问:
about Cartan's theorem
theorem : Let $SL(n,\Bbb{C})$ be the group of matrices of complex entries and determinant $1$ then  $SL(n,\Bbb{C})$ is a regular submanifold of $GL(n,\Bbb{C})$.

proof : $GL(n,\mathbb{C})$ is a Lie group and $SL(n,\mathbb{C})$ a closed subgroup. with  Cartan's theorem  $SL(n,\mathbb{C})$ is a regular Lie subgroup, in particular a regular submanifold.

Cartan's theorem : if $H$ is a closed subgroup of a Lie group G, then H is an embedded Lie group with the smooth structure (and hence the group topology) agreeing with the embedding.

why $SL(n,\mathbb{C})$ is  a closed subgroup ? what is dimension of manifold $SL(n,\mathbb{C})$ ?

答:
Consider the map $\det\colon GL(n,\Bbb C)\longrightarrow\Bbb C$. It is continuous and therefore $\det^{-1}\bigl(\{1\}\bigr)$ is closed. But $\det^{-1}\bigl(\{1\}\bigr)=SL(n,\Bbb C)$.

And its dimension is$$\dim\mathfrak{sl}(n,\Bbb C)=\dim\left\{M\in\Bbb C^{n\times n}\,\middle|\,\operatorname{tr}M=0\right\}=n^2-1.$$

Title: about Cartan's theorem


问:
How many arrangements? $(a, b, c, d, x, y, z, w)$
I would appreciate if somebody could help me with the following problem:

Q: How many arrangements?

