问:
Diameters of state spaces

The authors mention that if $M$ is not a factor, the diameter of $S_0(M)/Int(M)$ is equal to 2.

When $M$ is not a factor, it is easy to check that the $d(,)\leq 2$, how to check that $d(,)\geq2$?

答:
The idea used by Jonathan Hole can be implemented a bit more concisely.

If $M\cap M'$ is non trivial, then it has a non-trivial projection $p$. Define $v=2p-1$. Then $v$ is a selfadjoint unitary with spectrum $\{-1,1\}$ (this is trivial to see, as $$p=1\times p + (-1)\times (1-p)$$ is the spectral decomposition).

Assuming $M\subset B(H)$, choose unit vectors $\xi,\eta\in H$ with $p\xi=\xi$, $p\eta=0$. Define
$$
\omega(x)=\langle x\xi,\xi\rangle,\qquad \phi(x)=\langle x\eta,\eta\rangle.
$$
Since $pv=p$, $(1-p)v=p-1$,
$$
\omega(v)=\langle v\xi,\xi\rangle=\langle vp\xi,\xi\rangle=\langle \xi,\xi\rangle=1,
$$
and
$$
\phi(v)=\langle v\eta,\eta\rangle=\langle v(1-p)\eta,\eta\rangle=-\langle (1-p)\eta,\eta\rangle=-1.
$$
Then, using that $v$ is central, we have for any unitary $u$ that
$$
\|\omega-u\phi u^*\|\geq|\omega(v)-\phi(uvu^*)|=|\omega(v)-\phi(v)|=1-(-1)=2.
$$

Title: Diameters of state spaces


问:
How to solve $\int \frac{x^3}{(1+x)^5}\ dx$?
I am stuck at this integral. Can anybody help or give me a hint what to do with the denominator?

$$\int \frac{x^3}{(1+x)^5} \, dx$$

答:
Let $$1+x = 1/t$$
$$dx = \frac {-1} {t^2} dt$$
The integral transforms into $$\int \frac {(\frac 1 t -1)^3 } {(\frac 1 t)^5} \frac {-dt} {t^2}$$ $$\int (t-1)^3 dt$$
which can be easily solved further

Tip : Whenever the integral contains linear terms with higher powers in the denominator, substituting the linear term with 1/t becomes useful for solving the integral

ContentLicense: CC BY-SA 4.0


问:
Harmonic Function bounded by a linear function
Let $u$ be a harmonic function on $\mathbb C$. Suppose that for each $\epsilon > 0$, there is a constant $C_\epsilon$ such that

$$u(z) \leq C_\epsilon + \epsilon |z| .$$

I am trying to show that $u$ is constant. I tried the mean value property, as well as replacing $u$ by a related function, but was unsuccessful. Does anyone have any ideas?

答:
Without loss of generality, we may assume that $u(0)=0$, and it suffices to show that $u\equiv 0$.

Let us fix an arbitrary $\epsilon>0$ first. Choose $R_0>0$ such that $C_\epsilon <\epsilon R_0$. From the assumption on $u$ we know that for every $R\ge R_0$,
$$\max_{|z|\le R}u(z)\le C_\epsilon+\epsilon\cdot\max_{|z|\le R}|z|<2\epsilon R.\tag{$*$}$$

Proof 1(using Harnack's inequality) $(*)$ implies that $2\epsilon R-u(z)$ is a positive harmonic function on $|z|<R$. Then by Harnack's inequality, when $R\ge R_0$,
$$\frac{R-|z|}{R+|z|}\cdot 2\epsilon R\le 2\epsilon R-u(z)\le\frac{R+|z|}{R-|z|}\cdot 2\epsilon R,\quad \text{ if } |z|<R.$$
It is equivalent to say when $R\ge R_0$,
$$-\frac{4\epsilon R|z|}{R-|z|}\le u(z)\le\frac{4\epsilon R|z|}{R+|z|},\quad \text{ if } |z|<R.$$

Fixing $z$, letting $R\to\infty$ and $\epsilon\to 0$, the conclusion follows.

Remark: A similar argument using Harnack's inequality still works when $u$ is a harmonic function on $\mathbb R^n$($n\ge 3$) satisfying the same condition.

Proof 2(using Schwarz lemma)
Since $u$ is harmonic on $\Bbb C$ and $u(0)=0$, there exists a unique entire function $f=u+iv$ such that $f(0)=0$. Then it suffices to show that $f\equiv 0$.

Note that if $a>0$, the Möbius transformation $z\mapsto \frac{z}{2a-z}$ maps the half plane $\{x+iy\mid x<a\}$ conformally onto the unit disk $\Bbb D$, so by $(*)$, for $a=2\epsilon R$ with $R\ge R_0$,  the following map
$$g:\Bbb D\to \Bbb D,\quad w\mapsto \frac{f(Rw)}{4\epsilon R-f(Rw)}$$
is well defined and holomorphic; moreover, $g(0)=0$. Then by Schwarz lemma, when $R\ge R_0$ and $|z|<R$, for $w=\frac{z}{R}\in \Bbb D$,  $g(z)\le |z|$, i.e.
$$\frac{|f(z)|}{|4\epsilon R-f(z)|} \le \frac{|z|}{R}\Longrightarrow |f(z)|\le |z| \cdot |4\epsilon -\frac{f(z)}{R}|, \quad \text{ if } |z|<R.$$
Fixing $z$, letting $R\to\infty$ and $\epsilon\to 0$, the conclusion follows.

ContentLicense: CC BY-SA 3.0


问:
$\gcd(n,m)=10$, and $\text{lcm}(n,m)=100$
I'm having some trouble with the following question:

Determine all natural numbers $n,m$ such that: $\gcd(n,m)=10$, and $\text{lcm}(n,m)=100$.

Using this I arrived at the following conditions:

$\gcd(n,m)=10$

$nm=1000$

But I don't know how to proceed. How can I do this?

答:
Hint:$$mn = 1000, \ \ 10 \le m,n \le 100$$
So what you need to do is just to check $10$ numbers that are divisible by $10$ and not exceeding $100$.


Body:
I'm having some trouble with the following question:

Determine all natural numbers $n,m$ such that: $\gcd(n,m)=10$, and $\text{lcm}(n,m)=100$.

Using this I arrived at the following conditions:

$\gcd(n,m)=10$

$nm=1000$

But I don't know how to proceed. How can I do this?

Body:
Let $n = 2^a\cdot 5^b, m=2^c\cdot5^d$

Since $\gcd(n, m)=10$, we know that $\min(a,c)=1$ and $\min(b,d)=1$.

Since $lcm(n, m) = 100=10^2,$ we know that $\max(a,c)=2$ and $\max(b,d)=2$.

Hence, $(a, c), (b,d) \in \{(1,2), (2,1)\}$.


问:
Can you prove this implication between $\gcd$ equalities?
Given six positive integers $\{a_1,b_1,a_2,b_2,a_3,b_3\}$ such that
$$ \gcd(a_1,b_1) = \gcd(a_2,b_2) = \gcd(a_3,b_3) $$
then the following equality also holds,
$$ \gcd(d_{31},d_{32}) = \gcd(d_{31},d_{21}) = \gcd(d_{32},d_{21}) $$
being $d_{ij}=|a_ib_j - a_jb_i| $.

答: By dividing through by the common gcd, it suffices to assume that all the input pairs are relatively prime.  By symmetry, it suffices to show that, if $d \mid d_{31}$ and $d \mid d_{32}$, then $d \mid d_{21}$.  We have that $d \mid a_2 d_{31} - a_1 d_{32} = a_3 d_{21}$ and $d \mid b_2 d_{31} - b_1 d_{32} = b_3 d_{21}$.  Since $a_3$ and $b_3$ are relatively prime, this means that $d \mid d_{21}$.


问:
Elements of the same order in the same conjugacy class that commute are contained in the same cyclic subgroup
I have a conjecture, that I would like to prove.

EDIT: My first idea was not true, as pointed out by Derek Holt.

Statement:
Let $G$ be a (finite) group and $m$ the maximal order of an element of $G$, such that all elements of maximal order are conjugated.

If two commuting elements $x,y \in G$ have order $m$, then $x^n=y$ for some $n \in \mathbb{N}$, i.e. $y$ is contained in the cyclic subgroup generated by $x$

I looked for some examples:

$(1,2,3)=(1,3,2)^2\in S_3$

$(1,2,3,4)=(1,4,3,2)^3 \in S_4$, but for non-commuting elements of this conjugacy-class it is not true, see: $(1,2,3,4)\not=(1,3,2,4)^n \forall n \in \mathbb{N}$.

Is this true in general? Does anyone know a proof?

Thanks in advance.

答:
There is a counterexample to your revised question, which is the Frobenius group with structure $11^2:{\rm SL}_2(5)$. The highest order of an element is $11$, and all elements of order $11$ are conjugate. You can access this group in GAP as $\mathtt{PrimitiveGroup}(121,56)$.

I would guess that there are not too many couterexamples, and I wonder whether this might be essentially the only one. (By ``essentially'' I mean that you can get other counterexamples by taking direct products with other groups.)


问:
Balls in boxes - proof correction
I have been thinking about the following problem:

How many ways can $50$ different balls be placed into $50$ different sized boxes if exactly $2$ boxes have to remain empty?

My approach was the following:

There are two possible cases which satisfy the conditions:

There are $2$ boxes with $2$ balls and $46$ boxes with $1$ ball;

There is $1$ box with $3$ balls and $47$ boxes with $1$ ball.

In the first case, you can choose the $2$ empty boxes first ($\binom{50}{2}$ possibilities), then you can choose the boxes which have $2$ balls in them ($\binom{48}{2}$ possibilities). After that, you can choose $2$ balls which are placed in the bigger box chosen in the previous step ($\binom{50}{2}$ possibilities), then $2$ balls which are placed in the smaller box ($\binom{48}{2}$ possibilities). Finally, there are only $46$ balls left that have to be placed in the $46$ boxes remained, $1$ ball in each of them ($46!$ possibilities). Totally, there are $\binom{50}{2}* \binom{48}{2}* \binom{50}{2}* \binom{48}{2}* 46!$ possibilities.

In the second case, following a similar logic, there are $\binom{50}{2}* 48* \binom{50}{3}* 47!$ possibilities totally. The final answer to the question can be given adding these two numbers.

However, using the classic inclusion-exclusion formula, the following number of possibilities arise: $\binom{50}{2}* \sum_{k=0}^{48} (-1)^k \binom{48}{k} * (48-k)^{50}$.

I am quite sure that the number derived from the inclusion-exclusion formula is correct, however, I haven't got any clue what is wrong with my first solution.

I would appreciate any help or explanation why my first solution is not correct.

答: Both solutions seem to be the same.</p>

Title: Balls in boxes - proof correction

Body: Both solutions seem to be the same.


问:
Show $ g(s):=\int_{0}^{\infty}f(t)t^{s-1} dt$ is well defined and analytic.
Let f be a real-valued function as following : for fixed $z \in \mathbb{C}$
$$f(t) = \sum_{a,b \in \mathbb{N} }e^{-\lvert az +b \rvert^2 t}$$

Let's define $$ g(s):=\int_{0}^{\infty}f(t)t^{s-1} dt$$

Then, show that $g(s)$ is a well-defined and analytic for all $s \in \mathbb{C}$ except $s=0, 1$

[ My attempt ]

First, I showed that $\int_{t_0}^{\infty}f(t)t^{s-1} dt$ is well-defined for $t_0 >0$.

We know that for each $a>0$ , $b>0$ $$ a^2+b^2 \geq 2ab $$

Therefore,  $$ \lvert az+b \rvert^2 = a^2\lvert z \rvert^2+b^2+2\text{Re}(z)ab \geq 2\lvert z \rvert ab+2\text{Re}(z)ab = (2\lvert z \rvert + 2\text{Re}(z)) ab =:k_1ab$$

so

$$f(t) = \sum_{a,b \in \mathbb{N}}e^{-\lvert az +b \rvert^2 t} \leq \sum_{a,b \in \mathbb{N} }e^{-k_1ab t}$$

Therefore, we can check that $f(t)t^{s-1}$ has rapidly decaying at $\infty$

However i'm stuck here. I can't prove the integral of $g$ is well defined at $0$ and analytic part.

How to prove this?

Thank you for your attention.

Body:
You meant for $t >0,\Im(z)\ne 0$ $$f(t) = \sum_{a,b \in \mathbb{Z}^2 }e^{-\lvert az +b \rvert^2 t}$$ for $\Re(s) >1$ $$g(s):=\int_{0}^{\infty}(f(t)-1)t^{s-1} dt$$ is analytic, and it
extends to a meromorphic function on $\Bbb{C}$ with two simple poles at $0,1$.

For $v\in R^2$ look at
$$F_{M,t}(v)= \sum_{n\in Z^2} e^{-\pi ((v+n)^\top M (v+n))t}=\sum_{n\in Z^2} e^{-\pi \| M^{1/2} (v+n)\|^2 t}, \qquad M=\pmatrix{|z|^2&\Im(z)\\ \Im(z)& 1}$$
So that $$f(t)=F_{M,t/\pi}(0)$$

$F_{M,t}$ is $Z^2$ periodic, it has a Fourier series $$F_{M,t}(v)= \sum_{k\in Z^2} c_{M,t}(k) e^{2i\pi k^\top v}$$ $$ c_{M,t}(k)=\iint_{R^2} e^{-\pi \| M^{1/2} v\|^2 t}e^{-2i\pi k^\top v}dv= |\det(M)|^{-1/2} t^{-1} e^{-\pi \| M^{-\top/2} k\|^2/ t}$$

ie. $$F_{M,t}(0)= |\det(M)|^{-1/2} t^{-1} F_{M^{-\top},1/t}(0)$$
whence as $t\to 0$ $$F_{M,t}(0)  = |\det(M)|^{-1/2} t^{-1}+O(e^{-r/t})$$ so that
$$\int_0^\infty (F_{M,t}(0)-1+1_{t<1}-|\det(M)|^{-1/2} t^{-1} 1_{t<1}) t^{s-1}dt$$ $$=\pi^{-s} g(s)+\frac1s-|\det(M)|^{-1/2} \frac1{s-1}$$ is entire.


