问:
Holomorphic but not the uniform limit of polynomials
For any compact set $K\subset\mathbb C$ with $\mathbb C\setminus K$ disconnected, construct a holomorphic $f:K\to\mathbb C$ such that $f$ is not the uniform limit of polynomials.

Mergelyan’s theorem states that: if $K\subset\mathbb C$ is compact, $\mathbb C\setminus K$ is connected, then any continuous function $f:K\to\mathbb C$ can be approximated uniformly on $K$ by polynomials.

So I am looking for a counterexample when we allow $\mathbb C\setminus K$ to be disconnected, and require $f$ to be holomorphic.

I have tried to exploit the fact that polynomials always integrate around a loop to $0$, but this needn't be the case for holomorphic functions. However, I haven't been able to construct a concrete example.

答:
Pick $w \in U$ bounded component of $\mathbb C\setminus K$ which must exist since $K$ compact mean $\mathbb C\setminus K$ contains the exterior of a large closed disc so the unbounded component is unique. Let $0<d=\max_{z \in K}|z-w|$ which is finite since $K$ compact.

Then $f(z)=\frac{1}{z-w}$ is holomorphic in a neighborhood of $K$ and assuming there are $P_n \to f$ uniformly on $K$ we can pick $m$ st $|P_m(z)-f(z)|<\frac{1}{2d}, z \in K$

Then $|(z-w)P_m(z)-1| < \frac{|z-w|}{2d} \le \frac{1}{2}$ for all $z \in K$. But now $g(z)=(z-w)P_m(z)-1$ is a polynomial so an entire function and $\partial U \subset K$ by the defintion of $U$ hence restricting $g$ to $\bar U$ (which is compact since $U$ bounded - here we use the hypothesis), we get that it attains its maximum modulus on $K$ and by the above that maximum is less than $1/2$, while $g(w)=1$ and that is a contradiction!


问:
Showing that $x^5 - 5 a^4 x + a$ is irreducible.
A problem of finding a Galois group of $x^5 - 5 a^4 x + a$ appeared in a previous prelim at my school for $a \in \mathbb N$, and the only hard part seems like showing that the polynomial is irreducible, since it has 2 complex roots.

However, I do not know about any algebraic way to show that above is irreducible—other person from my school showed it by using rational root theorem and Rouche’s theorem.

So I was wondering if there is any way to show that above polynomial is irreducible in a purely algebraic way.

答:
Too long for a comment. For $a\in \Bbb{Z}$

If $a=0$ then it is reducible, if $a=\pm 1$ then it is irreducible $\bmod 3$.

If $a$ is not a 5th power then $f(x)=x^5+5a^4x+a$ is irreducible over $\Bbb{Q}_p$ whenever $5\nmid v_p(a)$, because $f(\gamma)=0$ gives that $v_5(\gamma)=v_5(a)/5$ so that $[\Bbb{Q}_p(\gamma):\Bbb{Q}_p]\ge 5$.

Otherwise $a= b^5$. Let $g(x)=a^{-1} f(xb)=x^5+5b^{16}x+1$, $h(x)=g(x-1)=x^5 - 5x^4 + 10x^3 - 10x^2 + 5xb^{16} + 5x- 5b^{16}$

If $5\nmid b$ then it is Eisenstein at $5$.

It remains to check the case $a=(5c)^5$.

ContentLicense: CC BY-SA 4.0


问:
Connected Sum of Riemannian Manifolds
As topological spaces, one can take connected sums of riemannian manifolds. Is there a way to give a Riemannian metric to the connected sum of two Riemannian Manifolds? If so is there a way to take this metric "canonical" in the sense that it extends the metrics of the manifolds used?

My intuiton tells me that when one takes a ball out of a manifold, the metric is not necessarily complete anymore. Is the metric on the connected sum complete?

答:
Completeness is not a problem. Removing an open ball out of a complete metric space, one obtains another complete metric space (manifold with boundary, in our situation). Then two manifolds with boundary are glued along the boundary. In the topological setting that would be the end of the story: the half-space charts on both sides of the cut stick together well enough for topological structure.

But if we want smooth gluing, one has to either change the metric on original manifolds or add a cylinder connecting the manifolds. Indeed, imagine glueing two $2$-spheres by removing small disks from each. The total curvature of the two spheres is almost $8\pi$, but the Gauss-Bonnet theorem says the connected sum must have total curvature $4\pi$, since  it is again a sphere. Therefore, negative curvature must appear somewhere.

One can argue as follows: pick a point $p\in M$ and two small radii $r<R$. The region between the spheres $S(p,r)$ and $S(p,R)$ is diffeomorphic to a cylinder (product of sphere with a line segment). Push forward the metric on $M$ to this cylinder. Then smoothly modify it (using partition of unity) so that near the end that corresponds to $S(p,r)$ it becomes the standard metric of the right circular cylinder. You can do the same on the other manifold $M'$. The ends corresponding to $S(p,r)$ and $S(p',r')$  glue together smoothly.

ContentLicense: CC BY-SA 3.0

Title: Connected Sum of Riemannian Manifolds


问:
Help with further simplification
I am attempting to further simplify 3^3^3^3^3^3^3 mod 1000. The period of 3^n is 100, which I have derived from doing the LCM of mod 8 and mod 125. Now I am stuck, and I need some guidance. How is doing 3^3^3^3^3^3^3 mod (the period) supposed to help me?</p>

答:
Now that you know that the period of $3^n\bmod{1000}$ is $100$, consider 3^3^3^3^3^3 mod $100$. The period of $3^n\bmod{100}$ is $20$, so consider 3^3^3^3^3 mod $20$. The period of $3^n$ mod $20$ is $4$, so consider 3^3^3^3 mod $4$. The period of $3^n$ mod $4$ is $2$, so hence, 3^3^3^3 mod $4$ is $3$.

And, $3^3\bmod{20}$ is $7$, and $3^7\bmod{100}$ is $87$, and so, the answer is $3^{87}\bmod{1000}$, which is easy to calculate via repeated squaring.

Title: Help with further simplification


问:
Is there a Graph theory way to solve this iteration problem
This thought came when i was trying to find the number of functions $f(x)$ from $\{1,2,3,4,5\}$ to $\{1,2,3,4,5\}$ that satisfy $f(f(x)) = f(f(f(x)))$ for all $x$ in $\{1,2,3,4,5\}$.  What i did was took cases when only one , two...etc  values of x such that all those are identity functions from that i got the answer as 756 which is correct.  But i was thinking if there was some kind of GT use can be done here  which can solve it more easily than case bash?

答:
The best I can give is a generating function solution, which gives a way to count such graphs automatically with a computer.