My Study

I solved by mathematica11.0

```<code>f[n_] := {a, b, c, d, x, y, z, w} /.
  Solve[0 < a < b < c < d < n && 0 < x < y < z < w < n && a < x &&
    b < y && c < z && d < w, {a, b, c, d, x, y, z, w}, Integers]
f[9]=490
</code>```

答:
We are counting strictly order-preserving maps $P\to[8]$, where $[8]$ denotes the poset $\{1<2<\dots<8\}$ and $P=[2]\times[4]$ with $(a,b)\le(c,d)$ if and only if $a\le c$ and $b\le d$:
$$
P : \quad
\begin{array}{ccccccc}
(1,1) & < & (1,2) & < & (1,3) & < & (1,4) \\
\style{display: inline-block; transform: rotate(90deg)}{<}
&& \style{display: inline-block; transform: rotate(90deg)}{<}
&& \style{display: inline-block; transform: rotate(90deg)}{<}
&& \style{display: inline-block; transform: rotate(90deg)}{<}
\\
(2,1) & < & (2,2) & < & (2,3) & < & (2,4).
\end{array}
$$
Hence, the number we are looking for is $\Omega^\circ_P(4)$, where $\Omega^\circ_P$ is the strict order polynomial of $P$.

For the strict order polynomial, there is the formula
$$
\Omega_P(n) = \sum_{w\in\mathcal L(P)} \binom{n+\operatorname{des}(w)}{|P|},
$$
where $\mathcal L(P)$ is the set of linear extensions of $P$ and $\operatorname{des}(w)$ is the number of descents of the linear extension $w$ with respect to some fixed natural labeling of $P$.

In our case, we can pick the natural labeling $\begin{matrix}1234\\5678\end{matrix}$. As mentioned by @saulspatz, there are $14$ linear extensions of $P$ by the hook length formula for counting standard Young tableaux. A list of these with their descent numbers is:
$$
12345678, d=0,\\
12354678, d=1,\\
12534678, d=1,\\
15234678, d=1,\\
12356478, d=1,\\
12536478, d=2,\\
15236478, d=2,\\
12563478, d=1,\\
15263478, d=2,\\
12356748, d=1,\\
12536748, d=2,\\
15236748, d=2,\\
12563748, d=2,\\
15263748, d=3.
$$
Here a descent is any position where a number is followed by a smaller number, for example in $15263478$ the descents are $52$ and $63$.

Hence, we have
$$\Omega^\circ_P(n) = \binom{n}{8} + 6 \binom{n+1}{8} + 6 \binom{n+2}{8} + \binom{n+3}{8}$$

This finally yields
$$\Omega^\circ_P(8) = \binom 8 8 + 6 \binom 9 8 + 6 \binom {10} 8 + \binom {11} 8 = 490.$$

ContentLicense: CC BY-SA 4.0


问:
Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$
Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ if $s_1 = 0, s_2 = 0, s_3 = 1$

I have attempted to use $p_n = c2^{n-2} - d$ [where $h_n = A(3)^n$, but to no avail] - i ended up with $c=-1$ and $d=-\frac{1}{2}$, which is incorrect.

Any help is appreciated! Thanks.

Edit: solution I require is $\frac{1}{2} (3^{n-1}+1-2^n)$

Edit2: Solutions to the homogeneous equation would be of the form $h_n = A(\alpha)^n + B(\beta)^n$, and $p_n$ will exist such that $s_n = h_n + p_n$

答:
A general technique is taught by Wilf's "generatingfunctionology". Define $S(z) = \sum_{n \ge 0} s_n z^n$ and write ($s_0$ you get from the recurrence "backwards", mostly for not having to mess around with indices):
$$
s_{n + 1} = 3 s_n + 2^{n - 1} - 1 \qquad s_0 = \frac{1}{6}
$$
Multiply the recurrence by $z^n$, add over $n \ge 0$ to get:
$$
\begin{align*}
\frac{S(z) - s_0}{z}
  &= 3 S(z) + \frac{1}{2} \sum_{n \ge 0} 2^n z^n - \sum_{n \ge 0} z^n \\
  &= 3 S(z) + \frac{1}{2} \frac{1}{1 - 2 z} - \frac{1}{1 - z}
\end{align*}
$$
Solving for $S(z)$ and expanding in partial fractions:
$$
S(z)
   = \frac{1}{6} \frac{1}{1 - 3 z}
       - \frac{1}{2} \frac{1}{1 - 2 z}
       + \frac{1}{2} \frac{1}{1 - z}
$$
Everything in sight is geometric series:
$$
s_n = \frac{1}{6} \cdot 3^n - \frac{1}{2} \cdot 2^n + \frac{1}{2}
    = \frac{1}{2} \left( 3^{n - 1} - 2^{n - 1} + 1 \right)
$$</p>

ContentLicense: CC BY-SA 3.0


问:
For the discrete topology, why must the singletons be in every basis?
I know the set of all singletons are a basis for a discrete topology, but if we had another basis, why must it also include all the singletons?</p>

答:
That's the only way to get the singletons to be open.  That is, suppose we have a basis $\mathcal B$ such that for a certain $x$, we have $\{x\}\not\in\mathcal B$.  Then $\{x\}$ isn't open, and the topology isn't discrete.

The key seems to be as @HallaSurvivor points out:  a singleton can't be written as the union of proper subsets.


问:
Stability of convergence of series through multiplication
I am asking myself the following question. Can we characterize the set of all sequences $(s_n)_n$ tending to $0$ so that if $(x_n)_n$ is a sequence whose series converges then the series $\sum_{n}s_n x_n$ converges too? I easily proved that this set contains $\ell^1$ but that's all...

答: No. If $s_n$ is any sequence tending monotonically to $0$, for example, then Dirichlet's test together with the convergence of $\sum x_n$ implies the convergence of $\sum s_nx_n$.


问:
prove function in product space is measurable
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a measurable function, define a function $F: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ by $(x,y) \rightarrow f(x-y)$, how do I show that this function is measurable? What I have tried is:

$F^{-1}(-\infty,\alpha) = \{(x,y) \in \mathbb{R} \times \mathbb{R}|  f(x-y) <  \alpha \} = \bigcup\limits_{x \in \mathbb{R}} \{x\} \times \{y \in \mathbb{R} | f(x-y) < \alpha\}$. Since Lebesgue measure is translation invariant, the function $f_{x}: y \rightarrow f(x-y)$ is also measurable, so the set $\{y \in \mathbb{R} | f(x-y) < \alpha\}$ is measurable. The problem is that the union is not over countable set, and I am not sure how to proceed.

答:
This is not specific to $\Bbb{R}$, so let me provide a proof in the general case. Suppose $f:\Bbb{R}^n\to\Bbb{R}$ is Lebesgue measurable, and $F:\Bbb{R}^n\times\Bbb{R}^n\to\Bbb{R}$ is defined as $F(x,y):=f(x-y)$. Now, define the function $T:\Bbb{R}^n\times\Bbb{R}^n\to \Bbb{R}^n\times \Bbb{R}^n$ as $T(x,y):= (x,x-y)$ and $\pi_2:\Bbb{R}^n\times\Bbb{R}^n\to\Bbb{R}^n$ as $\pi_2(x,y):=y$. Then, $F=f\circ\pi_2\circ T$, so for any $A\subset\Bbb{R}$, we have
\begin{align}
F^{-1}(A)&= T^{-1}\bigg[\pi_2^{-1}[f^{-1}(A)]\bigg]= T^{-1}\bigg[\Bbb{R}^n\times f^{-1}(A)\bigg].
\end{align}
Now, if $A$ is actually a Borel set, then by definition of $f$ being Lebesgue-measurable, we have that $f^{-1}(A)$ is a Lebesgue-measurable subset of $\Bbb{R}^n$. Next, the product $\Bbb{R}^n\times f^{-1}(A)$ is a Lebesgue-measurable subset of $\Bbb{R}^n\times\Bbb{R}^n$ (this should be one of the first few things you discuss when talking about Lebesgue measure on a product space). Finally, $T$ is an invertible linear transformation, which in particular means that the function $T^{-1}$ is actually Lipschitz continuous. As a result, it maps Lebesgue-measurable sets to Lebesgue-measurable sets. Thus, $F^{-1}(A)$ is a Lebesgue-measurable subset of $\Bbb{R}^n\times\Bbb{R}^n$, hence $F$ is Lebesgue measurable.

Edit in response to comments regarding measurability

When talking about measurability of functions, one has to be very careful to specify the $\sigma$-algebras involved; not specifying them can easily cause a lot of confusions. If $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ are measurable spaces and $f:X\to Y$ is a function, we say $f$ is $\mathcal{A}$-$\mathcal{B}$-measurable (or measurable with respect to $\mathcal{A}$ and $\mathcal{B}$... however you want to phrase things) if the preimage of every set in $\mathcal{B}$ lies in $\mathcal{A}$, i.e $f^{-1}(\mathcal{B})\subset \mathcal{A}$.

When dealing with a topological space $(Y,\tau)$, a natural choice for a $\sigma$-algebra is the one generated by $\tau$; this is called called the Borel $\sigma$-algebra of $Y$, and we shall denote this as $\mathcal{B}_Y$. So, if you have $(X,\mathcal{A})$ a measurable space, and $(Y,\tau)$ a topological space, you can consider the measurable space $(Y,\mathcal{B}_Y)$ so given a function $f:X\to Y$, we can talk about $\mathcal{A}$-$\mathcal{B}_Y$ measurability as I've defined above. By the way, it is a theorem (not too hard to prove) that to check a function is $\mathcal{A}$-$\mathcal{B}$ measurable, it suffices to check that for some (or equivalently, for any) generating set $\mathcal{S}$ of the $\sigma$-algebra $\mathcal{B}$, we have $f^{-1}(\mathcal{S})\subset \mathcal{A}$. So, if you apply this to the situation of the Borel $\sigma$-algebra of a topological space, we see that saying $f:X\to Y$ is $\mathcal{A}$-$\mathcal{B}_Y$ measurable is equivalent to saying preimage of every open set lies in $\mathcal{A}$ (i.e $f^{-1}(\tau)\subset \mathcal{A}$). This is precisely the definition you cite from Rudin.

So far we've talking quite generally. Let's specialize to $\Bbb{R}^n$. Well, this is a topological space, with the usual toplogy generated by (for example) the Euclidean norm. So we can consider the Borel $\sigma$-algebra $\mathcal{B}_{\Bbb{R}^n}$. This is certainly a natural choice for a $\sigma$-algebra. On the other hand, it is also quite natural to consider the Lebesgue $\sigma$-algebra $\mathcal{L}_{\Bbb{R}^n}$ (which you can for example define using the Caratheodory construction). Hence, we now have two fairly natural choices for $\sigma$-algebras on $\Bbb{R}^n$ (for the purposes we usually have in mind).

It is a fact that we have a strict inclusion: $\mathcal{B}_{\Bbb{R}^n}\subsetneq \mathcal{L}_{\Bbb{R}^n}$. So, if we now have a topological space $(Y,\tau)$, with the Borel $\sigma$-algebra $\mathcal{B}_Y$, and a function $f:\Bbb{R}^n\to Y$, then clearly, we have two different notions of measurability. We have the notion of $\mathcal{B}_{\Bbb{R}^n}$-$\mathcal{B}_{Y}$ measurability (for which we say $f$ is a Borel (measurable) function). We also have the notion of $\mathcal{L}_{\Bbb{R}^n}$-$\mathcal{B}_Y$ measurabiltiy (for which we say $f$ is a Lebesgue-measurable function). So as you can see the definition of Lebesgue-measurable is slightly nuanced.

You may now wonder the following question: for functions $f:\Bbb{R}^n\to\Bbb{R}^m$, why not define Lebesgue-measurability to mean $\mathcal{L}_{\Bbb{R}^n}$-$\mathcal{L}_{\Bbb{R}^m}$ measurability. Well, there aren't many such functions, because there exist even continuous functions which are not $\mathcal{L}_{\Bbb{R}}$-$\mathcal{L}_{\Bbb{R}}$ measurable, whereas every continuous function is $\mathcal{L}_{\Bbb{R}}$-$\mathcal{B}_{\Bbb{R}}$ measurable.


问:
I am looking for an example of an open set in $\operatorname{Spec}(R)$ equipped with the Zariski topology which is not compact.
I have learned the basic open sets i.e. the sets of the form $\{P\in\operatorname{Spec}(R): x\notin P\}$ where $x$ is a fixed element of $R$ are compact. Now any open set is union of such basic open sets. For an example I was thinking of taking $R$ as $k[x_1,x_2,x_3,...]$ then letting $X$=the set of inderminates, the set $U=\{P\in\operatorname{Spec}(R): X\not\subset P\}$ is open in $\operatorname{Spec}(R)$. I am thinking that this open set will be non compact. Am I right with my example? Please help me with it by commenting on my example or by providing a suitable one.

答: Your example is correct. $U$ is equal to the union $\bigcup_{i=1}^{\infty} D(x_i)$ since $X \not\subset p$ means exactly that $\exists i: x_i \not\in p,$ i.e. $p \in D(x_i).$ These is no finite subcover for this cover since for each $x_i$ there is an element in $U$ which lies only in $D(x_i)$ - the ideal generated by $X \setminus \{x_i\}.$


问:
A trivial map from $f:\{0\} \to \Bbb{R}^k$ is always smooth
Let $X = \Bbb{R}^0 = \{0\}$ and $Y = \Bbb{R}^n$ ,prove any map from $f:X\to Y$ is smooth.

I was bit confusing how to show this map is smooth ,since it seems the definition of continuous differential need the point varies in the domain,but here point in $X$ can not vary?

答:
This is a strange question.

However, $\{0\}$ is a vector space (maybe not the most interesting one...). You can endow it with whatever norm. Let say the Euclidean one $\Vert \cdot \Vert_2$. $(X, \Vert \cdot \Vert_2$) is a topological space. The only open subsets are $\emptyset$ and $\{0\}$ itself.

Based on that it is easy to verify (using derivative definition) that any $f$ defined on $X$ is smooth as it is constant...


问:
Rolling 30 indistinguishable dice all at once, such that every die has $6$ sides, How many different results can we get?
I am still trying like to understand the question, so I took an example of rolling two dice together, each with $6$ sides, I can get $\{ 2,3,4,5,6,7,8,9,10,11,12 \} $ as results which is equal to $11$ different results. But I'm not really succeeding at understanding how can I generalize or if I really understood the question correctly.

I would appreciate any help or explanation, thanks in advance.

答:
Find the lowest possible total, and the highest total, and decide whether every sum in between is possible.

OTOH, the way I read the question was that an outcome might be 'three 1s, seven 2s, six 3s, four 4s, two 5s and eight 6s'.  That way needs 'stars and bars', which you can look up.


问:
Matrix derivative: $\frac{\partial}{\partial A_{ij}}(\mathbb{A}(\mathbb{A}^T\Sigma\mathbb{A})^{-1}\mathbb{A}^T)$
Let $\mathbb{A} = (A_{ij})$ be a rectangular, $m\times n$ real matrix, full-rank and with $m \le n$. Let $\mathbb{C}$ be a real $n\times n$ symmetric and pos-def matrix. With these assumptions, $\mathbb{A}^T\mathbb{C}\mathbb{A}$ is invertible. Finally, let $\mathbf{u}$ be an $m$-vector.

I need to compute the following derivative:

$$\frac{\partial}{\partial \mathbb{A}}(\mathbf{u}^T\mathbb{A}(\mathbb{A}^T\mathbb{C}\mathbb{A})^{-1}\mathbb{A}^T\mathbf{u})$$

where $\Sigma$ is considered a constant.

But I keep struggling with the algebra which gets really messy.

Any tricks I can use?

答:
After some algebra, I arrive at the result:

$$\frac{1}{2}\frac{\partial}{\partial \mathbb{A}}(\mathbf{u}^T\mathbb{A}(\mathbb{A}^T\mathbb{C}\mathbb{A})^{-1}\mathbb{A}^T\mathbf{u})
=
\mathbf{u}\mathbf{u}^{\top} \mathbb{A} (\mathbb{A}^{\top}
\mathbb{C}\mathbb{A})^{- 1} -\mathbb{C}\mathbb{A} (\mathbb{A}^{\top}
\mathbb{C}\mathbb{A})^{- 1} \mathbb{A}^{\top} \mathbf{u}\mathbf{u}^{\top}
\mathbb{A} (\mathbb{A}^{\top} \mathbb{C}\mathbb{A})^{- 1}$$

I also checked this result numerically, using automatic differentiation in Julia (Zygote).


问:
Fourier transform of $f$ is $f$
I want to prove that for $f(x)=e^{-\pi x^{2}}$ the Fourier transform of $f$ equals $f$ itselfs.
I want to do it in three given steps (I got the three steps as a hint):

$h=\hat f$. Prove that $h$ is derivative and $h'(t)=-2\pi th(t)$

prove that the function $t\to h(t) e^{\pi t^{2}}$ is constant.

prove $h(0)=1$ and conclude that $\hat f=f$

Now  for (1) I first tried to find $\hat f$. \begin{align}
\hat f(t)&=\int_{\mathbb{R}}f(x) e^{-2\pi itx} \\
      &= \int_{\mathbb{R}}e^{-\pi x^{2}}e^{-2\pi itx} \\
&=....=e^{-\pi t^{2}}
\end{align}
I couldn't solve the integral myself but I find on the internet that the equation should have this answer. Maybe someone can help me to find it?

Now I now that $h(t)=e^{-\pi t^{2}}$ I know that $h'(t)=-2\pi t e^{-\pi t^{2}}=-2\pi t h(t)$

For (2) I proved that the derivative of that function is zero.
\begin{align}
\frac{d(h(t)e^{\pi t^{2}})}{dt}&=\frac{d(h(t))}{dt}e^{\pi t^{2}}+\frac{d(e^{\pi t^{2}})}{dt}h(t)& \\
      &= -2\pi t e^{-\pi t^{2}}e^{\pi t^{2}}+2\pi t e^{\pi t^{2}}e^{-\pi t^{2}} \\
&=0
\end{align}

For (3) $h(0)=e^{-2\pi 0^{2}}=1$. But I don't know how to conlcude that $\hat f=f$.

So in conclusion, I need some help to find the integral in (1) and to make the conclusion in (3).

EDIT: (2) you can also prove without knowing exactly what h(t) is, so if someone has a suggestion to prove (1) in an other way that calculating $\hat f$, then that would be helpfull too.

EDIT2: my question doesn't make sense if I would already find what equals to $h(t)$ so I would like to find another methode for option (1)

EDIT3: can I use for (1) that $h(t)=\hat f (t)= \int_{\mathbb{R}}f(x)e^{-2\pi itx}dx$. So $\frac{d(h(t))}{dt}=\frac{d(\int_{\mathbb{R}}f(x)e^{-2\pi itx}dx)}{dt}=\int_{\mathbb{R}}f(x)\frac{d(e^{-2\pi itx})}{dt}dx)=-2\pi t\int_{\mathbb{R}}f(x)e^{-2\pi itx}dx=-2\pi t h(t) $

答:
As $t\to h(t) e^{\pi t^{2}}$ is constant, you have $h(t) = h(0) e^{-\pi t^{2}}$ (the constant value is $h(0)$).

And as $h(0) = 1$, $\hat f(t) =h(t) = e^{-\pi t^{2}} =f(t)$.

Title: Fourier transform of $f$ is $f$


问:
Closed form for $n$th derivative of exponential of $f$
What is the closed form for:

$$\frac{\partial^n}{\partial x^n}\exp(f(x))=\exp(f(x))\cdot[????]$$

答:
Related problem:(I), (II). Here is a formula for the $n$th derivative of the function $e^{\sin(x)}$

$$\left( e^{\sin(x)}\right)^{(n)}={{ i}^{{n} }}\ e^{\sin ( x )} \sum _{k=0}^{n}
\sum _{j=0}^{n-k}\sum _{m = 0
}^{k}  \frac{\left(-1 \right)^{n-k-m} {i}^{{ j + m}} }{{2}^{(j+m)}}  {n\choose k} \left\{\matrix{n-k\\j}\right\} \left\{\matrix{k\\m}\right\}  {\rm e}^{ i x (m - j)}\\  n \in \mathbb{N},\, i = \sqrt{-1}. $$

where $\left\{\matrix{n\\k}\right\}$ are Stirling numbers of the second kind.

Note: I'll appreciate if someone is interested in verifying this formula with Maple or Mathematica. I already tried it with Maple and it should be fine.


答:
Since $(e^{\pi t^{2}} h(t))'=0$ it follows that $e^{\pi t^{2}} h(t)$ is a consatnt . Since it has the value $1$ when $t=0$ we get $e^{\pi t^{2}} h(t)=1$ for all $t$. So $\hat {f} (t)=h (t)=e^{-\pi t^{2}} $.

For (1) search the net for characteristic function of Gaussian distribution.

Title: Fourier transform of $f$ is $f$

Body:
I want to prove that for $f(x)=e^{-\pi x^{2}}$ the Fourier transform of $f$ equals $f$ itselfs.
I want to do it in three given steps (I got the three steps as a hint):

$h=\hat f$. Prove that $h$ is derivative and $h'(t)=-2\pi th(t)$

prove that the function $t\to h(t) e^{\pi t^{2}}$ is constant.

prove $h(0)=1$ and conclude that $\hat f=f$

Now  for (1) I first tried to find $\hat f$. \begin{align}
\hat f(t)&=\int_{\mathbb{R}}f(x) e^{-2\pi itx} \\
      &= \int_{\mathbb{R}}e^{-\pi x^{2}}e^{-2\pi itx} \\
&=....=e^{-\pi t^{2}}
\end{align}
I couldn't solve the integral myself but I find on the internet that the equation should have this answer. Maybe someone can help me to find it?

Now I now that $h(t)=e^{-\pi t^{2}}$ I know that $h'(t)=-2\pi t e^{-\pi t^{2}}=-2\pi t h(t)$

For (2) I proved that the derivative of that function is zero.
\begin{align}
\frac{d(h(t)e^{\pi t^{2}})}{dt}&=\frac{d(h(t))}{dt}e^{\pi t^{2}}+\frac{d(e^{\pi t^{2}})}{dt}h(t)& \\
      &= -2\pi t e^{-\pi t^{2}}e^{\pi t^{2}}+2\pi t e^{\pi t^{2}}e^{-\pi t^{2}} \\
&=0
\end{align}

For (3) $h(0)=e^{-2\pi 0^{2}}=1$. But I don't know how to conlcude that $\hat f=f$.

So in conclusion, I need some help to find the integral in (1) and to make the conclusion in (3).

EDIT: (2) you can also prove without knowing exactly what h(t) is, so if someone has a suggestion to prove (1) in an other way that calculating $\hat f$, then that would be helpfull too.

EDIT2: my question doesn't make sense if I would already find what equals to $h(t)$ so I would like to find another methode for option (1)

EDIT3: can I use for (1) that $h(t)=\hat f (t)= \int_{\mathbb{R}}f(x)e^{-2\pi itx}dx$. So $\frac{d(h(t))}{dt}=\frac{d(\int_{\mathbb{R}}f(x)e^{-2\pi itx}dx)}{dt}=\int_{\mathbb{R}}f(x)\frac{d(e^{-2\pi itx})}{dt}dx)=-2\pi t\int_{\mathbb{R}}f(x)e^{-2\pi itx}dx=-2\pi t h(t) $


问:
Fit sum of multivariate exponentials
I am working on a curve fitting task for a function which is multivariate sum of two exponentials.
The function which I am trying to fit has below form:
$$z(x,y)=ae^{bx}y+ce^{dx}$$
and want to find a, b, c, d values given values of x, y and z from measurement data.
Has anyone already worked on such function?
It seems not possible to linearize the problem using function integrals as done for fitting of sum of exponentials with single variable because second variable y is present in only one exponential term.

答:
The method of using integral equation to linearize the problem is explained in https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales . The case of two exponentials is treated pp.71-74. More exponentials are possible as well.

But the theory is done only in case of one variable.

In this paper the case of two variables is passed over superficially.pp.75-80. It is shown that the method can be extended in case of two variables if the data has a specific structure as shown in the numerical example pp.81-84 .

The case of $z(x,y)=ae^{bx}y+ce^{dx}$ is not explicitly treated. This would need more studies. The structure of the data (How the experimental points are presented in a table) is very important in practical application. In order to give a preliminary  advice on the feasibility in your case, you should joint to your question an example of data.

Title: Fit sum of multivariate exponentials


问:
Complement of closed convex set in Hilbert space
Let $H$ be a real Hilbert space and $C$ be a closed convex subset of $H$.

Is it true that if $x$ is a boundary point of $C$ then there exists $y\in H\setminus C$ such that the set $\{x+t(y-x);\;t>0\}$ lies in $H\setminus C$ ?

Thank you for any hint.

答:
I will use the following fact: if $x\in C$ and $x_0 \in \operatorname{int} C$ then $\lambda x + (1-\lambda)x_0 \in \operatorname{int} C$ for all $\lambda\in (0,1)$.

If $C$ has interior points, then the following construction works: Let $x_0$ be an interior point of $C$. Set $y:=x-x_0$.
Then $x+ty=(1+t)x-tx_0 $ cannot be in $C$, otherwise $x = \frac1{1+t}(x+ty) + \frac t{1+t} x_0$ would be an interior point of $C$.

If $x$ is the projection of a point $y\not\in C$, then $x+t(y-x)\not\in C$ for all $t>0$ as well: For $t\in (0,1)$ it follows from the projection property, for $t>1$ it follows with the same argument above.

Still there are situation, where both assumptions are not valid: Take $H=L^2(0,1)$, $C$ the cone of non-negative functions. It has empty interior. The point $x\equiv 1$ is a boundary point of $C$ that is not the projection of an element outside. Still we can construct the desired $y$: $y(x) = -|x|^{1/4}$.


问:
Intersection of a family of compact sets in Hausdorff topological space is compact
Let $\{G_i\}_{i \in \Lambda}$ be a family of compact sets in a Hausdorff topological space. Show that the intersection $\bigcap_{i \in \Lambda} G_i$ is compact.

My attempt was:

Since $G_j$ for some $j \in \Lambda$ is compact, there exists an open, finite covering. And $\bigcap_{i \in \Lambda} G_i \subset G_j$. So I can take the same covering for the intersection as for $G_j.$

Is that correct?

答:
I see my mistakes and realised that I misunderstood the definition of compactness. I would correct the proof as following:

Since a compact set in a Hausdorff topological space is closed, it follows that $G_i \forall i \in \Lambda$ is closed. Hence, $\bigcap_{i \in \Lambda} G_i$ is closed. Now define $F:=(\bigcap_{i \in \Lambda} G_i)^{c}$ which is the complement. Note that $F$ is open.

Let $\bigcup_{i \in I} U_i$ be an open covering of $\bigcap_{i \in \Lambda} G_i$. My goal is to show that this covering is finite or has a finite sub-covering.

Note that $\bigcup_{i \in I} U_i \cup F$ is the whole space. In particular it is a covering of $G_{i_0}$ for $i_0 \in \Lambda$. By compactness of $G_{i_0}$ there exists finite $I' \subset I$ s.t. $G_{i_0} \subset $ $\bigcup_{i \in I'} U_i \cup F$.

Note that:

$\bigcap_{i \in \Lambda} G_i \subset G_{i_0} \subset  \bigcup_{i \in I'} U_i \cup F$. But since $\bigcap_{i \in \Lambda} G_i \cap F = \emptyset$ we have that:

$\bigcap_{i \in \Lambda} G_i \subset \bigcup_{i \in I'} U_i$. So we have found a finite open covering and hence $\bigcap_{i \in \Lambda} G_i$ is compact.

Is that correct?


问:
Prove that the power set of an $n$-element set contains $2^n$ elements

Theorem. Let $X$ denote an arbitrary set such that $|X|=n$. Then $|\mathcal P(X)|=2^n$.

Proof. The proof is by induction on the numbers of elements of $X$.

For the base case, suppose $|X|=0$. Clearly, $X=\emptyset$. But the empty set is the only subset of itself, so $|\mathcal P(X)|=1=2^0$.

Now, the induction step. Suppose $|X|=n$; by the induction hypothesis, we know that $|\mathcal P(X)|=2^n$. Let $Y$ be a set with $n+1$ elements, namely $Y=X\cup\{a\}$. There are two kinds of subsets of $Y$: those that include $a$ and those that don't. The first are exactly the subsets of $X$, and there are $2^n$ of them. The latter are sets of the form $Z\cup\{a\}$, where $Z\in\mathcal P(X)$; since there are $2^n$ possible choices for $Z$, there must be exactly $2^n$ subsets of $Y$ of which $a$ is an element. Therefore $|\mathcal P(Y)|=2^n+2^n=2^{n+1}$. $\square$

Image that replaced text.

From the above explanation, I don't understand why the set that contains $\{a\}$ will contain $2^{|n|}$ elements when it should clearly be $2^{|1|}$.

The construction of a new set $S$ is the union of the old set with cardinality $n$ and a new element $\{a\}$, therefore the set that does not contain $\{a\}$ still has cardinality $n$ and the set that contains $\{a\}$ is just $\{a\}$, one element.

Can someone please elucidate?

答:
Here is another one that uses the Binomial Theorem.

Take, for example, the set $A = \{1, 2, 3, 4\}$

Then, $P(A) = \{∅, \{1\}, \{2\}, \{3\}, \{4\}, \{1, 2\}, \{1, 3\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \{4, 1\}, \{1, 2, 3\}, \{2, 3, 4\}, \{3, 4, 1\}, \{4, 1, 2\}, \{1, 2, 3, 4\}\}$

In $P(A)$, see that the number of subsets with cardinality 0 is 1, there are 4 subsets with cardinality 1, 6 with cardinality 2, then 4 again with 3 and then 1 more with cardinality 4. $\pmb{1-4-6-4-1}$; $16$ total. Note that these are the coefficients of the terms when you expand the binomial theorem (binomial coefficients) with $n=4$.

Now we want the sum of these numbers. This is can be easily achieved by replacing $x=y=1$ in the binomial formula where $n$ is the cardinality of set A and the formula follows from there.

Why the pattern appears is very simple. Let $n$ be $|A|$. You want to fill subsets of cardinalities $0, 1, \cdots,\ n$. You have n elements to choose from. So to fill the first subset with cardinality 0, you use $\binom{n}{0}$, for the second with cardinality 1 we use $\binom{n}{1}$ and so on.


问:
What will be the support of the convolution of two test functions.
If $g\in C^{\infty}_c$ defined on $\Bbb R^n$ and K is the support of function $g$. I want to find the support of $g_\epsilon$. Where $g_\epsilon$ is regularization of $g$.

Regularization of $g$ is defined as:

$g_\epsilon$:=  $g*\omega_\epsilon$ (convolution of g and a test function) i.e. $g_\epsilon(x
)$=$\int_{\Bbb R^n}g(x-y).\omega_\epsilon(y)dy $

I think, this process is also known as mollification.

Here
$ \omega_\epsilon (x)=
\begin{cases}
\frac{C^{-1}}{\epsilon^n} e^{{\frac{-\epsilon^2}{{\epsilon^2}-|x|^2}}},  &\text{for |x|< $\epsilon$ } \\
0, & \text{otherwise}  \\
\end{cases} $

I only know that support of convolution of two compactly supported functions is again a compact set. And, as support of $\omega_\epsilon $ is a ball $B(0,\epsilon)$ and support of g is a compact set K(given) then support of their convolution should be the intersection of the support of $g(x-y)$ and $\omega_\epsilon(y)$.
Because for non zero function integration remains non zero...

答:
Definition: Let $f:\mathbb{R}^N\to\mathbb{R}$ be a function. Consider the family $\omega_i$ of all open sets on $\mathbb{R}^N$ such that for each $i$, $f=0$ a.e. in $\omega_i$, then $f=0$ a.e. on $\omega=\cup\omega_i$ and we define the support of $f$ by $$\operatorname{supp}(f)=\mathbb{R}^N\setminus\omega$$

Suppose that $f,g\in C^\infty(\mathbb{R})\cap L^1(\mathbb{R})$. We can prove that $\operatorname{supp}(f\ast g)\subset\overline{\operatorname{supp}(f)+\operatorname{supp}(g)}$.

Take any $x\in\mathbb{R}$ and note that $$(f\ast g)(x)=\int_\mathbb{R}f(x-y)g(y)=\int_{(x-\operatorname{supp}(f))\cap \operatorname{supp}(g)}f(x-y)g(y)$$

For $x\notin \operatorname{supp}(f)+\operatorname{supp}(g)$, we have that $(x-\operatorname{supp}(f))\cap \operatorname{supp}(g)=\emptyset$ and then $(f\ast g)(x)=0$. It follows that if $x\in (\operatorname{supp}(f) + \operatorname{supp}(g))^c$ then $$(f\ast g)(x)=0$$

Therefore $$\operatorname{supp}(f\ast g)\subset\overline{\operatorname{supp}(f)+\operatorname{supp}(g)}$$

Now let's prove that the equality is not necessary. Consider the functions in $\mathbb{R}$ defined by $f(x)=x\chi_{[-1,1]}(x)$ and $g(x)=\chi_{[-2,2]}(x)$, where $\chi_A$ is the characteristic function of the set $A$.

Note that in this case $\overline{\operatorname{supp}(f)+\operatorname{supp}( g)}=[-3,3]$, however, the interval $(-1,1)$ does not belong to the support of $f\star g$ (which is equal to $[-3,-1]\cup [1,3]$).

Remark 1: The proof I gave here can be found in Brezis's book of Functional Analysis.

Remark 2: I don't know a expliclty characterization of the support of the convolution, but by the given formula, you can see that if the two functions has compact support, then does the convolution.

Update: I have corrected some errors in the text.


问:
How to Solve Large Scale Linear Least Squares with Simplex Constraints
I need to solve a rather simple optimization problem, however the structure of the problem makes out-of-the-box commercial solvers fail. The problem is:\begin{aligned}
&\min_{x_i}\|\sum_{i=1}^{10}x_i\mathbf{a}_i - \mathbf{b}\|_2^2 \\
&s.t.: \sum_{i=1}^{10}x_i = 1,\; x_i\geq0
\end{aligned}
The decision variables are $x_i$, which are scalars, and typically the problem will have 10 variables. The complexity comes from the fact that $\mathbf{a}_i, \mathbf{b}$ are VERY LONG constant vectors, varying from 10,000 to 100,000 elements and possibly more. Without going too much into details, black-box solvers fail since they impute the variable over vectors, thus converting a 10-D problem to 100K-vars problem.

Since the problem has a simple structure (only complicated by the constant vectors), I think that interior point algorithms can perform well here. Also, the Hessian is very simple and easy to compute - it's $ij$-th entry is simply $\langle \mathbf{a}_i, \mathbf{a}_j \rangle$, with size $10\times 10$. I'm very familiar with optimization, however my experience is mostly with first-order algorithms. I'm almost clueless when it come to IPMs and not sure how to implement an appropriate algorithm. Does anyhow have a ready script I can use? Or any other tips that might help?

答:
Since you mention cvxpy I wrote quick code and it works without any problem. $A,b$ are like in the other answer.