问:
How can I see that the rank of $\mathrm{SU}(N)$ is $N-1$?
Is there a way to understand intuitively why this must be the case? Or is there a kind of easy-to-see proof of this result? Thank you in advance.</p>

答:
$\DeclareMathOperator\U{U}\DeclareMathOperator\diag{diag}\DeclareMathOperator\SU{SU}\DeclareMathOperator\Cent{C}\DeclareMathOperator\GL{GL}$Per @Max's suggestion, I add an approach very similar to theirs, but where things boil down slightly differently in the end.

It is clear that $T = \ker(\det : \U(1)^n \to \U(1))$ is an $(n - 1)$-dimensional torus.  (Explicitly, the isomorphism with $\U(1)^{n - 1}$ is given by $(z_1, \dotsc, z_n) \mapsto (z_1, \dotsc, z_{n - 1})$.)  It is straightforward to show that $\Cent_{\SU(n)}(T) = T$, but I'll put an explanation in the next paragraph anyway.  Granted this, we know that a maximal torus in $\SU(n)$ containing $T$ must be contained in $\Cent_{\SU(n)}(T) = T$, hence equal $T$, so that some (hence every, because they are all conjugate) maximal torus is ($n - 1$)-dimensional.

Now to show that $\Cent_{\SU(n)}(T) = T$.  I'll show more generally (since $T$ is the group of diagonal matrices in $\SU(n)$) that $\Cent_{\GL(n)}(T)$ consists of diagonal matrices.  In fact, even $\Cent_{\GL(n)}(t)$ consists of diagonal matrices for any single diagonal matrix $t = \diag(z_1, \dotsc, z_n)$ where the $z_i$ are all different and all have norm $1$.  (By working with the full torus rather than just a single element, this argument can be made to work even over fields that don't have enough scalars to produce such a diagonal matrix.)  If $g$ commutes with $t$ and $i \ne j$, then the $(i, j)$ entry of $g t$ is $g_{i j}z_j$, but the $(i, j)$ entry of $t g$ is $z_i g_{i j}$, and the fact that these are equal (since $g t = t g$) but that $z_i \ne z_j$ implies that $g_{i j}$ is $0$.

OwnerDisplayName: Polarized photon


答:
What you have done is correct.

You can apply Principle of Inclusion Exclusion and that leads to the same answer.

Using P.I.E, ${50 \choose 2} \times  \sum \limits_{i=0}^{48} {(-1)^i} {48 \choose i} (48-i)^{50}$

Another way to do it will be to choose $2$ empty boxes out of $50$ and then use Stirling Number of the second kind to make $48$ non-empty heaps from $50$ balls and then permute the heaps as we have distinct boxes. That again leads to the same answer.

$S2(50,48) \times 48! \times {50 \choose 2}$

Here is the answer when you calculate any of the above $3$ expressions using Mathematica - $10804606609908677549993379050994509131965157167373221888000000000000000$

Title: Balls in boxes - proof correction

Body:
I have been thinking about the following problem:

How many ways can $50$ different balls be placed into $50$ different sized boxes if exactly $2$ boxes have to remain empty?

My approach was the following:

There are two possible cases which satisfy the conditions:

There are $2$ boxes with $2$ balls and $46$ boxes with $1$ ball;

There is $1$ box with $3$ balls and $47$ boxes with $1$ ball.

In the first case, you can choose the $2$ empty boxes first ($\binom{50}{2}$ possibilities), then you can choose the boxes which have $2$ balls in them ($\binom{48}{2}$ possibilities). After that, you can choose $2$ balls which are placed in the bigger box chosen in the previous step ($\binom{50}{2}$ possibilities), then $2$ balls which are placed in the smaller box ($\binom{48}{2}$ possibilities). Finally, there are only $46$ balls left that have to be placed in the $46$ boxes remained, $1$ ball in each of them ($46!$ possibilities). Totally, there are $\binom{50}{2}* \binom{48}{2}* \binom{50}{2}* \binom{48}{2}* 46!$ possibilities.

In the second case, following a similar logic, there are $\binom{50}{2}* 48* \binom{50}{3}* 47!$ possibilities totally. The final answer to the question can be given adding these two numbers.

However, using the classic inclusion-exclusion formula, the following number of possibilities arise: $\binom{50}{2}* \sum_{k=0}^{48} (-1)^k \binom{48}{k} * (48-k)^{50}$.

I am quite sure that the number derived from the inclusion-exclusion formula is correct, however, I haven't got any clue what is wrong with my first solution.

I would appreciate any help or explanation why my first solution is not correct.


问:
Knowing $f(x) \to 0$ for $x \to 0$, why is $f(x)/x \to 0$ for $x \to 0$?
Let's say, I've found out, that $\lim\limits_{x\rightarrow 0} f(x)= 0$, why is then $\lim\limits_{x \to 0} f(x)/x = 0$? I want to prove, that the function is differentiable at $x=0$, for what I have to show that this limit exists.
$f:\mathbb{R} \to \mathbb{R}$ is defined as $0$ for $x \leq 0$ und $f(x) = e^{-1/x}/x^n$ for x>0, where I've proven, that in case $x<0$ and x>0 it is differentiable.

答:
Your claim is  not always true.

With $$ f(x)=\sin(x)\sin(\frac 1x)$$
you have
$$\lim_{x\to 0}f(x)=0$$
because $$|f(x)|\le |\sin(x)|$$
but
$$\lim_{x\to 0}\frac{f(x)}{x}=\lim_{x\to 0}\sin(\frac 1x)$$
does not exist.

For the second, put $ t=\frac 1x $,

So,
$$\lim_{x\to 0^+}\frac{f(x)}{x}=\lim_{t\to+\infty}t^{n+1}e^{-t}=0$$

In fact, taking logarithm, we get
$$\lim_{+\infty}\ln(t^{n+1}e^{-t})=$$
$$\lim_{+\infty}\Bigl(t((n+1)\frac{\ln(t)}{t}-1)\Bigr)=$$
$$+\infty(0-1)=-\infty$$


问:
How to evaluate the following importer intergal?
Evaluate the integral $$\int_0^{\infty}\frac{dx}{(x^2+25)(2+\cos x)}.$$
I know how to integrate
both the integrals
$$\int_0^{\infty}\frac{dx}{(x^2+25) }, \ \ \ \int_0^{\infty}\frac{dx}{2+\cos x}.$$
Is it any way to utilize the above integrals or please tell me how to proceed?

答:
Let $\alpha = -2 + \sqrt{3}$, $\beta = -2 - \sqrt{3}$ and $u = e^{ix}$. For real $x$, we have:

$$\begin{align}
\frac{1}{2 + \cos x}
&= \frac{1}{2 + \frac{u + u^{-1}}{2}} = \frac{2u}{u^2+4u+1}\\
&= \frac{2u}{(u-\alpha)(u-\beta)} = \frac{2}{\alpha-\beta}\left(\frac{\alpha}{u-\alpha} - \frac{\beta}{u-\beta}\right)\\
&= \frac{1}{\sqrt{3}}
\Re\left(\frac{\alpha}{u-\alpha} - \frac{\beta}{u-\beta}\right)
= \frac{1}{\sqrt{3}}
\Re\left(\frac{\alpha}{\bar{u}-\alpha} - \frac{\beta}{u-\beta}\right)\\
&= \frac{1}{\sqrt{3}}
\Re\left(\frac{\beta^{-1}}{u^{-1}-\beta^{-1}} - \frac{\beta}{u-\beta}\right)\\
&= \frac{1}{\sqrt{3}}\Re\left(\frac{\beta + e^{ix}}{\beta - e^{ix}}\right)
\end{align}
$$
Using this, we can transform the integral at hand as

$$\begin{align}
\mathcal{I} \stackrel{def}{=} &\;
\int_0^\infty \frac{dx}{(x^2 + 5^2)(2+\cos x)}
= \frac12 \int_{-\infty}^\infty \frac{dx}{(x^2 + 5^2)(2+\cos x)}\\
= &\;\frac{1}{2\sqrt{3}}\Re\int_{-\infty}^{\infty} \frac{1}{z^2+5^2}\frac{\beta + e^{iz}}{\beta - e^{iz}}dz
\end{align}
$$
Notice $|\beta| > 1$ and $|e^{iz}| \le 1$ for $\Im z \ge 0$. The factor
$\frac{\beta + e^{iz}}{\beta - e^{iz}}$ is holomorphic and bounded over upper half plane.
We can evaluate $\mathcal{I}$ by completing the contour in upper half plane and then take the residue at $5i$, the only pole of the integrand. The end result  is

$$\begin{align}
\mathcal{I} &= \frac{1}{2\sqrt{3}}
\Re\left(2\pi i \mathop{\rm Res}\limits_{z=5i}\left[\frac{1}{z^2+5^2}\frac{\beta + e^{iz}}{\beta-e^{iz}}\right]\right)
= \frac{1}{2\sqrt{3}}\Re\left(\frac{2\pi i}{10i}\frac{\beta + e^{-5}}{\beta - e^{-5}}\right)\\
&= \frac{\pi}{10\sqrt{3}}\frac{2+\sqrt{3} - e^{-5}}{2+\sqrt{3} + e^{-5}}
\approx 0.180726180098288\ldots
\end{align}
$$


答: HintL Let $m=10m'$ and $n=10n'$. Then $\gcd(m',n')=1$ and $m'n'=10$. There are only a few solutions now.

Title: $\gcd(n,m)=10$, and $\text{lcm}(n,m)=100$


问:
Math contest: Find number of roots of $F(x)=\frac{n}{2}$ involving a strange integral.

Edit summary: A good answer appeared. CW full answer added, based on given answers. Removing my ugly-looking attempts, as they still remain in the rev. history.

Here's a final-round calculus contest problem in our district in 2009. Test paper isn't available online. I originally proved $F(n/2)<n/2$ but failed to prove $F(n)>n/2$ and the monotonicity of $F(n)$.

Define
      $$F(x)=\int\limits_0^x \mathrm e^{-t} \left(1+t+\frac{t^2}{2!}+\cdots+\frac{t^n}{n!}\right) \mathrm d t$$
    where $n>1, n \in \mathbb N^+$; and an equation: $\displaystyle F(x)=\frac{n}{2}$.

Find the number of roots of this equation within the interval $\displaystyle I=\left(\frac{n}{2},n\right)$.

Tedious Previous Attempts Removed.

答:
Another way to show this result could be the one using integration by parts: it is indeed an easy computation (using integration by parts) the one which leads to
$$F(n) = n(1-e^{-n})-e^{-n}\sum_{j=1}^n\frac{n^j}{j!}.$$
I've written it in this way so that you can imagine how to get this result. Now let me use that
$$\sum_{j=1}^n\frac{n^j}{j!}=e^n-\sum_{j=n+1}^{+\infty}\frac{n^j}{j!}$$
so that we can substitute it in the previous expression to get
$$F(n) = n(1-e^{-n}) - 1 + \sum_{j=n+1}^{+\infty}\frac{n^j}{j!}>n(1-e^{-n}) - 1.$$
Thus finally
$$F(n)-\frac{n}{2}>n\biggl(\frac{1}{2}-e^{-n}\biggr)-1$$
If $n\geq 4$ we get
$$F(n)-\frac{n}{2} > \frac{n}{4}-1\geq 0$$
The other cases can be checked directly, moreover from these computations you can also deduce monotonicity of F(n).


问:
Fourier Transform of a compactly supported distribution : how to write the definition in a "better way"?
I am really troubled by the following Theorem :

*Theorem : If $u\in \mathcal{E}'(\mathbb R^d)$ (space of compactly supported distributions), then its Fourier transform is a function, and :
$\hat{u}(\xi)=\langle u,e^{-ix\cdot\xi}\rangle$

For a distribution of $u\in \mathcal{S}'(\mathbb R),$ we define the Fourier transform of $u$ as a distribution, defined for all $\phi \in \mathcal{S}(\mathbb R^d)$ by $\langle\hat u,\phi\rangle=\langle u, \hat\phi \rangle$

Moreover $\mathcal{E}'(\mathbb R^d) \subset \mathcal{S}'(\mathbb R^d)$.

So.. If I follow the definition that is given to me I'm supposed to have, for $u \in \mathcal{E}'(\mathbb R^d),$ $\hat{u}$ as a distribution.

Edit : Evaluating $\hat{u}$ in a real or complex number then doesn't make any sense to me. A distribution of this type has a Schwartz function as argument , and not a complex number to my opinion. So I don't understand how can the symbolism "$\hat{u}(\xi)"$ have a sense in the theorem ? It can also simply be a misunderstanding on my part of the rôle of $\hat{u}$,

A bit stuck with this. Also, if it's just a notation problem, I'd also be interested on how you can easily understand/read this theorem, as this makes a nonsense to me.

答:
The distributional Fourier transform of $u \in \mathcal{E}'(\mathbb{R}^d)$ is by definition a distribution. But there is a function $f \in L^1_{\text{loc}}(\mathbb{R}^d)$ such that $\hat{u}$ is given by integration against $f,$ i.e.
$$\langle \hat{u}, \varphi \rangle = \int f(\xi)\,\varphi(\xi)\,d\xi.$$
What the theorem says is that $f(\xi)=\langle u(x), e^{-ix\cdot\xi} \rangle.$


问:
Prove that integral converges
How can I prove that this integral converges?

$$\int\limits_1^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}\ dx$$

I tried to show that function $0 < \frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ is decreasing and is continious for $x$ in $(1,\infty)$. Furthermore the above integral converges when series $\sum_{1}^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ converges.

Are there better methods?