If you make a directed graph with an edge $x\to f(x)$ for each $x\in \{1,\dots,5\}$, and ignore any self loops $x\to x$, what remains is a rooted forest where each tree has depth at most $2$. If we let $F_{h,n}$ be the number of rooted forests on $n$ vertices with a depth of at most $h$, and $F_h(x)=\sum_{n\ge 0}F_{h,n}x^n$ be the exponential generating function, then
$$
F_h(x)=\exp(x F_{h-1}(x))
$$
This is because a rooted forest is given by partitioning the underlying set ($\exp$), choosing a root for each set $(x)$, and then putting a rooted forest structure with depth $h-1$ on the remaining elements of each set $F_{h-1}(x)$. In your case, we get
$$
F_2(x)=\exp(xF_1(x))=\exp(x\exp(xF_0(x)))=\exp(x\exp (x\exp(x)))
$$
Therefore, it follows that
$$
F_{2,n}=n![x^n]\exp(x\exp (x\exp(x)))
$$
where $[x^n]f(x)$ is the coefficient of $x^n$ in the power series $f(x)$. For example, the Mathematica code

```<code>5! * SeriesCoefficent[ Exp[x Exp[x Exp[x]]], {x, 0, 5}]
</code>```

verifies your count of $756$, as shown on Wolfram Alpha.

Body:
The best I can give is a generating function solution, which gives a way to count such graphs automatically with a computer.

If you make a directed graph with an edge $x\to f(x)$ for each $x\in \{1,\dots,5\}$, and ignore any self loops $x\to x$, what remains is a rooted forest where each tree has depth at most $2$. If we let $F_{h,n}$ be the number of rooted forests on $n$ vertices with a depth of at most $h$, and $F_h(x)=\sum_{n\ge 0}F_{h,n}x^n$ be the exponential generating function, then
$$
F_h(x)=\exp(x F_{h-1}(x))
$$
This is because a rooted forest is given by partitioning the underlying set ($\exp$), choosing a root for each set $(x)$, and then putting a rooted forest structure with depth $h-1$ on the remaining elements of each set $F_{h-1}(x)$. In your case, we get
$$
F_2(x)=\exp(xF_1(x))=\exp(x\exp(xF_0(x)))=\exp(x\exp (x\exp(x)))
$$
Therefore, it follows that
$$
F_{2,n}=n![x^n]\exp(x\exp (x\exp(x)))
$$
where $[x^n]f(x)$ is the coefficient of $x^n$ in the power series $f(x)$. For example, the Mathematica code

5! * SeriesCoefficent[ Exp[x Exp[x Exp[x]]], {x, 0, 5}]

verifies your count of $756$, as shown on Wolfram Alpha.


问:
First order logic: What is this formula saying?

$$\varphi(u, v) := \lnot E(u, v) \land \forall w (E(u, w) \lor E(v, w) \lor u = w \lor v = w)$$

If we interpret $E$ as an edge relation and $u, v$ and $w$ as vertices, I think the formula is looking for two vertices (two free variables) which do not share an edge and form a dominating set, but the last part ($..\lor u = w \lor v = w$) is throwing me off and I'm not sure why it's there or how to understand it. If we were indeed looking for a dominating set, shouldn't it be ( $..\land u \neq w \land v \neq w$)? Would really appreciate your thoughts.

答:
Remember $\lnot \phi \Rightarrow \psi$ is equivalent to $\phi \lor \psi$. Here take $\psi \equiv E(u, w) \lor E(v, w)$ and $\phi \equiv u = w \lor v = w$. I think you were expecting to see $\lnot \phi \Rightarrow \psi$, i.e., $(u \neq w \land v \neq w) \Rightarrow (E(u, w) \lor E(v, w))$, but what you were given was the equivalent formula $\psi \lor \phi$, i.e., $E(u, w) \lor E(v, w) \lor u = w \lor v = w$.

(And this does indeed mean that $\{u,  v\}$ is a dominating set.)


问:
Determining a line in $\Bbb{R}^4$ that meets a given line and plane

I have the equation for the first line in $\Bbb{R}^4$
$$q : [1,3,1,1] + t(-1,2,-1,1)$$ where $t$ is a parameter. Then I have my plane given by these two equations
$$\rho : x_1 + x_2 - x_3 -3x_4 = -2 \quad\text{and}\quad 2x_1 + x_2 -2x_4 = 5$$ I have to construct a line $p$ such that it intersects both the plane and the line $q$ and I know that it has the form of $p: A + s(1,1,2,0)$, where $A$ is some point.

Now I would know how to solve this if I was given the point instead of the vector but in this situation I have no idea.

Does anyone have any tips on how to proceed? Thank you for your help in advance.

答:
If we take any line with direction vector which is NOT parallel to the plane, then it will intersect the plane at some point. Try to visualize this, if you are having a hard time then visualize it in 3 dimensions instead. Now we simply need to ensure that this line will intersect q. To ensure this we can simply take any point on q and incorporate it into our line. The easiest point to take would be [1,3,1,1]. Now we just need to add to this any direction vector, not parallel to the plane, with a parameter.

In short, you already know how to solve this question if you were given a point instead of a line. So just do that! Take any point on q and solve as you usually would. Be sure to double check that it works


问:
Convergence of $1+\frac{1}{2}\frac{1}{3}+\frac{1\cdot 3}{2\cdot 4}\frac{1}{5}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{1}{7}+\cdots$
Is it possible to test the convergence of $1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$ by Gauss test?

If I remove the first term I can see $\dfrac{u_n}{u_{n+1}}=\dfrac{(2n+2)(2n+3)}{(2n+1)^2}
\\=\dfrac{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{3}{2n}\right)}{\left(1+\dfrac{1}{2n}\right)^2}
\\={\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{3}{2n}\right)}{\left(1+\dfrac{1}{2n}\right)^{-2}}\\=\left(1+\dfrac{5}{2n}+\dfrac{3}{2n^2}\right)\left(1-\dfrac{1}{n}\ldots\right)\\=1+\dfrac{3}{2n}+O\left(\dfrac{1}{n^2}\right)$

So the series is convergent.

Is it a correcct attempt?

答:
I'm not sure about your result; the ratio test is inconclusive anyway.  The way I look at this, I express each term as

$$b_k = \frac{a_k}{2 k+1}$$

where

$$a_k = \frac{1}{2^{2 k}} \binom{2 k}{k}$$

Using, e.g., Stirling's formula, you may show that

$$a_k \sim \frac{1}{\sqrt{\pi k}} \quad (k \to \infty)$$

so that the ratio test on the coefficients provides

$$\frac{b_{k+1}}{b_k} \sim \left(1+\frac{1}{k}\right)^{-1/2} \left (1+\frac{2}{2 k+1}\right) \sim 1+\frac{1}{2 k}\quad (k \to \infty)$$

The limit test is inconclusive, as the limit if the ratio is $1$.  However, we may see that the series converges by the comparison test because the terms in the sum behave as

$$b_k \sim \frac{1}{2 \sqrt{\pi}} k^{-3/2} \quad (k \to \infty)$$

That said, we know that the series converges to $\pi/2$ as follows: consider the generating function

$$f(x) = \sum_{k=0}^{\infty} a_k \frac{x^{2 k+1}}{2 k+1}$$

Then for all $x$ within the radius of convergence of the series on the right, we may differentiate to get

$$f'(x) = \sum_{k=0}^{\infty} a_k \, x^{2 k} = \frac{1}{\sqrt{1-x^2}}$$

Therefore

$$f(x) = \arcsin{x}$$

and the series has value $f(1) = \pi/2$.


问:
Intertwining relations and operator decomposability
The following proposition is made in a functional analysis paper without proof. I assume that the argument is a very simple one-liner, but I can't seem to come up with it. Any help would be appreciated.

Consider separable Hilbert spaces $\mathcal{H}$, $\mathcal{K}$ and $\mathcal{K}_1$, where the dimension of $\mathcal{K}_1$ is not lower than the dimension $\mathcal{K}$. If an isometry $U: \mathcal{H} \otimes \mathcal{K} \to \mathcal{H} \otimes \mathcal{K}_1$ satisfies the following relation
$$
 U \left( a \otimes 1_K \right) = \left( a \otimes 1_{{K}_1}\right)U,
$$
for all bounded operators $a \in \mathcal{B}(\mathcal{H})$, it is of the form $U = 1 \otimes U_1$ with $U_1: \mathcal{K} \to \mathcal{K}_1$ an isometry.

答:
Let $E$,  $F$,  and $G$  be
orthonormal bases for $H$, $K$,  and $K_1$,  respectively.

Fix any vector $e_0$ in $E$ and consider the operators
$$
  W:\xi \in K\mapsto e_0\otimes \xi \in H\otimes K,
  $$
and
$$
  W_1:\eta \in K_1\mapsto e_0\otimes \eta \in H\otimes K_1.
  $$
Finally, put $U_1=W_1^*UW$, so that $U_1$ is an operator from $K$ to $K_1$.

We then  claim that $U=1\otimes U_1$.

To see this
notice that, given $e_1,e_2\in E$, $f\in  F$, and $g\in G$, we have that
$$
  \big \langle (1\otimes U_1)(e_1\otimes f),e_2\otimes g\big \rangle  =
  \big \langle e_1\otimes U_1(f),e_2\otimes g\big \rangle ,
  \tag 1
  $$
which clearly vanishes when $e_1\neq e_2$. Otherwise we have that (1) equals
$$
  \big \langle U_1(f),g\big \rangle  =
  \big \langle UW(f),W_1(g)\big \rangle  =
  \big \langle U(e_0\otimes f),e_0\otimes g\big \rangle .
  $$

In order to prove the claim it is enough to show that
$\big \langle U(e_1\otimes f), e_2\otimes g\big \rangle $ coincides with (1).

Assuming first that $e_1=e_2$, let $\sigma $ be any permutation of $E$
such that $\sigma (e_0)=e_1$, and consider the unitary operator $a$ on $H$ determined by its values on $E$ by
$$
  a(e) = \sigma (e),\quad\forall e\in  E.
  $$
Then,
$$
  \big \langle U(e_1\otimes f), e_2\otimes g\big \rangle  =
  \big \langle (a\otimes 1)^{-1} U(a\otimes 1)(e_1\otimes f), e_2\otimes g\big \rangle  =$$$$=
  \big \langle U(\sigma (e_1)\otimes f), \sigma (e_2)\otimes g\big \rangle  =
  \big \langle U(e_0\otimes f), e_0\otimes g\big \rangle .
  $$

In case $e_1\neq e_2$, let $a$ be the orthogonal projection from $H$ to ${\mathbb C}e_1$, so
$$
  \big \langle U(e_1\otimes f), e_2\otimes g\big \rangle  =
  \big \langle U(p\otimes 1)(e_1\otimes f), e_2\otimes g\big \rangle  = $$ $$ =
  \big \langle (p\otimes 1)U(e_1\otimes f), e_2\otimes g\big \rangle  =
  \big \langle U(e_1\otimes f), (p\otimes 1)(e_2\otimes g)\big \rangle  = 0.
  $$

This concludes the proof.


答:
I'll give a strategy that generalizes using generating functions. As Mike Earnest points out, we want to count rooted labelled forests of size $5$ with depth at most $2$. Let's find the EGF of such forests of size $n$.

A tree is a root with a set of trees attached. Let $\mathcal T^{(2)}$ be the class of trees of depth at most $2$, and let $\mathcal Z$ denote a node. Then using the symbolic method we have
$$
\mathcal T^{(2)} = \mathcal Z\star SET(\mathcal Z\star SET(\mathcal Z))
$$
(Each use of $SET$ correspond to a possible layer). A forest is just a set of trees, so we get the following EGF
$$
\mathcal F^{(2)} = SET(\mathcal T^{(2)})
\implies F^{(2)}(z) = \exp(z\exp(z\exp(z)))
$$
The number of these forests that have $n$ nodes is then given by $F_n^{(2)} = n![z^n]F^{(2)}(z)$, where $[z^n]$ means the coefficient at $z^n$ in the series expansion of the EGF. We find
$$ \begin{split}
F^{(2)}(z) &= \exp(z\exp(z\exp(z))) \\
&= \sum_{m\ge0}\frac{z^me^{mze^z}}{m!}\\
&= \sum_{m\ge0}\frac{z^m}{m!} \sum_{k\ge0}\frac{m^kz^ke^{kz}}{k!}\\
&= \sum_{m\ge0}\frac{z^m}{m!} \sum_{k\ge0}\frac{m^kz^k}{k!}  \sum_{j\ge0}\frac{k^jz^j}{j!}\\
&= \sum_{m,k,j\ge0}\frac{m^kk^j}{m!k!j!}z^{m+k+j}\\
&= \sum_{n\ge0} \left(\sum_{m=0}^n \sum_{k=0}^{n-m} \frac{m^kk^{n-m-k}}{m!k!(n-m-k)!}\right)z^n
\end{split}$$
So our final answer is
$$
F_n^{(2)} = n! \sum_{m=0}^n \sum_{k=0}^{n-m} \frac{m^kk^{n-m-k}}{m!k!(n-m-k)!}
$$
and we can verify that $F_5^{(2)}$ is indeed $756$ using this formula. You can find the sequence at A949 on OEIS. I don't know if we can do much more than this, other than cleaning up the edge cases and perhaps rewriting with a trinomial coefficient:
$$
F_n^{(2)} = 1 + \sum_{m=1}^{n-1} \sum_{k=1}^{n-m}\binom{n}{m,k,n-m-k} m^kk^{n-m-k}
$$

Title: Is there a Graph theory way to solve this iteration problem

Body: This thought came when i was trying to find the number of functions $f(x)$ from $\{1,2,3,4,5\}$ to $\{1,2,3,4,5\}$ that satisfy $f(f(x)) = f(f(f(x)))$ for all $x$ in $\{1,2,3,4,5\}$.  What i did was took cases when only one , two...etc  values of x such that all those are identity functions from that i got the answer as 756 which is correct.  But i was thinking if there was some kind of GT use can be done here  which can solve it more easily than case bash?


问:
Is simplification of $(t^2)(t^2-1^2)(t^2-2^2)\ldots(t^2-a^2) \bmod (2t-1)$ possible?
Is it possible to further simplify or reduce the following equations,

$$(t^2)(t^2-1^2)(t^2-2^2)(t^2-3^2)\ldots(t^2-a^2) \bmod (2t-1)$$
Or
$$(t-a) \ldots (t-3)(t-2)(t-1)(t^2)(t+1)(t+2)(t+3) \ldots (t+a) \bmod (2t-1)$$

assuming $a,t \in \mathbb Z_+$ and $a < t-1$

答: Assuming that $a,t \in \mathbb Z_+$ and $1\le a \le t-1$ inclusive, then the term $t^2-(t-1)^2=2t-1\equiv 0 \bmod (2t-1)$, and the product $\equiv 0 \bmod (2t-1)$


问:
How is this proof that $a+0=a$ correct?
I found this in the 4th edition of Spivak's Calculus (which I have just begun reading) on page 4:

If $a+x=a$ then $(-a)+(a+x)=(-a)+a=0$.

Hence $((-a) + a) + x = 0$
[from $(a + b) + c = a + (b + c)$ and $(-a) + a = 0$, both of them had been previously mentioned]

Hnce $0 + x = 0$. Finally $x = 0$.

My problem is on the last line; we're proving that $a + 0 = a$, but aren't we using the fact that $0 + a = 0$, to conclude from $0 + x = 0$ that $x = 0$?

答:
I assume you're working in a field, as I think Spivak deals with them first.

The field axioms are the defining properties of a field and its operations. One of the axioms states that in a field $\mathbb{F}$, there exists some element in the field (which we call $0$, the "additive identity") such that for every element $a\in\mathbb{F}$,

$$a+0=0+a=a.$$

Axioms are something that you simply state as true without proof. All fields must satisfy the axioms - if it doesn't then it's not a field (and we don't really care about it). In this case, we define $0$ to be the additive identity, and because we are working in a field we know it must exist (the axioms say so).

In the proof you are referring to, we are not proving that $a+0=a$. It would be pointless to prove this, as it is true by definition. What we are proving is that $0$ is unique. That is, the only element in $\mathbb{F}$ that is an additive identity is the $0$ that's defined in the axiom. To do this, we first suppose that $x\in\mathbb{F}$ satisfies $a+x=a$ for all $a\in\mathbb{F}$. With a little bit of work (and using the other field axioms), we show that this implies $x=0$. Basically,

$$a+x=a\implies x=0.$$

This means that $0$ is the only element that is an additive identity, and we are done.

Title: How is this proof that $a+0=a$ correct?


问:
Using dimensional analysis to ‘get’ planck’s ${\lambda}$ equation with avogadro’s constant
This is for when you calculate bond energy in: ${j\cdot mol(n)^{-1}}$ and want to use planck’s equation for photon energy to calculate ${\lambda}$

Planck’s equation for photon energy:
$${\lambda=\frac{h\cdot c}{E}}$$