```<code>import cvxpy as cp
import numpy as np

N = 500000
K = 10

A = np.random.uniform(size=(N,K))
b = np.random.uniform(size=(N))

x = cp.Variable(K)
prob = cp.Problem(cp.Minimize(cp.norm(A@x-b)), [cp.sum(x)==1,x>=0])

prob.solve(solver=cp.MOSEK,verbose=True)
</code>```

Body:
Since you mention cvxpy I wrote quick code and it works without any problem. $A,b$ are like in the other answer.

import cvxpy as cp
import numpy as np

N = 500000
K = 10

A = np.random.uniform(size=(N,K))
b = np.random.uniform(size=(N))

x = cp.Variable(K)
prob = cp.Problem(cp.Minimize(cp.norm(A@x-b)), [cp.sum(x)==1,x>=0])

prob.solve(solver=cp.MOSEK,verbose=True)


问:
Do universally connected spaces exist?
Let's call a topological space $X$ connected iff it can not be represented as a disjoint union of two non-empty open sets.

Now, suppose $X$ and $Y$ are both topological spaces. Let's say $X$ is $Y$-connected iff $\forall x_0, x_1 \in X$ $\exists y_0, y_1 \in Y$ and a continuous function $f: Y \to X$ such that $f(y_0) = x_0$ and $f(y_1) = x_1$.

Those two variants of connectivity are connected in the following ways:

If $X$ is $Y$-connected and $Y$ is connected then $X$ is also connected.

Proof: Suppose, $X$ is not connected. Then it is a disjoint union of two non-empty open sets $U$ and $V$. Let's take $x_0 \in U$ and $x_1 \in V$. Then $Y$ will not be connected as a disjoint union of two non-empty open sets $f^{-1}(U)$ and $f^{-1}(V)$, Q.E.D.

Converse is however generally false - for example, $\{(t, \sin(\frac{1}{t})|t \in \mathbb{R})\} \cup \{(0, 0)\}$ is connected, but not $[0;1]$-connected.

If $Y$ is not connected, then all topological spaces are $Y$-connected.

Proof: Suppose, $Y$ is not connected. Then it is a disjoint union of two non-empty open sets $U$ and $V$. Now, suppose $X$ is an arbitrary topological space and $x_0, x_1 \in X$. Now we take $y_0 \in U$, $y_1 \in V$ and define $f$ as

$$f(y) = \begin{cases} x_0 & \quad y \in U \\ x_1 & \quad y \in V \end{cases}$$

Those $y_0, y_1$ and $f$ satisfy our conditions. That means, $X$ is $Y$-connected, Q.E.D.

If $X, Y$ and $Z$ are such topological spaces, that $X$ is $Y$-connected and $Y$ is $Z$-connected then $X$ is $Z$-connected.

Suppose $x_0, x_1 \in X$. Thes $\exists y_0, y_1 \in Y$ and and a continuous function $f: Y \to X$ such that $f(y_0) = x_0$ and $f(y_1) = x_1$. There also $\exists z_0, z_1 \in Z$ and a continuous function $g: Z \to Y$ such that $f(z_0) = y_0$ and $f(z_1) = y_1$. Then $f(g(z_0)) = x_0$ and $f(g(z_1)) = x_1$, which means $X$ is $Z$-connected, Q.E.D.

Now let's call a topological space $Y$ universally connected iff a topological space being $Y$-connected is equivalent to it being connected (equivalently, $Y$ is connected and all connected spaces are $Y$-connected)

My question is:

Do universally connected spaces exist?

答:
No such space exists.

Assume $Y$ is universally connected. Now consider $\lambda$ to be any ordinal with the corresponding cardinality greater then that of $Y$. Let $X=\lambda\times[0,1)\sqcup\{\lambda\}$ with the lexicographical order/topology. The extended extremely long line. ;) The space $X$ is connected and in fact connected subsets of $X$ are intervals. In particular if $A\subseteq X$ is a connected subset containing the beginning $x_0$ and the end $x_1$ then $A=X$. This means that if $f:Y\to X$ is any continuous map with $\{x_0,x_1\}$ in the image then $f(Y)=X$. But this is impossible due to cardinality.

Title: Do universally connected spaces exist?


问:
Analytic solution to $\alpha , \beta \in \mathbb{R}$ such that $\cos \alpha \cdot \sin \beta = \cos \bigl(\sin (\alpha \cdot \beta)\bigr)$?
Are there any $\alpha , \beta \in \mathbb{R}$ such that $$\cos \alpha \cdot \sin \beta = \cos \bigl(\sin (\alpha \cdot \beta)\bigr)?$$

The trivial solutions are $(\alpha, \beta)=(0, \dfrac{\pi}{2})$. But are there more? I dont see an obvious way of tackling this problem but I keep coming back to it because it seems interesting to me (look at the graph of the solutions in desmos: https://www.desmos.com/calculator/ms8ad8cxqt).

I tried simplifying matters by trying out the case where $\alpha = \beta$ where we get that $$\cos \alpha \cdot \sin \alpha = \cos \bigl(\sin (\alpha ^{2})\bigr)$$ and using the fact that $2\sin \alpha \cos \alpha = \sin 2\alpha$ we really want to find $\alpha \in \mathbb{R}$ such that $$\dfrac{1}{2}\sin 2\alpha = \cos \bigl(\sin (\alpha ^{2}) \bigr),$$ though I have to admit that this does not make the problem easier (it seems to me at least).

Has anyone looked at this problem before and have a solution or maybe some hints or suggestions on how to go further in solving the problem?