Title: Prove that integral converges

Body:
First, note that
$$
\int_1^\infty \frac{\ln x\,dx}{(x-1)\sqrt{x^2-1}}=
\int_0^\infty \frac{\ln (1+t)\,dt}{t\sqrt{t(t+2)}}
=\int_0^1 \frac{\ln (1+t)\,dt}{t\sqrt{t(t+2)}\,dt}+\int_1^\infty \frac{\ln (1+t)\,dt}{t\sqrt{t(t+2)}}
$$
Now, due to Mean Value Theorem, for $f(t)=\ln (1+t)$,
$$
\frac{\ln (1+t)}{t}=\frac{\ln(1+t)-\ln 1}{t-0}=(\ln (1+s))'=\frac{1}{1+s}<1, \quad s\in (0,t),
$$
and so
$$
0\le \int_0^1 \frac{\ln (1+t)\,dt}{t\sqrt{t(t+2)}}\le
\int_0^1 \frac{dt}{\sqrt{t(t+2)}}\le \int_0^1 \frac{dt}{\sqrt{t}}=2.
$$
Meanwhile, for $t\ge 0$,
$$
\ln(1+t)=\int_0^t\frac{ds}{1+s}\le \int_0^t\frac{ds}{\sqrt{1+s}}=2\sqrt{1+t}-2<2\sqrt{1+t}
<2\sqrt{t+2},
$$
and hence
$$
0\le \int_1^\infty \frac{\ln (1+t)\,dt}{t\sqrt{t(t+2)}}\le
\int_1^\infty \frac{2\sqrt{t+2}\,dt}{t\sqrt{t(t+2)}}=
\int_1^\infty \frac{2\,dt}{t\sqrt{t}}=\left.-t^{-1/2}\right|_1^\infty=1
$$
Altogether
$$
0\le \int_1^\infty \frac{\ln x\,dx}{(x-1)\sqrt{x^2-1}}\le 3
$$


问:
Parameterizing Geodesics on the Sphere in Polar Coordinates
I seem to have this seemingly trivial Problem, but can't figure it out.

Situation:
I have my Unit Sphere, $S^2$ defined as a Riemannian Manifold. Parameterizing Geodesics (Great circles) on this sphere is absolutely no Problem, IF I embed it in $\mathbb{R}^3$. BUT, I don't want to do this.

What I really want to do, is use riemannian polar coordinates. Now genereally, this isn't a problem, as I could just take any point on the geodesic and call it $(0,0)$ (with $(r,\varphi)$ being my riemannian polar coordinates). And all the lines going through the origin would be my great circles.

However, The origin of the exponential map I'm using does not lie on the geodesic of interest. I do however know 2 points (say $P_1, P_2$) and their polar coordinates.

So basically, I have $\text{exp}_p^{-1} : S^2\to \mathbb{R}^2$ and my two points in $\mathbb{R}^2$ defined by $(r_1,\varphi_1)$ and $(r_2,\varphi_2)$.

I'm looking for some parameterisation to connect these two points via a great circle. And I just can't think of anything sensible.

Any help would be highly appreciated.

Edit: To clarify, I have a working method using an $\mathbb{R}^3$ embedding, but it's very bulky and "unpretty".

Further, I'm using this geodesic to "bound" one of my integrals and shifting the Origin is therefore not an option

答:
Neither approach I propose here are pleasant, but it demonstrates that it can be done in principle.

You can write down an explicit expression for the exponential map in coordinates $(r, \theta) \mapsto (x, y, z)$. If we put the origin of Riemannian polar coordinates at $O = (0, 0, 1)$, then we have $(x, y, z) = (\cos \theta \sin r, \sin \theta \sin r, \cos r)$. You can then try to invert this to obtain expressions for great circles in $(r, \theta)$.

A slightly more clever approach would be to use stereographic projection $\pi : S^2 \to \mathbb{R}^2 \cup \{ \infty \}$:
$$\displaystyle (u, v) = \left( \frac{x}{1-z}, \frac{y}{1-z} \right)$$
So changing to Riemannian polar coordinates,
$$\displaystyle (u, v) = \left( \frac{\sin r \cos \theta}{1 - \cos r}, \frac{\sin r \sin \theta}{1 - \cos r} \right)$$
we see that
$$\displaystyle \sqrt{u^2 + v^2} = \frac{\sin r}{1 - \cos r}$$
and the polar angle in the $(u, v)$ plane is equal to $\theta$. We know what the action of $\text{SO}(3)$ on the sphere corresponds to in the $(u, v)$ plane (namely certain Möbius transformations), so we can write down expressions for rotations of the sphere in terms of $(r, \theta)$. (These will probably be quite ugly.) This then allows us to reduce to the case where $P_1 = O$, so we can apply the inverse rotation to get the expression for a great circle through $P_1$ and $P_2$ in Riemannian polar coordinates centred at $O$.


问:
In a Discussion Involving Formal Languages, Can a Word Be Regarded as an "Expression"?
In order to avoid sounding repetitious by too frequently using the word "word," I would like to be able use the word "expression," writing, e.g., "Applying the function $f$ to the lang$_{1}$ word $ab$ yields the expression $abba$, which is a word in the lang$_{2}$ language."

Is that an acceptable usage of the term expression?

Also, if I were to instead write "Applying the function $f$ to the lang$_{1}$ word $ab$ yields the word $abba$, which is a member of the lang$_2$ language," could I be criticized for first identifying $abba$ as a word without specifying the set in which it is considered such until I reach my sentence's dependent clause to reveal that crucial piece of information?  Or is it okay to call something a "word" in the general sense that there exists some language in which it is a word?  (I know I could rephrase using "...the lang$_{2}$ word $abba$...", but such rephrasing would prove extremely awkward, given the context of my discussion.)

答:
Expression isn’t actually wrong, but it’s awfully general; when you’re talking about a word, it’s generally better to use the more precise term word (or sometimes string). For the example in your first paragraph I would write simply:

Applying $f$ to $ab\in\text{lang}_1$ yields $abba\in\text{lang}_2$.

Your version is unnecessarily wordy. There’s no real need here for function, word, or expression: all are perfectly clear from the context.


问:
Map positive real line to $[0,1]$
I am looking for a function (ideally a bijection) $f(x)$ that maps the unit interval to the positive real numbers. The function should fulfill the following restrictions

$\text{lim}_{x\rightarrow0} f(x) \rightarrow 0$

$f(0.5) = 1$

$\text{lim}_{x\rightarrow1} f(x) \rightarrow \infty$

I've been playing around with variants of $\frac{x}{1+x}$ without success. It's easy to get restrictions 1 and 3 right, but I couldn't manage to find a solution that works without a ''replacement'' solution like the top answer here.

答:
Hint:

If $f(0)=0$ and $\lim_{x\to 1} f(x)=\infty$, then $g_k(x)=f(x)+kx$ (where $k\geq 0$) also satisfies those constraints. You can then just massage $k$ to ensure that $g_k(\frac 12)=1$

Coincidentally, I also stumbled upon $f(x)=-(1+\frac{1}{x-1})$ which fits the bill without needing that extra step.

Title: Map positive real line to $[0,1]$


答:
I suppose the limits you are talking about are deleted limits (otherwise, this is just impossible).

With that in mind, here are some hints:

You can just have $f$ be identity on $(0,1/2)$; this deals with the first requirement.

You have no choice with $f(1/2)=1$. You might as well keep going and have $f(0)=2$, $f(1)=3$. You are left with finding a bijection $(1/2,1)\to (1/2,+\infty)\setminus \{1,2,3\}$.

Start with a bijection $(1/2,1)\to (1/2,+\infty)$, and then compose it with a bijection $(1/2,+\infty)\to (1/2,+\infty)\setminus \{1,2,3\}$.

Title: Map positive real line to $[0,1]$

Body:
I am looking for a function (ideally a bijection) $f(x)$ that maps the unit interval to the positive real numbers. The function should fulfill the following restrictions

$\text{lim}_{x\rightarrow0} f(x) \rightarrow 0$

$f(0.5) = 1$

$\text{lim}_{x\rightarrow1} f(x) \rightarrow \infty$

I've been playing around with variants of $\frac{x}{1+x}$ without success. It's easy to get restrictions 1 and 3 right, but I couldn't manage to find a solution that works without a ''replacement'' solution like the top answer here.


答: Set $u:=\ln x$. The reason $\int_0^\infty\frac{ue^udu}{(e^u-1)\sqrt{e^{2u}-1}}$ converges is that its integrand is asymptotic to $\tfrac{1}{\sqrt{2}}u^{-1/2}$ for small $u>0$ and to $ue^{-u}$ for large $u$. Note $\int_0^\epsilon u^{-1/2}du=2\sqrt{\epsilon}$ for small $\epsilon>0$, and $\int_0^\infty ue^{-u}du=1$.

Title: Prove that integral converges

Body:
How can I prove that this integral converges?

$$\int\limits_1^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}\ dx$$

I tried to show that function $0 < \frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ is decreasing and is continious for $x$ in $(1,\infty)$. Furthermore the above integral converges when series $\sum_{1}^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ converges.

Are there better methods?


问:
$\sum \frac{\ln(n)}{\sqrt{n^5}}$ test for convergence
Let $\sum a_{n}=\sum \frac{\ln(n)}{\sqrt{n^5}}$. To find if the serie is convergence or not, I had some difficult on finding the proper serie to test the given one.

After some work around, I found this sequence $b_{n}=\frac{1}{\sqrt{n^3}}$, whose serie converges!

Then I applyed the limit test comparasion.

$$\lim_{n \to \infty} \frac{\ln(n)}{\sqrt{n^5}}\cdot \sqrt{n^3}=\lim_{n \to \infty} \frac{\ln(n)}{\sqrt{n^{2}}}=\lim_{n \to \infty} \frac{\ln(n)}{n}=0$$

If the limit is equal $0$ and the $\sum b_{n}$ converges, then $\sum a_{n}$ converges to.

My main difficult was to fint out the proper $b_{n}$. Is there an easy way to find it? Thanks

答: You should have in your repository that exponential functions grow faster than any power - and conversely logarithms grow slower than any power, including fractionals powers. That is, $\ln n <\sqrt n$ for sufficiently big $n$. This leaves you simply with $\frac 1{n^2}$ (or even $\frac1{n^{5/2-\epsilon}}$ for any positive $\epsilon$).</p>


问:
What is the probability that at least 2 professors (out of 100) pick the same course (out of 200)?
Suppose each of 100 professors in a large mathematics department picks at random one of 200 courses. What is the probability that at least two professors pick the same course?

The answer given in 1 - 6.6 x 10^-14.
I'm pretty sure you can get this answer by 1-(200!/100! * 1/(200 ^ 100)), but I'm having trouble understanding the basic concepts behind getting the answer.

I don't fully understand why the total possible outcomes would be 200^100 instead of 100^200, if not all courses could possibly be picked by at least one of the professors.

And finally, I don't understand how 200!/100! would be the possible outcomes of no courses being chosen by at least two professors.

Please illuminate me! I'm clearly missing a link.

答:
Its the birthday problem in disguise. Just compute the probability of the complement which is that no two professors pick the same course. This means the first professor picks a course with probability 200/200. Second one picks a course with p = 199/200, third one picks a course with p = 198/200 ... and the 100th professor picks a course with p = 101/200.
Let A = the event that at least two professors pick the same course
Let B = the event that no two professors pick the same course
So P(A) = 1 - P(B) = 1 - (200 * 199 * 198 .. 101)(200 ^ 100).

This is likely very close to 1.


问:
Find all the ideals of a ring
Let $(S,+,\cdot)$ be the commutative and unitary ring given by

$$S=\left\{\left(\begin{array}{cc}a&b\\0&a \end{array}\right)\mid a,b\in\mathbb{R}\right\}$$

Find all the ideals.

Are they all prime ideals?

I found that one ideal is
$$I=\left(\begin{array}{cc}0&b\\0&0 \end{array}\right)$$ and it's not a prime ideal.
How can you know that there are or are not more ideals?

答:
Note that the elements of the ideal $I$ you found are exactly the non invertible elements of the ring. It follows that if $J$ is a proper ideal then $J\subseteq I$. Also, if $J$ contains a nonzero element $\left(\begin{array}{cc}0&b\\0&0 \end{array}\right)$, then for each $c\in\mathbb{R}$ we have:

$\left(\begin{array}{cc}0&c\\0&0 \end{array}\right)=\left(\begin{array}{cc}0&b\\0&0 \end{array}\right)\left(\begin{array}{cc}\frac{c}{b}&1\\0&\frac{c}{b} \end{array}\right)\in J$

And so $J=I$ in that case. So it follows that $I$ is the only nontrivial ideal.

And by the way, $I$ is a maximal ideal, so in particular it is prime.

Title: Find all the ideals of a ring


问:
Show that $-(\lambda_2\lambda_4-\lambda_2\lambda_6-\lambda_4\lambda_5)/(\lambda_1\lambda_4\lambda_5+\lambda_2\lambda_3\lambda_6)=A$
Let $A,B,C$ be three pairiwise coprime positive integers, i.e., there exist six integers $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5},\lambda_{6}$ such that

$$\lambda_{1}A+\lambda_{2}B=1$$

$$\lambda_{3}A+\lambda_{4}C=1$$

$$\lambda_{5}B+\lambda_{6}C=1$$

Solving with respect to $A,B,C$, we get

$$A=-(\lambda_{2}\lambda_{4}-\lambda_{2}\lambda_{6}-\lambda_{4}\lambda_{5})/(\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6})$$

$$B= (\lambda_{1}\lambda_{4}-\lambda_{1}\lambda_{6}+\lambda_{3}\lambda_{6})/(\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6})$$

$$C= (\lambda_{2}\lambda_{3}+\lambda_{1}\lambda_{5}-\lambda_{3}\lambda_{5})/(\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6})$$

Generally, it is not possible to decide if $\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6}$ is not zero and one can find examples about this.

My question:

If there exist some $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5},\lambda_{6}$ verifying $(\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6})$ is not zero and $(\lambda_{2}\lambda_{4}-\lambda_{2}\lambda_{6}-\lambda_{4}\lambda_{5})(\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6})<0$, then show that the expression $-(\lambda_{2}\lambda_{4}-\lambda_{2}\lambda_{6}-\lambda_{4}\lambda_{5})/(\lambda_{1}\lambda_{4}\lambda_{5}+\lambda_{2}\lambda_{3}\lambda_{6})$ must be a representation of the positive integer $A$.