Planck’s equation for photon energy with avogadro’s constant:

$${\lambda=\frac{h\cdot c\cdot N_A}{E}}$$

I don’t understand how ${N_A}$ (${\frac{1}{n}}$) is at the top as:

${n=\frac{g}{g\cdot n^{-1}}\Rightarrow{n=\frac{1}{n}}}$

Energy in this case:

${E=\frac{j}{mol\ or\ n \ (1/n)}}$

I can’t rearrange ${1/n}$ to get it to the to the top in order to get rid of the ${n}$ at the bottom so that:

${E=j}$

Can someone show me verbosely with dimensional analysis how to do this?

Units:

${\lambda=m}$

${E=j}$

${h=js}$

${c=ms^{-1}}$

${N_A=n^{-1}}$

Dimensional analysis:

${\lambda=js\cdot ms^{-1}\cdot j^{-1}(1^{-1}n^{-2})}$

${\lambda=\frac{j\cdot s\cdot m}{j\cdot s(\frac{1}{n})}}\Rightarrow{\lambda=\frac{j\cdot s\cdot m}{\dfrac {j\cdot s}{n}}}\Rightarrow{\lambda=\frac{m}{\dfrac{1}{n}}}$

Then I can’t get ${\lambda=m}$

Is this correct, and if not then why?:

${\frac{j}{mol(1/n)}=\frac{j}{\dfrac{1}{n}}}$

Bonus question:

I have been using dimensional analysis to ‘get’ equations before I even knew that word, as my memory is bad and I couldn’t remember equations, I used to think it was a trick

Are there more/better ways to ‘get’ equations and cases where dimensional analysis won’t work?

I really need help here and would appreciate it very much

答:
$\require{cancel}$

Let's assume for the following identity to hold:

$$\star:\left\{\lambda=\frac{hc N_A}{E}\right\}$$

Which expressed in units of measurement is:

$$\lambda=\frac{hc N_A}{E}=\frac{[\text J\cdot\text s][\text m\cdot \text s^{-1}][\text{mol}^{-1}]}{[\text J\cdot\text{mol}^{-1}]}=[\text m]$$

where $$E=[\text{J}\cdot\text{mol}^{-1}]$$

Where the 'naked' variables are quantities, and the square bracket-ed variables are units of measurements.

Let's divide both sides of the equation$(\star)$ by $\color{blue}{N_A}$ as per the rules of algebra.

$$\frac{\lambda}{\color{blue}{N_A}}=\frac{hc {N_A}}{E{\color{blue}{N_A}}}$$

Then we can simplify by $N_A$

$$\frac{\lambda}{\color{blue}{N_A}}=\frac{hc \cancel{N_A}}{E\cancel{\color{blue}{N_A}}} \Leftrightarrow \frac{\overbrace{[\text J\cdot\text s]}^h   \overbrace{[\text m\cdot \text s^{-1}]}^c    \overbrace{\cancel{[\text{mol}^{-1}]}}^{N_A}}{\underbrace{[\text J\cdot\text{mol}^{-1}]}_{E}    \underbrace{\cancel{[\text{mol}^{-1}]}}_{\color{blue}{N_A}}}=\frac{[\text m]}{[\text{mol}^{-1}]}$$


问:
Periodic functions summation order
It can be shown, if I am not mistaken, that for $g(n) \geq n,\; n \in \mathbb{N}$ holds:
\begin{equation}
    F(n,g(n)) = \sum_{d=1}^{n}\sum_{k=1}^{g(n)} f(k,d) \sim \sum_{d=1}^{n} \frac{g(n)}{d}\sum_{k=1}^{d} f(k,d)
\end{equation}
where $ f(k, d) \geq 0 $ is any nonnegative periodic function of the argument $k$ and period $d$, for example: $k \bmod d$; $|\sin(\frac{d k}{\pi})|$. Similarity sign means the functions are in the same order.

Question: Does the next expression follow directly?:
\begin{equation}\label{eq2}
    F(n,g(n)+1) - F(n,g(n)) \sim \sum_{d=1}^{n} \frac{1}{d}\sum_{k=1}^{d} f(k,d)
\end{equation}
If not, how can prove or disprove it?

答:
No, this does not follow from the previous expression, because in general, $f
_1 \sim g_1$ and $f_2 \sim g_2$ does not imply $f_2 - f_1 \sim g_2 - g_1$.  The $\sim$ relation can hide differences that can have quite large variance, for instance:

$$ n^2 \sim n^2, \qquad n^2 + n \sim n^2 + 1,$$
but $ n \not \sim 1$.

In your specific case, the LHS admits a trivial
simplification:

$$F(n,g(n)+1) - F(n,g(n)) = \sum_{d=1}^n f(g(n)+1, d).$$

Whether this is of the same order as the sum of the first $d$ average values of $f(\cdot,d)$ depends somewhat on your quantifiers (can the
constants depend on $f$ and $g$ or are they uniform in one or both?).  But it does not seem true in a strict sense as one can choose $f, g$ so that LHS is exactly 0 for all $n$ but the RHS is non-zero for any non-trivial $f$.

Title: Periodic functions summation order


问:
Energy norm. Why is it called that way?
Let $\Omega$ be an open subset of $\mathbb{R}^n$. The following

$$\lVert u \rVert_{1, 2}^2=\int_{\Omega} \lvert u(x)\rvert^2\, dx + \int_{\Omega} \lvert \nabla u(x)\rvert^2\, dx$$

defines a norm on $H^1(\Omega)$ space, that is sometimes called energy norm.

I don't feel easy with the physical meaning this name suggests. In particular, I see two non-homogeneous quantities, $\lvert u(x)\rvert^2$ and $\lvert \nabla u(x)\rvert^2$, being summed together. How can this be physically consistent?

Maybe some example could help me here. Thank you.

答:
To expand on my comment (see e.g. http://online.itp.ucsb.edu/online/lnotes/balents/node10.html):

The expression which you give can be interpreted as the energy of a $n$-dimensional elastic manifold being elongated in the $n+1$ dimension (e.g. for $n=2$, membrane in three dimension); $u$ is the displacement field.

Let me put back the units
$$E[u]= \frac{a}{2}\int_{\Omega} \lvert u(x)\rvert^2\, dx + \frac{b}{2} \int_{\Omega} \lvert \nabla u(x)\rvert^2\, dx.$$ The first term tries to bring the manifold back to equilibrium (with $u=0$), the second term penalizes fast changes in the displacement. The energy is not homogenous and involves a characteristic length scale
$$\ell_\text{char} = \sqrt{\frac{b}{a}}.$$
This is the scale over which the manifold returns back to equilibrium (in space) if elongated at some point. With $b=0$, the manifold would return immediately, you elongate it at some point and infinitely close the manifold is back at $u=0$. With $a=0$ the manifold would never return to $u=0$. Only the competition between $a$ and $b$ leads to the physics which we expect for elastic manifold. This competition is intimately related to the fact that there is a characteristic length scale appearing.

It is important that physical laws are not homogenous, in order to have characteristic length scales (like $\ell$ in your example, the Bohr radius for the hydrogen problem, $\sqrt{\hbar/m\omega}$ for the quantum harmonic oscillator, ...). The energy of systems only become scale invariant in the vicinity of second order phase transitions. This is a strong condition on energy functionals to the extend that people classify all possible second order phase transitions.

Title: Energy norm. Why is it called that way?


问:
If fundamental group of a compact path connected space is finite then its universal cover is compact
Let $X$ be a compact and path connected space and let $p:\tilde X\rightarrow X$ be its universal cover. I can show that if $\tilde X$ is compact then $\pi_{1}(X)$ is finite:

$\pi_{1}(X, x_{0})$ acts on $p^{-1}(x_{0})$  via $[\gamma].y=\tilde\gamma(1)$ where $\tilde\gamma$ is the unique lifting strating at y. Another observation is that
Stab(y)={$[\gamma]$$\in\pi_{1}(X, x_{0})$ : $\tilde\gamma$ is a closed loop}=$p_{*}(\pi_{1}(\tilde X, \tilde x_{0}))$ where $p_{*}$ is the induced homomorphism of $p$. Since this action is transitive,  there is a bijection between  $p^{-1}(x_{0})$ and $\pi_{1}(X, x_{0})$/$p_{*}(\pi_{1}(\tilde X, \tilde x_{0}))$. In our case $\tilde X$ is simply connected. So $p^{-1}(x_{0})$ and $\pi_{1}(X, x_{0})$ has the same cardinality. But $p^{-1}(x_{0})$ is closed and discrete subspace of compact space $\tilde X$. so it is finite, as desired.

But I couldn't show the converse. I am new in algebraic topology and I follow Hatcher's book chapter 1. Can any one give a hint or answer? Thanks!

答:
I find the following a useful way of thinking about it.

We may as well prove the statement that a covering map with finite fibers is a proper map (this means that the pre-image of any compact set is compact). (This argument is nice because it easily generalizes to prove the statement that a locally trivial fiber bundle with compact fibers is proper).

First note that when the covering map is just the projection $U \times F \rightarrow U$ then this map is proper because the preimage of a compact set $K \subset U$ is just $K \times F$ which is compact (being the product of compact spaces). In general, suppose that $\widetilde{X} \rightarrow X$ is a cover with finite fiber, $F$, then we may take a cover $\{U_\alpha\}$ of $X$ so that the restriction of the map to the cover just looks like $U_\alpha \times F \rightarrow U_\alpha$, and we've already seen that this is a proper map.

It is an easy exercise to see that properness (as I've defined it above) is local on the target, i.e. that the map $\widetilde{X} \rightarrow X$ is proper if and only if there is a cover of $X$ where all the restrictions are proper. (Indeed, any compact subset of $X$ is covered by finitely many $U_\alpha$, and so the preimage of $X$ is the union of the preimages of those specific $X \cap U_\alpha$. A finite union of compact sets is compact, QED.)

It follows that $\widetilde{X} \rightarrow X$ is a proper map. But now, if $X$ is compact, then the preimage of $X$, which is $\widetilde{X}$, must be compact, which was to be shown.


问:
Some (trivial?) doubts on the proof of chromatic number of any planar graph is at most 6
I am trying to show that chromatic number of any planar graph is at most 6. This is a weaker statement of the Four-Colour Theorem.

I have a vague idea about the proof but not sure how to convince myself that each statement of the proof really implies the next ones.

I used the fact that any simple planar graph has a vertex of degree at most 5. We then remove the vertex, then the remaining graph can be coloured with at most 5 colours (I am not sure on this part). If we add back the vertex, the graph can be coloured by at most 6 colours.

It seems to me that I am a bit confused how removing a vertex of degree at most 5 really does the job. Do we need to use induction somewhere? And how are we supposed to use induction?

I think about it like this: The fact that I used only says that a simple planar graph has a vertex of degree at most 5. But it may have a vertex of degree 6 or above right? The statement does not say all vertices have degree at most 5. Let say it has one vertex of degree 100. Then how can the chromatic number be at most 6?

I am still not very confident in proving graph theoretical statements.

Thanks.

答:
Kempe’s proof is actually fairly straightforward! We give it here.

Proof. We proceed by contradiction. Assume not: that there are planar graphs on finitely many vertices that need at least 5 colors to be colored properly. Consequently, there must
be some smallest planar graph G, in terms of the number of its vertices, that needs at
least five colors to color its vertices! Pick such a graph G. Notice that if we remove any
vertex v from G, we have a graph on a smaller number of vertices than G. Consequently, the graph G \ {v} can be colored with four colors!
Let v be the vertex in G with degree at most 5. Delete v from G: this leaves us a graph that we can four-color. Do so.


问:
Spliting Field over $\mathbb{F}_3$
How to find the splitting field of $f(x)=x^3-x+1$ and $g(x)=x^3-x-1$ over $\mathbb{F}_3$ and how to construct a isomorphism between them?</p>

答:
The polynomials $x^3-x-1$ and $x^3-x+1$ are irreducible over $\mathbb F_3$ as pointed out
in the comment by @JyrkiLahtonen.

Now suppose that $\alpha$ is a root of $x^3-x-1$,
that is, $\alpha^3 = \alpha+1$.
Then the set of $27$ elements of the form $a_0 + a_1\alpha + a_2\alpha^2$,
$a_i \in \mathbb F_3$ is a three-dimensional vector space over $\mathbb F_3$.
Also, the set is closed under multiplication since the product of two elements,
being a polynomial of degree at most $4$ can be expressed as a polynomial
of degree at most $2$ by setting $\alpha^3 = \alpha+1$ and $\alpha^4 = \alpha^2+\alpha$
and collecting coefficients.  The proof of the
existence of unique multiplicative inverses takes
a little longer, but can be done, and the end result is that these $27$ elements
constitute a field of $27$ elements.

Now, $\alpha^3 -\alpha = 1 \Rightarrow -(\alpha^3 -\alpha)=-1
\Rightarrow (-\alpha)^3 -(-\alpha) = -1$, that is, $-\alpha$ is a
a root of the other polynomial $x^x-x+1$. Writing $\beta$ for $-\alpha$,
we have, as before, that the elements $b_0 + b_1\beta+b_2\beta^2$, $b_i \in \mathbb F_3$
constitute a field of $27$ elements. But these two different-seeming
fields are in fact the same field since for all choices of $3$
elements $a_0,a_1,a_2 \in\mathbb F_3$, the element
$a_0 + a_1\alpha + a_2\alpha^2$ in the first field is the same as
the element $a_0 -a_1 \beta + a_2\beta^2$ in the second field:
$$a_0 + a_1\alpha + a_2\alpha^2 = a_0 -a_1(-\alpha) + a_2(-\alpha)^2
= a_0 -a_1 \beta + a_2\beta^2$$

Title: Spliting Field over $\mathbb{F}_3$


问:
Section 4.3 Page 26 in Loring Tu's Differential Geometry
On Section 4.3, Page 26 in Loring Tu's Differential Geometry:

The underlined part is what I cannot understand. Why is $\dfrac{dV}{dt}$ in general not defined? What does "a canonical frame of vector fields" exactly mean?

I got confused at this point, and I hope someone could explain it explicitly for me.

答:
A frame field is a collection of vector fields that form a basis of the tangent space at every point of your manifold. These need not exist for an arbitrary manifold. So, you may not be able to represent $V$ as given, hence, you may not be able to define the derivative in that manner, either.

A smooth vector bundle has a smooth frame if and only if the vector bundle is trivial. Here is another useful situation, where you can get an orthonormal one (summary: on a surface with a Riemannian metric, you get it iff the surface is orientable and admits a nowhere vanishing vector field): Existence of orthonormal frame.


问:
Group Actions of $S_n$ and $O(n)$
I have been reading up a bit on group actions and whether they are faithful, free and transitive. A question I found states:

Consider the following group actions

1) The symmetric group $S_n$ acting on an n-element set.

2) The orthogonal group $O(n)$ acting on $\mathbb{R}^n$.

3) The orthogonal group $O(n)$ acting on the $(n-1)$-sphere $S^{n-1}$.

Which of these actions is faithful, transitive or free and what are the group orbits?

An explicit action is not specified, but I assumed that we could take the 'obvious' action: 1) I took $S_n$ as permuting the elements of the set and for 2) and 3) I took $O(n)$ as rotations. Is this correct, or can groups only act on a set in one way anyway?

