the domain to $x\geq-1$ this should not be a problem (derivative is positive for $x>-1$ so function is strictly increasing). So now my question becomes: "Why can't we find an inverse function for $f$ over the interval $[-1,\infty)$?"

Perhaps it is because $e^x$ is transcendental (not algebraic).
However, we can find an inverse for $g(x) = e^x$, which is also transcendental. Is that because we're "cheating" by defining another transcendental function, namely $\ln(x)$, to be its inverse? In other words, would it be fundamentally no different to define a new function, call it $\text{lnx}(x)$ (if that's not already something else), to be the inverse of $xe^x$ over $[-1,\infty)$ and then say that $f$ has a "closed form" / "symbolic" / ??? inverse function $f^{-1}(x)=\text{lnx}(x)$ over the interval $[-1, \infty)$?

SageMath source to generate plot

xs = (x,-5,2)
ys = (y,-1,5)
p1 = implicit_plot(x*exp(x)-y,xs,ys, color='blue', legend_label='y=x*e^x')
p2 = implicit_plot(x-y,xs,ys, color='orange', linestyle='dashed', legend_label='y=x')
p3 = implicit_plot(exp(x)-y,xs,ys, color='green', linestyle='dotted', legend_label='y=e^x')
combined = p1 + p2 + p3
combined.axes_labels(['x', 'y'])
combined.legend(True)
combined.show(title='Transcendental Stuff', frame=True, axes=True, legend_loc='lower right')

See also

How to know if I can't solve an equation with "standard" methods? (this had some helpful answers & comments)

If this equation can't be solved algebraically, can it be solved some other way?

Is there a theory of transcendental functions?

Is there a proof that there is no general method to solve transcendental equations?

答:
$f\colon [-1,\infty)\to\mathbb{R}\colon x\mapsto f(x)=xe^x$ is bijective and has an inverse therefore. The relation Lambert W consists of the branches of the inverse relation of the function $\mathbb{C}\to\mathbb{C}\colon z\mapsto ze^z$.

$f$ is an elementary function (see differential algebra), but Lambert W not.

There are some proofs proving that Lambert W is not an elementary function. See e.g. the following references.

Liouville, J.: Mémoire sur la classification des transcendantes et sur l'impossibilité d'exprimer les racines de certaines équations en fonction finie explicite des coefficients. Journal de mathématiques pures et appliquées 2 (1837) 56–105, 3 (1838) 5233–547

The theorem in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 is also proved in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math 101 (1979) (4) 743-759.

Ritt, J. F.: Integration in finite terms. Liouville's theory of elementary methods. Columbia University Press, New York, 1948

Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22

Bronstein, M.; Corless, R. M.; Davenport, J. H.; Jeffrey, D. J.: Algebraic properties of the Lambert W function from a result of Rosenlicht and of Liouville. Integral Transforms and Special Functions 19 (2008) (10) 709-712

Title: Why can't $y=xe^x$ be solved for $x$?


答:
Apply the Frullani integral formula below

$$\int_0^\infty \frac{f(ax)-f(bx )}x =(f(0)-f(\infty))\ln\frac ba $$

with $f(x)=\frac {\ln(1+x)}x$ to obtain

$$I
=ab \int_0^{\infty}\frac1x\left(  \frac{\ln(1+ax)}{ax} -\frac{\ln(1+bx)}{bx}\right)  \,dx
=ab \ln \frac ba
$$

Title: Evaluate the integral $\int_0^{\infty} \frac{b\ln{(1+ax)}-a\ln{(1+bx)}}{x^2} \,dx$

Body:
Evaluate the following integral

$$I=\int_0^{\infty} \frac{b\ln(1+ax)-a\ln(1+bx)}{x^2} \,dx$$ with $\ a,b\in\mathbb{R},\ 0<a<b$.

My first attempt was to write $b\ln(1+ax)-a\ln(1+bx)$ as another integral, so I could substitute in the initial integral and then, by reversing the order of integration, try to calculate $I$.

I've tried writing $$b\ln(1+ax)-a\ln(1+bx)=\frac{ab}{t}\cdot\ln(1+tx)\Biggr|_{t=b}^{t=a}=ab\int_b^a\left(\frac{x}{t(1+tx)}-\frac{\ln(1+tx)}{t^2}\right)\,dt$$ but it doesn't seem to help me evaluating $I$.

What else could I try?


问:
Modus ponens and left-nested conditionals (part 2)
Someone told me that (a ⊃ b) ⊃ a and ((a ⊃ b) ⊃ a) ⊃ a together with modus ponens entail a and subsequently b. Is this true? And if yes, could please someone show me how to prove it? (I haven't found the proof yet).

答:
hint

Modus Ponens argument insure that
$$(p\implies q) \wedge p \therefore q$$

Now, take
$$p=(a\implies b)\implies a$$
and
$$q=a$$


答:
Let's use differentiation under the integral to find the integral.