答:
$$
M =
\left(
\begin{array}{ccc}
   0 & \lambda_2 &  \lambda_1 \\
   \lambda_4 & 0 &  \lambda_3 \\
   \lambda_6 &  \lambda_5 &  0 \\
\end{array}
\right)
$$
in
$$
\left(
\begin{array}{ccc}
   0 & \lambda_2 &  \lambda_1 \\
   \lambda_4 & 0 &  \lambda_3 \\
   \lambda_6 &  \lambda_5 &  0 \\
\end{array}
\right)
\left(
\begin{array}{c}
   C\\
   B \\
   A \\
\end{array}
\right) =
\left(
\begin{array}{c}
   1\\
   1 \\
   1 \\
\end{array}
\right)
$$

$$
M^{-1} =
\frac{1}{ \lambda_6  \lambda_3  \lambda_2 +  \lambda_5  \lambda_4  \lambda_1} \hspace{2mm}
\left(
\begin{array}{ccc}
  - \lambda_5  \lambda_3 &   \lambda_5  \lambda_1 &  \lambda_3  \lambda_2 \\
    \lambda_6  \lambda_3 & - \lambda_6  \lambda_1 &  \lambda_4  \lambda_1 \\
    \lambda_5  \lambda_4 &  \lambda_6  \lambda_2 & - \lambda_4  \lambda_2 \\
\end{array}
\right)
$$

Indeed, it is routine to confirm the matrix product
$$
\left(
\begin{array}{ccc}
   0 & \lambda_2 &  \lambda_1 \\
   \lambda_4 & 0 &  \lambda_3 \\
   \lambda_6 &  \lambda_5 &  0 \\
\end{array}
\right)
\left(
\begin{array}{ccc}
  - \lambda_5  \lambda_3 &   \lambda_5  \lambda_1 &  \lambda_3  \lambda_2 \\
    \lambda_6  \lambda_3 & - \lambda_6  \lambda_1 &  \lambda_4  \lambda_1 \\
    \lambda_5  \lambda_4 &  \lambda_6  \lambda_2 & - \lambda_4  \lambda_2 \\
\end{array}
\right) =
\left(
\begin{array}{ccc}
     \lambda_6  \lambda_3  \lambda_2 +  \lambda_5  \lambda_4  \lambda_1  & 0 &  0 \\
   0 &   \lambda_6  \lambda_3  \lambda_2 +  \lambda_5  \lambda_4  \lambda_1  &  0 \\
   0 &  0 &    \lambda_6  \lambda_3  \lambda_2 +  \lambda_5  \lambda_4  \lambda_1  \\
\end{array}
\right)
$$

This tells us that
$$
\left(
\begin{array}{c}
   C\\
   B \\
   A \\
\end{array}
\right) =
\frac{1}{ \lambda_6  \lambda_3  \lambda_2 +  \lambda_5  \lambda_4  \lambda_1} \hspace{2mm}
\left(
\begin{array}{ccc}
  - \lambda_5  \lambda_3 &   \lambda_5  \lambda_1 &  \lambda_3  \lambda_2 \\
    \lambda_6  \lambda_3 & - \lambda_6  \lambda_1 &  \lambda_4  \lambda_1 \\
    \lambda_5  \lambda_4 &  \lambda_6  \lambda_2 & - \lambda_4  \lambda_2 \\
\end{array}
\right)
\left(
\begin{array}{c}
   1\\
   1 \\
   1 \\
\end{array}
\right)
$$

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

hello world


答: Consider $f(x)=\tan(\frac{\pi}{2}x)$? This function maps the interval $(0,1)$ to $(0,\infty)$ and satisfies your three requirements.

Title: Map positive real line to $[0,1]$


问:
Spanning Theory of Linear Algebra
I am not really sure how we can decide that some linearly independent set spans $\mathbb R^n$? Can you explain what are the criteria when we say it spans $\mathbb R^n$?

答:
If it spans a space, then the elements of that space can be written as the linear combination of its elements.

Lets assume that $<1,2>,<3,4>$ are two vector. If we can the values of $c_1,c_2$ in terms of $a,b$ when we wrote it in the form of $c_1<1,2>+c_2<3,4>=$ ,then it spans the space.

In other words, when you write a linear equation in the form of augmented matrix, if you have infinity many solutions then it spans the space.

When we comes to your question , if we cannot find $b$ in the system of $Ax=b$, then it does not span the space

Title: Spanning Theory of Linear Algebra

Body:
If it spans a space, then the elements of that space can be written as the linear combination of its elements.

Lets assume that $<1,2>,<3,4>$ are two vector. If we can the values of $c_1,c_2$ in terms of $a,b$ when we wrote it in the form of $c_1<1,2>+c_2<3,4>=<a,b>$ ,then it spans the space.

In other words, when you write a linear equation in the form of augmented matrix, if you have infinity many solutions then it spans the space.

When we comes to your question , if we cannot find $b$ in the system of $Ax=b$, then it does not span the space


问:
Prime Distribution Linked To Shannon Entropy
I've recently produced a result that links the distribution of primes to Shannon Entropy. The result uses The Prime Number Theorem. I would like to know if my result is valid and if so how can it be interpreted or extended?

The result is here:
http://www.math-math.com/2015/05/a-prime-number-surprise.html

答:
Gregory Chaitin suggested some years ago that there is a link between the primes and disorder or entropy in Chaitin, G.J.; Schwartz, J.T. A note on Monte Carlo primality tests and algorithmic information theory. Pap. Algorithmic Inf. Theory 1990, 8, 197.

Seven years ago I used simple function called BiEntropy which I had designed to investigate the order and disorder of the e.g. prime constant - the 1's and 0's corresponding to the primality of the natural numbers. ie we encode 0,1,2,3,4,5,6,7,8,9.... as 0011010100....  The BiEntropy function returns a real number between 0.0 and 1.0 reflecting the order and disorder of the binary series (of arbitrary length) using a weighted average of the Shannon entropies of the string and its first n-2 binary derivatives. This series shows that the prime constant (in binary) is disordered. In fact it is irrational as proven by Hardy & Wright in the 1930's. That work is documented here: https://arxiv.org/abs/1305.0954

A couple of years ago, I finally got round to investigating the BiEntropy of the primes themselves. It turns out that there is empirically a very close relationship between BiEntropy and Primality - the higher the BiEntropy, the more likely a number is prime. The result is repeated in Trinary. Plus there is a related simple theory of periodic numbers that shows how these empirical results generalise to all the Natural numbers. The paper is available here: https://www.mdpi.com/1099-4300/22/3/311

To answer your question directly, there is a very close relationship between entropy, specifically Shannon's Entropy, and the distribution of primes.


答:
This ring is isomorphic to $\mathbb R[X]/\langle X^2\rangle$, which is essentially $\mathbb R[\varepsilon]$, where $\varepsilon$ is an element with the property that $\varepsilon^2=0$. You can show this using the fundamental theorem on homomorphisms: the homomorphism $\mathbb R[X]\longrightarrow S$ which maps $r\mapsto rE_2$ for all $r\in\mathbb R$ and $X\mapsto\begin{pmatrix}0&1\\0&0\end{pmatrix}$ is surjective and has kernel $\langle X^2\rangle$. The fundamental theorem on homomorphisms (I think sometimes known as first isomorphism theorem) then gives us the isomorphism I claimed in the beginning.

Now you can use the correspondence theorem, which says that the ideals of $S$ are in a one-to-one correspondence with the ideals of $\mathbb R[X]$ containing $\langle X^2\rangle$. Which are pretty easy to find, since $\mathbb R[X]$ is a principal ideal domain. So you're looking for all ideals of the form $\langle p\rangle$ with $p\in\mathbb R[X]$, which contain $\langle X^2\rangle$. That's exactly the ideals generated by a divisor of $X^2$, of which there are three, up to association: $\langle 1\rangle,~\langle X\rangle$, and $\langle X^2\rangle$ itself. So there are only three ideals of $\mathbb R[X]$ containing $\langle X^2\rangle$, and thus only three ideals of $S$. Two of those are the trivial ideals, and you found the third one by yourself.

Title: Find all the ideals of a ring

Body:
Let $(S,+,\cdot)$ be the commutative and unitary ring given by

$$S=\left\{\left(\begin{array}{cc}a&b\\0&a \end{array}\right)\mid a,b\in\mathbb{R}\right\}$$

Find all the ideals.

Are they all prime ideals?

I found that one ideal is
$$I=\left(\begin{array}{cc}0&b\\0&0 \end{array}\right)$$ and it's not a prime ideal.
How can you know that there are or are not more ideals?


问:
Almost Everywhere Conservative Vector Field
$$F(x,y) = \left \langle\frac{-y}{\sqrt{x^2 + y^2}}, \frac{x}{\sqrt{x^2 + y^2}},0 \right \rangle $$

Is a famous example of a vector field whose curl is zero away from the origin but does not have potential function associated with it. However, can a potential function be defined almost everywhere that works? I think this function can be increasing as it goes - say starting at zero on the x-axis. When it comes back, it will be at a higher value. But since I have specified that it be defined a.e., the value along the x-axis won't matter. I don't know how to explicitly write this function though.

答:
For simplicity, we ignore the last $0$ component of $F$ and just consider it as a vector field in $\mathbb R^2$.

A potential function can always be defined on a simply connected open set $\Omega \subseteq \mathbb R^2\backslash \{0\}$ by picking any $x_0\in \Omega$ and write
$$
\phi(x) =\int_{x_0}^x -F(y)dy.
$$
The integral is, of course, path-independent, by Green's formula, as long as the path lie in $\Omega.$

An explicit expression can be obtained by evaluation this integral explicitly, which isn't that difficult. To make sure that almost every point in $\mathbb R^2$ lies in $\Omega,$ just take $\Omega = \mathbb R^2 \backslash \Gamma,$ where $\Gamma$ is any simple curve connecting $0$ to $\infty.$

Of course, all things above are in classical sense. If we work in distributional sense things can be a little more interesting.


问:
Positive and negative correlation -what’s the relationship?-covariance and correlation
Question - show that if y=ax+b with a>0 then corr(x,y)=1 . What relationship between X and Y would give Cov(x,y)=-1?

I got from subbing in y=ax+b that it is equal to
$$a/√a^2$$
Which I think will give you $$ a/mod a$$

Body: If $y=-x+b$  (i.e. $a=-1$), then $Cov(y,x)=corr(y,x)=-1$.


问:
Let $G$ be a group that acts on $P(G)$ via conjugation.

Let $G$ be a group that acts on a set $X.$ Let $\alpha, \beta \in X$ be in the same orbit. If $G$ acts on $P(G)$ via conjugation, then $G_{\alpha}$ and $G_{\beta}$ are in the same orbit.

$P(G)$ is the power set of $G$ i.e the set of all subset of $G$

$G_{\alpha}$ = {$g\in G: \alpha .g=\alpha$}

In the same way, $G_{\beta}$ = {$g\in G: \beta .g=\beta$}

How can I prove this statement?

Recall that the orbit of $G_{\alpha}$ is equal to $\mathcal O_{G_{\alpha}} = \{g^{-1} h g \,|\, g \in G \text{ and } \alpha \cdot h = \alpha \}.$ To show $G_{\alpha}$ and $G_{\beta}$ are in the same orbit, then for some $ g \in G,$ we have that $G_{\alpha} \cdot g = G_{\beta}.$ Any hints?

答:
If $\alpha$ and $\beta$ are in the same orbit of some action $\cdot$ , then $G_\alpha$ and $G_\beta$ (stabilizers) are conjugate in $G$. In fact, $\beta\in O(\alpha)\Rightarrow\exists \tilde g\in G\mid \beta=\tilde g\cdot \alpha$; but then, by action properties:

\begin{alignat}{1}
G_\beta &= \{g\in G\mid g\cdot \beta=\beta\} \\
&= \{g\in G\mid g\cdot (\tilde g\cdot \alpha) =\tilde g\cdot \alpha\} \\
&= \{g\in G\mid (g\tilde g)\cdot \alpha =\tilde g\cdot \alpha\} \\
&= \{g\in G\mid \tilde g^{-1}\cdot((g\tilde g)\cdot \alpha) =\tilde g^{-1}\cdot(\tilde g\cdot \alpha\}) \\
&= \{g\in G\mid (\tilde g^{-1}(g\tilde g))\cdot \alpha =(\tilde g^{-1}\tilde g)\cdot \alpha\} \\
&= \{g\in G\mid (\tilde g^{-1}g\tilde g)\cdot \alpha =\alpha\} \\
&= \{\tilde gg'\tilde g^{-1}\in G\mid g'\cdot \alpha =\alpha\} \\
&= \tilde g\{g'\in G\mid g'\cdot \alpha =\alpha\}\tilde g^{-1} \\
&= \tilde gG_\alpha\tilde g^{-1} \\
\tag 1
\end{alignat}

So, $G_\alpha$ and $G_\beta$ are coniugate in $G$. This precisely means that they lie on the same orbit of the action of $G$ by conjugation on $\mathcal{P}(G)$, say $\star$ . In fact: $\tilde g\star G_\alpha=\tilde g G_\alpha\tilde g^{-1}\stackrel{(1)}{=}G_\beta$.


问:
If $0<a<1, 0<b<1$, $a+b=1$, then prove that $a^{2b}+ b^{2a} \le 1$
I have been really struggling with this problem ... please help!
Let a,b be real numbers. If $0<a<1, 0<b<1, a+b=1$, then prove that $a^{2b} + b^{2a}  \le 1$

What I have thought so far:
without loss of generality we can assume that $a \le b$, since $a^{2b} + b^{2a}$ is symmetric in $a$ and $b$. This gives us $0\le 1/2, 1/2 \le b<1$. But then I am stuck.
I also thought of solving by Lagrange's multiplier method, but it produces huge calculations.

Any help is welcome :)

答:
Not having a good day with websites. I have downloaded what seems to be the source of the question, a 2009 paper by  Vasile Cirtoaje which is about 14 pages. Then a short answer, in a four page document by Yin Li, probably from the same time or not much later. The question was posted on MO by a selfish guy who knew the status of the problem but was hoping for a better answer, a complete answer was also given there in 2010 by fedja, https://mathoverflow.net/questions/17189/is-there-a-good-reason-why-a2b-b2a-1-when-ab1

I have both pdfs by Cirtoaje and Li, email me if you cannot find them yourself.

This is not a reasonable homework question, so I would like to know more of the story, what course for example.

========================

Yin Li of Binzhou University, Shandong, 2009 or 2010, excerpts  I believe he just spelled Jensen's incorrectly, see JENSEN

========================


问:
Let $m$ be arbitrary positive rational number. Show that the sequence $(mn)_{n=1}^{\infty}$ is not $\epsilon$-steady for any $\epsilon>0$
The proposition I would like to prove:

Proposition 1. For arbitrary positive rational $m$, the sequence $(nm)_{n=1}^{\infty}$ is not $\epsilon$-steady for any positive rational $\epsilon$.