Going with these actions, I thought that 1) would be faithful as the only permutation which leaves a set unchanged is the identity permutation; transitive as you can always find a bijection to send one permutation to another; and free as the only bijection which leaves a permutation unchanged is the identity permutation. The orbit would be all possible permutations of $n$ elements. My problem with this though is that my reasonings for faithful and free seem identical, so I don't think my train of thinking is correct.

For 2) and 3), I also don't see how what they are acting on seems to affect the characteristics of the group action. What kind of things do I need to think about?

答:
Let us remember the definition of faithful, transitive and free:

The action of a group $G$ on a set $X$ is called:

Transitive: if $X$ is non-empty and if for any $x,y\in X$ there exists $g\in G$ such that $g\cdot x = y$.

Faithful (or effective): if for any two distinct $g,h\in G$ there exists $x\in X$ such that $g\cdot x \neq h\cdot x$; or equivalently, if for any $g\neq e\in G$ there exists $x\in X$ such that $g\cdot x \neq x$. Intuitively, in a faithful group action, different elements of $G$ induce different permutations of $X$.

Free: if, given $g,h\in G$, the existence of  $x\in X$ with $g\cdot x = h\cdot x$ implies $g = h$. Equivalently: if $g$ is a group element and there exists an $x\in X$ with $g\cdot x = x$ (that is, if $g$ has at least one fixed point), then $g$ is the identity.

Now we can determine your three group actions:

The action of $S_n$ on a set of $n$-elements is as you showed transitive. It is also faithful as you explained, but it is not free, because there are elements on $S_n$ that do have fixed points that are not the identity, for example $1\to 1, 2\to 3, 3\to 4, n-1\to n, n\to 2$. Since the action is transitive the group of orbits is the group of one element.

The action of $O(n)$ on $S^{n-1}$, it is transitive, since there is always a rotation that can send any point of $S^{n-1}$ onto the north pole. For $n>2$, It is not free since for any point on $S^{n-1}$ there are many rotations different from the identity, whose axis of rotation is the line that passes through that point and the origin,  and all of these fix that point. It is not faithful neither since any element of $O(n)$ that is not the identity moves $S^{n-1}$, therefore there exists a point on $S^{n-1}$ that is not fixed by the action. Since the action is transitive the group of orbits is the group of one element. The case $n=2$ the action it is transitive, free but not faithful.

The action of $O(n)$ on $\mathbb{R}^n$ is not free since every element of $O(n)$ fixes the origin but is not necessarily the identity. It is not transitive either, since for two elements of $\mathbb{R}^n$ whose length are not equal, you cannot find an element of $O(n)$ that sends one on the other (the elements of $O(n)$ preserve norm). It is not faithful neither since when you restrict the action to $S^{n-1}$ it is not faithful either. Since $\mathbb{R}^n=[0,\infty)\times S^{n-1}$ the set of orbits can be identified with $[0,\infty)$ since when $O(n)$ is restricted to the second factor it is transitive.

Title: Group Actions of $S_n$ and $O(n)$


问:
Infinite union of closed intervals, is the result opened or closed?
Consider the union  $\space\bigcup\limits_{n \geq 1} \left[\dfrac{1}{n}; 1 - \dfrac{1}{n}\right] $. I considered $\inf\left(\dfrac{1}{n}\right )$ and $\sup\left(1 - \dfrac{1}{n}\right )$ and got the obtained the result
$$
\space\bigcup\limits_{n \geq 1} \left[\dfrac{1}{n}; 1 - \dfrac{1}{n}\right] = \left(0, 1\right).
$$

I'm wondering whether it should be $\left(0, 1\right)$ or $[0;1] $ since one takes the infinite union. Could someone clarify this for me?

Body: You can consider its complement in $[0,1]$. Let $A^c$ denote the complement of $A$. $(\cup_{n>1} [\frac{1}{n}, 1-\frac{1}{n}])^c=\cap_{n>1} [\frac{1}{n}, 1-\frac{1}{n}]^c$. Then $x \in \cap_{n>1} [\frac{1}{n}, 1-\frac{1}{n}]^c$ only if $0\leq x< \frac{1}{n}$ for all $n$ or $1-\frac{1}{n}<x \leq 1$ for all $n$, then either $x=0$ or $x=1$. Thus $(\cup_{n>1} [\frac{1}{n}, 1-\frac{1}{n}])^c=\{0,1\}$. So $\cup_{n>1} [\frac{1}{n}, 1-\frac{1}{n}]=(0,1)$.


问:
Make a matrix symmetric and positive semi-definite if it already approximately is
I have a matrix that ought to be symmetric and positive semi-definite (with real entries). However, due to small rounding errors that can accumulate over time, that might not be strictly true. For example, the smallest eigenvalue might be $-10^{-10}$ even though it ought to be $0$ if we had been using exact calculations. This isn't particularly important overall, except that some computations will fail altogether with even this small negative number.

Is there a simple expression that will fix this? As an example of what I'm looking for:

If I were asking "how do I make a real number non-negative if it already nearly is?" I'd like the answer "take the absolute value". This leaves non-negative numbers precisely unchanged, and only has a small effect on small strictly negative numbers.

If I were asking "how do I make a matrix $M$ symmetric" I'd like the answer "use $M + M^t$". Actually that is one of the things I want so that will be my first step!

So what I'm asking, analogously, is: "how do I make the matrix positive semidefinite?"

答:
If all you care about is positive semidefinite, the bluntest instrument you can use is:

First, symmetrize, as you suggest, to produce a new matrix $S.$

Second, take the pseudo-inverse $P$ of S (np.linalg.pinv if you are using python, with a reasonable cutoff (say $1.0e-6$), and then the pseudo-inverse $Q$ of $P.$
Then, Q will be the matrix you seek. Or at least A matrix you seek.

A more direct (though not faster) method is write your $S = P^t D P,$ where $P$ is orthogonal, D diagonal, and clip the negative entries of $D$ to $0.$ It will produce the same result as above.


问:
Distance between opposite vertices of 2 joint (by base) tetrahedrons
I have two tetrahedrons joint by base (let's call it ABC triangle):

tetrahedron ABCD

tetrahedron ABCE

I know lengths (read as distances) of all edges. Is there any reasonably simple way to find length (distance) of DE?

Also, if you know how to find height of the tetrahedron by the lengths of edges, please comment or add answer - this I feel would be 90% of the solution.

PS: those figures are general and are not of special case.

Image just to clarify how it looks like

Any input would be helpful! Thanks!

答:
The question can be solved using the formulas for the $n$-dimensional volume of an $n$-dimensional simplex. It can be found here.

For five points $P_0,\ldots,P_4$ in 4-dimensional space and their mutual distances $d_{ij}=\overline{P_iP_j}$, $i\neq j$ the 4-dimensional volume or content $C$ is given by
$$
\begin{align}
C^2 & = -\frac{1}{(4!)^2 2^4}
\begin{vmatrix}
0 & d_{01}^2 & d_{02}^2 & d_{03}^2 & d_{04}^2 & 1 \\
d_{01}^2 & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2 & 1 \\
d_{02}^2 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2 & 1 \\
d_{03}^2 & d_{13}^2 & d_{23}^2 & 0  &d_{34}^2 & 1 \\
d_{04}^2 & d_{14}^2 & d_{24}^2 & d_{34}^2& 0 & 1 \\
1 & 1 & 1 & 1 & 1 & 0
\end{vmatrix}.
\end{align}$$
In the question, we have five points. We rename them $P_1=A$, $P_2=B$, $P_3=C$, $P_4=D$ and finally $P_0=E$. The key obervation is that they are contained in a 3-dimensional subspace if and only if $C=0$ (and $P_1$,..,$P_4$ and $P_0$,$P_2$,$P_3$,$P_4$, respectively, form non-degenerate tetrahedra).

This gives an equation for $x=d_{01}^2$:
$$
\begin{align}
\begin{vmatrix}
0 & x & d_{02}^2 & d_{03}^2 & d_{04}^2 & 1 \\
x & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2 & 1 \\
d_{02}^2 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2 & 1 \\
d_{03}^2 & d_{13}^2 & d_{23}^2 & 0  &d_{34}^2 & 1 \\
d_{04}^2 & d_{14}^2 & d_{24}^2 & d_{34}^2& 0 & 1 \\
1 & 1 & 1 & 1 & 1 & 0
\end{vmatrix}=0.
\end{align}$$
Using Laplace's expansion, it can be seen that this is a quadratic equation for $x$. The larger solution is $x_1=\overline{DE}^2$, the smaller solution is $x_2=\overline{DE'}^2$, where $E'$ is the symmetric to $E$ with respect to the plane containing $A,B,C$. Observe that $\overline{AE'}=\overline{AE}$, $\overline{BE'}=\overline{BE}$ and $\overline{CE'}=\overline{CE}$.

The coefficient $M$ of $x^2$ in this equation actually is the negative of a subdeterminant:
$$M=-
\begin{align}
\begin{vmatrix}
0 & d_{23}^2 & d_{24}^2 & 1 \\
 d_{23}^2 & 0  &d_{34}^2 & 1 \\
 d_{24}^2 & d_{34}^2& 0 & 1 \\
 1 & 1 & 1 & 0
\end{vmatrix}.
\end{align}$$
Again by the determinant formula, this is $M=16A^2$, where $A$ is the area of the triangle $P_2P_3P_4$.

The heights mentioned in the question can also be expressed using determinants.
The volume $V$ of the tetrahedron $P_1P_2P_3P_4$ is given by
$$V^2=\frac1{3!^2 2^3}
\begin{align}
\begin{vmatrix}
0 & d_{12}^2 & d_{13}^2 & d_{14}^2 & 1 \\
 d_{12}^2 & 0 & d_{23}^2 & d_{24}^2 & 1 \\
 d_{13}^2 & d_{23}^2 & 0  &d_{34}^2 & 1 \\
 d_{14}^2 & d_{24}^2 & d_{34}^2& 0 & 1 \\
 1 & 1 & 1 & 1 & 0
\end{vmatrix}.
\end{align}$$
One  height of this tetrahedron, i.e. the distance $h$ of $P_1$ from $P_2P_3P_4$ is
then obtained from the formula
$$V=\frac13 Ah$$
where $A$ is the area of the triangle $P_2P_3P_4$ expressed above.


问:
Recovering a group from its quotient group
Suppose I've a group $G$ and a normal subgroup $H$. I know the structure of $G/H$ as well as the structure of $H$. Is it possible to recover the original structure of G from this?</p>

答:
There is one important piece of data that you have omitted: namely, if $H$ is normal in $G$, then $G$ acts on $H$ by conjugation, and so there is a homomorphism
$G/H \to Out(H)$ (the group of outer automorphisms of $H$); without remembering this, there would be no chance of reconstructing $G$.

Now in general giving $G/H$, $H$, and a map from $G/H$ to $Out(H)$, there is
still no chance of reconstructing $G$.  E.g. even if $G$ and $H$ are abelian,
so that the homomorphism $G/H \to Out(H)$ is trivial, Zev's answer gives
a counterexample.

However, if $G/H$ and $H$ are finite of coprime order, then in fact any extension
of $G/H$ by $H$ is necessarily a semidirect product, and so in this case your question (suitably modified by remembering the conjugation action) has a positive answer.  This is the Schur--Zassenhaus theorem.

This is why counterexamples (such as in Zev's answer) tend to focus on $p$-groups.


问:
Representation of translation group on $\mathbb{R}^n$.
I am having the following question:

If we consider the group $G_{\alpha}$, $\alpha \in \mathbb{R}^n$ that acts on a point $x \in \mathbb{R}^n$, by $g_{\alpha}x = (x_{1}+a_{1}, ..., x_{n}+a_{n})$, is there some way of getting a representation for an element $g_{\alpha}$ that is given by a matrix?

Is this possible if we restrict to $\alpha$ being one-dimensional, i.e. the action is $g_{\alpha}x = (x_{1}+a, ..., x_{n}+a)$ of a further restriction that $G$ is finite, i.e. we only pick a finite number of real numbers for $\alpha$.

Thank you

答:
Matrix multiplication is linear. That is, for every $n\times n$ matrix $A$ the map $\mathbb R^n\to\mathbb R^n$ given by $x\mapsto Ax$ is a linear map. All the maps $x\mapsto x+\alpha$ for $\alpha\in\mathbb R^n$ are non-linear, with the obvious exception for $\alpha=0$ — they are translations.

Hence, you can't describe those translations by a matrix in the standard meaning of "description by a matrix".

However, by embedding $\mathbb R^n$ into the projective space $\mathbb R\mathbb P^n$ (using homogeneous coordinates), you can indeed describe translation by matrices:
$$
\begin{pmatrix}
1 & 0 & \cdots & 0 & a_1 \\
0 & 1 & \ddots & \vdots & \vdots \\
\vdots & \ddots & \ddots & 0 & a_{n-1}\\
\vdots & & \ddots & 1 & a_n \\
0 & \cdots & \cdots & 0 & 1
\end{pmatrix}
\begin{pmatrix} x_1\\\vdots\\x_n\\ 1\end{pmatrix}
=
\begin{pmatrix} x_1+a_1\\\vdots\\x_n+a_n\\ 1\end{pmatrix}
$$


问:
What is meant by : The sum over one period of the values of a periodic complex exponential is zero unless that complex exponential is a constant.
What is meant by : The sum over one period of the values of a periodic complex exponential is zero unless that complex exponential is a constant.

$$\sum_{n=0}^{N-1} e^{\frac{j2 \pi (p-k)n}{N}}$$
$$ =N; (p-k)=0, \pm N, \pm 2N,...$$
$$ =0; (p-k) \neq N$$

Why is the above summation equal to $0$ if $(p-k)\neq N$

答: This has nothing to do with L'Hôpital's rule.  That's for taking limits, and there's no limit here.  Once you get to $$\frac{1-e^{\frac{j2\pi (p-k) N}{N}}}{1- e^{\frac {j2\pi (p-k)}{N}}},$$ note that the numerator is $$1-e^{2\pi j(p-k)}=1-1=0,$$ since $p-k$ is an integer.


问:
Finding the smallest $\alpha>0$ for which $\exists\beta(\alpha)>0$ so that $\sqrt{1+x}+\sqrt{1-x}\le 2-\frac{x^\alpha}\beta,\forall x\in[0,1]$.

Find the smallest $\alpha>0$ for which there is $\beta(\alpha)>0$ so that the following inequality holds $\forall x\in[0,1]$: $$\sqrt{1+x}+\sqrt{1-x}\le 2-\frac{x^\alpha}\beta$$ and find $\min\beta(a)$ for that particular $a$.

Source: Elezović, N., Odabrani zadatci elementarne matematike, Zagreb, 1992

EDIT: By $\beta(\alpha)$ I denoted a positive $\beta$ depends on $\alpha$. If $\beta>0$ exists for some $\alpha>0$, it doesn't have to be unique.

My thoughts:

Function $f(x)=\sqrt{1+x}+\sqrt{1-x}$ is concave on $(-1,1)$, and particulary on $(0,1)$, which follows from the fact that $$f''(x)=-\frac14\left(\frac1{\sqrt{(1+x)^3}}+\frac1{\sqrt{(1-x)^3}}\right)<0,\space\forall x\in(0,1).$$

Since $f$ is concave, each of its tangents is above the graph $\Gamma(f)$.

My first idea was to observe the polynomial $g(x)=f'(c)(x-c)+f(c),c\in (0,1),\deg g=1,$ which made me suspect that $\alpha\ge 1$.

I noticed that a function $f_2(x)=-\frac{x^\alpha}\beta$ is concave for $\alpha>1,$ convex for $0<\alpha<1$ and both convex and concave for $\alpha=1,$ when $f_2$ is linear. Then,
$$\alpha\ge 1\implies f_3(x)=2-\frac{x^\alpha}\beta\text{ is also concave}.$$

and this polynomial $f_3$ and the function $f$ should have at most one intersection and at $x=0$.

However, I'm not able to justify my claim that this is the only possibility.

Also, $f'$ is strictly decreasing on an open interval, and hence, my attempt to express $\min\beta(\alpha)$ in terms of $f'$ in the second part of the task failed.