答:
Here's one thing i did with this problem. Please let me know if it is legitimate or maybe trivial, hehe.

Suppose $\alpha , \beta \in \mathbb{R}$ where $\alpha \approx 0$ and $\beta \approx 0$, then it immediately follows that $\alpha \cdot \beta \approx 0$.

Now what I was thinking is that we can use the small angle approximations of $\cos \theta \approx 1-\dfrac{\theta ^{2}}{2}$ and $\sin \theta \approx \theta$ and get that $$\cos \alpha \cdot \sin \beta \approx \Bigl(1-\dfrac{\alpha ^{2}}{2}  \Bigr)\cdot \beta = \beta - \dfrac{\alpha ^{2}\beta}{2}$$ and $$\cos \Bigl(\sin \alpha \cdot \beta \Bigr)\approx \cos \alpha \cdot \beta \approx 1-\dfrac{\alpha^{2}\beta^{2}}{2}.$$ So now we may ask: for which $\alpha , \beta \in \mathbb{R}$ does $$\beta - \dfrac{\alpha ^{2}\beta}{2}=1-\dfrac{\alpha^{2}\beta^{2}}{2}?$$

After some algebraic manipluations we get that $$\dfrac{(\beta -1)(\beta \cdot \alpha^{2}+2)}{\alpha^{2}}= 0 \Leftrightarrow \boxed{\beta = 1 \: \text{or} \: \beta = -\dfrac{2}{\alpha^{2}}}.$$ Since it is assumed that $\beta \approx 0$ I dont know if the solution $\beta = -\dfrac{2}{\alpha^{2}}$ is even legitimate since if $\alpha \approx 0$, then $\Bigl|-\dfrac{2}{\alpha^{2}}\Bigr|>>0.$ Anyhow, if we let $\beta = 1$ (can this be considered $\beta \approx 0$?) and let $\alpha$ be something $\approx 0$, say $\alpha = 0.001$, then we get $$\cos \alpha \cdot \sin \beta =\cos 0.001 \cdot \sin 1 \approx 0.\bar{9}\cdot 0.84147\approx 0.84147$$ and $$\cos \Bigl( \sin \alpha \cdot \beta \Bigr)\approx \cos \Bigl( \sin 0.001\cdot 1 \Bigr)\approx \cos 0.\bar{9}\approx 0.\bar{9}.$$ Thus $$\cos \alpha \cdot \sin \beta \approx \cos \Bigl( \sin \alpha \cdot \beta \Bigr),$$ whenever $\beta =1$ and $\alpha \approx 0$.


问:
Conditioning a summand on a sum of independent random variables
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space

Denote by $\mathcal{B}$ the Borel $\sigma$-algebra on $\mathbb{R}$, denote by $\lambda$ the Lebesgue measure on $(\mathbb{R}, \mathcal{B})$. We will use the term random variable to refer to a function $\varphi : \Omega\rightarrow\mathbb{R}$ that is $\mathcal{F}/\mathcal{B}$-measurable. For every random variable $\varphi$, we will write $R_{\varphi}$ to denote $\varphi$'s range, and we will write $\mathbb{P}_{\varphi}$ to denote $\varphi$'s distribution, in other words $\mathbb{P}_{\varphi}:\mathcal{B}\rightarrow[0,1]$ is the function satisfying $\mathbb{P}_{\varphi}(B) = \mathbb{P}\big(\varphi^{-1}(B)\big)$ for every Borel set $B \in \mathcal{B}$. For every random variable $\varphi$ we will denote by $\mathbb{E}(\varphi)$ $\varphi$'s expectation w.r.t. $\mathbb{P}$, and for every pair of random variables $\varphi, \psi$ we will denote by $\mathbb{E}(\varphi|\psi)$ the conditional expectation of $\varphi$ given $\psi$ w.r.t. $\mathbb{P}$.

Now, let $X, Y$ be random variables such that

$X$ and $Y$ are mutually independent w.r.t. $\mathbb{P}$,

$R_X$ is countable (i.e. $R_X$'s cardinality is $\leq \aleph_0$),

$\mathbb{P}_Y \ll \lambda$,

$X$ is integrable in $(\Omega, \mathcal{F}, \mathbb{P})$.

I'd like to find $\mathbb{E}(X|X+Y)$, up to a $\mathbb{P}$-null set.

I managed to solve this problem for a special case (see below), but I need help with the general case.

Solution for a special case

I will solve the question for the special case that

$\inf\ \big\{|x - y|\ :\ x, y \in R_X,\ x \neq y \big\} \in (0,\infty)$.

$R_Y \subseteq \big(-\frac{a}{2}, \frac{a}{2}\big)$, where $a$ is the above infimum.

Define $g$ to be the function $g:\mathbb{R}\rightarrow\big(R_X\cup\{0\}\big)$ satisfying for every $r \in \mathbb{R}$ that $g(r) = x$ if $r \in \big(x-\frac{a}{2}, x+\frac{a}{2}\big)$ for some $x \in R_X$ and $g(r) = 0$ otherwise.

$g$ is $\mathcal{B}/\mathcal{B}$-measurable and satisfies: $g\circ(X+Y) = X$. Therefore $g\circ(X+Y)$ is $\sigma(X+Y)/\mathcal{B}$-measurable ($\sigma(X+Y)$ being the $\sigma$-algebra on $\Omega$ generated by $X+Y$) and we have for every $A \in \sigma(X+Y)$,
$$
\mathbb{E}\Big(\big(g\circ(X+Y)\big)\mathbb{1}_A\Big) = \mathbb{E}(X\mathbb{1}_A).
$$

This shows that $g\circ(X+Y) = \mathbb{E}(X|X+Y)$, up to a $\mathbb{P}$-null set. QED

Remark

Note that my special-case solution makes no use of the fact that $X$ and $Y$ are independent, nor of the fact that $\mathbb{P}_Y \ll \lambda$. Also, the fact that $R_X$ is countable can be deduced from the assumption that $\inf\ \big\{|x - y|\ :\ x, y \in R_X,\ x \neq y \big\} \in (0,\infty)$.

答:
Let $p$ and $f$ denote the pmf and pdf of $X$ and $Y$, respectively. Then for $\epsilon>0$,
$$
h_z(\epsilon):=\mathsf{P}(|X+Y-z|\le \epsilon)=\sum_{x\in R}p(x)\int_{z-x-\epsilon}^{z-x+\epsilon}f(y)\,dy
$$
and
$$
g_z(\epsilon):=\mathsf{E}[X1\{|X+Y-z|\le \epsilon\}]=\sum_{x\in R}xp(x)\int_{z-x-\epsilon}^{z-x+\epsilon}f(y)\,dy.
$$
Finally,
$$
\mathsf{E}[X\mid X+Y=z]=\lim_{\epsilon\to 0}\frac{g_z(\epsilon)}{h_z(\epsilon)}.
$$


问:
Find partial sums formula of series: $\sum_{n=1}^{\infty}\frac{n}{1\cdot3\cdot5\cdot...\cdot(2n+1)}$
$$\sum_{n=1}^{\infty}\frac{n}{1\cdot3\cdot5\cdot...\cdot(2n+1)}=\sum_{n=1}^{\infty}a_n$$
For $n>1$: $$a_n=\frac{1}{2}(\frac{(2n+1)-1}{1\cdot3\cdot5\cdot...\cdot(2n+1)})\\=\frac{1}{2}(\frac{1}{3\cdot5\cdot...\cdot(2n-1)}-\frac{1}{3\cdot5\cdot...\cdot(2n+1)})$$
and $a_1=\frac1{3}$, hence
$$S_n=\frac1{3}+\frac1{2}(\frac1{3}-\frac1{3\cdot5\cdot\cdot...\cdot(2n+1)}).$$
I'm not sure why is there $\frac1{3}$inside the bracket and would be glad if someone clarified it.
(copied from a workbook)

Body:
It's a telescoping series.

Let $u_n=\frac{1}{3\cdot5\cdots(2n+1)}$, then $u_1=\frac13$, $a_n=\frac12(u_{n-1}-u_n)$ and

$$S_2=a_1+a_2=a_1+\frac12(u_1-u_2)$$

$$S_3=a_1+a_2+a_3=a_1+\frac12(u_1-u_2+u_2-u_3)=a_1+\frac12(u_1-u_3)$$

$$S_n=a_1+\frac12(u_1-u_2+u_2-\dots+u_{n-1}-u_n)=a_1+\frac12(u_1-u_n)$$

You can also prove this by induction.


问:
Ring Inside an Algebraic Field Extension
Let $E|F$ be an algebraic field extension and a ring $K$ such that $F\subseteq K\subseteq E$. It is true that $K$ is a field?</p>

答: Yes. Suppose $0\ne k\in K$. Since $k\in E$ and $E/F$ is algebraic, we have some minimal polynomial $x^n+a_{n-1}x^{n-1}+\cdots+a_0$ with coefficients in $F$ that is satisfied by $k$. By minimality the coefficient $a_0$ must be nonzero, so it has an inverse $a_0^{-1}$ in $F$. Then $k(-a_0^{-1})(k^{n-1}+a_{n-1}k^{n-2}+\cdots+a_1)=1$, so $k^{-1}=(-a_0^{-1})(k^{n-1}+a_{n-1}k^{n-2}+\cdots+a_1)$. Since $k$ and each $a_i$ are in $K$, it follows that $k^{-1}$ is in $K$. Thus $K$ is a field.</p>


问:
How to proof squareroot inequality by induction?
Hello so I'm making an exercise where you need to proof the following:
$$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\leqq2\sqrt{n}\text{ with }n\in\mathbb{N}$$
But I seem to be stuck, this is what I have:

Base Case

Take $n$=1, then $\frac{1}{\sqrt{1}}\leqq 2\sqrt{1}$, which is true

Induction case

take $k$ $\in\mathbb{N}$ and assume that: $$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}\leqq2\sqrt{k}$$
add k+1, then:
$$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\leqq2\sqrt{k+1}$$
This is where I'm stuck, normally you could replace $1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}$ with $2\sqrt{k}$ and then prove that the lefthand side equals the righthandside. But because its less than or equal I (at least I think)  can't replace it, so now I can't show if its less than or equal. So I can't seem to find a way to proof it by induction.

If you know the answer please share your wisdom

Thanks in advance.

答:
We know $(\sqrt {k+1} - \sqrt k)^2 \gt 0$

$2k + 1 - 2 \sqrt {k+1} \sqrt k \gt 0$

$2(k + 1) - 2 \sqrt {k+1} \sqrt k \gt 1$

$2(\sqrt{k + 1} - \sqrt k) \gt \frac{1}{\sqrt {k+1}}$

This should help.


答:
You know that
$$1+\frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{k}} \leq 2\sqrt{k}$$
from the induction hypothesis. Add $\frac{1}{\sqrt{k+1}}$ on both sides to get
$$1+\frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{k}} + \frac{1}{\sqrt{k+1}} \leq 2\sqrt{k} + \frac{1}{\sqrt{k+1}}.$$
Now, it only remains for you to show that
$$2\sqrt{k} + \frac{1}{\sqrt{k+1}} \leq 2\sqrt{k+1}.$$
To do so, try to collect all the terms with a square-root on one side of the inequality, then square both sides. You should eventually get to something like $0 \leq 1$ which is trivially fulfilled.

Title: How to proof squareroot inequality by induction?

Body:
Hello so I'm making an exercise where you need to proof the following:
$$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\leqq2\sqrt{n}\text{ with }n\in\mathbb{N}$$
But I seem to be stuck, this is what I have:

Base Case

Take $n$=1, then $\frac{1}{\sqrt{1}}\leqq 2\sqrt{1}$, which is true

Induction case

take $k$ $\in\mathbb{N}$ and assume that: $$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}\leqq2\sqrt{k}$$
add k+1, then:
$$1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+1}}\leqq2\sqrt{k+1}$$
This is where I'm stuck, normally you could replace $1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{k}}$ with $2\sqrt{k}$ and then prove that the lefthand side equals the righthandside. But because its less than or equal I (at least I think)  can't replace it, so now I can't show if its less than or equal. So I can't seem to find a way to proof it by induction.

If you know the answer please share your wisdom

Thanks in advance.


问:
Proof regarding the differentiability of arccos.

Okay so this is my attempt of proving a, this was probably one of the most confusing things I did in my 2 month career in calculus with all the $x$'s and $y$'s and figuring out which is which so I'm not confident at all about the solution…

Regarding b, I just couldn't figure it out with the tool we currently have… we cannot use L'Hopital and we've just only started messing around with differential functions, so I'm stuck with picking the right variable to switch to in the one sided limits…

Question 1.

a.

Consider the function $g: (-1,1) -> \Bbb R, g(x) = \cos(x)$.

Note that we've seen in class that $g$ is both continuous and differential on $(-1,1)$.

We also know that $g'(x) = -\sin(x) \ne 0$, and $g^{-1}(y) = \arccos(y)$ for every $y \in (-1,1)$.

Also, by the algebra of continuous functions $g^{-1}$ is continuous in $(-1,1)$.

Therefore, we have

$f'(y) =\left(g^{-1}\right)^{'}\left(y\right)=\frac{1}{g'\left(x\right)}=\frac{1}{-\sin\left(x\right)}=\frac{1}{-\sin\left(\arccos\left(y\right)\right)}$

Note that we know $\sin^{2}\left(\arccos y\right)+\cos^{2}\left(\arccos y\right)=1$, then we have

$\sin^{2}\left(\arccos y\right)=1-\cos^{2}\left(\arccos y\right)$

$\sin^{2}\left(\arccos y\right)=1-y^{2}$

$\sin\left(\arccos y\right)=\pm\sqrt{1-y^{2}}$

Note that for every $y \in (-1,1)$, $\sin\left(\arccos y\right) > 0$, Therefore

$\sin\left(\arccos y\right)=\sqrt{1-y^{2}}$, which means

$f'(x) = \frac{1}{-\sin\left(\arccos x\right)}=\frac{-1}{\sqrt{1-x^{2}}}$

b.