Couple of notes:

I will use notation $d(x,y)$ to refer to the distance between rational numbers $x,y$, i.e., $d(x,y) = |x-y|$.

The definition of $\epsilon$-steadiness presented in the book: For some $\epsilon > 0$, the sequence $(a_{n})_{n=m}^{\infty}$ is $\epsilon$-steady if and only if for all natural numbers $j,k$ larger or equal to $m$, $d(a_{j},a_k) ≤ \epsilon$

My attempt:

We will first prove the following lemma:

Lemma 1. For arbitrary positive rational $m$, the sequence $(nm)_{n=1}^{\infty}$ is not $\epsilon$-steady for any positive natural number $\epsilon$.

We will use induction. For the base case, consider $\epsilon := 1$ and suppose that the sequence is $\epsilon$-steady. If $m ≥ 1$, then $d(3m,m) = 2m > 1$, a contradiction. Suppose $m < 1$. Since $m$ is non-zero, there must be some natural number $k$ such that $\frac{1}{m} < k$, and so we have  $1 < mk$. But then $d(m(k+1),m)= mk > 1$, a contradiction. So for any positive rational $m$, the sequence cannot be $1$-steady. Now inductively suppose that for some positive natural number $\epsilon$, the sequence is not $\epsilon$-steady. We want to show that the sequence cannot be $(\epsilon+1)$-steady. Since the sequence is not $\epsilon$-steady, it follows that the sequence must contain some elements $a_j,a_k$ (without loss of generality, let's suppose that $a_j > a_k$), such that $|a_j - a_k| = a_j - a_k > \epsilon$. Since $a_j,a_k$ are multiples of $m$, the rationals $3a_{j},3a_{k}$ will also be multiples of $m$, and thus $3a_{j},3a_{k}$ are also elements of the sequence $(nm)_{n=1}^{\infty}$. Furthermore, we have $d(3a_j,3a_k) = |3a_j - 3a_k| = 3a_j - 3a_k > 3\epsilon$, and since $\epsilon ≥ 1$, $3\epsilon > \epsilon + 1$, which implies that $d(3a_j,3a_k) > \epsilon + 1$, but that means that the sequence is not $(\epsilon+1)$-steady. $\Box$

Now back to the proposition 1:

Take arbitrary positive rationals $m,\epsilon$ and suppose that the sequence $(nm)_{n=1}^{\infty}$ is $\epsilon$-steady. Since $\epsilon$ is rational, we know that there must exist some natural number $k$ such that $\epsilon < k$. By lemma 1, we know that the sequence is not $k$-steady, implying that there are some elements $a_i, a_j$ such that $d(a_i,a_j) > k$. But then clearly,  $d(a_i,a_j) > \epsilon$, a contradiction. $\Box$

Question 1.

Is the proof correct?

Question 2.

Are there other alternatives to proving the proposition?

答: Possibly an easier approach: If $(a_n)_{n=m}^\infty$ is $\epsilon$-steady, then $(a_n)_{n=m}^\infty$ is bounded. That implies $(a_n)_{n=1}^\infty$ is bounded.  But in your situation, $(nm)_{n=1}^\infty$ is unbounded.


问:
How to draw things like commutative diagram in PDE theory
The question maybe a little vague. I am working on PDE theory, I am trying to make my mind more clear by drawing some graph like commutative diagram. But the problem comes since in PDE theory u may need to put your object in different space. For example, for elliptic PDE, we consider the operator $\sum_{ij}a^{ij}\partial_{ij}$ with domain $H^1$ and codomain $L^2$, though we set our domain as $H^1$, but the precise domain is actually $H^2$! But we will need some property from $H^1$. Also, we may require to put the operator in different space. That could be some examples how things get messy, could anyone help with that?

答:
I think there are two slightly different issues here, and the fact that it's not just one issue did confuse me for a long time.

First, there is a useful body of abstraction about "unbounded operators on Hilbert space", abstracting differential operators (and also unbounded multiplication operators). As noted in your question, and looming larger than might be anticipated, there's the issue of "domain" of unbounded operators. They map to the Hilbert space (concretely, $L^2$), but cannot be reasonably defined on all of $L^2$ (and still reasonably map to $L^2$). This can be discussed in a mildly abstract setting without mentioning Sobolev spaces of distributions...

On another hand, for specific operators such as diff ops, especially Laplacians and relatives, we can define Sobolev spaces, families of Hilbert (or Banach) spaces, exactly designed so that $\Delta$ maps $H^s$ to $H^{s-2}$, and so on. (And usually there is some version of "Sobolev imbedding/inequality", comparing Sobolev spaces' $L^2$-differentiability with classical, pointwise differentiability.) This viewpoint also can be abstracted, to "Gelfand triples" and "rigged Hilbert spaces" (meaning families like the $H^s$).

Is this bit of discussion heading in a helpful direction?

EDIT: As in the original asker's comments, indeed, often one discovers that the "original operator" (which in naive mathematics is not mutable), has domain problems. Then, yes, unless one has been around this block before, it is probably wise to give various restrictions and extensions different names (even while realizing they're different incarnations of the same intuitive thing).

For example, in my own experience, various incarnations of Laplace-Beltrami operators play a role, and from a naive viewpoint dredge up serious muddles. Although my own interest is in such operators on modular curves and other "arithmetic quotients", which have various bits of number-theoretic significance, most of the mechanisms already make complete sense in far simpler situations, like on the circle or on the line, not to mention $\mathbb R^n$.

A typical parade of incarnations of "the Laplacian" is: well, first, sure, we have the distributional Laplacian, which can act on nearly anything, and map it to a distribution. Trying to control this a little, one orthodox approach is to restrict the Laplacian to test functions, where it is continuous (after all), and then think about extensions... ideally to $L^2$, but we know this is not literally possible.

One way to adapt is via Sobolev spaces ("tuned" to the Laplacian, most often), to refine the extreme range between test functions and distributions, with $L^2$ "in the middle".

Also, in the case of "semi-bounded" operators such as Laplace-Beltrami operators, there is a canonical, and relatively simple to describe, self-adjoint extension of the merely-symmetric operator Laplacian-restricted-to-test-functions, the Friedrichs extension. (This did already appear as a comment in von Neumann, a few years earlier, as K. Klinger-Logan pointed out to me.) But/and as von Neumann (and many others since) observed, in general there is not a unique self-adjoint extension of a symmetric (but unbounded, hence only densely defined...) operator.

The latter ambiguity has more consequences than many people seem to realize. E.g., "the adjoint", easily mistakable for the original operator in terms of computations, will often have eigenvalues consisting of all complex numbers, or at least an upper or lower half-plane of them. In particular, in violent contrast to one's expectations about "self-adjoint" operators, they're not necessarily real...

Although I do not claim to understand the context of general PDE discussion of this, it is already clear to me that the seemingly fiddly/silly little issues about "domains" and $H^2$ versus $H^1$ are not actually silly, but embody some essential issues.

Put otherwise, the "same" operator but with different domains should be seriously treated as a different thing, and not obstinately "collapsed" to being the same.


问:
Finding the Frenet frame field of a curve $\vec{C}(t)$ = $(\frac{t^3}{3}, 2t-1,t^2+2)$
Working out the unit tangent $T$

$\vec{r} = \frac{t^3}{3}i + (2t-1)j + (t^2+2)k$

$\frac{dr}{dt} = t^2 +2j+2tk$

$\frac{ds}{dt} = |\frac{dr}{dt}| = \sqrt{t^4+4+2t^2}  = t^2+2$

$T = \frac{dr}{ds} = \frac{dr}{dt}/\frac{ds}{dt} = \frac{t^2i+2j+2tk}{t^2+2}$ = $\frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k$

Now working out the Principal Normal N,

$\frac{dT}{dt} = \frac{4t}{t^4+4}i-\frac{4t}{t^4+4}j-\frac{2t^2+4}{t^4+4}$ By the Quotient Rule

$\frac{dT}{ds} = \frac{dT}{dt}/\frac{ds}{dt} = \frac{4t}{t^6+8}i-\frac{4t}{t^6+8}j-\frac{2t^2+4}{t^6+8}k$

Now $N$ is $\frac{1}{k}$$\frac{dT}{ds}$ so in order to find $N$, I must first find $k$.

$k$ = $|\frac{dT}{ds}$| and this is where is becomes problematic. Since $|\frac{dT}{ds}$| = $\sqrt{(\frac{4t}{t^6+8})^2i-(\frac{4t}{t^6+8})^2j-(\frac{2t^2+4}{t^6+8})^2k}$

=$\sqrt{(\frac{4t^2}{t^12+64})i-(\frac{4t^2}{t^12+64})j-(\frac{2t^4+16}{t^12+64})k}$

So $N$ = $\frac{1}{$\sqrt{(\frac{4t^2}{t^12+64})i-(\frac{4t^2}{t^12+64})j-(\frac{2t^4+16}{t^12+64})k}} \times \frac{4t}{t^6+8}i-\frac{4t}{t^6+8}j-\frac{2t^2+4}{t^6+8}k$

Which I don't know what it equals to.

I wonder how I would be able to work out the Binormal B, knowing that $B = T \times N$.

Maybe I'm making a mistake... Any tips?

Body:
You've missed an $i$ in your calculations for $\frac{d\vec{r}}{dt}$.
$$\frac{d\vec{r}}{dt} = t^2i +2j+2tk$$
Even though the final result for $\frac{ds}{dt}$ is good, the rightmost term in the radical sign is incorrect.
$$\frac{ds}{dt} = \left| \frac{dr}{dt} \right| = \sqrt{t^4+4+(2t)^2}  = t^2+2$$
Luckily, you put back the $i$ in the expression for $\vec{T}$.
However, you've incorrectly squared the denominator $t^2 +2$ as $t^4 + 4$.
The correct $\frac{d\vec{T}}{dt}$ should be
$$\frac{d\vec{T}}{dt} = \frac{4t}{(t^2+2)^2}i-\frac{4t}{(t^2+2)^2}j-\frac{2(t^2-2)}{(t^2+2)^2}k$$
because $$\frac{d}{dt} \frac{2t}{t^2+2} = \frac{2(t^2+2)-2t(2t)}{(t^2+2)^2} = \frac{4-2t^2}{(t^2+2)^2}.$$
You've got the denominator wrong in the same way.
$$\frac{d\vec{T}}{ds} = \frac{d\vec{T}}{dt}/\frac{ds}{dt} = \frac{4t}{(t^2+2)^3}i-\frac{4t}{(t^2+2)^3}j-\frac{2(t^2-2)}{(t^2+2)^3}k$$
\begin{aligned}
\kappa &= \left| \frac{d\vec{T}}{ds} \right| \\
&= \frac{\sqrt{(4t)^2+(4t)^2 + 4(t^2-2)^2}}{(t^2+2)^3} \\
&= \frac{\sqrt{4(4t^2+4t^2+t^4-4t^2+4)}}{(t^2+2)^3} \\
&= \frac{\sqrt{4(t^4+4t^2+4)}}{(t^2+2)^3} \\
&= \frac{2}{(t^2+2)^2}
\end{aligned}
So
\begin{aligned}
\vec{N} &= \frac1\kappa \frac{d\vec{T}}{ds} \\
&= \frac{(t^2+2)^2}{2} \left[ \frac{4t}{(t^2+2)^3}i-\frac{4t}{(t^2+2)^3}j-\frac{2(t^2-2)}{(t^2+2)^3}k \right] \\
&= \frac{1}{2(t^2+2)} [4ti-4tj-2(t^2-2)k] \\
&= \frac{2ti-2tj-(t^2-2)k}{t^2+2}
\end{aligned}
Finish the exercise with $\vec{B} = \vec{T} \times \vec{N}$.
\begin{aligned}
\vec{B} &= \vec{T} \times \vec{N} \\
&= \left( \frac{t^2}{t^2+2}i+\frac{2}{t^2+2}j+\frac{2t}{t^2+2}k \right) \times \frac{2ti-2tj-(t^2-2)k}{t^2+2} \\
&= \frac{1}{(t^2+2)^2} [(-2(t^2-2)-(-2t)(2t))i - ((t^2)(-(t^2-2))-(2t)(2t))j + ((t^2)(-2t)-(2t)(2))k] \\
&= \frac{2(t^2+2)i+t^2(t^2+2)j-2t(t^2+2)k}{(t^2+2)^2} \\
&= \frac{2i+t^2j-2tk}{t^2+2}
\end{aligned}


答:
A "spanning set" or "a set that spans the vector space" is a set such that any member of the vector space can be written as a linear combination of vectors in the set.

A spanning set is NOT necessarily a basis.  A basis is a set that both spans the vector space AND such that the vectors in the set are linearly independent.  It easy to see that a set consisting of a single (non-zero) vector is linearly independent but probably wont span the space!  As we add more vectors to the set they are less likely to be independent but more likely to span the space.  On the other hand, the set of all vectors certainly spans the space but can't be linearly independent.  As we remove vectors from the set they become more likely to be independent but less likely to span the space.

In Linear Algebra, dealing with finite dimensional vector spaces, there is one number, n, where they "meet in the middle".  That is, there exist a set of vectors that both spans the space and is linearly independent.  That set is a basis for the space and the specific number "n" is the "dimension" of the vector space.

Title: Spanning Theory of Linear Algebra

Body: I am not really sure how we can decide that some linearly independent set spans $\mathbb R^n$? Can you explain what are the criteria when we say it spans $\mathbb R^n$?


问:
Shortest distance for cuboid by taking a certain path in 3D

I am stuck on part (b) of the question. Please see image above.
How would I work out the shortest distance with reducing the amount of cheese the ant has to crawl through?

答:
The Pythagorean theorem works in unlimited dimensions so, if she love cheese and would eat her way through, the distance $$d=\sqrt{20^2+20^2+30^2}=\sqrt{1700}\approx 41.23$$

If she hates cheese, the way along the edges is $d=20+20+30=70$

OTOH, if she walks half-way down a 20-side while walking horizontally the other full 20-side, she can the continue the rest of the way down the 20-side while walking the full length of the 30-side.

$$\sqrt{20^2+10^2}+\sqrt{10^2+30^2}=\sqrt{500}+\sqrt{1000}\approx
22.36 + 31.62\approx 53.98$$

Another way is to think of the 20X30 and a 20X20 sections as a single rectangle.

$$\sqrt{(30+20)^2+20^2}=\sqrt{2900}\approx 53.85$$


问:
blood investigation probability
A large number $N$ of people are subjected to a blood investigation. This investigation can be organized in two
ways.