May I ask for advice on solving this task? Thank you very much in advance!

答:
Clearly, $2 - \sqrt{1 + x} - \sqrt{1 - x} > 0$ for all $x\in (0, 1]$.

We have, for all $x \in (0, 1]$,

\begin{align}
\beta &\ge \frac{x^\alpha}{2 - \sqrt{1 + x} - \sqrt{1 - x}}\\
&= \frac{x^{\alpha - 2}}{4}(2 + \sqrt{1 + x} + \sqrt{1 - x})
(2 + 2\sqrt{1 - x^2})\\
&\triangleq f(x).
\end{align}
If $0 < \alpha < 2$, then $f(x) \to \infty$ as $x\to 0^{+}$.

If $\alpha \ge 2$, then $f(x) \le \frac{1}{4}(2 + \sqrt{2} + 1)(2 + 2)$ for all $x \in (0, 1]$.

Thus, $\alpha = 2$ is the smallest possible $\alpha$, with the corresponding smallest possible
$\beta = 4$ (not difficult to obtain).


问:
A and B are subsets of the reals; neither A nor B is an open set, but  ∪  is an open set.
Either give an example that satisfies the statement or explain why no such example is possible. The context is the reals.

A and B are subsets of the reals; neither A nor B is an open set, but  ∪  is an open set.

So for this one If you have 2 non open sets, I feel like then I would assume these are both closed sets. I don't think 2 closed sets can form an open set. I just don't know how to go about proving it.

答:
It is not necessarily the case that all non-open sets are closed. Take half-open intervals such as $(a,b]$ for example. That might help motivate an example.

Recall that $\Omega\subseteq\mathbb{R}$ is closed if $\Omega^c=\mathbb{R}\setminus\Omega$ is open. What must the complement of a finite subset of $\mathbb{R}$ be?

Remember that for $\Omega\subseteq\mathbb{R}$ to be open, $\text{Int}(\Omega)=\Omega$. What does that tell you about the non-empty intersection of two open sets (that also must be open)?

Your intuition is good! Closed sets including their boundary is what lets this happen.

I hope this pushes you in the right direction!


问:
Topologist's Sine Curve not a regular submanifold of $\mathbb{R^2}$?
I am trying to work out the details of following example from page 101 of Tu's An Introduction to Manifolds:

Example 9.3. Let $\Gamma$ be the graph of the function $f(x) =
 \sin(1/x)$ on the interval $]0, 1[$, and let $S$ be the union of
  $\Gamma$ and the open interval $I=\{(0,y)\in \mathbb{R^2} |−1<y<1\}$.

The subset $S$ of $\mathbb{R^2}$ is not a regular submanifold for the
  following reason: if $p$ is in the interval $I$, then there is no
  adapted chart containing $p$, since any sufficiently small
  neighborhood $U$ of $p$ in $\mathbb{R^2}$ intersects $S$ in infinitely
  many components.

Using Tu's definitions, I need to show that given a neighborhood $U$ of $p$, there exists no homeomorphism $\phi$ from $U$ onto an open set of $V \subset\mathbb{R^2}$ with the property that $U \cap S$ is the pre-image with respect to $\phi$ of the $x$ or $y$ axes intersected with $V$.

I am not sure where the fact that $U \cap S$ has infinitely many components comes into play. I tried to derive a contradiction using this fact; but even if restrict my attention to connected neighborhoods $U$ of $p$, the intersection of the connected set $\phi(U)$ with the $x$ or $y$ axes might have infinitely many components (I think), so there's no contradiction with the fact that homeomorphism preserves components.

I would appreciate any help!

答:
One simple way to see that $S$ is not a regular submanifold around $p$ is that it is not locally Euclidean: while there exist open subsets of $\mathbb{R}$ with as many connected components as you like, there are no points like $p$ in $\mathbb{R}$, that is, with every neighborhood not connected.
So, in the sentence "since any sufficiently small neighborhood $U$ of $p$ in $\mathbb{R}^2$ intersects $S$ in infinitely many components" the important part (besides "many components") is "any neighborhood".


答:
We can construct a tetrahedron with vertices $(0,0,0),(1,1,0), (1,0,1), (0,1,1)$

The center face opposite the vertex at the origin is $(\frac 23,\frac 23,\frac 23)$

From this it is simple enough to find the height and length of an edge.
The corresponding vertex opposite the origin $(\frac 43,\frac 43,\frac 43)$

Edge length: $\sqrt 2$

Height: $\frac {2}{\sqrt 3}$

And the distance between the two opposite vertices: $\frac {4}{\sqrt 3}$

We can scale all of this by $\frac {s}{\sqrt 2}$ to suit a different edge length.

Title: Distance between opposite vertices of 2 joint (by base) tetrahedrons

Body:
I have two tetrahedrons joint by base (let's call it ABC triangle):

tetrahedron ABCD

tetrahedron ABCE

I know lengths (read as distances) of all edges. Is there any reasonably simple way to find length (distance) of DE?

Also, if you know how to find height of the tetrahedron by the lengths of edges, please comment or add answer - this I feel would be 90% of the solution.

PS: those figures are general and are not of special case.

Image just to clarify how it looks like

Any input would be helpful! Thanks!


问:
How can a manifold be Hausdorff and have an atlas where coordinate charts intersect?
I'm sorry if this question seems "stupid", but I am having trouble visualizing a manifold that is Hausdorff with an atlas. Manifolds that are Hausdorff have non-intersecting neighborhoods and an atlas tells us a manifold is smooth. So let us say we are on a manifold, maybe the earth, if i have a atlas, I am able to transition between 1 coordinate chart to the other. Maybe outside of my country But then at the same time Hausdorff tells me something on this manifold needs to be non-intersecting. I don't understand what that is. Maybe i am going about this the wrong way?</p>

答:
This is not a stupid question (not that there is no argument to be made for asking stupid questions). It took humanity a long time to formalize all this local-global stuff. Also as a disclaimer, there is a lot of fine print I ignore below.

To build on your example, consider the surface of the earth. We have atlasses (as in geography) with certain specified information (so an atlas could only contain the information of the borders of countries, or the distribution of sugarcane farms, or the distribution of infected people in the case of a pandemic). Considering an atlas in book form, we have different maps for different regions of the earth. As far as I remember some regions depicted in different pages overlap, so that the user always has some idea where the region depicted in a specific page is located. (I'll leave it to you to think about the modern day version of this.)

When you abstract away this idea into mathematics, being able to flip the pages of an atlas and being able to remember the maps of which two regions you want to use in tandem correspond to the transition cocycle you mentioned. Observe that just as there are different kinds of geographical atlasses that depict different information there are also different kinds of mathematical atlasses that depict different information. A smooth atlas like you mention is one that allows you to do calculus without worrying too much about running out of derivatives.

Let me also mention that a mathematical atlas bypasses any issues related to the so-called map-territory relation. As far as calculus is concerned, the surface of the earth is its depiction in an atlas (at least as a first approximation).

So far all I've mentioned was about the existence of an atlas. Let us add the Hausdorff assumption into the story. Hausdorffness guarantees like you said that given any two distinct points there are non-intersecting open sets separating them. This means that, returning to our example of a geographical atlas, given any two different locales, there is a page in the atlas that includes a depiction of the first one but does not include a depiction of any part of the second one. In other words you have the ability to localize to anywhere you wish. (I'll leave it to you to think of what this corresponds to in terms of geographical atlasses. In this case it's much easier to think of the modern day versions.) (As another little thought experiment, what does a "locale" mean here? One is not be able to do this with two neighboring countries, or two neighboring houses, for instance.)

This in turn allows you to localize to different regions, build a thing on each one of the regions without worrying about making a mess at regions you are not looking at, and then patch the thing you constructed at different locales up to get a global thing that works all across the earth. This is the so-called partition of unity (for which in general Hausdorffness is not sufficient but necessary, at least in classical contexts).

To sum up, both the notion of an atlas and Hausdorffness is precisations of the idea of being able to split a big thing into smaller things. An atlas tells you that as long as you know how to put down one small piece and take up another you have the whole big thing (and vice versa) (What does the previous paren mean?). Hausdorffness tells you that if you want to think about one specific part of the big thing and simultaneously ignore some other specific part of the big thing, there is a small enough piece you can pick up that does the job. This (under certain conditions) means that if you want to build a thing on the big thing you can build things on each small piece and put them together. (For this last paragraph it might be useful to think of a puzzle of the earth, with somehow the sizes of each puzzle piece changing at your will.)

Finally here are two natural exercises that I hope will be useful to think about:

Exercise 1: What is an example of a topological space that is Hausdorff but does not have an atlas (in the sense we consider here)?

Exercise 2: What is an example of a topological space that is not Hausdorff but does have an atlas?


问:
Determine the convergence of $\sum\limits_{n=2}^\infty \frac{\sqrt{n+1}}{n(n-1)}$
So I need to know the convergence of this sum. The integral test does not seem to work (WolframAlpha gives a confusing answer which implies it's not the correct method). I also tried the limit comparison test for various harmonic series, but they are all inconclusive. I'm also not getting very far with the ratio test.
$$
\sum_{n=2}^\infty \dfrac{\sqrt{n+1}}{n(n-1)}
$$</p>

答:
$\dfrac{\sqrt{n+1}}{n(n-1)}\le\dfrac{{n+1}}{n-1}=(1+\dfrac{1}{n})(1-\dfrac{1}{n})^{-1}=(1+\dfrac{1}{n})(1+\dfrac{1}{n}-\cdots)=1+\dfrac{2}{n}+O(\dfrac{1}{n^2}).$

$\displaystyle\sum\dfrac{{n+1}}{n-1}$ is convergent and so is the given series.


问:
Derive Equation for Dual Curve to Ellipse not at Origin Given Implicitly
This is a second follow-up to this question. For the ellipse centered at $(h,k)$ defined implicitly as

$$
\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}-1=0 \tag{1}
$$

Please show the steps for calculating its dual curve and provide the equation for the dual curve. Do not solve by redefining the function parametrically. I know how to solve for parametric functions, but need to see the general algorithm for implicit functions.

Wikipedia provides a trivial example that does not help me.

My Attempt

At point $(p,q)$,

$$
X = \lambda \frac{2(p-h)}{a^2} \implies p = h + \left( \frac{a^2}{2\lambda} \right) X
$$

$$
Y = \lambda \frac{2(q-k)}{b^2} \implies q = k + \left( \frac{b^2}{2\lambda} \right) Y
$$

For incidence to occur, the inner product of the line $[X,Y]$ and point $(p,q)$ must be zero:

$$
\frac{a^2}{2\lambda}X^2 + hX + \frac{b^2}{2\lambda}Y^2 + kY = 0
$$

$$
a^2X^2 + b^2Y^2 + 2\lambda (hX + kY) = 0 \tag{2}
$$

How should $\lambda$ be eliminated at this point?

Thought 1:

Can we homogenize the linear terms by $\frac{\left( hX + kY \right)}{-2\lambda}$? If so, we get

$$
a^2X^2 + b^2Y^2 - (hX+kY)^2 = 0
$$

$$
a^2X^2 + b^2Y^2 - (h^2X^2 + 2hkXY + k^2Y^2) = 0
$$

$$
(a^2-h^2)X^2 + (b^2-k^2)Y^2 - 2hkXY = 0
$$

but I'm not sure if this is the correct equation for the dual. Is there a way to check?

Thought 2:

Alternatively, can I just say that $\lambda = 1$? In this case, I get:

$$
a^2X^2+b^2Y^2+2(hX+kY)=0
$$

However, now my answer in Thought 1 is different from Thought 2.

答:
I see now. So the general quadratic form in the projective plane is

$$
\begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} a & b & d \\ b & c & e \\ d & e & f \end{pmatrix} \begin{pmatrix} x \\ y \\ z\end{pmatrix} = 0
$$

which yields

$$
ax^2+cy^2+fz^2+2bxy+2dxz+2eyz=0,
$$

but when we restrict ourselves to the affine plane,

$$
\begin{pmatrix} x & y & 1 \end{pmatrix} \begin{pmatrix} a & b & d \\ b & c & e \\ d & e & f \end{pmatrix} \begin{pmatrix} x \\ y \\ 1\end{pmatrix} = 0
$$

becomes

$$
ax^2+cy^2+f+2bxy+2dx+2ey=0.
$$

The adjoint (signed cofactor matrix) is denoted $A^\Delta$:

$$
\begin{pmatrix} a & b & d \\ b & c & e \\ d & e & f \end{pmatrix} ^ \Delta = \begin{pmatrix} + \begin{vmatrix} c & e \\ e & f \end{vmatrix} & - \begin{vmatrix} b & e \\ d & f \end{vmatrix} & + \begin{vmatrix} b & c \\ d & e \end{vmatrix} \\ - \begin{vmatrix} b & d \\ e & f \end{vmatrix} & + \begin{vmatrix} a & d \\ d & f \end{vmatrix} & - \begin{vmatrix} a & b \\ d & e \end{vmatrix} \\ + \begin{vmatrix} b & d \\ c & e \end{vmatrix}& - \begin{vmatrix} a & d \\ b & e \end{vmatrix} & + \begin{vmatrix} a & b \\ b & c \end{vmatrix} \end{pmatrix}
$$


问:
average proportions over time
Let us say I have a number of events per week over some time periods. During each week a proportion of these events are important (important_events_per_week/number_of_events_per_week). The number_of_events_per_week also vary over time. How does one calculate the average important number of events per week over time please? I would think it is too naive to simply calculate the proportion per week and than take the arithmetic mean, so I thought I ask proper mathematicians? Should one use a weighted average to account for the varying number of events? Thanks!</p>

答:
Maybe the number $X$ of events per week has $X \sim \mathsf{Pois}(\lambda = 10),$ so that on average there are ten events a week. If $1/5$ of these events are "important," then the number $Y$ of important events per week has
$Y \sim \mathsf{Pois}(\lambda_I = \lambda/5 = 2),$ so that there are two important events in a week on average.

Then $P(Y = k) = e^{-2}2^k/k!$, for $k = 0, 1, 2, \dots.$

In particular $P(Y = 5) = 0.0361$ and $P(Y \le 3) = 0.8571,$ (rounded to four places) can be
found from the PDF formula above or in R, as follows:

```<code>dpois(5, 2)
[1] 0.03608941
exp(-2)*2^5/factorial(5)
[1] 0.03608941

sum(dpois(0:3, 2))
[1] 0.8571235
ppois(3, 2)
[1] 0.8571235
</code>```

In R a Poisson PDF is denoted dpois and a Poisson CDF
is denoted ppois, each with appropriate parameters.

If you don't know the average number of important events per week, you might look at numbers of
important events over the last 52 weeks.

```<code>y
 [1] 2 3 3 2 3 2 0 2 0 4
[11] 0 1 2 1 3 1 2 3 2 1
[21] 2 2 3 1 4 0 1 4 2 3
[31] 3 0 2 1 1 1 1 3 1 1
[41] 1 3 2 1 3 1 0 2 2 1
[51] 2 2

summary(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  0.000   1.000   2.000   1.788   3.000   4.000
</code>```

So a reasonable estimate of $\lambda_i$ as $\hat \lambda_I = 1.788.$

Here is one way to find a 95% confidence interval (CI) $(1.46, 2.19)$ for $\lambda_I,$ based on the total number of important events in a year:

```<code>sum(y)
[1] 93
CI.52 = 93 + 2 + qnorm(c(.025,.975))*sqrt(93+1)
CI.52/52
[1] 1.461489 2.192357
</code>```