Assume toward contradiction that $f$ is differentiable at $1$. Then we have

$\lim _{x\to 1^-}\frac{\left(\arccos\left(x\right)-\arccos\left(1\right)\right)}{x-1}=\lim_{x\to \:1^-}\frac{\left(\arccos\left(x\right)-0\right)}{x-1}$

We use a change of variable, $t = \arccos x$.

$\lim _{t\to 0^+}\frac{\left(t\right)}{\cos\left(t\right)-1}=\lim _{t\to 0^+}\frac{\left(t\right)}{\cos\left(t\right)-1}\cdot \frac{t}{t}=\lim _{t\to 0^+}\frac{\left(t\right)^2}{\cos\left(t\right)-1}\cdot \frac{1}{t}$

$=\lim _{t\to 0^+}\frac{1}{\frac{\left(\cos\left(t\right)-1\right)}{t^2}}\cdot \frac{1}{t}=\lim_{t\to0^+}-2\cdot \frac{1}{0^+}=-\infty$

Since the limit doesn't exist, we have a contradiction.

Assume toward contradiction f is differentiable at $-1$. Then we have

$\lim _{x\to -1^+}\frac{\left(\arccos\left(x\right)-\arccos\left(-1\right)\right)}{x+1}=\lim \:_{x\to \:-1^+}\frac{\left(\arccos\left(x\right)-\pi \right)}{x+1}$

We use a change of variable, $t = \arccos x$.

$\lim _{t\to \pi^+}\frac{\left(t\right)-\pi }{\cos\left(t\right)+1}$

We use a change of variable, $u = t - \pi$.

Note that $\cos(x + \pi) = -\cos(x)$.

$\lim _{u\to 0^-}\left(\frac{u}{cos\left(u+\pi \right)+1}\right)=\lim _{u\to 0^-}\left(\frac{u}{-cos\left(u\:\right)+1}\right)\cdot \frac{\left(cos\left(u\right)\:+1\right)}{cos\left(u\right)+1}=\lim _{u\to 0^-}\left(\frac{u\cdot \left(cos\left(u\right)+1\right)}{sin^2\left(u\right)}\right)=\lim _{u\to 0^-}\frac{\left(u\right)}{sin\left(u\right)}\cdot \frac{\left(u\right)}{sin\left(u\right)}\cdot \frac{1}{u}\cdot \left(cos\left(u\right)+1\right)=-\infty $

Since the limit doesn't exist, we have a contradiction.

答: Suppose that $\arccos$ is differentiable at $1$. The $\cos\circ\arccos$ is also differentiable at that point. But\begin{align}(\cos\circ\arccos)'(1)&=\cos'(\arccos(1))\arccos'(1)\\&=-\sin(0)\arccos'(1)\\&=0.\end{align}However, $\left(\forall x\in[-1,1]\right):\cos\bigl(\arccos(x)\bigr)=x$, and therefore $(\cos\circ\arccos)'(1)=1$. The same argument works for $-1$


问:
$g(x) = 0$ if $f(x)$ is rational measurability
Say $f$ is a measurable real-valued function on $X$, and put $g(x) = 0$ if $f(x)$ is rational and $g(x)=1$ if $f(x)$ is irrational. Is $g$ measurable?</p>

答: Hint: The Dirichlet function is in fact measurable (it is Borel). Note that the composition of measurable functions is measurable.


Title: Find partial sums formula of series: $\sum_{n=1}^{\infty}\frac{n}{1\cdot3\cdot5\cdot...\cdot(2n+1)}$

Body:
$$\sum_{n=1}^{\infty}\frac{n}{1\cdot3\cdot5\cdot...\cdot(2n+1)}=\sum_{n=1}^{\infty}a_n$$
For $n>1$: $$a_n=\frac{1}{2}(\frac{(2n+1)-1}{1\cdot3\cdot5\cdot...\cdot(2n+1)})\\=\frac{1}{2}(\frac{1}{3\cdot5\cdot...\cdot(2n-1)}-\frac{1}{3\cdot5\cdot...\cdot(2n+1)})$$
and $a_1=\frac1{3}$, hence
$$S_n=\frac1{3}+\frac1{2}(\frac1{3}-\frac1{3\cdot5\cdot\cdot...\cdot(2n+1)}).$$
I'm not sure why is there $\frac1{3}$inside the bracket and would be glad if someone clarified it.
(copied from a workbook)

Body:
Simpliy from OP's 3rd step:
$$A_k=\frac{1}{2}[B_k-B_{k+1}],~~ B_k=\frac{1}{1.3.5.7....(2k-1)}$$
By telescopic summation we get
$$S_n=\sum_{k=1}^{n} A_k= \frac{1}{2}[B_1-B_{n+1}]=\frac{1}{2}\left[ 1-\frac{1}{1.3.5.7....(2n+1)}\right]$$
So $$S_{\infty}=\frac{1}{2}$$


问:
Difference between Ornstein-Uhlenbeck, Vasicek and Geometric Mean Reversion
I am currently reading the following 3 paper:

Paper 1: Decision Support for IT Investment Projects

Paper 2: A Stochastic Processes Toolkit for Risk Management

Paper 3: Calibration of the exponential Ornstein–Uhlenbeck process when spot prices are visible through the maximum log-likelihood method. Example with gold prices

Each of them is terming what to me seems the same stochastic process differently. Paper 1 p. 387 Equation 4 is terming the process Geometric Mean Reversion (GMR) and Equation 5 the exponential version Exponential Mean Reversion (EMR), Paper 2 p. 27 Equation 47 is terming it Vasicek Model and on p. 31/32 Equation 62 Exponential Vasicek Model and Paper 3 p. 4 Equation 3 is terming it Exponential Ornstein-Uhlenbeck process.

Am I correct, that all of these 3 essentially are the same stoachstic process?

Also, for the exponential version two paper seem to offer a closed form solution, where I spotted a small difference. Paper 1 p. 387 Equation 6 has the following solution:

$$S_{T,i}=\exp\left(\ln(S_0)e^{-\alpha T}+\left(\theta-\frac{\sigma^2}{4\alpha}\right)(1-e^{-\alpha T})+\sqrt{(1-e^{-2\alpha T})\frac{\sigma^2}{2\alpha}}\epsilon_i\right)$$

Paper 3 p. 5 Equation 9 on the other hand has the following solution:

$$\ln(X_{t})=\ln(X_{t-1})e^{-\theta \Delta t}+\left(\mu-\frac{\sigma^2}{2\alpha}\right)(1-e^{-\theta \Delta t})+\sigma\sqrt{\frac{1}{2\alpha}(1-e^{-2\theta \Delta t})}\epsilon_i$$

Now different variables aside, the first equation has $\frac{\sigma^2}{4\alpha}$ and the second equation has $\frac{\sigma^2}{2\alpha}$. Can anyone tell me which one is correct?

答:
Short Answer:

Ornstein-Uhlebeck = Vasicek (only difference being that Vasicek can sometimes have (deterministic) non-constant reversion speed parameter, whilst Ornstein-Uhlenbeck has all parameters constant by definition).

Exponetial Vasicek = EMR.

The term $4 \alpha$ in the denominator looks like a typo, should be $2 \alpha$

Long Answer:

Ornstein-Uhlenbeck process is defined as (I use $\theta$ instead of your $\alpha$, otherwise the notation is identical):

$$\tag{1} X_t:=X_0+\int_{h=0}^{h=t}\theta(\mu- X_h)dh + \int_{h=0}^{h=t}\sigma dW_h$$

Or in "short-hand" notation we can write:

$$dX_t=\theta (\mu-X_t)dt+\sigma dW_t$$

In an answer to another question on Math SE here, I show how the Ornstein-Uhlenbeck can be solved step-by-step (the bottom section of my answer titled "Why not to use Short-hand notation")

The final step I don't show is to complete the Riemann integral which solves as follows:

$$\int_{h=0}^{h=t}\left(e^{\theta(h-t)}\theta\mu\right)dh=\mu(1-e^{\theta t})$$

So the full solution is:

$$X_t=X_0e^{-\theta t}+\mu(1-e^{\theta t})+\sigma\int_{h=0}^{h=t} e^{\theta(h-t)} dW_h$$

Ornstein-Uhlenbeck and Vasicek (and GMR) processes are the same thing (except that Vasicek process can have the parameter $\theta$ non-constant, it can be a deterministic function of $t$).

In the first reference you provide, the author introduces the EMR process (or Exponential Vasicek) for $S_t$ (i.e. what I call $X_t$) as:

$$dS_t=\alpha(\theta - ln(S_t))S_tdt + \sigma_t S_t dW_t$$

Or in long-hand notation:

$$S_t=S_0+\int_{h=0}^{h=t}\alpha(\theta - ln(S_h))S_hdh+\int_{h=0}^{h=t}\sigma_hS_s dW_h$$

We can see right away that this is not the same process as the Ornstein-Uhlenbeck defined in (1).

As for your last question, the $4\alpha$ in the denominator looks like a typo, should be $2\alpha$ I believe (indeed the author who has $4 \alpha$ doesn't derive the solution, he just quotes Brigo. Also see the Wikipedia link at the bottom for why it should be $2 \alpha$).

Last but not least, note that Ito integral of any deterministic function is a Normally distributed random variable. Variance of the term $\sigma\int_{h=0}^{h=t} e^{\theta(h-t)} dW_h$ is derived here on Wikipedia, and this variance and the Normal distribution property of the Ito Integral are used to rewrite the solution to the EMR process in the form you show in the two equations you quote.


问:
Pullback of locally free sheaf over the associated vector bundle
Let $X$ be a scheme over $k$, $\mathcal{F}$ be a locally free sheaf on $X$ of rank $r$, $F$ be the vector bundle associated to $\mathcal{F}$ i.e. $F=\operatorname{Spec}\operatorname{Sym}(\mathcal{F}^{*})$, and $\pi\colon F\rightarrow X$ be a natural projection.

There is a canonical locally free sheaf $\pi^{*}\mathcal{F}$ on F, and
I want an explicit description of $\pi^{*}\mathcal{F}$.

Take affine open set $U=\operatorname{Spec}A$ of $X$, and assume $V=F\mid_{U}=U\otimes k[t_{1},\dots, t_{r}]$.

Then $\pi^{*}\mathcal{F}(V)=\pi^{-1}\mathcal{F}(V)\otimes_{\pi^{-1}\mathcal{O}_{X}(V)} \mathcal{O}_{F}(V)=\mathcal{F}(U)\otimes_{\mathcal{O}_{X}(U)} \operatorname{Sym}(\mathcal{F}^{*})(U)$.

But what is $\operatorname{Sym}(\mathcal{F}^{*})(U)$ ? I'm stuck here. Thanks in advance.

答:
Let $U:=Spec(A)$ and $F:=A\{e_1,..,e_1\}$ the free $A$-module on $e_i$ with $F^*:=A\{x_1,..,x_n\}$. Let $\mathcal{F}$ be the sheafification of $F$.

Question: "But what is $Sym(F∗)(U)$?"

Answer:  $Sym_A^*(F^*)(U)\cong A[x_1,..,x_n]$ is the polynomial ring on $x_i$ over $A$.

There is a canonical map

V2. $\pi: \mathbb{V}(F^*):=Spec(Sym_A^*(F^*)):=\mathbb{A}^n_X \rightarrow X$

defined by the canonical map

V3. $\phi: A \rightarrow A[x_1,..,x_n]:=B$

and by definition $\pi^*\mathcal{F}$ is the sheafification of the $B$-module $F\otimes_A B:=F\otimes_A A[x_1,..,x_n]$.


问:
Finding maximum radius of inscribed circle
I am working through a pure maths text book as a hobby. I am having difficulty with the second half of this problem:

The equal sides AB and AC of an isosceles triangle have length a, and the angle A is denoted by $2 \theta$. Show that the radius of the inscribed circle of the triangle is given by $\frac {a \sin\theta \cos \theta}{1 + \sin \theta}$

Show that, if a is constant and $\theta$ varies between $0^\circ$ and $90^\circ$, the maximum value of r occurs when $\sin \theta = \frac{(\sqrt 5 - 1)}{2}$

I have solved the first part of the question. Now I need to differentiate $\frac {a \sin\theta \cos \theta}{1 + \sin \theta}$

Using the quotient formula I get this to be:

$\frac{dr}{d\theta} = \frac{(1 + \sin \theta)(a\cos^2\theta - a\sin^2\theta) - (a\sin \theta \cos\theta)\cos \theta}{(1 + \sin\theta)^2}$

But I cannot get this to equal the answer given.

答:
HINT:

Let $$\frac{dr}{d\theta}=0$$
and then rewrite the numerator completely in terms of $\sin \theta$.

If you need any more help please don't hesitate to ask :)

EDIT:

As you correctly found, the numerator of the derivative is equal to
$$a(1-2\sin^2\theta-\sin^3\theta)$$
Hence, $$a(1-2\sin^2\theta-\sin^3\theta)=0$$
as when a fraction is equal to $0$ the denomiantor plays no part in making the fraction equal to $0$; the numerator must be equal to $0$. Now, assuming $a\ne0$, we can divide through by $a$ to obtain
$$1-2\sin^2\theta-\sin^3\theta=0\implies \sin^3\theta+2\sin^2\theta-1=0$$
Now we can simply verify that $\sin\theta=\frac{\sqrt5 -1}{2}$ is a solution the equation above, and we are done.

Alternatively, we can find this solution directly. Notice that substituting $\sin\theta=-1$ gives us $0$; hence by the factor theorem $(\sin\theta+1)$ is a factor of $\sin^3\theta+2\sin^2\theta-1$. Hence,
$$\sin^3\theta+2\sin^2\theta-1=(\sin\theta+1)(\sin^2\theta+\sin\theta-1)=0$$
Solving for the second factor we obtain the required solution. Notice that the other $2$ solutions are outside of the range of allowed solutions for $\theta$.


问:
Closed form of $x_{n+1}=x_n-1/x_n$ with $x \neq 0,1$
I have the sequence

$x_{n+1}=x_n-1/x_n$ with $x \neq 0,1$

For certain values of $x_0$ this sequence will be periodic. To find these values one can look at the root of:
$x_{p} = x_0$ where $p$ is the period. For $p=2$ this can be solved easily.