$(1)$ The blood of each person is investigated separately. In this case $N$ analyses are needed.

$(2)$ The blood of $k$ people are mixed and the mixture is analysed. If the result is negative, then this single analysis is sufficient for $k$ persons.
But if it is positive, then the blood of each one must be subsequently investigated separately, and in toto for $k$ people, $k + 1$ analysis are needed. It is assumed that the probability of a positive result ($p$) is the same for all people and that the results of the analysis are independent in the probabilistic sense.

• What is the probability that the analysis of the mixed blood of $k$ people is positive?

• What is the expectation of the number of analysis necessary in the second method of testing?

• For what $k$ is the minimum expected number of necessary analysis obtained?

答:
Nice question.

What is the probability that the analysis of the mixed blood of $k$ people is positive?

Let $X_i = 1$ iff the $i$-th person is positive and $X_i=0$ otherwise. We have
$$\mathbb{P}(\text{mixed blood of $k$ people is positive}) \ = \ \mathbb{P}(X_1 \ \text{or } \ X_2 \ \text{or} \ \dots \ \text{or} \ X_k \ \text{is} \ 1)$$
$$= \ 1-\mathbb{P}(X_1, \dots, X_k \ \text{are all} \ 0 ) \ = \ 1-(1-p)^k.$$

What is the expectation of the number of analysis necessary in the second method of testing?

Let as assume that $k$ is a divisor of $N$, i.e. $N=k\cdot m$. Then the answer is
$$m\cdot \Big(\text{expected number of tests needed to check a group of $k$ people}\Big)$$
$$= \ m\cdot\Big[1\cdot(1-p)^k \ + \ (k+1)\cdot\Big(1-(1-p)^k\Big)\Big].  $$

I will live You with further calculations. Please note, that the independence assumption is crucial.

Title: blood investigation probability


问:
Trigonometric problem
I'm trying to get the roots for a complex number $x^2+1$

$x^2+1=0\rightarrow x^2=-1 \rightarrow x = \sqrt{-1} \rightarrow i$

So, $w^2 = 0 + 1i$

$p = \sqrt{0^2+1^2} = 1$

$\theta = \tan^{-1} \left( \frac{1}{0} \right )$

But I don't know what I can do to get the $\tan^{-1}(\frac10)$

答: $\theta = \tan^{-1}(\infty) =\dfrac{\pi}2$ because the real part and the imajinary part are positive and if the real part and the imajinary part are negative also we have $\theta=\dfrac{\pi}{2}$, if one of this side is negative and the other is positive then $\theta=-\dfrac{\pi}{2}$.</p>

Title: Trigonometric problem


问:
Inflection points with natural logs.
Where are the inflection points, if any exist, for $f(x)=10\frac{\ln(\ln(x))}{\ln(x)}$?
I don't think any exist. But could use a little help. My guess would be that with the second derivative, if $x = e^{-2}$ you get 0, but this is not in the domain of the function. Is this correct, or completely off? Any help would be appreciated.</p>

答:
Inflection point must exist.

Indeed we have a maximum at $x=e^e$ and $f(x)\to 0$ as $x\to +\infty$, therefore concavity must change because $f(x)$ has no intersection with $x$-axis after $x=e$.

We have $$f''(x)=10\,\frac{ (\log x+2) \log (\log x)-\log x-3}{x^2 \log ^3x}$$

$f''(x)=0$ if $$(\log x+2) \log (\log x)-\log x-3=0$$
substitute $\log x=u$
$$(u+2)\log u-u-3=0\to u=3.28455$$
Finally we have $x=e^{3.28455}\approx 26.7$

the graph below shows the curve $y=f(x)$, the inflection point and the tangent

$$...$$

Title: Inflection points with natural logs.


问:
The relations between the second fundamental forms of conformal immersions
Let $f:M^{n}\rightarrow \tilde{M}^{m}$ be an isometric immersion, that is, $\langle\cdot, \cdot\rangle_{f}=\langle\cdot, \cdot \rangle$, where $\langle \cdot, \cdot \rangle$ is the metric with respect to $M$. Also denote $\langle \cdot,\cdot\rangle_{\tilde{M}}$ by the metric associeted with $\tilde{M}$.

Denote $f^{*}T\tilde{M}$ the induced bundle over $M$ whose fiber at $x \in M$ is $T_{f(x)}\tilde{M}$. The orthogonal complement of $f_{*}T_{x}M$ in $T_{f(x)}\tilde{M}$ is called the normal space of $f$ at $x$ and is denoted by $N_{f}M(x)$. The normal bundle $N_{f}M$ of $f$ is the vector subbundle of $f^{*}T\tilde{M}$ whose fiber at a point $x\in M$ is $N_{f}M(x)$. Then we have the orthogonal decomposition $$ f^{*}T\tilde{M}=f_{*}TM\oplus N_{f}M.$$ Given $X, Y$ vector fields in $M$, decompose $$ \tilde{\nabla}_{X}f_{*}Y=(\tilde{\nabla}_{X}f_{*}Y)^{T}+(\tilde{\nabla}_{X}f_{*}Y)^{\perp}.$$ As $$ \nabla_{X}Y:=f_{*}^{-1}(\tilde{\nabla}_{X}f_{*}Y)^{T}$$ defines a compatible torsion-free connection on $TM$ we have by uniqueness that $\nabla_{X}Y$ is the Levi-Civita connection of $M^{n}$. The map $\alpha:\Gamma(TM) \times \Gamma(TM) \rightarrow \Gamma(N_{f}M$) defined by $$ \alpha^{f}(X,Y)=(\tilde{\nabla}_{X}f_{*}Y)^{\perp}$$ is called the second fundamental form of $f$. The shape operator $A_{\xi}$ of $f$ at $x \in M$ with repect to $\xi \in N_{f}M(x)$ is defined by $$ \langle A_{\xi}X,Y\rangle=\langle \alpha(X,Y), \xi \rangle_{\tilde{M}}.$$

Definition: Let $f, g:M^{n}\rightarrow \mathbb{R}^{m}$ be immersions. We say that $f$ and $g$ are conformal, if there exists $\varphi \in C^{\infty}(M)$ such that the metrics induced by $f$ and $g$ are related by $$ \langle \cdot , \cdot\rangle_{g}=\varphi^{2}\langle \cdot, \cdot\rangle_{f}$$

Problem: I would like to know if someone could show me the relations between the second fundamental forms of $f$ and $g$.

I thought of using the relation $$ \nabla_{X}^{g}{Y}=\nabla_{X}^{f}Y+\frac{1}{\varphi}(Y(\varphi)X+X(\varphi)Y-\langle X,Y \rangle_{f}\mbox{grad}_{f}\varphi)$$ where $\mbox{grad}_{f}\varphi$ is defined by $$ X(\varphi)=\langle \mbox{grad}_{f}\varphi, X \rangle_{f}.$$

答:
There is no relation. Take the simplest case when $n=1$ and $m=2$. That is, $f, g$ are two immersed curves in $\mathbb R^2$. One can give both $f, g$ the arc-length parametrization, which in your notations means $\langle \cdot, \cdot\rangle_g = \langle \cdot, \cdot\rangle_f$, or $\varphi = 1$. But there is no restrictions on $f$ and $g$, and so no restrictions on the curvature $\kappa$.

On the other extreme, let $f$ be any isometric immersion and let $g = f\circ h$, where $h : M \to M$ is any conformal immersion. They have the same second fundamental forms, and it has nothing to do with the conformal factor.


问:
Why can any inner product in $\Bbb{R}^n$ be represented in a general form?
Why can any inner $\langle\mathbf{x}, \mathbf{y}\rangle$ Product in $\Bbb{R}^{n}$ be rewritten as $x^{T}Ay$ (so it can be rewritten as the standard euclidean product)?  I know that $$
\langle\mathbf{x}, \mathbf{y}\rangle=\left\langle\sum_{i=1}^{n} x_{i} \mathbf{e}_{i}, \sum_{j=1}^{n} y_{j} \mathbf{e}_{j}\right\rangle=\sum_{i, j=1}^{n} x_{i} y_{j}\left\langle\mathbf{e}_{i}, \mathbf{e}_{j}\right\rangle
$$
where $\bf{e}$ denotes the canonical basis. But doesn't that only say that matrix $A$ can be the identity matrix, since the inner product of the orthonormal basis vectors is always $0$ or $1$?

答: The basis vectors $\mathbf e_1, \mathbf e_2, \dots, \mathbf e_n$ are only orthonormal with respect to the standard inner product. If $\langle \cdot, \cdot\rangle$ is some other inner product, then you don't know what $\langle \mathbf e_i, \mathbf e_j\rangle$ is. All you can do is define $A_{ij} = \langle \mathbf e_i, \mathbf e_j\rangle$, and then what you've done shows that $$\langle \mathbf x , \mathbf y \rangle = \sum_{i=1}^n \sum_{j=1}^n x_i y_j A_{ij} = \mathbf x^{\mathsf T}\!A \mathbf y.$$ (From properties of inner products, you can deduce other things. For instance, if you know that $\langle \mathbf x , \mathbf y \rangle = \langle \mathbf y, \mathbf x\rangle$ for all $\mathbf x, \mathbf y$, then you know that $\langle \mathbf e_i, \mathbf e_j\rangle = \langle \mathbf e_j, \mathbf e_i\rangle$. Therefore $A_{ij}=A_{ji}$, so $A$ is symmetric.)


问:
Is 3-SAT useful for anything practical?
If all NP-complete problems can be converted to 3-SAT problems I had an idea that would not solve the NP problem but might be a practical solution.

What you could do is simply, make a huge table of 3-SAT problems and whether they are solvable or not. Store this on a giant server.

Then it takes less than polynomial time to look at a 3-SAT problem and look it up in a table.

Two questions, for practical purposes, how big a table would be useful? Does such a table already exist?

答:
I think your approach would be completely infeasible and I am sure that no such table exists. The good news is that SAT solvers using clever algorithms and heuristics are able to solve formulas involving thousands of variables with acceptable running times. If you try to estimate how many entries there would be in your table if it contained all formulas with up to 10,000 variables and 1,000,000 symbols, you will see why I say your approach is infeasible.

The moral is that just because a problem is NP-hard does not mean that algorithms to solve that problem are useless in practice.

Title: Is 3-SAT useful for anything practical?


问:
Let X and Y be random variable with correlation coefficient  prove : ((|)) ≤ (1 − ^2 ) ()
Let X and Y be random variable with correlation coefficient  prove that :

$((|)) ≤ (1 − ^2 ) ()$

My attempt:

We know that $Var(Y|X) = E(Y^2|X) - E(Y|X)^2$

$E[Var(Y|X)] = E[E(Y^2|X)] - E[E(Y|X)^2] = E[Y^2] - E[E(Y|X)^2]$

I'm stuck here.

答:
Wlog $\mathbb E X = \mathbb E Y = 0$ and $Var(X) = Var(Y)=1$.

Then, $\mathbb E[Y-\rho X]=0$ and $Var(Y-\rho X) = 1+\rho^2-2\rho^2 = 1-\rho^2$. Moreover, $Cov(X,Y-\rho X) = 0$.

Let's pose $Z:=Y-\rho X$, then $Y = Z+\rho X$ and $Var(Y|X) = Var(Z|X)$. We need $$ \mathbb E[Var(Z|X)] \stackrel ?\leq 1-\rho^2 = Var(Z)$$

However,
$$\mathbb E[Var(Z|X)] = \mathbb E[Z^2] - \mathbb E[\mathbb E[Z|X]^2] = Var(Z) +\mathbb E[Z]^2- \mathbb E[\mathbb E[Z|X]^2] = Var(Z)+\mathbb E[\mathbb E[Z|X]]^2- \mathbb E[\mathbb E[Z|X]^2] = Var(Z)-Var(\mathbb E[Z|X]) \leq Var(Z)$$


问:
Expectancy of a RV
I have a question. I know that a RV $X$ has the hyperbolic secant distribution (with density function $f_X(x)=\frac{2}{\pi}\frac{1}{e^x+e^{-x}}$). I have a few questions about calculating the expectancy (mean value):
Is $\int^{\infty}_{0}f_{X}(x)dx=\mathbb{E}[X\space{1}_{\{x>0\}}]$? (where $1$ indicates the indicator function)
And why does the fact that $X$ and $-X$ has the same distribution imply that $2\mathbb{E}[X\space{1}_{\{x>0\}}]=\mathbb{E}[X\space{1}_{\{x>0\}}]+\mathbb{E}[-X\space{1}_{\{x<0\}}]=\mathbb{E}[|X|]$?

Thanks in advance

答:
No, it should be $\int_0^\infty x f_X(x) \, dx = E[X 1_{X > 0}]$.

If $X$ has the same distribution as $-X$, then $X1_{X > 0}$ has the same distribution as $(-X) 1_{(-X) > 0} = -X1_{X < 0}$, so $E[X1_{X>0}] = E[-X1_{X<0}]$

In general for any real number $x$ we have $|x| = x 1_{x>0} + (-x)1_{x<0}$, so  $E[|X|] = E[X1_{X > 0} + (-X) 1_{X < 0}] = E[X 1_{X > 0}] + E[-X1_{X<0}]$

Title: Expectancy of a RV


问:
Given first fundamental form, provide an example of its surface

How to find surface of this first fundamental form (not a plane)? I know it may sound stupid, but my few attempts got me $(u,v,0)$, but it is plane. Is there any algorithm, I could not find any.

答:
If we want a surface embedded in $\mathbb{R}^3$, we can write a partial differential equation to find such surfaces.