That's $(93+2 \pm 1.96\sqrt{93+1})/52.$

Note: My 'data' above were sampled from
$\mathsf{Pois}(2)$ as shown below. In a real
application you wouldn't know the exact true
value of $\lambda_I.$

```<code>set.seed(315)
y = rpois(52, 2)
</code>```

Title: average proportions over time

Body:
Maybe the number $X$ of events per week has $X \sim \mathsf{Pois}(\lambda = 10),$ so that on average there are ten events a week. If $1/5$ of these events are "important," then the number $Y$ of important events per week has
$Y \sim \mathsf{Pois}(\lambda_I = \lambda/5 = 2),$ so that there are two important events in a week on average.

Then $P(Y = k) = e^{-2}2^k/k!$, for $k = 0, 1, 2, \dots.$

In particular $P(Y = 5) = 0.0361$ and $P(Y \le 3) = 0.8571,$ (rounded to four places) can be
found from the PDF formula above or in R, as follows:

dpois(5, 2)
[1] 0.03608941
exp(-2)*2^5/factorial(5)
[1] 0.03608941

sum(dpois(0:3, 2))
[1] 0.8571235
ppois(3, 2)
[1] 0.8571235

In R a Poisson PDF is denoted dpois and a Poisson CDF
is denoted ppois, each with appropriate parameters.

If you don't know the average number of important events per week, you might look at numbers of
important events over the last 52 weeks.

y
 [1] 2 3 3 2 3 2 0 2 0 4
[11] 0 1 2 1 3 1 2 3 2 1
[21] 2 2 3 1 4 0 1 4 2 3
[31] 3 0 2 1 1 1 1 3 1 1
[41] 1 3 2 1 3 1 0 2 2 1
[51] 2 2

summary(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  0.000   1.000   2.000   1.788   3.000   4.000

So a reasonable estimate of $\lambda_i$ as $\hat \lambda_I = 1.788.$

Here is one way to find a 95% confidence interval (CI) $(1.46, 2.19)$ for $\lambda_I,$ based on the total number of important events in a year:

sum(y)
[1] 93
CI.52 = 93 + 2 + qnorm(c(.025,.975))*sqrt(93+1)
CI.52/52
[1] 1.461489 2.192357

That's $(93+2 \pm 1.96\sqrt{93+1})/52.$

Note: My 'data' above were sampled from
$\mathsf{Pois}(2)$ as shown below. In a real
application you wouldn't know the exact true
value of $\lambda_I.$

set.seed(315)
y = rpois(52, 2)


问:
Delta function equivalent : limit $\sin(t/e)/(\pi t)$ tends to delta(t) as $e$ approaches $0$
I plotted the above function in MATLAB and observed the delta function, but I can't theoretically prove that limit of $$\frac{\sin(t/\varepsilon)}{\pi t}$$ tends to delta when $\varepsilon$ approaches $0$.
As we choose $t$ near $0$ we get infinity as the result, but I can't prove where ever except $t=0$ we will have zero output.

答:
A sequence $(\delta_\alpha)_{\alpha\in\mathbb{N}}$ of integrable functions $\delta_\alpha\in L^1(\mathbb{R}^n)$ is called a nascent delta function if

for all $x\in\mathbb{R}$ and $\alpha\in\mathbb{N}$, $(\delta_\alpha)_{\alpha\in\mathbb{N}}\ge0,$

for all $\alpha\in\mathbb{N}$, $$\int_{\mathbb{R}^n}\delta_{\alpha}(x)\,\mathrm{d}x=1,$$

for all $\varepsilon$, $$\lim_{\alpha \to \infty} \int_{\mathbb{R}^n \setminus B_\varepsilon (0)} \delta_{\alpha}(x) \mathrm{d}x = 0.$$

The function
$$\delta_{\epsilon}(x)=\frac{1}{\pi x}\sin\left(\frac{x}{\epsilon}\right)$$
is not a nascent Delta-function since it is not positive for all $\varepsilon\ge0$. Still the other two conditions are met so it tends to the Delta-function.


问:
Need help or tips on how to find inverse on groups
i need some tips on how to start searching for inverses on a group structure, I seem to have a difficulty with actually finding an inverse when the operation is specified
for example
$$(Q\backslash \{-1\},*), \ \ a*b=a+b+ab$$
the first thing i think is the algebraic inverse or rather $\frac{-p}{q}$ but in cases like this that the operation is specified and "new" i am lost on how to find it, i know that $\frac{-p}{q}$ in this case wouldn't work for the exclusion of -1.

Are there any tips on how to proceed when proving inverse?

I know that in this example that 0 is the identity element but i don't know how to show this in more abstract terms:
$$a*a^{-1}=\frac{p}{q}+\frac{r}{s}+\frac{pr}{qs}=0$$
i've tried putting them togheter like this
$$\frac{ps+rq+pr}{qs}=0$$ or $$\frac{ps+rq}{qs}=-\frac{pr}{qs}$$

but i still am lost

If this question is inappropiate please let me know so i can delete it

Thanks in advance!

答:
I don't know what $Q/[-1]$ means, so I'll just call it $G$.

Step 1: find the identity element $-$ look for an element $e\in G$ such that $a*e = e*a = a$ for all $a$.

Step 2: given $a\in G$, look for $b\in G$ such that $a*b=e$ (for the right inverse) or $b*a=e$ (for the left inverse). Your group is commutative, so these are the same.

That's it!


问:
How to solve this non-autonomous differential equation?
I have a question about the resolution of a differential equation. I have the answer but i don't understand how to find this solution.

Consider the following system :

$$\frac{d\omega}{dt}= \begin{pmatrix} \frac{\beta_v(t)}{\lambda}-\mu_v(t) & 0 \\ \frac{\beta_h(t)}{\lambda} & -(\mu_h(t)+\gamma(t))
\end{pmatrix}\omega$$

which has fundamental matrix :

$$\Psi(t) = \begin{pmatrix} exp(\frac{1}{\lambda} \int_{0}^{t} \beta_v(s)ds-\int_0^t \mu_v(s)ds) & 0 \\f(t) & exp(-\int_0^t(\mu_h(s)+\gamma(s))ds\end{pmatrix}$$

where $$f(t)=exp(-\int_{0}^{t}(\mu_h(s)+\gamma(s))ds)(\frac{1}{\lambda}\int_{0}^{t}\beta_h(s)exp(\frac{1}{\lambda} \int_{0}^{t} \beta_v(\tau)d\tau+\int_{0}^{t}(\mu_h(\tau)+\gamma(\tau)-\mu_v(\tau))d\tau)ds)$$

I can't calculate this fundamental matrix.. I tried the diagonalisation to calculate a matrix exponential but i don't find the same result..

Can you help me ?

Thank you,

Lehinig

答:
I do not believe you need any special results to find the solution for this differential system under pretty reasonable assumptions. Observe that the first component is independent, i.e. we can simply write that $\omega_1(t)$ solves

$$\dot{\omega}_1(t) = \left[ \frac{1}{\lambda}\beta_v(t) - \mu_v(t) \right] \omega_1(t).$$

Assuming the integrability of the coefficient, the solution must be

$$\omega_1(t) = \omega_1(0)\;\mathrm{exp}\left( \int_{0}^t \frac{1}{\lambda} \beta_v(s) - \mu_v(s)\, ds \right).$$

We now have a known solution for $\omega_1(t)$ so we can treat that as known.
Therefore we may treat $\omega_2(t)$ as a solution to the non-homogenous linear differential equation

$$\dot{\omega}_2(t) + (\mu_h(t)+\gamma(t)) \omega_2(t) = \frac{1}{\lambda} \beta_h(t) \omega_1(t).$$

Assuming the integrability of the coefficient (once again) we now use variation of parameters. Define

$$\omega_2(t) = v(t)\; \mathrm{exp}\left(-\int_0^{t} \mu_h(s)+\gamma(s)\,ds \right).$$

Substituting this in to DE for $\omega_2(t)$ gives

$$\dot{v}(t)\; \mathrm{exp}\left(-\int_0^{t} \mu_h(s)+\gamma(s)\,ds \right) = \frac{1}{\lambda} \beta_h(t) \omega_1(t).$$

At which point you can resolve $v(t)$ by direct, painful integration with initial condition $v(0) = \omega_2(0)$ and complete the solution. Specifically

$$v(t) = \omega_2(0) + \int_0^t \mathrm{exp}\left(-\int_0^{\tau} \mu_h(s)+\gamma(s)\,ds \right) \frac{1}{\lambda} \beta_h(\tau) \omega_1(\tau)\, d\tau.$$

Substitute our expression for $\omega_1$

$$v(t) = \omega_2(0) + \omega_1(0) \int_0^t \frac{1}{\lambda} \beta_h(\tau) \mathrm{exp}\left(-\int_0^{\tau} \mu_h(s)+\gamma(s)\,ds \right)\mathrm{exp}\left( \int_{0}^\tau \frac{1}{\lambda} \beta_v(s) - \mu_v(s)\, ds \right) \, d\tau.$$

and simplify

$$v(t) = \omega_2(0) + \omega_1(0) \int_0^t \frac{1}{\lambda} \beta_h(\tau) \mathrm{exp}\left(\int_0^{\tau} \frac{1}{\lambda} \beta_v(s) -\mu_h(s)-\gamma(s)-\mu_v(s)\,ds \right)\, d\tau.$$

I think my $\tau$ and $s$ have flipped roles in your integral, but same idea. Backsubstitute $v(t)$ into $\omega_2(t)$ and you are done.


问:
How to show that this is a bijection?
I am trying to prove that the Cartesian product of two finite sets is also finite.

These are the steps:

Let S and T be finite sets:

$$S×T=\{(s,t):s∈S,t∈T\}$$by definition of set union:$$S×T=\bigcup\limits_{s∈ S}^{}( \{s\}×T)$$also the mapping:$$g_s:\{s\}×T→T$$defined by :$$g_s(s,t)=t$$ must be a bijection. But Im not able to prove the bijection.

This is my try (bijection proof) :

First I must proof thats injection:

If $t_1=t_2$ then $g_s(s_1,t_1)=g_s(s_2,t_2)$

Question 1:

I dont know how to show $(s_1,t_1)=(s_2,t_2)$?

Question 2:

I know the definition of surjective function but I dont know how to apply that to this function?

答:
Your definition of "onto" doesn't really make sense. What are $s_1, s_2$? Where did they come from? I'll explain what they must be below, but before you know this they seem to come from nowhere. And the style of that statement is not what it typically means to be onto. Being onto means that your function reaches every element of the codomain $T$, but your definition is more akin to saying $g_s$ is well defined, i.e. that the same input yields the same output. This is important, but immediately true as your definition of $g_s$ was unambiguous. Anyways, let's proceed to the questions at hand.

We're trying to prove that $g_s$ is injective so let $g_s(s_1, t_1) = g_s(s_2, t_2)$. First and foremost, by the very definition of $g_s$, to define $g_s(s_1, t_1)$ we need $(s_1, t_1) \in \{s_1\} \times T$. That is, $s_1 \in \{s\}$. But the only element $\{s\}$ is $s$, so we must have $s_1 = s$. The same argument applies to $s_2$, so $s_1 = s = s_2$. This is what I meant above by showing what $s_1, s_2$ must be. They must be $s$ to even try and apply $g_s$ on them. Thus, $g_s(s_1, t_1) = g_s(s, t_1) = t_1$. Similarly, $g_s(s_2, t_2) = g_s(s, t_2) = t_2$. You assumed that $g_s(s_1, t_1) = g_s(s_2, t_2)$. Hence, $t_1 = t_2$. We have already shown that $s_1 = s_2$, so $(s_1, t_1) = (s_2, t_2)$. Thus, $g_s$ is injective.

Take a $t \in T$. We want to show that $g_s$ reaches $t$ in its image. Indeed, we see that $g_s(s, t) = t$ so every element of $t$ is realized as an output of $g_s$. Thus, $g_s$ is surjective.

Title: How to show that this is a bijection?


答: A function $f:A \rightarrow B$ being "onto" aka surjective means that it "hits" every element in it's "target set". Formally this means that for every $b\in B$, you can find an $a\in A$ such that you "hit" $b$ with $f$ aka $f(a)=b$. In your example $T$ is the target set so you have to show that for every $t\in T$ you can find an element $a \in \{s\} \times T$ such that $g_s(a)=t$. Note that here $a$ is of the form $a=(s,t')$ for some $t'\in T$. What could $t'$ be?

Title: How to show that this is a bijection?

Body:
I am trying to prove that the Cartesian product of two finite sets is also finite.

These are the steps:

Let S and T be finite sets:

$$S×T=\{(s,t):s∈S,t∈T\}$$by definition of set union:$$S×T=\bigcup\limits_{s∈ S}^{}( \{s\}×T)$$also the mapping:$$g_s:\{s\}×T→T$$defined by :$$g_s(s,t)=t$$ must be a bijection. But Im not able to prove the bijection.

This is my try (bijection proof) :

First I must proof thats injection:

If $t_1=t_2$ then $g_s(s_1,t_1)=g_s(s_2,t_2)$

Question 1:

I dont know how to show $(s_1,t_1)=(s_2,t_2)$?

Question 2:

I know the definition of surjective function but I dont know how to apply that to this function?


问:
Show that $\lim_{n\to+\infty} \int_{0}^{1} f(x)\sin(nx) \,dx = 0$ for a continuous $f(x)$.
Let $f : [0, 1] \to \mathbb{R}$ be a continuous function.

Prove that $\lim_{n\to+\infty} \int_{0}^{1} f(x)\sin(nx) \,dx = 0$

Seems like I can say that f has its maximum and minimum (Let's call them $m_1$ and $m_2$) and than say that:

$m_1\int_{0}^{1} \sin(nx) \,dx  \leq \int_{0}^{1} f(x)\sin(nx) \,dx \leq m_2\int_{0}^{1} \sin(nx) \,dx $

And the question is about to prove that $\int_{0}^{1} \sin(nx) \,dx$ tends to 0 for big $n's$. But how can I prove it?

答:
Hint

Prove the result for a constant map.

Then for a step function.

Conclude by approximating $f$ with step functions. Any continuous map is the uniform limit of step functions.


答: Hint: Consider the value of $\int_{\epsilon_0}^{\epsilon_1} f(x)\sin(nx)dx$ for $\epsilon_0$ and $\epsilon_1$ so close to each other that $f(x)$ is approximately constant on $[\epsilon_0, \epsilon_1]$ and $n$ so large that $\sin(nx)$ makes at least one full cycle over the interval.

Title: Show that $\lim_{n\to+\infty} \int_{0}^{1} f(x)\sin(nx) \,dx = 0$ for a continuous $f(x)$.

Body:
Let $f : [0, 1] \to \mathbb{R}$ be a continuous function.

Prove that $\lim_{n\to+\infty} \int_{0}^{1} f(x)\sin(nx) \,dx = 0$

Seems like I can say that f has its maximum and minimum (Let's call them $m_1$ and $m_2$) and than say that:

$m_1\int_{0}^{1} \sin(nx) \,dx  \leq \int_{0}^{1} f(x)\sin(nx) \,dx \leq m_2\int_{0}^{1} \sin(nx) \,dx $

And the question is about to prove that $\int_{0}^{1} \sin(nx) \,dx$ tends to 0 for big $n's$. But how can I prove it?


问:
Let f(x) be a non constant thrice differentiable function such $f(x)=f(6-x)$ and $f’(0) =0 = f’(2)=f’(5)$. Find minimum number of roots

Let f(x) be a non constant thrice differentiable function such $f(x)=f(6-x)$ and $f’(0) =0 = f’(2)=f’(5)$. Find minimum number of roots of $(f’’(x))^2 +f’(x)f’’’(x)=0$ in the interval $x\in [0,6]$

Clearly $f’(x)=-f’(6-x)$

So $f’(1)=f’(5)=f’(4) = f’(2) = f’(6)=f’(0)=0$

Also $f’(3) =0$

This all I was able to find out. How should I proceed after this?

答:
Use the fact that $(f’’(x))^2 +f’(x)f’’’(x)=\dfrac{\mathrm{d}}{\mathrm dx}(f'(x)f''(x))$.

Since, you have found out that $f'(x)$ has atleast $7$ roots, it implies $f''(x)$ will have atleast $6$ roots, thus $f'(x)f''(x)$ will atleast have $13$ roots and thus it's derivative will atleast have $12$ roots.


问:
Determinant of sandwiched square matrix
Suppose that $A$ is an $m\times m$ matrix, and $S$ a $m\times n$ matrix. Is it possible for the relation
$$ \det S^{\dagger}A S= \det A \det S^{\dagger}S$$ to hold also for $n\neq m$?

For instace, if $\det A = 0$ is it true that also $\det S^{\dagger}AS = 0$?

答:
The equality does not hold in general. Consider the following counter-example:

$A=\begin{bmatrix}
1 & 0\\
0 & x
\end{bmatrix}$ and $S=\begin{bmatrix}
y \\
0
\end{bmatrix} .$ Then note that $$
\det (S^{\dagger}AS) = y^2,\quad \det(A)\det(S^{\dagger}S)=xy^2.
$$Picking suitable $x$ and $y$ you can see that the equality does not hold. You can also take $x=0$ if you need a non-invertible example.

Title: Determinant of sandwiched square matrix


答:
If you want the average proportion of the number of interesting events to the number of total events over some period of multiple weeks,
one simple method is to add up the total number of events during that period
(which you can do because you have the total number of events in each week of the period),
then add up the total number of interesting events during that period
(which you can do because you have the total number of interesting events in each week of the period).
Finally, divide the total number of interesting events by the total number of events.

This is how many such averages are computed in real life.
For example, to find the average speed for a trip of $20$ miles, you don't take the speed on each mile and do some kind of fancy weighted average of those $20$ speeds to find the average speed, you just take the total trip time and divide by the total distance (which is $20$ miles).

Title: average proportions over time

Body: Let us say I have a number of events per week over some time periods. During each week a proportion of these events are important (important_events_per_week/number_of_events_per_week). The number_of_events_per_week also vary over time. How does one calculate the average important number of events per week over time please? I would think it is too naive to simply calculate the proportion per week and than take the arithmetic mean, so I thought I ask proper mathematicians? Should one use a weighted average to account for the varying number of events? Thanks!</p>


问:
Limit function of Taylor series of $x\ln(x)$
The question is closely connected to another question of mine (see Show $\lim\limits_{x\to0}x\ln(x)=0$ by using Taylor series) but it emphasizes rather on the Taylor series itself.

Let's assume we already know that $\ln(x)=\sum\limits_{k=1}^{\infty}(-1)^{k+1}\frac{(x-1)^k}{k}$ for all $x\in(0,2]$. In other words $\ln(x)$ is the limit of its Taylor series expanded at point $a=1$.

How do I show that $x\ln(x)$ is the limit of its Taylor series expanded at $a=1$, $x\ln(x)\overset{??}{=}(x-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}\frac{(x-1)^k}{(k-1)k}$ for all $x\in[0,2]$?

My approach:

Let be
$$
T_n(x):=(x-1)+\sum\limits_{k=2}^{n}(-1)^{k}\frac{(x-1)^k}{(k-1)k},\\
S_n(x):=1+\sum\limits_{k=1}^{n}(-1)^{k+1}\frac{(x-1)^k}{k}.
$$
Then we see quickly that $T(x):=\lim\limits_{n\to\infty}T_n(x)$ exists for all $x\in[0,2]$ and that $(x-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}\frac{(x-1)^k}{(k-1)k}$ converges uniformly on $[0,2]$ (I won't go into further details because this is clear to me). Moreover we know that $1+\sum\limits_{k=1}^{\infty}(-1)^{k+1}\frac{(x-1)^k}{k}$ converges uniformly on $[\delta,2]$, where $0<\delta$.