For $p>10$ it's not viable to define the function recursively and search for roots numerically.
To look for the roots at higher values of $p$  I'm searching for a closed form of this formula.

I already found an answer to a similar problem, but failed to transfer the solution to my different sequence. I tried using the same approach.

By $x=\cot(\theta)$ you get:
$x_1=2\cot(2\theta)=\cot(\theta)-\frac{1}{\cot(\theta)}$

I calculated $x_2$ by hand which is:
$x_2=\cot(4\theta)+\frac{3}{2}\cot(2\theta)$

But I wasn't able to find a simple formula for $x_3$

Any help with finding a closed form (or an idea how to find the roots more easily) would be appreciated!

EDIT:

I moved my follow up questions to a seperate topic

答:
The equations $x_{n+1} = x_n - \frac 1{x_n}$ and $x_{n+1} = \frac 12 \left( x_n - \frac 1{x_n} \right)$ are maybe similar in appearance but very different qualitatively. For instance, the first one has no fixed points while the second one does (in the complex numbers):
$$
x = \frac 12 \left( x - \frac 1x \right) = \frac{x^2-1}{2x} \quad \Longrightarrow \quad 2x^2 = x^2 - 1 \quad \Longrightarrow \quad x^2 + 1 = 0
$$
and so it has the solutions $\pm i$. If the first one had fixed points, it would imply $x = x - \frac 1x$ has a solution, but this clearly has no solution.

Working on our original problem again but considering $x_{n+2}$ as a function of $x_n$ instead, we get
$$
x_{n+2} = x_{n+1} - \frac 1{x_{n+1}} = \frac{x_{n+1}^2 - 1}{x_{n+1}} = \frac{\left(\frac{x_n^2-1}{x_n}\right)^2 - 1}{\frac{x_n^2 -1}{x_n}} = \frac{x_n^4 - 3x_n^2 + 1}{x_n(x_n^2-1)}
$$
and if the recurrence equation $y_n = x_{2n}$ has a fixed point, then it satisfies $y_{n+1} = x_{2n+2} = x_{2n} = y_n$, so we can use the above equation and deduce
$$
-\frac 14 = y^4 - y^2 = y^2(y^2-1) = y^4 - 3y^2 + 1 = \frac 14 - \frac 32 + 1 \quad \Longrightarrow \quad 2y^2 - 1 = 0
$$
which means $y = \pm \frac 1{\sqrt 2}$ are the possible values of $x_0$ for which the sequence $x_{2n}$ will be periodic.

You might wonder, if starting with $x_0 = \frac 1{\sqrt 2}$ means $x_0 = x_2 = \cdots$, what happens to $x_1,x_3,\cdots$? Well, since $x_{2n+1} = f(x_{2n})$, and $x_{2n} = x_0$ for all $n \ge 0$, it follows that $x_{2n+1} = x_1$ for all $n \ge 0$. This means that if $x$ is a solution to $f(f(x)) = x$, then $f(x)$ is also a solution to $f(f(x)) = x$, and because we know there are no solution to $f(x) = x$, it has to be a distinct one! We deduce the following result:

Let $p$ be a positive integer. If $x$ is a solution to the equation $f^{(p)}(x) = x$ (where $f^{(p)}(x)$ is obtained by applying $f$ on $x$ precisely $p$ times), then $x, f(x), \cdots, f^{(k)}(x), \cdots, f^{(p-1)}(x)$ are all solutions to the equation $f^{(p)}(x)$. Furthermore,

if $q > 1$ is a positive integer that divides $p$, then $f^{(p/q)}(x)$ is a solution to the equation $f^{(q)}(x) = x$, so that if $x$ has period $p$, $f^{(p/q)}(x)$ has period $q$

if $p$ is prime, then the set $\{x, f(x), \cdots, f^{(k)}(x), \cdots, f^{(p-1)}(x)\}$ are distinct solutions to the equation $f^{(p)}(x)$, and therefore are periodic solutions of period $p$

To be able to say a little bit more about the periods of order $p$, consider the following general pattern. Write $x_{n+p} = \frac{f_p(x_n)}{g_p(x_n)}$ where $f_p$ and $g_p$ are the polynomials in $x_n$ we obtain by iterating the above idea $p$ times instead of $2$ times. We saw that
$$
\deg f_1 = 2, \quad \deg g_1 = 1, \quad \deg f_2 = 4, \quad \deg g_2 = 3.
$$
Let us prove by induction on $n$ that $\deg f_p = 2^p$ and $\deg g_p = 2^p-1$. The base case is proved already, and
$$
\frac{f_{p+1}(x_n)}{g_{p+1}(x_n)} = \frac{f_p(x_n)}{g_p(x_n)} - \frac{g_p(x_n)}{f_p(x_n)} = \frac{f_p(x_n)^2 - g_p(x_n)^2}{f_p(x_n)g_p(x_n)}.
$$
Since $\deg g_p < \deg f_p$ by induction, no cancellation occurs in the numerator, so $\deg f_{p+1} = 2 \deg f_p = 2^{p+1}$. Then $\deg g_{p+1} = \deg f_p + \deg g_p = 2^p + 2^p - 1 = 2^{p+1}-1$.

This proves our induction hypothesis, so we can deduce the following result:

Let $p > 1$ be a positive integer. There are at most $2^p - 2$ solutions to the equation $f^{(p)}(x) = x$.

The proof is simple; the solutions to that equation are solution to the polynomial equation $f^p(x) - xg_p(x) = 0$, and this polynomial has degree at most $2^p$.

We can slightly improve our result by noticing that there is some cancellation involved in that subtraction. By induction on $n$, one easily sees that the leading coefficients of $f_p$ and $g_p$ are both $1$, so the leading term of $f_p(x)$ and $xg_p(x)$ are the same.

We can also see by induction on $n$ that $f_p(x)$ is an even polynomial function and $g_p(x)$ is an odd polynomial function, so $x g_p(x)$ is even. Even polynomial functions can be written as polynomials in $x^2$, meaning that the terms of odd degree of $f_p(x)$ and $xg_p(x)$ are all zero. This brings our upper bound down on the number of solutions to $2^p-2$.

I believe this upper bound is sharp (but this is conjectural) since using sympy, I computed

$$
f_3(x) = x^8 - 7x^6 + 13x^4 - 7x^2 + 1, \quad xg_3(x) = x^8 - 4x^6 + 4x^4 - x^2
$$
and so we see that there is no cancellation. I repeated the reasoning computationally and it seems to hold for $p \le 15$, where the polynomial has degree $2^p-2$.

Note that nothing says the polynomial doesn't have double roots, so we can't claim that there are always exactly $2^p-2$ solutions. Some of them might be repeated roots of that polynomial.

If I find more interesting results, I'll add to this answer.

EDIT: Adding here my sympy code to generate the polynomials $f_p$ and $g_p$.

```<code>from sympy.abc import x
def recurrence_relation():
    f = x
    g = 1
    while True:
            yield (f,g)
            f,g = f**2 - g**2, f*g

def first_n_terms(n):
    for (f,g), i in zip(recurrence_relation(), range(n)):
            if i <= 1:
                    continue
            print(f"Polynomial f_p: {f}")
            print(f"Polynomial xg_p: {x*g}")
            print(f"Degree of f_p - xg_p: {poly(f-x*g, x).degree()}")
</code>```

I hope this tells you roughly how to operate with sympy for your own experiments.

Hope that helps,

Body:
The equations $x_{n+1} = x_n - \frac 1{x_n}$ and $x_{n+1} = \frac 12 \left( x_n - \frac 1{x_n} \right)$ are maybe similar in appearance but very different qualitatively. For instance, the first one has no fixed points while the second one does (in the complex numbers):
$$
x = \frac 12 \left( x - \frac 1x \right) = \frac{x^2-1}{2x} \quad \Longrightarrow \quad 2x^2 = x^2 - 1 \quad \Longrightarrow \quad x^2 + 1 = 0
$$
and so it has the solutions $\pm i$. If the first one had fixed points, it would imply $x = x - \frac 1x$ has a solution, but this clearly has no solution.

Working on our original problem again but considering $x_{n+2}$ as a function of $x_n$ instead, we get
$$
x_{n+2} = x_{n+1} - \frac 1{x_{n+1}} = \frac{x_{n+1}^2 - 1}{x_{n+1}} = \frac{\left(\frac{x_n^2-1}{x_n}\right)^2 - 1}{\frac{x_n^2 -1}{x_n}} = \frac{x_n^4 - 3x_n^2 + 1}{x_n(x_n^2-1)}
$$
and if the recurrence equation $y_n = x_{2n}$ has a fixed point, then it satisfies $y_{n+1} = x_{2n+2} = x_{2n} = y_n$, so we can use the above equation and deduce
$$
-\frac 14 = y^4 - y^2 = y^2(y^2-1) = y^4 - 3y^2 + 1 = \frac 14 - \frac 32 + 1 \quad \Longrightarrow \quad 2y^2 - 1 = 0
$$
which means $y = \pm \frac 1{\sqrt 2}$ are the possible values of $x_0$ for which the sequence $x_{2n}$ will be periodic.

You might wonder, if starting with $x_0 = \frac 1{\sqrt 2}$ means $x_0 = x_2 = \cdots$, what happens to $x_1,x_3,\cdots$? Well, since $x_{2n+1} = f(x_{2n})$, and $x_{2n} = x_0$ for all $n \ge 0$, it follows that $x_{2n+1} = x_1$ for all $n \ge 0$. This means that if $x$ is a solution to $f(f(x)) = x$, then $f(x)$ is also a solution to $f(f(x)) = x$, and because we know there are no solution to $f(x) = x$, it has to be a distinct one! We deduce the following result:

Let $p$ be a positive integer. If $x$ is a solution to the equation $f^{(p)}(x) = x$ (where $f^{(p)}(x)$ is obtained by applying $f$ on $x$ precisely $p$ times), then $x, f(x), \cdots, f^{(k)}(x), \cdots, f^{(p-1)}(x)$ are all solutions to the equation $f^{(p)}(x)$. Furthermore,

if $q > 1$ is a positive integer that divides $p$, then $f^{(p/q)}(x)$ is a solution to the equation $f^{(q)}(x) = x$, so that if $x$ has period $p$, $f^{(p/q)}(x)$ has period $q$

if $p$ is prime, then the set $\{x, f(x), \cdots, f^{(k)}(x), \cdots, f^{(p-1)}(x)\}$ are distinct solutions to the equation $f^{(p)}(x)$, and therefore are periodic solutions of period $p$

To be able to say a little bit more about the periods of order $p$, consider the following general pattern. Write $x_{n+p} = \frac{f_p(x_n)}{g_p(x_n)}$ where $f_p$ and $g_p$ are the polynomials in $x_n$ we obtain by iterating the above idea $p$ times instead of $2$ times. We saw that
$$
\deg f_1 = 2, \quad \deg g_1 = 1, \quad \deg f_2 = 4, \quad \deg g_2 = 3.
$$
Let us prove by induction on $n$ that $\deg f_p = 2^p$ and $\deg g_p = 2^p-1$. The base case is proved already, and
$$
\frac{f_{p+1}(x_n)}{g_{p+1}(x_n)} = \frac{f_p(x_n)}{g_p(x_n)} - \frac{g_p(x_n)}{f_p(x_n)} = \frac{f_p(x_n)^2 - g_p(x_n)^2}{f_p(x_n)g_p(x_n)}.
$$
Since $\deg g_p < \deg f_p$ by induction, no cancellation occurs in the numerator, so $\deg f_{p+1} = 2 \deg f_p = 2^{p+1}$. Then $\deg g_{p+1} = \deg f_p + \deg g_p = 2^p + 2^p - 1 = 2^{p+1}-1$.

This proves our induction hypothesis, so we can deduce the following result:

Let $p > 1$ be a positive integer. There are at most $2^p - 2$ solutions to the equation $f^{(p)}(x) = x$.

The proof is simple; the solutions to that equation are solution to the polynomial equation $f^p(x) - xg_p(x) = 0$, and this polynomial has degree at most $2^p$.

We can slightly improve our result by noticing that there is some cancellation involved in that subtraction. By induction on $n$, one easily sees that the leading coefficients of $f_p$ and $g_p$ are both $1$, so the leading term of $f_p(x)$ and $xg_p(x)$ are the same.

We can also see by induction on $n$ that $f_p(x)$ is an even polynomial function and $g_p(x)$ is an odd polynomial function, so $x g_p(x)$ is even. Even polynomial functions can be written as polynomials in $x^2$, meaning that the terms of odd degree of $f_p(x)$ and $xg_p(x)$ are all zero. This brings our upper bound down on the number of solutions to $2^p-2$.

I believe this upper bound is sharp (but this is conjectural) since using sympy, I computed

$$
f_3(x) = x^8 - 7x^6 + 13x^4 - 7x^2 + 1, \quad xg_3(x) = x^8 - 4x^6 + 4x^4 - x^2
$$
and so we see that there is no cancellation. I repeated the reasoning computationally and it seems to hold for $p \le 15$, where the polynomial has degree $2^p-2$.

Note that nothing says the polynomial doesn't have double roots, so we can't claim that there are always exactly $2^p-2$ solutions. Some of them might be repeated roots of that polynomial.

If I find more interesting results, I'll add to this answer.

EDIT: Adding here my sympy code to generate the polynomials $f_p$ and $g_p$.

from sympy.abc import x
def recurrence_relation():
    f = x
    g = 1
    while True:
            yield (f,g)
            f,g = f**2 - g**2, f*g

def first_n_terms(n):
    for (f,g), i in zip(recurrence_relation(), range(n)):
            if i <= 1:
                    continue
            print(f"Polynomial f_p: {f}")
            print(f"Polynomial xg_p: {x*g}")
            print(f"Degree of f_p - xg_p: {poly(f-x*g, x).degree()}")

I hope this tells you roughly how to operate with sympy for your own experiments.

Hope that helps,


答: Notice that $ 1 + \frac{1}{ \sqrt{2}} + ... + \frac{1}{ \sqrt{k}} \leq 2 \sqrt{k}$ implies that $ 1 + \frac{1}{ \sqrt{2}} + ... + \frac{1}{ \sqrt{k}} + \frac{1}{ \sqrt{1 + k}} \leq 2 \sqrt{k} + \frac{1}{ \sqrt{1 + k}} $ and that $2 \sqrt{k} + \frac{1}{ \sqrt{1 + k}} = \frac{2 \sqrt{k(1+k)} +1}{ \sqrt{1+k}} $ which is bounded above by $ 2 \sqrt{k+1} $. To prove the last statement , take a look at $ 2 \sqrt{k(1+k)} +1  \leq 2(k+1) $ ...


问:
Quotients and extensions of scalars in polynomial algebras
Let $K$ be a ring, $L$ an extension of $K$, and $I$ an ideal of $K[X_1,...,X_n]$. Is it true that

$$ L[X_1,...,X_n]/(IL[X_1,...,X_n]) \cong (K[X_1,...,X_n]/I) \otimes_K L \ \ ?$$

(additional question: does something similar hold more generally, for instance when $K$ is a ring and $L$ a $K$-algebra (and $K$ does not necessarily inject into $L$)?)

I have troubles applying general commutative algebra machinery because the situation is very specific: we have an identification $K[X_1,...,X_n] \otimes_K L \cong L[X_1,...,X_n]$, into which $K[X_1,...,X_n]$ injects.

答:
Hint: As a scalar extension, the first step is that
$L[X] \simeq K[X]\otimes_K L$.

Moreover, for the ideal, $I L[X] \simeq I\otimes_K L$.

Thus $L[X]/(IL[X]) \simeq (K[X]\otimes_K L) / (I\otimes_KL)\simeq (K[X]/I)\otimes_K L$.


问:
Fit sum of multivariate exponentials
I am working on a curve fitting task for a function which is multivariate sum of two exponentials.
The function which I am trying to fit has below form:
$$z(x,y)=ae^{bx}y+ce^{dx}$$
and want to find a, b, c, d values given values of x, y and z from measurement data.
Has anyone already worked on such function?
It seems not possible to linearize the problem using function integrals as done for fitting of sum of exponentials with single variable because second variable y is present in only one exponential term.

答:
I came up with a solution exactly at 4:20. Your specific case is a lucky one because the term that depends on the independent variable $y$ affects linearly the non-linear exponentials that depend on $x$. So the problem remains linear and you can still apply the method I proposed in this other answer (with some minor changes).

First you have to store your data $z(x,y)$ in a (tall) column vector, putting each $y$ sample head-to-tail with each other. Assuming you have $nx$ number of samples in $x$, and $ny$ number of samples in $y$:

$$
z = \left[
\begin{array}{c}
z(x, y_1) \\
z(x, y_2) \\
\vdots \\
z(x, y_{ny})
\end{array}
\right]
$$

Now we need to equate the right hand side of the equation according to the proposed method. For all $ny$ samples in $y$, the exponental terms ("lambdas") are the same, but the terms that multiply $x$ and the constant are not, so the matrix should have the form:

$$
Z = \left[
\begin{array}{c}
\int{z(x, y_1)} & \int^2{z(x, y_1)} & x & 1 & 0 & 0 & \dots & 0 & 0 \\
\int{z(x, y_2)} & \int^2{z(x, y_2)} & 0 & 0 & x & 1 & \dots & 0 & 0 \\
\vdots \\
\int{z(x, y_{ny})} & \int^2{z(x, y_{ny})} & 0 & 0 & 0 & 0 & \dots & x & 1
\end{array}
\right]
$$

So now we have:

$$
z = Z A
$$

So:

$$
A = (Z^T Z)^{-1}Z^T z
$$

$$
\lambda = eig\left(
\left[
\begin{array}{c}
A(1) & A(2) \\
1 & 0
\end{array}
\right]
\right)
$$

The rest follows by the proposed method. The respective Matlab code is:

```<code>clear all;
clc;

%% Create test data

% independent variable x
dx = 0.02;
x  = (dx:dx:1.5)';
nx = length(x);
% independent variable y
dy = 0.05;
y  = (dy:dy:1.5)';
ny = length(y);
% dependent variable z(x, y)
% stored in column vector head-to-tail for each y[k]
z = zeros(nx*ny, 1);
for k = 1:1:ny
    z_start = nx*(k-1)+1;
    z_end   = nx*k;
    % calculate z(y[k])
    z(z_start:z_end) = 5*exp(0.5*x)*y(k) + 4*exp(-3*x);
end

%% Estimate exponential contants ("lambdas")

% build least squares matrix Z to obtain lambdas
Z = zeros(nx*ny, 2+2*ny);
for k = 1:1:ny
    z_start = nx*(k-1)+1;
    z_end   = nx*k;
    % calculate integrals of z(y[k]) wrt x, for each kth sample in y
    iz1 = cumtrapz(x, z(z_start:z_end));
    iz2 = cumtrapz(x, iz1);
    % store in appropriate location
    Z(z_start:z_end, 1:2) = [iz1, iz2];
    Z(z_start:z_end, 2+2*(k-1)+1:2+2*(k-1)+2) = [x, ones(size(x))];
end
% get exponentials lambdas
A = pinv(Z)*z;
lambdas = eig([A(1), A(2); 1, 0]);
lambdas
%lambdas =
%  -2.9991
%   0.5000

%z_hat = Z*A;
%figure();
%plot(z, 'b-'); hold on;
%plot(z_hat, 'r--'); hold on;

%% Estimate exponentials multipliers

% build least squares matrix X to obtain multipliers
X = zeros(nx*ny, 2);
for k = 1:1:ny
    x_start = nx*(k-1)+1;
    x_end   = nx*k;
    X(x_start:x_end, :) = [exp(lambdas(2)*x)*y(k), exp(lambdas(1)*x)];
end
P = pinv(X)*z;
P
%P =
% 4.9999
% 3.9995
</code>```

You can try it online here:

https://octave-online.net/bucket~UfZSeFmySresJFMLNy4Y34

Please upvote my proposed method in the other question if you found it useful.

Edit:

@PayasVartak, the lambdas computation using the eigenvalues is indeed the optimal solution for the two lambdas (there only two values). The reason why you you might not be getting the best fit, might be because of the second least squares problem, in the "Estimate exponentials multipliers" section of the code.

Note that I used the equation:

```<code>[exp(lambdas(2)*x)*y(k), exp(lambdas(1)*x)]
</code>```

for the fit, because I already knew in advance which exponential gets multiplied by y(k). But actually we don't know at this point which of the two exp(lambda(?)*x) is the one that gets multiplied by y(k). So actually you need to try also:

```<code>[exp(lambdas(2)*x), exp(lambdas(1)*x)*y(k)]
</code>```

And see which one makes the best fit.

As @JJacquelin proposed, it might be a good idea to post some of your experimental data so we can help you better.

Title: Fit sum of multivariate exponentials

Body:
I came up with a solution exactly at 4:20. Your specific case is a lucky one because the term that depends on the independent variable $y$ affects linearly the non-linear exponentials that depend on $x$. So the problem remains linear and you can still apply the method I proposed in this other answer (with some minor changes).

First you have to store your data $z(x,y)$ in a (tall) column vector, putting each $y$ sample head-to-tail with each other. Assuming you have $nx$ number of samples in $x$, and $ny$ number of samples in $y$:

$$
z = \left[
\begin{array}{c}
z(x, y_1) \\
z(x, y_2) \\
\vdots \\
z(x, y_{ny})
\end{array}
\right]
$$

Now we need to equate the right hand side of the equation according to the proposed method. For all $ny$ samples in $y$, the exponental terms ("lambdas") are the same, but the terms that multiply $x$ and the constant are not, so the matrix should have the form:

$$
Z = \left[
\begin{array}{c}
\int{z(x, y_1)} & \int^2{z(x, y_1)} & x & 1 & 0 & 0 & \dots & 0 & 0 \\
\int{z(x, y_2)} & \int^2{z(x, y_2)} & 0 & 0 & x & 1 & \dots & 0 & 0 \\
\vdots \\
\int{z(x, y_{ny})} & \int^2{z(x, y_{ny})} & 0 & 0 & 0 & 0 & \dots & x & 1
\end{array}
\right]
$$

So now we have:

$$
z = Z A
$$

So:

$$
A = (Z^T Z)^{-1}Z^T z
$$

$$
\lambda = eig\left(
\left[
\begin{array}{c}
A(1) & A(2) \\
1 & 0
\end{array}
\right]
\right)
$$

The rest follows by the proposed method. The respective Matlab code is:

clear all;
clc;

%% Create test data

% independent variable x
dx = 0.02;
x  = (dx:dx:1.5)';
nx = length(x);
% independent variable y
dy = 0.05;
y  = (dy:dy:1.5)';
ny = length(y);
% dependent variable z(x, y)
% stored in column vector head-to-tail for each y[k]
z = zeros(nx*ny, 1);
for k = 1:1:ny
    z_start = nx*(k-1)+1;
    z_end   = nx*k;
    % calculate z(y[k])
    z(z_start:z_end) = 5*exp(0.5*x)*y(k) + 4*exp(-3*x);
end

%% Estimate exponential contants ("lambdas")

% build least squares matrix Z to obtain lambdas
Z = zeros(nx*ny, 2+2*ny);
for k = 1:1:ny
    z_start = nx*(k-1)+1;
    z_end   = nx*k;
    % calculate integrals of z(y[k]) wrt x, for each kth sample in y
    iz1 = cumtrapz(x, z(z_start:z_end));
    iz2 = cumtrapz(x, iz1);
    % store in appropriate location
    Z(z_start:z_end, 1:2) = [iz1, iz2];
    Z(z_start:z_end, 2+2*(k-1)+1:2+2*(k-1)+2) = [x, ones(size(x))];
end
% get exponentials lambdas
A = pinv(Z)*z;
lambdas = eig([A(1), A(2); 1, 0]);
lambdas
%lambdas =
%  -2.9991
%   0.5000

%z_hat = Z*A;
%figure();
%plot(z, 'b-'); hold on;
%plot(z_hat, 'r--'); hold on;

%% Estimate exponentials multipliers

% build least squares matrix X to obtain multipliers
X = zeros(nx*ny, 2);
for k = 1:1:ny
    x_start = nx*(k-1)+1;
    x_end   = nx*k;
    X(x_start:x_end, :) = [exp(lambdas(2)*x)*y(k), exp(lambdas(1)*x)];
end
P = pinv(X)*z;
P
%P =
% 4.9999
% 3.9995

You can try it online here:

https://octave-online.net/bucket~UfZSeFmySresJFMLNy4Y34

Please upvote my proposed method in the other question if you found it useful.

Edit:

@PayasVartak, the lambdas computation using the eigenvalues is indeed the optimal solution for the two lambdas (there only two values). The reason why you you might not be getting the best fit, might be because of the second least squares problem, in the "Estimate exponentials multipliers" section of the code.

Note that I used the equation:

[exp(lambdas(2)*x)*y(k), exp(lambdas(1)*x)]

for the fit, because I already knew in advance which exponential gets multiplied by y(k). But actually we don't know at this point which of the two exp(lambda(?)*x) is the one that gets multiplied by y(k). So actually you need to try also:

[exp(lambdas(2)*x), exp(lambdas(1)*x)*y(k)]

And see which one makes the best fit.

As @JJacquelin proposed, it might be a good idea to post some of your experimental data so we can help you better.


问:
Question regarding $\lim_{x \to 0} \left(\exp(\sin (x)) + \exp \left(\frac{1}{\sin (x)}\right)\right)$
I wanted to find out whether the following limit exists, and find the value if it does.
$$\lim_{x \to 0} \left(\exp(\sin (x)) + \exp \left(\frac{1}{\sin (x)}\right)\right).$$

Attempt

After many attempt to prove that the limit exists, I looked up Wolfram Alpha, and it turns out that it doesn't. So I set out to prove that. By the algebra of limits we have
$$\lim_{x \to 0} \left(\exp(\sin (x)) + \exp \left(\frac{1}{\sin (x)}\right)\right)= \lim_{x \to 0} \exp(\sin (x)) + \lim_{x \to 0} \exp \left(\frac{1}{\sin (x)}\right).$$

Now, let $$f(x)=\frac{1}{\sin (x)} \ \text{and} \ g(x)=\exp (x),$$ and consider the sequences
$$(y_n)=\dfrac{1}{\frac{\pi}{2}+ 2n\pi} \ \text{and} \ (z_n)=\dfrac{1}{-\frac{\pi}{2}+ 2n\pi }\text{.}$$

Then $(y_n) \rightarrow 0$ and $(z_n) \rightarrow 0$ as $n \rightarrow 0$, however $f(y_n)=1$ and  $f(z_n)=-1$ for all $n \in \mathbb{N}$.

Hence $\lim_{n \to \infty}f(y_n) \neq \lim_{n \to \infty}f(z_n)$ and $\lim_{x \to 0} \frac{1}{\sin (x)}$ does not exist.

If the working above is correct, I then would like to show that

since $\lim_{x \to 0} \frac{1}{\sin (x)}$, then $\lim_{x \to 0} \exp \left(\frac{1}{\sin (x)}\right)$ does not exist,

if $\lim_{x \to 0} \exp \left(\frac{1}{\sin (x)}\right)$ does not exist, then $\lim_{x \to 0} \exp(\sin (x)) + \lim_{x \to 0} \exp \left(\frac{1}{\sin (x)}\right)$ does not exist.

Question

A. Is my working above correct?

B. How do I show (1) and (2)? I know the rule for finding the limit of the composition of two functions and the addition of limits, if the limit exists, but not if the limit doesn't exist.

C. Is there any other way to tackle this question?

D. Could you give me some tips to identify if the limit of a function does not exist, without any aid of computers? (In order to not waste time trying to prove the existence of the limit when it actually doesn't.)

Thank you for your time.

答:
Note that $\frac{1}{\sin x}$ blows up as $x$ approaches $0$ from the right. So the limit does not exist. Even if we allow infinite limits, the limit does not exist, for our function approaches $1$ as $x$ approaches $0$ from the left.

Remark: If the question is changed to $\lim_{x\to 0^-}$ (limit as $x$ approaches $0$ from the left), then the limit does exist, and is equal to $1$.