Let $x: U\to\mathbb{R}^3$ be a patch with $ds^2=\lambda(du^2+dv^2)$. We can write this out in components as $$x(u,v)=(f(u,v),g(u,v),h(u,v))$$ with $f,g,h$ smooth functions from $\mathbb{R}^2\to\mathbb{R}$. Then $$x_u=(f_u,g_u,h_u) \text{ and } x_v=(f_v,g_v,h_v)$$ Our coordinate functions must satisfy $x_u\cdot x_u=\lambda$, $x_v\cdot x_u=0$ and $x_v\cdot x_v=\lambda$ giving us the following set of partial differential equations. \begin{align}
x_u\cdot x_u&= f_u^2+g_u^2+h_u^2=\lambda\\
x_v\cdot x_u&=f_uf_v+g_ug_v+h_uh_v=0\\
x_v\cdot x_v&= f_v^2+g_v^2+h_v^2=\lambda
\end{align}
Solving these equations will give us a surface with the desired property.


答: You can't just brute force 3-sat. That would take too much computational resources. I am pretty sure all the computers on earth would not be enough. This is typical of all problems considered infeasable in computational complexity theory. The doubling of computational power that is happening due to Moores law still isn't good enough to solve many problems</p>

Title: Is 3-SAT useful for anything practical?

Body:
If all NP-complete problems can be converted to 3-SAT problems I had an idea that would not solve the NP problem but might be a practical solution.

What you could do is simply, make a huge table of 3-SAT problems and whether they are solvable or not. Store this on a giant server.

Then it takes less than polynomial time to look at a 3-SAT problem and look it up in a table.

Two questions, for practical purposes, how big a table would be useful? Does such a table already exist?


问:
What does $\vdash s \rightarrow (\neg s\rightarrow t)$ mean?
What does this statement mean $\vdash s \rightarrow (\neg s\rightarrow t)$?

And how can I prove it?

答:
Writing $T\vdash\varphi$ means that if we assume $T$ then we can prove $\varphi$. If $T$ is omitted then this means that without any assumptions we can prove $\varphi$, that is to say that $\varphi$ is logically true.

In this case it means that $s\rightarrow(\lnot s\rightarrow t)$ is true, regardless to the truth values of $s$ and $t$.

I'll leave you with the task that this statement is indeed true, but I'm going to give you a small hint:

Hint: Use the truth table of $\rightarrow$ and calculate the possible truth values of the statement based on the possible combinations of truth values of $s$ and $t$.


答:
Firstly, by definition
$$
\mathbb E(\phi(X)) = \int_{\mathbb R} f_X(t) \phi(t) dt.
$$
(This can be proven from the somehow more basic definition $\mathbb E(x) =\int x f_X(x)$.) So if $\phi(x) =x 1_{\mathbb R^+},$ then
$$
\mathbb E(\phi(X)) = \int_{\mathbb R} f_X(t) \phi(t) dt=\int_{\mathbb R^+} f_X(t)tdt.
$$

For the second question, let me add some details: since $X$ and $-X$ have the same distribution, we have
$$
\mathbb E(\phi(X)) = \mathbb E (\phi(-X)).
$$
Now, what is $\mathbb E (\phi(-X))?$ Well, $\phi(-X) = (-x)1_{\mathbb R^+}(-x) = -x1_{\mathbb R^-} (x)$, which is what you have in the question.

Title: Expectancy of a RV

Body:
I have a question. I know that a RV $X$ has the hyperbolic secant distribution (with density function $f_X(x)=\frac{2}{\pi}\frac{1}{e^x+e^{-x}}$). I have a few questions about calculating the expectancy (mean value):
Is $\int^{\infty}_{0}f_{X}(x)dx=\mathbb{E}[X\space{1}_{\{x>0\}}]$? (where $1$ indicates the indicator function)
And why does the fact that $X$ and $-X$ has the same distribution imply that $2\mathbb{E}[X\space{1}_{\{x>0\}}]=\mathbb{E}[X\space{1}_{\{x>0\}}]+\mathbb{E}[-X\space{1}_{\{x<0\}}]=\mathbb{E}[|X|]$?

Thanks in advance


问:
Density of linear transformation of random vector
Let $x \in \mathbb{R}^n$ be a random vector with density $f : \mathbb{R}^n \rightarrow \mathbb{R}$. Let $A \in \mathbb{R}^{m \times n}$ with $m < n$. What can we say about the density of $Ax$?

If $A \in \mathbb{R}^{n \times n}$ is invertible then by a change of variables, we know that the density of $Ax$ is given by $f(A^{-1}x)/|\text{det} A|$. However, I can't seem to find a concrete reference for a general case.

答:
I will focus on the case where $A$ has full row rank. The key is figuring out how to handle the case where $A$ has a non-trivial kernel.

First, consider the case where $A$ is "diagonal". If we have $A = [\Sigma\ \ 0]$ with
$$
\Sigma = \pmatrix{\sigma_1\\ & \ddots\\ && \sigma_m},
$$
then for any set $S \subset \Bbb R^m$, the vector $y = Ax$ is an element of $S$ if and only if the truncated vector $\hat x = (x_1,\dots,x_m)$ is such that $\hat x = \Sigma^{-1} y$. With that, we deduce that the distribution of $y = Ax$ is given by
$$
f(y) = f_{1,\dots,m}(\Sigma^{-1} y)/|\det(\Sigma)| = ,
$$
where $f_{1,\dots,m}$ denotes the marginal distribution over $x_1,\dots,x_m$.

For an arbitrary full row-rank matrix $A$, $A$ necessarily has a singular value decomposition
$$
A = U [\Sigma \ \ 0] V^T,
$$
where both $U$ and $V$ are orthogonal matrices (which in particular implies that $|\det(U)| = |\det(V)| = 1$). The density of $V^Tx$ is given by $g(x) = f(Vx)$, the density of $[\Sigma \ \ 0] V^T x$ is given by $h(y) = g_{1,\dots,m}(\Sigma^{-1}y)/|\det\Sigma|$, and the density of $U[\Sigma \ \ 0]V^Tx = Ax$ is therefore given by
$$
p(y) = h(U^Ty).
$$
If we take $R$ to denote the row-space of $A$ and $N$ to denote the nullspace, then this amounts to the statement
$$
p(y) = f_R(A^+y)/\sqrt{\det(AA^T)},
$$
where we define
$$
f_R(\vec x) = \int_{\vec z \in N} f(\vec x + \vec z)\,d\vec z.
$$


问:
How many $8$-digit numbers do not contain the string $55555$?
I'm having a little trouble figuring out this exercise.

Consider an $8$-digit number. How many numbers do not contain even numbers? How many numbers do not contain the string $55555$?

For the first part I've worked out that since there are only $5$ odd numbers $\{1,3,5,7,9\}$ and a total of $8$ digits the answer is $5^8$, but I don't know how to tackle the second part.

答:
If $55555$ appears as a (consecutive) substring, its earliest appearance begins with one of the first four digits. If that earliest appearance begins with the first digit, then the final three digits can be anything. If it begins with the second, third, or fourth digit, then the digit immediately preceding it cannot be a $5$ but can be anything else, and the remaining two digits, wherever they are, can be anything. Thus the total numbero of eight-digit strings containing $55555$ is

$$10^3+3(9\cdot10^2)=3700$$

so the number of eight-digit strings without $55555$ as a substring is

$$10^8-3700$$


问:
Existence of divisors in a finite set.
Is it true that however we pick 1011 numbers alongside of $\{1,2,...,2020\}$ then for at least two of them we'll have that one of those two numbers is a divisor of the other?

Meaning that however we pick $A\subseteq ${$1,2,...,2020$} with $|A|=1011$, then $\exists x,y\in A$ s.t. $x|y$ or $y|x$.

What I thought of is somehow using the Pidgeonhole principle, but I've no constructive thoughts from that on. I'll appreciate any help. Thanks.

答: Every integer is a power of two times an odd number. There are only 1010 candidates for the odd part. Apply pigeon hole</p>

Title: Existence of divisors in a finite set.


问:
Equivalent formulas and equivalent statements in logic.
I am reading a book titled "How to prove it", and I am learning about equivalences. I could be completely wrong here, so thanks for the patience and the help.

From the book, the absorption law looks like this:

Absorption law:

P ∨ (P ∧ Q) is equivalent to P.

My question regarding this.

Say I have two statements:

P: It is dark.

Q: It is cold.

P ∨ (P ∧ Q) would mean:

It is either dark, or dark and cold, or both.

Which seems different from simply P:

It is dark.

Equivalent seems to mean the following here:

The validity of the original statement (P ∨ (P ∧ Q)) depends on P alone, all other parts of the statement (like Q) don't change the validity.

And I would add, that by applying the absorption law, the resulting (simplified) form shouldn't be mistaken for having the same meaning as the original statement.

However, this seems to be a bit different in case of other equivalences:

¬¬P is equivalent to P in a sense that it is a different (simplified) form, without loosing the meaning of the original statement.

Or in case of De Morgan's law:

¬(P ∧ Q) is equivalent to ¬P ∨ ¬Q. The latter from still produces a statement that can be interpreted as having the same meaning:

It is not both dark and cold.

And

It is either not dark, or not cold, or neither dark nor cold.

Is it the case, that some laws or modifications can be used to simplify statements, whereas some can only be used to analyze certain relationships within the original statement?

答:
"$\varphi$ is logically equivalent to $\psi$" means that in any model, $\varphi$ is true if and only if $\psi$ is true. In the case of propositional logic, a model is just an assignment of a truth value to each basic proposition (i.e. a row of a truth table).

In the case of $P$ and $P\lor(P \land Q)$, it is easy to check that these are logically equivalent. If $P$ is true, then by the meaning of $\lor$, $P\lor (P\land Q)$ is true. If $P$ is false, then $(P\land Q)$ is false by the meaning of $\land$. Since also $P$ is false, $P\lor (P\land Q)$ is false by the meaning of $\lor$. We've shown that $P\lor (P\land Q)$ is true whenever $P$ is true and false whenever $P$ is false, so they are logically equivalent.

You should try to avoid allowing intuitions from ordinary language to creep into your understanding of logical words in mathematics. Words like "and", "or", "implies", and "equivalent" all have precise meanings in mathematics, which do not always match the informal and imprecise ways we use these words in ordinary language.


问:
Stopping time + constant
Let $\tau$ be a stopping time with respect to some filtration $(\mathcal{F}_t)_{t\geq 0}$. Then we know that $\{ \tau \leq t \} \in \mathcal{F}_t$ for all $t\geq 0$.

But how do I prove that $\tau+s$ for $s \geq 0$ is a stopping time aswell?

答:
Suppose that $t \ge s$.

$$ \{ \tau + s   \le t \}  =  \{ \tau  \le t -s \} \in \mathcal{F}_{t-s}$$
because $\tau$ is stopping time.

Thus
$$ \{ \tau + s   \le t \} \in \mathcal{F}_{t-s} \subset  \mathcal{F}_{t}$$
by definition of filtration.

Hence, $$ \{ \tau + s   \le t \} \in \mathcal{F}_{t}$$
and $\tau + s$ is stopping time.

Now suppose that  $t < s$. Thus $\{ \tau + s   \le t \} = \{ \tau  \le t -s \} = \varnothing \in  \mathcal{F}_{t}.$

Is there any questions?

Title: Stopping time + constant


问:
Proof verification that series $\sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{10^{\frac{1}{n+1}}} $ converges
Proof that the series converges

$$\sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{10^{1/(n+1)}} $$

My proof:
$$
\left(\frac{(-1)^{n+1}}{10^{1/(n+1)}}\right)^{1/n} = \frac{(-1)^{1+1/n}}{10^{1/(n^2+n)}}
$$
when n goes to infinity we get:

$$\frac{-1\cdot 1}{1} = -1 < 1 \implies\text{series is convergent}$$

After retry with absolute value:

$$\left|\frac{-1\cdot 1}{1}\right| = 1 \implies \text{we don't know if it converges}  $$

答:
First of all,
$10^{1/(n+1)}
=e^{\ln(10)/(n+1)}
\approx 1+\ln(10)/(n+1)
\lt 2$
for $n > \ln(10)$
so the fractions are between
$\frac12$ and $1$ for
$n > 3$.

Therefore the sum is neither
absolutely nor conditionally
convergent.


问:
Validity of manipulating infinite (Fourier) series?
I found a problem that goes like this:

Suppose $f\in C^1(\mathbb{R}/\mathbb{Z})$ and $\int_0^1 fdx=0$. Show that $\int_0^1 (f')^2dx\ge4\pi^2\int_0^1 f^2dx$.

My first thought was to use Fourier series, by writing
$$ f(x) = \sum_{n=1}^\infty \left(a_n\sin(2\pi nx)+b_n\cos(2\pi nx)\right) $$

Then we have:

$f^2(x) = \left(\sum_{n=1}^\infty \left(a_n\sin(2\pi nx)+b_n\cos(2\pi nx)\right) \right)^2$

$\int_0^1 f^2dx=\sum_{n=1}^\infty \frac{a_n^2+b_n^2}{2\pi n}$

$f'(x)=\sum_{n=1}^\infty \left(2\pi na_n\cos(2\pi nx)-2\pi nb_n\sin(2\pi nx)\right)$

$(f')^2(x)=\left(\sum_{n=1}^\infty \left(2\pi na_n\cos(2\pi nx)-2\pi nb_n\sin(2\pi nx)\right)\right)^2$

$\int_0^1 (f')^2dx=\sum_{n=1}^\infty 2\pi n(a_n^2+b_n^2)$

And then the inequality of the problem follows easily. My problem is I'm not sure if each step above is valid for the infinite series. I'm pretty sure the first two are fine since $f$ is continuously differentiable and the domain is compact, which implies the Fourier series converges uniformly. However, I'm not sure if differentiating term-by-term in step 3 is valid, and I have no idea how to justify the last two. Are these infinite series manipulations valid?

Finally, although I'm more interested in knowing whether the above approach is mathematically sound, I'm also interested to hear about other ways to solve the original problem.

答:
I would be nice if series $3$ converges uniformly; however, this is not always the case. Examples of continuous function with divergent Fourier series are not difficult to find. To tackle this, note that whether or not we can differentiate term by term (in classical sense), the Fourier series of $f'$ is always given by series $3$. Since $f'\in C^0,$ the Fourier series converges almost everywhere to $f'$. This is sufficient for us to perform Riemann or Lebesgue integral on $f'$, which is what we need to prove.

It is then a simple application of Plancherel theorem to go from 3 to 5. Or, if you want to square the series directly, then you can use Mertens' theorem (see also Cauchy product).