If we take the derivative of $T_n(x)$ then we see that $T'_n(x)=S_{n-1}(x)$ and it follows that (see https://en.wikipedia.org/wiki/Uniform_convergence#To_differentiability) $T'(x):=\lim\limits_{n\to\infty}T'_n(x)=\lim\limits_{n\to\infty}S_n(x)=1+\ln(x)$ for all $x\in(\delta,2]$. Additionally, it follows that $T_n(x)$ converges to the antiderivative of $\lim\limits_{n\to\infty}T'_n(x)=S_n(x)=1+\ln(x)$ which is equal to $x\ln(x)$. So $x\ln(x)=(x-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}\frac{(x-1)^k}{(k-1)k}$ on $[\delta,2]$. The case when $x=0$ needs to be handled separately: $(0-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}\frac{(0-1)^k}{(k-1)k}=0=0\ln(0)$. Hence, $x\ln(x)=(x-1)+\sum\limits_{k=2}^{\infty}(-1)^{k}\frac{(x-1)^k}{(k-1)k}$ on the whole intervall $[0,2]$.

Is this correct? Any comments or suggestions are welcome :)

答: Your question was answered, already (and you gave a link to the answer): for $|x|<1$, you can get the Taylor series of $x\,\ln x$ as the product of the Taylor series $x=1+(x-1)$ and $\ln(1+(x-1)))$, or by integration of $1+\ln(1+(x-1)))$. For $|x|=1$, that follows from Abel's theorem. If you have problems with those arguments, you should address them specifically, and not just restate the question, risking to receive equally unintelligible (for you) answers.


问:
Unknowns in backward, forward, and centered difference methods
for a second-order time dependent PDE, like
$\frac{\partial^2 u}{\partial t^2}=L(u)$.
Centered difference approximation for time is
$\frac{u^{n+1}-2u+u^{n-1}}{\Delta t^2}$.
Backward difference approximation for time is
$\frac{2u^{n}-5n^{n-1}+4n^{n-2}-u^{n-3}}{\Delta t^2}$.
Forward difference approximation for time is
$\frac{2u^{n}-5n^{n+1}+4n^{n+2}-u^{n+3}}{\Delta t^2}$.

I am confused, what are the unknowns in these methods?

For example, in the Centered difference method, is $u^{n+1}$ the unknown or $u^{n}$ the unknown in the current timestep $n$?

Formulas are from
https://www.dam.brown.edu/people/alcyew/handouts/numdiff.pdf

答:
The basic formula for the second derivative is
$$
u''(t)=u''(t-h)+O(h)=\frac{u(t)-2u(t-h)+u(t-2h)}{h^2}+O(h)
$$
But we can add terms from the Taylor expansion and their low-order difference-quotient over the available sample points
\begin{align}
u''(t)&=u''(t-h)+u'''(t-h)h+O(h^2)\\
&=\frac{u(t)-2u(t-h)+u(x-2h)}{h^2}+\frac{u(t)-3u(t-h)+3u(t-2h)-u(t-3h)}{h^{3-1}}+O(h^2)\\
&=\frac{2u(t)-5u(t-h)+5u(t-2h)-u(t-3h)}{h^2}+O(h^2)
\end{align}
In the next order higher one would have to consider that the third order divided difference has its minimal error as approximation of $u(t-1.5h)$, the shift to that position gives another Taylor term. Then that the second order derivative has a second order error term involving the 4th order derivative. And for higher orders it gets more complex, which is why a more systematic approach is needed, see Numerical Method Forward Finite Difference Coefficient

In any case you want $u''(t)=L(u(t))$ with matching times on both sides. One gets more flexibility if more values of the $L$ function are involved in the formula, the easiest is the 4th order Numerov method
$$
\frac{u(t+h)-2u(t)+u(t-h)}{h^2}=\frac{L(u(t+h))+10L(u(t))+L(u(t-h))}{12}+O(h^4)
$$
$u(t+h)$ is the new state that is computed in a Numerov step.


问:
$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}$, what does it mean?
I would like to know what these fractional exponents means in a derivative $\frac{d}{dx}$ operator. Like, I've seen $\frac{d^2}{dx^2}$ but I don't know what $\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}}$ means.</p>

答: $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$ is the operator such that $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$$\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$=$\frac{d}{dx}$, ie. $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$$\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}f$=$\frac{d}{dx}f$ for differentiable functions $f$, much like how $\frac{d^2}{dx^2}=\frac{d}{dx}\frac{d}{dx}$. It is basically the square root of $\frac{d}{dx}$. For example, $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}e^{ax}=\sqrt{a}e^{ax}$, since then $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}\sqrt{a}e^{ax}=ae^{ax}=\frac{d}{dx}e^{ax}$.</p>


问:
Maximum value of $|\sum_{n=1}^\infty \frac{g\left(\frac1n\right)}{2^n}|$
Let's consider expression:

$$ L = \left|\sum_{n=1}^\infty \frac{g\left(\frac1n\right)}{2^n}\right|$$

for $$g \in C([0,1]) \quad\text{such that}\quad \int_0^1|g(t)| dt \le 1.$$

I want to find supremum of $L$ under such conditions on $g$. For me the biggest value of $L$ which statisfies my conditions will be taken for $g = 1$. Then $$L = \sum_{n=1}^\infty\frac{1}{2^n} = 1.$$
However, I'm not sure how to prove that there is no other function that will exceed value of $L$.

Could you give me a hand with justifying so?

答:
Note that the p-norm on $C([0,1])$ is defined as $||f||_p:=(\int_0^1|f(t)|^pdt)^{1/p}$ for $p \in [1, \infty)$ and $||f||_\infty=\sup_{t \in [0,1]}|f(t)|$ for $p= \infty$. So basically you know that $||g||_1 \leq 1$.

EDIT: This is wrong in general: (I mixed something up)

You can prove that: $||f||_r \leq ||f||_s $ if $s \leq r$. This implies $||g||_\infty \leq ||g||_1 \leq 1$ . So:
$$L=|\sum_{n=1}^\infty \frac{g(1/n)}{2^n}|\leq \sum_{n=1}^\infty |\frac{g(1/n)}{2^n}| \leq  \sum_{n=1}^\infty\frac{||g||_\infty}{2^n} \leq \sum_{n=1}^\infty\frac{||g||_1}{2^n} \leq \sum_{n=1}^\infty\frac{1}{2^n} =1$$
This doesn't follow:  This means that $L$ can never be higher than 1

EDIT 2: Adding to @Thomas idea here is an example:
Let $M>0$ and consider: $$f(x)=\frac{M}{1+((x-1/2)M \pi)^2}$$
$f\in C([0,1]), f \geq 0, f(1/2)=M$ and
$$\int_0^1|f(t)|dt=\int_0^1f(t)dt=\frac{2}{\pi}\arctan(M\pi/2) \leq 1 $$
If you plot $f$ you can see that the graph gets very large at 1/2 and drops off very fast. This means for $g=f$, $L\geq M/4$, so $L$ can get as big as you want.


问:
$A$ and $B$ are nonempty open subsets of the reals with nonempty intersection, and that intersection is a finite set.
$A$ and $B$ are nonempty open subsets of the reals with nonempty intersection, and that intersection is a finite set. Give an example if possible, prove if not.

The wording of this confuses me. I feel like this is not possible because something like $(0,1)$ and $(1,10)$ would not have any intersection and once you move one a little like $(0,1.1)$ and $(1,10)$ the area between $1$ and $1.1$ is infinite, no matter how small you make it. I just don't know how to prove this mathematically.

Would the set $\{1,2,3,4,5\}$ be considered open? Or is that one of the ones that is considered neither? Because if it is considered open, then something like $A=(0,10)$ and $B=\{1,2,3,4,5\}$ would work.

答:
Providing that $\mathbb R$ is endowed with the usual topology

The intersection of two open subsets $A, B$ of the reals is an open subset $C$. Either $C$ is empty... and indeed $C$ is finite. Or $C$ is not empty, but then it contains a nonempty interval and $C$ is therefore infinite. It even has the power of the continuum.


问:
Does this equation also holds for complex parameters?
I know that the following integral equation holds

$$\int_{-\infty}^{\infty}\frac{\log(a^2+x^2)}{b^2+x^2}dx=\frac{2\pi}{b}\log(a+b)$$

for $a,b>0\in\mathbb{R}$. Does it also hold for $a,b\in\mathbb{C}$ ?

答:
Let's investigate $I(a,b)=\int_{-\infty}^{\infty}\frac{\log(a^2+x^2)}{b^2+x^2}dx$ in more details. We want to know whether the formula $\int_{-\infty}^{\infty}\frac{\log(a^2+x^2)}{b^2+x^2}dx=\frac{2\pi}{b}\log(a+b)$ holds for all $a,b\in\mathbb{C}$, or there are some limitations of its usage.

The idea of analytic continuation is that we would like to get a continuous function $I(a,b)$, if we switch from  real $a,b$ to complex ones by adding phases, i.e $a\to{a}e^{i\phi_a}$ and $a\to{b}e^{i\phi_b}$

Please also notice that the integrand is expressed in terms of $a^2$ and $b^2$ - integral "does not know" anything about $a$ and $b$. We could expect that $I(a,b)$ is in fact $I(a^2,b^2)$

Making change of the variable $t=x^2$ we get

$$I(a^2,b^2)=\int_{-\infty}^{\infty}\frac{\log(a^2+x^2)}{b^2+x^2}dx=\int_0^{\infty}\frac{\log(a^2+t)}{b^2+t}\frac{dt}{\sqrt{t}}=\int_0^{\infty}\frac{\log(A+t)}{B+t}\frac{dt}{\sqrt{t}}$$

where $A=a^2$ and $B=b^2$

To make $A$ and $B$ complex we add phases: $A=|A|e^{i\phi_A}$ and $B=|A|e^{i\phi_B}$. We  suppose that $\phi_B>\phi_A>0$

$$I(A,B)=\int_0^{\infty}\frac{\log(e^{i\phi_A}(|A|+te^{-i\phi_A}))}{|B|e^{i\phi_B}+t}\frac{dt}{\sqrt{t}}\overset{t\to{t}e^{i\phi_A}}=e^{-\frac{i\phi_A}{2}}\int_0^{\infty}\frac{\log(e^{i\phi_A}(|A|+t))}{|B|e^{i(\phi_B-\phi_A)}+t}\frac{dt}{\sqrt{t}}$$

In fact by the change $t\to{t}e^{i\phi_A}$ we also change the integration path in the complex plane. But $I(A,B)$ does not change - due to the fact that both integrals along a part of a circle of big radius $R\to\infty$ and along a part of small circle near zero (radius $r\to0$) $\to0$ and there are no singularities (residuals) inside the closed contour (point $B$ is outside this contour due to our choice $\phi_B>\phi_A$).

Introducing $C=|B|e^{i(\phi_B-\phi_A)}$ we get

$$I(A,B)=e^{-\frac{i\phi_A}{2}}i\phi_A\int_0^{\infty}\frac{dt}{\sqrt{t}(t+C)}+e^{-\frac{i\phi_A}{2}}\int_0^{\infty}\frac{\log(t+|A|)}{\sqrt{t}(t+C)}dt$$

To evaluate $\int_0^{\infty}\frac{dt}{\sqrt{t}(t+C)}$ we integrate along the contour (below), on the both sides 0f the cut from $0$ to $\infty$

$$\int_0^{\infty}\frac{dt}{\sqrt{t}(t+C)}=\frac{\pi{i}}{\sqrt{Ce^{\pi{i}}}},\,\text{if}\,\,\phi_C= \phi_B-\phi_A\neq\pi$$
If $\phi_C=\pi\,$ integral is zero.

To evaluate the second integral $\int_0^{\infty}\frac{\log(t+|A|)}{\sqrt{t}(t+C)}dt$ we use the contour with an additional cut from $-|A|$ to $-\infty$ - to make logarithm a sing-valued function.

$$\int_0^{\infty}\frac{\log(t+|A|)}{\sqrt{t}(t+C)}dt=\pi{i}\frac{\log(|A|+Ce^{\pi{i}})}{\sqrt{Ce^{\pi{i}}}}+\pi{i}\int_{|A|}^{\infty}\frac{dt}{\sqrt{te^{\pi{i}}}(t-C)}dt=$$ $$=\pi{i}\frac{\log(|A|+Ce^{\pi{i}})}{\sqrt{Ce^{\pi{i}}}}+\frac{\pi{i}}{\sqrt{Ce^{\pi{i}}}}\log({\frac{\sqrt{|A|}+\sqrt{C}}{\sqrt{|A|}-\sqrt{C}}})=$$$$=2\pi\frac{\log({\sqrt{|A|}+\sqrt{C}})}{\sqrt{C}},\, \text{if}\, \phi_C\neq\pi$$
$$=\frac{2\pi{e}^{-\frac{i\phi_A}{2}}}{\sqrt{|B|}}\arctan{\sqrt{\frac{|B|}{|A|}}}, \, \text{if}\,\, \phi_C=\phi_B-\phi_A=\pi$$

Taking all together and switching to $a^2=|A|e^{i\phi_A}=|a^2|e^{i\phi_{a^2}}$ and $b^2=|B|e^{i\phi_B}=|b^2|e^{i\phi_{b^2}}$ we get finally:

$$I(a^2,b^2)=\frac{2\pi}{\sqrt{b^2}}\log(\sqrt{a^2}+\sqrt{b^2}), \, \phi_{a^2}+\pi>\phi_{b^2}>\phi_{a^2}$$
$$I(a^2,b^2)=\frac{2\pi}{\sqrt{b^2e^{-\pi{i}}}}\arctan\sqrt{\frac{b^2e^{-\pi{i}}}{a^2}}, \, \text{if} \, \,\phi_{b^2}=\phi_{a^2}+\pi$$


答:
Given the condition on $g$ it is not difficult to find, for every $M>0$,  a $g>0$ such that:

$g(1/2) > M$

$\int_0^1 g(t)dt=1$

Just think of a delta function centered on 1/2. Told another way, the fact that the integral of $g$ is limited does not say much about its maximum value.

Now it is clear that the addend at $n=1$ is larger than $M/2$. Therefore $L$ can take arbitrarily large values.

Title: Maximum value of $|\sum_{n=1}^\infty \frac{g\left(\frac1n\right)}{2^n}|$

Body:
Let's consider expression:

$$ L = \left|\sum_{n=1}^\infty \frac{g\left(\frac1n\right)}{2^n}\right|$$

for $$g \in C([0,1]) \quad\text{such that}\quad \int_0^1|g(t)| dt \le 1.$$

I want to find supremum of $L$ under such conditions on $g$. For me the biggest value of $L$ which statisfies my conditions will be taken for $g = 1$. Then $$L = \sum_{n=1}^\infty\frac{1}{2^n} = 1.$$
However, I'm not sure how to prove that there is no other function that will exceed value of $L$.

Could you give me a hand with justifying so?


问:
Are there mini-mandelbrots inside the julia set?
I've seen a julia set zoom but it is not nearly as interesting as a mandelbrot zoom. I also have not seen corresponding julia sets for zooms in the mandelbrot deeper than the original image. I'm wondering if it is possible to locate some form of a mini mandelbrot inside the julia set.</p>

答: The Mandelbrot set has nonempty interior, so if a Julia set contained a Mandelbrot set, then necessarily that Julia set would have nonempty interior. As soon as a Julia set has nonempty interior, however, it must be all of the Riemann sphere. </p>


问:
Convexity of the product of two convex sets of non-square stochastic matrices
Suppose $X\subseteq\mathbb{R}^{l\times m}$ and $Y\subseteq\mathbb{R}^{m\times n}$ are sets of matrices with real entries. Moreover, $X$ is the set of all $l\times m$ stochastic matrices (each column sums up to 1, and each entry is non-negative), so $X$ is convex. Moreover, $Y$ is a (possibly strict) subset of the set of all $m\times n$ stochastic matrices, and assume $Y$ is also convex.

Is it true that $Z:=\{xy\mid x\in X\text{ and }y\in Y\}$ is also a convex set?

Intuitively I think this may not be true in general, but I had no luck to produce a counter example. Any help would be appreciated.

答:
The proposition is likely not true in general. Here is a way to produce a numerical counterexample:

Let $X\subseteq\mathbb{R}^{2\times2}$ and $Y\subseteq\mathbb{R}^{2\times5}$. Moreover, let $Y=\{\lambda Y_u+(1-\lambda)Y_{v}\,|\,0\leq\lambda\leq1, Y_u\in\mathbb{R}^{2\times5} \text{ and }Y_v\in\mathbb{R}^{2\times5} \text{ are some column-stochastic matrices}\}$.

The question then becomes, for any $X_1,X_2\in X$, $Y_1,Y_2\in Y$ and $0\leq\mu_0\leq1$, can we find an $X_3\in X$ and $0\leq\mu_1\leq1$ such that
\begin{equation}\tag{1}
\mu_0X_1Y_1+(1-\mu_0)X_2Y_2 = X_3(\mu_1Y_u+(1-\mu_1)Y_v).
\end{equation}

Since we just need a counterexample, we can fix $Y_1=Y_u$, $Y_2=Y_v$, $\mu_0=0.2$ (for example) and pick $X_1\in X$ and $X_2\in X$ arbitrarily. In this case, (1) becomes
\begin{equation}
0.2X_1Y_u+0.8X_2Y_v = X_3(\mu_1Y_u+(1-\mu_1)Y_v).
\end{equation}
In the above equation, $X_1$, $X_2$, $Y_u$ and $Y_v$ are all known (fixed or picked). There are $10$ equations in total (because $X\subseteq\mathbb{R}^{2\times2}$ and $Y\subseteq\mathbb{R}^{2\times5}$), but there are only $5$ variables ($X_3$ and $\mu_1$), this give a hint that the solution might not exist.


问:
Visually stunning math concepts which are easy to explain
Since I'm not that good at (as I like to call it) 'die-hard-mathematics', I've always liked concepts like the golden ratio or the dragon curve, which are easy to understand and explain but are mathematically beautiful at the same time.

Do you know of any other concepts like these?

答:
This is a proof of the Pythagorean by US President J. A Garfield.

As you can see, when you line up the triangles like this, it forms a trapezoid. One way to find the area of the trapezoid is by adding up the areas of all the triangles that make it up, so we get $A = \frac{ab}{2} + \frac{ab}{2}+ \frac{c^2}{2}$. Also, we can find the area using the formula for a trapezoid, like this $A = \frac{a+b}{2}*(a+b)$. Now when we set the areas equal to each other, we get $\frac{ab}{2} + \frac{ab}{2}+ \frac{c^2}{2} = \frac{a+b}{2}*(a+b) \implies ab + \frac{c^2}2 = \frac{(a+b)^2}2 \implies 2ab + c^2 = (a+b)^2 \implies a^2+b^2=c^2$


问:
For any i.i.d. $X_n$ it follows $\lim_{n\to\infty}X_n/n=0$ pointwise?
Let $X_n$ denote a sequence of i.i.d. random variables and $Y_n=X_n/n$.

$Y_n$ converges pointwise towards Y if

$~~~{\displaystyle \lim _{n\to \infty }Y_{n}(\omega )=Y(\omega ),\,\,\forall \omega \in \Omega}$,

where $\Omega$ is the sample space of the underlying probability space over which the random variables are defined.

It seems trivial that $\lim_{n\to\infty}X_n(\omega)/n=0$, $\forall \omega \in \Omega$ and i.i.d. distributions of $X_n$.
Then, also all other kinds of convergence are true. What do I miss?

I'm confused by the comments of experienced guys in this post:
Proving almost sure convergence

答:
Here is an example for iid variables where $X_n(\omega)/n$ does not converge to zero.

Suppose that $X_n$ are iid copies of the St. Petersburg random variable. That is, you repeatedly flip a coin until the result is heads, and let $X=2^{\text{# flips}}$.

I claim that with probability one, $X_n(\omega)/n$ will not converge to zero. To show this, I will show that $X_n(\omega)/n\ge 1$ will occur infinitely often with a probability of one.

For any $n$ in the range $2^k$ to $2^{k+1}-1$, we have
$$
P(X_n/n\ge 1)=P(X_n\ge n)= P(X_n\ge 2^k)=P(\hbox{# flips}\ge k)=2^{-(k-1)}
$$
Therefore, the probability that all of the numbers $X_n/n$ in the range $2^k$ to $2^{k+1}-1$ are less than one is at most
$$
P\left(\bigcap_{n=2^k}^{2^{k+1}-1}\{X_n/n<1\}\right)\le (1-2^{-(k-1)})^{2^k}=\left(1-\frac{2}{2^k}\right)^{2^k}\approx e^{-2}.
$$
so the probability that at least one of the $X_n/n\ge 1$ occurs is $1-e^{-2}\approx 86\%$. Therefore, in each of then intervals $[2^k,2^{k+1})$, there is at least an $86\%$ chance that there will be a variable $X_n/n\ge 1$. This these events are all independent, in the long run, it will occur infinitely many times.

It turns out that the same thing will happen whenever $X$ has infinite expectation. The intuition is that having a large expectation increases the chance of $X$ being large, so it turns out infinite expectation implies $X_n$ will actually be larger than $n$ infinitely often.