Let $I(a,b)=\int_0^\infty\frac{b\ln(1+ax)-a\ln(1+bx)}{x^2}dx$. Then, let's take the derivative with respect to $b$:
$$\frac{\partial I}{\partial b}=\int_0^\infty\frac{\partial}{\partial b}\frac{b\ln(1+ax)-a\ln(1+bx)}{x^2}dx=\int_0^\infty \frac{\ln(1+ax)-\frac{ax}{1+bx}}{x^2}dx$$
Now, take the derivative with respect to $a$:
\begin{align*}\frac{\partial^2 I}{\partial a \partial b}&=\int_0^\infty\frac{\partial}{\partial a} \frac{\ln(1+ax)-\frac{ax}{1+bx}}{x^2}dx\\&=\int_0^\infty\frac{\frac x{1+ax}-\frac x{1+bx}}{x^2}\\&=\int_0^\infty\frac 1x\cdot\frac{1+bx-1-ax}{(1+ax)(1+bx)}dx\\&=\int_0^\infty\frac{b-a}{(1+ax)(1+bx)}dx\\&=\int_0^\infty\frac{b}{1+bx}-\frac a{1+ax}dx\text{ by partial fractions}\\&=\ln{(1+bx)}-\ln{(1+ax)}\Big|_0^\infty\\&=\ln b - \ln a\end{align*}
Where the last equality is left as an exercise for the reader :)

So, let's move backwards now, integrating with respect to $a$:
$$\frac{\partial I}{\partial b}=a\ln b-a(\ln a -1)+c_1$$
for some real constant $c_1$. Then we integrate again with respect to $b$:
$$I(a,b)=ab(\ln b-\ln a)+c_1b+c_2$$
for real constants $c_1,c_2$. Our goal now is to find those constants: Note that for non-negative real $a$, we have $I(a,a)=\int_0^\infty\frac{a\ln(ax+1)-a\ln(ax+1)}{x^2}dx=\int_0^\infty\frac 0{x^2}dx=0$. So, consider $I(1,1)$ and $I(2,2)$:

\begin{align*}
I(1,1)&=0&=(1)(1)(\ln 1-\ln1)+c_1(1)+c_2&=c_1+c_2\\
I(2,2)&=0&=(2)(2)(\ln 2-\ln2)+c_1(2)+c_2&=2c_1+c_2
\end{align*}
So $c_1=c_2=0$ and our final result is:
$$I(a,b)=\int_0^\infty\frac{b\ln(1+ax)-a\ln(1+bx)}{x^2}dx=ab(\ln b-\ln a)$$


问:
Discuss whether or not the following binary operations are commutative, associative, ...
Discuss whether or not the following binary operations are commutative, associtive, have neutral elements and for which elements there are inverse elements.
In between are what I have said, but if it doesn't seem right please let me know.

$1. (R, \circ ) \text{ with }  x \circ y := |x \cdot y|;$

1: Commutative, Associative, no neutral element (unsure about this), and inverses for R+

$2. (N_0 ,\circ ) \text{ with } x \circ  y := |x - y|; $

2: Commutative, Not Associative, neutral element = 0, and elements are own inverses

$3. (R, \circ ) \text{ with } x \circ  y := 2 \cdot x - 3 \cdot y ;$

3: Not Commutative, Not Associative, no neutral element, and no inverses

$4. (R, \circ ) \text{ with } x \circ  y := x + y^2;$

4: Not Commutative, Not Associative, no neutral element, and no inverses

$5. (R_{2 \times 2} , \circ ) \text{ with } A \circ  B := A . B - B . A  , (A \cdot B \text{ is matrix multiplication});$

5: Not Commutative, Not Associative, no neutral element, and no inverses

$6. (R, \circ ) \text{ with } x  y := x + y - 2x^2y^2;$

6: Commutative, Not Associative, neutral element = 0, not sure about inverse.

$7. (\{\pm 1\} \times Z, \circ ) \text{ with } (\epsilon, a) \circ (\delta,  b) := (\epsilon \cdot \delta,  a \cdot\delta +  b).$

I am unsure about this one.

答:
This is indeed commutative and associative, and it has no neutral element, since there is no $x\in\Bbb R$ such that $x\circ-1=-1$, for instance. Since it has no neutral element, the question of inverses doesn’t even arise: inverses are defined in terms of a neutral element.

You’re fine here, though it would be good to provide a counterexample to associativity; $(3\circ 3)\circ 4=0\circ 4=4\ne 2=3\circ 1=3\circ(3\circ 4)$ is one.

Looks good.

This is also correct, though you might want to notice that the operation does have a one-sided neutral element on the right: $x\circ 0=x$ for all $\in\Bbb R$.

And again correct, though a counterexample to associativity would be a good idea.

For inverses fix $a\in\Bbb R$; to find $a^{-1}$, you must solve $a+x+2a^2x^2=0$ for $x$. That’s just a quadratic, so you know that $$x=\frac{-1\pm\sqrt{1-8a^3}}{4a^2}\;;$$ for which $a$ does this have a real solution?

Compare $\langle -1,0\rangle\circ\langle -1,1\rangle$ with $\langle -1,1\rangle\circ\langle -1,0\rangle$. Then multiply out and compare $$\big(\langle\epsilon_1,a_1\rangle\circ\langle\epsilon_2,a_2\rangle\big)\circ\langle\epsilon_3,a_3\rangle$$ and $$\langle\epsilon_1,a_1\rangle\circ\big(\langle\epsilon_2,a_2\rangle\circ\langle\epsilon_3,a_3\rangle\big)\;.$$ For a neutral element you want $\langle\epsilon_0,a_0\rangle$ such that $\langle\epsilon_0,a_0\rangle\circ\langle\epsilon,a\rangle$ for each $\langle\epsilon,a\rangle$, i.e., $\langle\epsilon_0\epsilon,\epsilon a_0+a\rangle=\langle\epsilon,a\rangle$. It’s clear that $\epsilon_0$ has to be $1$, not $-1$; is there an $a_0\in\Bbb Z$ that works?