As for more ways of proving this, search for Poincaré inequality. The thing you prove generalises to higher dimensions. This is a very interesting topic.


问:
Precalculus in reverse?
In real analysis, I am aware of James Propp's "Real Analysis in Reverse" which does "naive reverse mathematics", showing the equivalence of various theorems of analysis to the completeness property of the real line.

Recently I've been thinking about the various geometric results that are typically covered in a precalculus course, such as:

The Pythagorean theorem

The law of cosines

The equivalence between algebraic and geometric dot product ($u\cdot v = \|u\|\|v\|\cos \theta$)

The formula for projections: $\operatorname{proj}_u(v) = \frac{v\cdot u}{\|u\|^2}u$

The sine and cosine angle addition formulas

The geometric interpretation of complex multiplication ("add the angles and multiply the lengths")

Determinant of a $2\times 2$ matrix as the signed area of the parallelogram spanned by its columns

...and so on. It seems to me that these results are all saying "basically the same thing" (whatever that means); in particular it's often possible to prove one of them using some of the others.

I am wondering if there is a textbook or paper that does something like "naive reverse mathematics of precalculus". What would the "base axioms" be in this setting? What is the relationship between these theorems? Why are they so closely related? Are there other "equivalent" theorems that are missing from this list? Are there any keywords I should be using in my search?

答:
From what I have seen here
(and I have contributed to this),
just about about any inequality
is a consequence of
that most elementary inequality,
Bernoulli's inequality.</p>

Title: Precalculus in reverse?


问:
Convolution with functions and two Dirac deltas
I have to calculate the following convolution $\mathcal{J}$:
$$
\mathcal{J} = f(\omega)\delta(\omega-\xi)\ast g(\omega)\delta(\omega-\zeta),
$$
where $f$ and $g$ are functions well behaved to do this operation and $\delta$ is the Dirac delta. According to the definition of convolution, we have
$$
\mathcal{J} = \int_{-\infty}^{\infty}f(\omega-\eta)\delta(\omega-\xi-\eta)g(\eta)\delta(\eta-\zeta)d\eta.
$$
However, the result of $\mathcal{J}$ depends on which $\delta$ I consider to do the integration. I look for answers on this site, and found out this question where it is used (what I think ) the following property:
$$
f(x)\delta(x-a)\ast g(x)\delta(x-b) = f(a)g(b)\delta(x-a-b).
$$
Is this a definition for such type of convolution? If not, how can I calculate $\mathcal{J}?$

答:
Forgive me for using notation I'm more used to. Instead of $\omega$, I'll use $x$ as the free variable and instead of $\delta(x-x_0)$, I'll just write $\delta_{x_0}$. The convolution you're asking for is then:
$$
  f \, \delta_\xi \; * \; g \, \delta_\zeta
$$

Solution using Fourier transform

Initially, I would approach this problem using the properties of Fourier transform. Specifically these two properties will be useful:
$$
  \def\R{\mathbb{R}}
  \def\F{\mathcal{F}}
  \def\e{\mathrm{e}}
  \def\i{\mathrm{i}}
  \def\d{\mathrm{d}}
  \F(fg) = \F(f) * \F(g) \\
  \F(f * g) = \F(f) \, \F(g)
$$
As it turns out, there's a much simpler solution (presented down below), but I'll keep it here just for completeness. Applying the rules to the distributions in question, we get:
$$
\begin{aligned}
  &\F( \; f \, \delta_\xi \; * \; g \, \delta_\zeta \; ) \\
  &= \F( f \, \delta_\xi) \; \F(g \, \delta_\zeta)
  = \Big( \F(f) * \F(\delta_\xi) \Big) \; \Big( \F(g) * \F(\delta_\zeta) \Big)
  = ( \tilde f * \e^{-2\pi\i\xi x} )( \tilde g * \e^{-2\pi\i\zeta x} ) \\
  &= \Bigg( \int \tilde f(y) \; \e^{-2\pi\i\xi \, (x-y)} \d{y} \Bigg)
  \Bigg( \int \tilde g(y) \; \e^{-2\pi\i\zeta \, (x-y)} \d{y} \Bigg) \\
  &= \e^{-2\pi\i\xi x} \;\Bigg( \int \tilde f(y) \; \e^{2\pi\i\xi y} \d{y} \Bigg) \;
  \e^{-2\pi\i\zeta x} \;\Bigg( \int \tilde g(y) \; \e^{2\pi\i\zeta y} \d{y} \Bigg) \\
  &= \e^{-2\pi\i\xi x} \; \F^{-1}(\tilde f)(\xi) \; \e^{-2\pi\i\zeta x} \; \F^{-1}(\tilde g)(\zeta) \\[5pt]
  &= \e^{-2\pi\i \, (\xi + \zeta) \, x} \; f(\xi) \; g(\zeta)
\end{aligned}
$$
Finally we will do the inverse Fourier transform:
$$
\begin{aligned}
  f \, \delta_\xi \; * \; g \, \delta_\zeta
  &\;=\;
  \F^{-1}(\; \e^{-2\pi\i \, (\xi + \zeta) \, x} \; f(\xi) \; g(\zeta) \;)
  \\[5pt]
  &\;=\;
  f(\xi) \; g(\zeta) \; \F^{-1}(\; \e^{-2\pi\i \, (\xi + \zeta) \, x} \;)
  \\[5pt]
  &\;=\;
  f(\xi) \; g(\zeta) \; \delta_{\xi + \zeta}
\end{aligned}
$$
Keep in mind that all the refinements I did were only formal, I didn't check whether all the expressions were well-defined. You should check that yourself.

Simpler solution using properties of delta

Alternatively, we can observe that $f \, \delta_\xi = f(\xi) \, \delta_\xi$ holds for any continuous function $f$ and that $\left\langle T * \delta_\xi, \; \varphi(x) \right\rangle = \left\langle T, \; \varphi(x - \xi) \right\rangle$ holds for any distribution $T$. Specifically, it is true that $\delta_\xi * \delta_\zeta = \delta_{\xi + \zeta}$. Then the solution can be found by application of these rules.
$$
  f \, \delta_\xi \; * \; g \, \delta_\zeta
  \;=\; f(\xi) \, \delta_\xi \; * \; g(\zeta) \, \delta_\zeta
  \;=\; f(\xi) \; g(\zeta) \; \delta_{\xi + \zeta}
$$


问:
Elements of the same order in the same conjugacy class that commute are contained in the same cyclic subgroup
I have a conjecture, that I would like to prove.

EDIT: My first idea was not true, as pointed out by Derek Holt.

Statement:
Let $G$ be a (finite) group and $m$ the maximal order of an element of $G$, such that all elements of maximal order are conjugated.

If two commuting elements $x,y \in G$ have order $m$, then $x^n=y$ for some $n \in \mathbb{N}$, i.e. $y$ is contained in the cyclic subgroup generated by $x$

I looked for some examples:

$(1,2,3)=(1,3,2)^2\in S_3$

$(1,2,3,4)=(1,4,3,2)^3 \in S_4$, but for non-commuting elements of this conjugacy-class it is not true, see: $(1,2,3,4)\not=(1,3,2,4)^n \forall n \in \mathbb{N}$.

Is this true in general? Does anyone know a proof?

Thanks in advance.

答:
Thanks again for your contribution!

If you are interested, I explain my motivation. In a paper(https://doi.org/10.1080/00029890.2019.1528826) it is proven that a arbitrary group $G$ with a finite number of elements of maximal order has bounded size. Namely: $|G|\leq\frac{mk^2}{\varphi(m)}$, where $m$ is the maximal order and $k$ the number of elements that have order $m$. I wanted to characterize for which groups $G$ the bound is sharp, i.e. $|G|=\frac{mk^2}{\varphi(m)}$. Using GAP I found all groups with this property up to order 1023 and was able to state a conjecture.
It is easy to see in the paper, that a group has the property only if all elements of maximal order are conjugated. So we need this as as a requirement.

Conjecture. Let $G$ be a group with $k<\infty$ elements of maximal order $m$, in which all elements of maximal order are conjugated. Then the following are equivalent.

i) $|G|=\frac{mk^2}{\varphi(m)}$

ii) $k=\varphi(m)$

iii) $G$ has a unique subgroup of order $m$

iv) $C_m \cong C_G(x)=C_G(y)\trianglelefteq G$ for all $x,y\in G$ with maximal order

I already proved the equivalence of ii), iii), iv) and ii) $\implies$ i). What I am missing is i) $\implies$ ii). I already proved, that i) implies, that all elements of maximal order commute, but I could not finish till now (this is where my assumption of this post would have helped).


问:
Second order homogeneous differential equations: why do repeated roots modify the solution set?
Consider the following differential equation with constant coefficients:
$$
y'' + 2ky' + k^2y = 0 \, .
$$
The auxiliary equation is
\begin{align}
m^2 + 2km + k^2 &= 0 \\[4pt]
(m+k)^2 &= 0 \\[4pt]
m &= -k \, .
\end{align}
This means that the solutions of the differential equation are $Ae^{-kx}+B\color{red}{x}e^{-kx}$, where $A$ and $B$ are arbitrary constants. The $x$-term highlighted in red is what confuses me. Generally speaking, if the roots of the auxiliary equation are $\alpha$ and $\beta$, then the solutions to the differential equation are $Ae^{\alpha x} + Be^{\beta x}$. However, if $\alpha=\beta$, then the solutions are not $Ae^{\alpha x} + Be^{\alpha x}$—or at least, this solution is incomplete. I can verify that $y=xe^{-kx}$ is a solution of the differential equation by plugging it back in to the equation, but I'm looking for a deeper reason for why this occurs.

答:
As the commenters point out, we know that there must be two independent solutions, even without knowing what they are. It is a sensible guess to try something like $xe^{-kx}$. An algorithmic way to find that this is the correct factor is to use variation of parameters.

Indeed, you may find it illuminating to transform the equation by making the substitution $y = e^{-kx}v$ and deriving the equation for $v$.

There are many direct ways of interpreting this change in the solution set. One way is by making small perturbations. Indeed, suppose we had solutions $k_\pm = -k \pm \varepsilon$. Then we get independent solutions $y = Ae^{(-k-\varepsilon)x} + Be^{(-k+\varepsilon)x} = e^{-kx}(Ae^{-\varepsilon x} + Be^{\varepsilon x})$. Then, with some initial data we have $$y = e^{-kx}\bigg(y(0) \cosh(\varepsilon x) + (y'(0) + ky(0)) \frac{\sinh(\varepsilon x)}{\varepsilon}\bigg) \to e^{-kx}\bigg(y(0) + (y'(0)+ky(0))x\bigg).$$

There are other ways. You can rewrite the equation in matrix form:

$$\frac{d}{dx}\pmatrix{y' \\ y} = \pmatrix{-2ky' - k^2 y  \\ y'} = \pmatrix{-2k & -k^2 \\ 1 & 0}\pmatrix{y' \\ y}$$

In the two-root case the matrix is diagonalisable, and the solutions given by the matrix exponential. The matrix above is not diagonalisable, and the $x$ factor arises from its Jordan Normal Form. There are details in this talk if you're interested.


问:
Making Sense of Independent Increments Proving The Martingale Property
I've got a question about a proof about using the independent increment property of a stochastic process when it comes to showing the Martingale Property of the said process.

Suppose $\{B_t\}_{t\geq0}$ is a standard Brownian motion with some $\sigma\in(0,\infty)$ sampled at the times $0=t_0<t_1<\cdots<t_n$. That is we are working in discret time. We are to show the Martingale property for this process holds, and that arguments goes as follows (in my textbook):

$$ \mathbb{E}[B_t \vert B_{t-1}, \ldots,B_0] = \mathbb{E}[B_t - B_{t-1} + B_{t-1}\vert B_{t-1}, \ldots,B_0] =^* \mathbb{E}[B_t-B_{t-1}]+ \mathbb{E}[B_{t-1} \vert B_{t-1}, \ldots,B_0] = B_{t-1}$$

My question is about the * part. The argument is that Brownian motion has indepdent increments. But how do we make sense of saying that $B_t-B_{t-1}$ is independent of every single one of the $B_{t-1}, \ldots, B_0$. Why can we just throw away the conditioning by claiming "independent increments"?

Note, I understand that the unconditional expectation is $0$ and that the other conditional will be $B_{t-1}$. My question is solely about how $\mathbb{E}[B_t-B_{t-1} \vert B_{t-1}, \ldots,B_0]=\mathbb{E}[B_t-B_{t-1}]$.

答:
Let us consider vector $(B_t, B_{t-1}, \ldots, B_0)$. It has normal distribution (a propery of Brownian motion). Consider the vector $Z = (B_t - B_{t-1}, B_{t-1}, B_{t-2}, \ldots, B_1, B_0)$. It's a linear transformation of $(B_t, B_{t-1}, \ldots, B_0)$ and hence is has normal distribution too. We know that $cov(B_{i}, B_j) = \min(i,j)$. Hence the covariance matrix $\Sigma$ of vector $Z$ is such that $\Sigma_{1i} = \Sigma_{i1} = 0$ for $i \ge 2$. We know the form of characteristic function of $Z$, because $Z = (Z_t, \ldots, Z_0)$ is normal. It's easy to see that $Ee^{i (\vec{s}, \vec{Z})} = Ee^{i s_t Z_t} \cdot Ee^{i (\sum_{k=0}^{t-1} s_k Z_k)}$. Hence $Z_t$ is independent of $(Z_{t-1}, \ldots, Z_0)$. Q.e.d.

Is there any questions?


问:
How do you determine the particular solution to a non-homogeneous DE by undetermined coefficients?
I am asked to solve

$y'' +2y' = 2x + 5 -e^{-2x}$

I can find the general solution easily, but the particular solution in this case is hard to find. Here's the answer. I don't know why they got $Ax^2 + Bx + Cxe^{-2x}$

I know the $x$ in $Cxe^{-2x}$ is due to the $e^{-2x}$ being found in the general solution, but I don't know where the $Ax^2 + Bx$ part comes from.

Shouldn't it be $Ax + B + Cxe^{-2x}$, instead?

答: It's Ax^2 and Bx, because there is a constant C in the general solution and because of Bx we can't have Ax, so it's Ax^2.</p>