The calculations in the post for values of $x$ of special form are not useful for this problem. The function $\sin x$ behaves very smoothly for $x$ near $0$, and it makes no difference if $x$ is of special shape such as $\frac{1}{2n\pi+\frac{\pi}{2}}$.  Such values of $x$ would be very important if we were studying expressions that involve $\sin(1/x)$ for $x$ near $0$.

As to your question about ways other than using computers, it is a matter of visualizing the behaviour of these fairly familiar functions. For positive $x$ near $0$, $\sin x$ is close to $0$ and positive, so $\exp(1/\sin x)$ is really really big. For negative $x$ near $0$, $\frac{1}{\sin x}$ is big negative, so $\exp(1/\sin x)$ is close to $0$.


问:
How to construct an RSA crypto based on prime numbers?
Discrete Mathematics - Modular Arithmetic:

Say we have the prime numbers 7 and 17 and with encryption exponent 5. How to construct an RSA crypto based on these prime numbers and encrypt for example the number 9?

Thanks!

答:
You have $p=7 ,q= 17 $ and so $ n = 7*17 = 119$ and $e = 5$ and $ \phi(n) = (p-1)(q-1) = 96$ you have the encryption key and you need the decryption key to fulfill the congruence $ e d = 1 \mod \phi(n) = 1 \mod 96$ which is $d = 77$.

We know from Euler that $a^{\phi(n)} = 1 \mod n$ when $gcd(a,n)=1$ and since $e d =1 \mod \phi(n)$ we will have that $ a^{e d} = a \mod n$ and so $ (a^e)^{d} = a \mod n$.

Now you encrypt $9$ to $ 9^{5} = 25 \mod 119$.

If you want to decrypt $25$ just compute $ 25^{77} = 9 \mod 119$


问:
Basis unit vectors in Orthogonal Curvilinear Coordinates, how to calculate?
Bipolar cylindrical coordinates (u, v, w) are defined by the transformation equations

x =a sinh v/(cosh v − cos u), y =a sin u/(cosh v − cos u), z = w

where a is a positive constant.

(a) Find the basis unit vectors eu, ev and ew for this coordinate system

Answer is:eu =(− sin u sinh v, cos u cosh v − 1, 0)/cosh v − cos u
,ev =(1 − cos u cosh v, − sin u sinh v, 0)/cosh v − cos u
,ew = (0, 0, 1) But not sure how to get these.

答:
If $\underline{r} = \underline{r}(u_1, u_2, u_3)$ then If we change $u_1$ by a small amount, $du_1$, then $\underline{r}$ moves to position $(\underline{r} + d\underline{r})$, where

\begin{align}
d\underline{r} &= \frac{\partial \underline{r}}{\partial u_1}\,d u_1 \\
&= h_1 \,e_1 \,du_1
\end{align}
Where scale factor $h_1$ is given by
$$h_1 = \bigg|\frac{\partial \underline{r}}{\partial u_1}\bigg|$$

and the base unit vector $e_1$ is given by
$$e_1 = \frac{1}{h_1} \frac{\partial \underline{r}}{\partial u_1}$$

As an example (from Spherical Polar Coordinates), let $u_1 = r, u_2 = \theta, u_3 = \phi$ with

$$\underline{r} = r \sin \theta \cos \phi \textbf{i} + r\sin \theta \sin \phi \textbf{j}+r \cos \theta \textbf{k}$$

It's easy to show that
\begin{align}
\frac{\partial \underline{r}}{\partial r}&= \sin \theta \cos \phi \textbf{i} + \sin \theta \sin \phi \textbf{j}+\cos \theta \textbf{k} \implies h_r=1 \\
\frac{\partial \underline{r}}{\partial \theta}&= r \cos \theta \cos \phi \textbf{i} + r \cos  \theta \sin \phi \textbf{j}-r \sin \theta \textbf{k} \implies h_\theta=r \\
\frac{\partial \underline{r}}{\partial \phi}&= -r \sin \theta \sin \phi \textbf{i} + r \sin \theta \cos \phi \textbf{j}\implies h_\phi= r \sin \theta
\end{align}

Thus
\begin{align}
e_r &= \sin \theta \cos \phi \textbf{i} + \sin \theta \sin \phi \textbf{j}+\cos \theta \textbf{k} \\
e_\theta &= \cos \theta \cos \phi \textbf{i} + \cos \theta \sin \phi \textbf{j}-\sin \theta \textbf{k} \\
e_\phi &= -\sin \phi \textbf{i} + \cos \phi \textbf{j}
\end{align}

This method can then be applied to your example above.


问:
$u_t+(u(1-u))_x=a(1-2u)$, method of characteristics for traffic flow equation with riemann initial data
We consider the non conserving equation
$$u_t+(f(u))_x=af'(u)$$
where $a$ is a constant and $f(u)=u(1-u)$.

I am trying to solve this equation by method of characteristics with the initial condition
$$u(x,0)=\begin{cases}
u_l & x\leq0 \\
u_r & x>0 \\
\end{cases}
$$
By method of characteristics, I have $\displaystyle \frac{dt}{1}=\frac{dx}{1-2u}=\frac{du}{a(1-2u)}$, this means that the characteristics equation is $$\displaystyle \frac{dx}{dt}=1-2u$$ along with $\displaystyle \frac{du}{dx}=a, \displaystyle \frac{du}{dt}=a (1-2u).$

Solving these equations, I reached upto $u(x,t)=ax+ g(t)$ where $g$ is some function of $t$ alone. I don't know how to proceed further.

I was able to solve this when we had the equation $$u_t+(f(u))_x=0$$ as there $u$ was constant along the line of characteristics.
Thanks in advance for any help.

答:
From $\mathrm{d}u/\mathrm{d}x = a$ you get $u - ax = c_1$, and from $a\mathrm{d}t = \mathrm{d}u/(1-2u)$ you obtain $u = \frac{1}{2}(1-c_2 \mathrm{e}^{-2 at})$. Let $c_2 = f(c_1)$ to derive an implicit solution for $u$, determined by the equation

$$ u = \frac{1}{2}\left[1-f(u - ax) \, \mathrm{e}^{-2 at}\right]$$

The task at hand now is to determine $f$ from the initial condition and eventually solve for $u$. Can you take it from here?


问:
For continuous distribution $P(X_3<X_2<\max(X_1,X_4))$ equals?

Let $X_1,X_2,X_3,X_4$ be i.i.d random  variables have a continuous
distribution. Then $P(X_3<X_2<\max(X_1,X_4))$ equals?

$A=\dfrac{1}{2}$

$B=\dfrac{1}{3}$

$C=\dfrac{1}{4}$

$D=\dfrac{1}{6}$

My attempt:

$P(X_3<X_2<\max(X_1,X_4))=F_{X_{2}}(\max(X_1,X4))-F_{X_{2}}(X_3)=P(X_{2}<\max(X_1,X_{4}))-P(X_{2}<X_{3}))=1-P(\max(X_1,X_{4})<X_2)-P(X_{2}<X_{3}))=1-P(X_1<X_2)P(X_{4})<X_2)-P(X_{2}<X_{3}))=1-P(X_1<X_2)^2-P(X_{2}<X_{3}))$

I am blank now. No distribution is provided I don't know what I am missing here.

I know the fact that $Z=F_{X}(x) \sim U(0,1)$ but how do I use that here?

答:
Ignoring equalities (which happen with probability $0$ since this is a continuous distribution) there are $4!=24$ equally likely orders

Of these, half $(12)$ have $X_3>X_2$ and can be excluded, while a quarter $(6)$ have $X_2$ as the largest value and can also be excluded.

Every other case has $X_3<X_2<\max(X_1,X_4)$ and so $$\mathbb P(X_3<X_2<\max(X_1,X_4)) = 1-\frac12-\frac14 = \frac14.$$

The six orders which meet the condition are

$X_4 < X_3 < X_2 < X_1$

$X_3 < X_4 < X_2 < X_1$

$X_3 < X_2 < X_4 < X_1$

$X_3 < X_2 < X_1 < X_4$

$X_3 < X_1 < X_2 < X_4$

$X_1 < X_3 < X_2 < X_4$

so we can also say $$\mathbb P(X_3<X_2<\max(X_1,X_4)) = \frac{6}{24}= \frac14.$$


问:
Create a number by multi by $2$ and divide by $3$ (integer part)
How can I create a given postitive integer $N$ by multi by $2$ and divide by $3$ (integer part) ?

(Write a computer program is allowed)
For example:
$$100 = 2*2*2*2*2*2*2*2*2*2*2/3/3/3/3*2*2$$

(The numbers of $2, 3$ need to be minimum)

($/3$ mean $\text{div}\ 3$, eg: $2/3 = 0, 4/3 = 1...$)

Is solving equation

$$\left \lfloor{\frac {2^a}{3^b} }\right \rfloor *2^c = N$$
help ?

If it help then, how to solve
$$\left \lfloor{\frac {2^a}{3^b} }\right \rfloor *2^c = N$$

such that $a > c, a+b+c$ is minimum.

答:
What I'm trying is this.

We go the inverse way; starting on value $w_0=N$ and then $w_1=w_0/2^{\nu_2(w_0) }$.

After that, possibly one of $\{3w_1,3w_1+1,3w_1+2\}$ is a perfect power of $2$. If it is, we've solved the problem.

If not, possibly one of $ 3^2 \cdot w_1 +\{0,1,2,3,4,5,6,7,8\}$ is a perfect power of $2$ . If yes, then we've solved the problem.

If not, possibly one of $ 3^3 \cdot w_1 +\{0,1,\ldots,3^3-1\}$ is a perfect power of $2$.

We can proceed the obvious way.

For $N=w_0=100$ I got
$2^{A_1}=4$ so $A_1=2$ and $w_1=25$

Then in the interval $25\cdot 3 \to 75+\{0,1,2\}$ there is no perfect power of $2$

Then in the interval $25 \cdot 3^2 \to 225+\{0,1,...,8\}$ there is no perfect power of $2$

Then in the interval $25 \cdot 3^3 \to 675+\{0,1,...,26\}$ there is no perfect power of $2$

Then in the interval $25 \cdot 3^4 \to 2025+\{0,1,...,80\}$ there is a perfect power of $2$ (namely $2^{11}$and we have the solution (which was already given by the OP):

$ \qquad \qquad 1 \cdot 2^{11} \backslash 3 \backslash 3\backslash 3\backslash 3 \cdot 2^2 = N (= 100)$

where "$\backslash$" means integer division and we proceed from left to right without compacting the integer-divisions.

However, I didn't attempt yet to find out, whether this is an optimal solution in any way...

For instance, using $w_1=25$ gives $4$ steps of multiplication by $3$ (inlcusive the residue correction) but it we use $w_1=w_0=100$ then I needed $57$ multiplications by $3$ (inclusive the residue corrections). So there are clearly optimality criteria to define/satisfy.

update1

I found it interesting, that a simple restriction is then equivalent with the Collatz-transformation. If we throw away the requirement that in the listed example we do not increase the exponent at 3 until a perfect power of 2 occurs in the interval $25*3^k + \{0,1,2,...,3^k-1\}$ but instead at a occuring even number (at odd residue) we remove the even factor and then proceed taking that value for iteration, we simply have the $(3a+1)/2^A$-transformation, and we could use the orbit of that transformation as upper limit for the optimal length, giving a basic normalization of the representation of a number.

As a perhaps more interesting option I looked at another restriction:

In the example check the intervals using $N=25$ again:

$$ 25\cdot 3 \to w_{1,k}=75+\{1\}_k \\
 25 \cdot 3^2 \to w_{2,k}=225+\{1,3,...,7\}_k \\
  25 \cdot 3^3 \to w_{3,k}=675+\{1,3,...,25\}_k \\
 \vdots \\
 25 \cdot 3^e \to w_{e,k}=25 \cdot 3^e +\{1,3,...,3^e-2\}_k
$$
Test the even numbers $w_{e,k}=2^j \cdot r_k$ (= remove the even prime factor $2^j$) along increasing $e$ and $k$ until first time $r_{k}<25$. And iterate, until we arrive at $r_k=1$.

This gives a trajectory of strictly decreasing intermediate values, being another candidate for the normalization.
Unfortunately this way needs sometimes even more steps of the basic operation than using the $3a+1$-rule, so again this method does not always produce an optimal (shortest) representation of some $N$.
However, it seems an interesting question to prove, that such a rule is always applicable at all, leading towards $1$.

To get a better overview in the possibilities of finding an optimal representation the following picture :

Here we may start at the left column, with some power of 2, and walk right for the $ /3$ steps, until $2$ or $1$ occurs. If we assume, that this process covers all natural numbers, it might be the best strategy for getting a shortest trajectory for any number $N$. If we count the steps down and then steps right towards some number $N$  as length of the trajectory, it seems  that  lengthes towards some duplicates (for some small $N$ visible in the more rightpart and colored orange) are always longer. To help intuition I placed  duplicate indicator lines for the values $N=8$ and $N=42$  in the picture.

For the question of whether all numbers are covered by this scheme I brute force sorted all occuring intermediate values smaller than $1000$ and using the first $120$ rows of the table I got the following list of occuring numbers $N$ (the smaller part of them occur multiple times):

```<code>    0    1    2    3    4    5    6    7    8    9
   10   11   12   13   14   15   16   17   18   19
   20   21   22   23   24   25   26   27   28   29
   30   31   32   33   34   35   36   37   38   39
   40   41   42   43   44   45   46   47   48   49
   50   51   52   53   54   55   56   57   58   59
   60   61   62   63   64   65   66   67   68   69
   70   71   72   73   74   75   76   77   78   79
   80   81   83   84   85   86   87   88   89   90
   92   93   94   95   97   98   99  100  101  102
  103  105  106  107  109  110  112  113  115  116
  118  119  121  122  124  126  127  128  129  131
  132  133  134  136  138  140  141  142  143  145
  147  149  151  153  155  157  159  161  163  166
  168  170  172  174  177  179  181  184  186  189
  191  194  196  199  201  202  204  207  210  212
</code>```

But I think it is not too difficult to prove that the infinite table indeed covers all natural numbers.

Looking at the solutions in @HandeBruijns's answer, for instance $N=11$, he has a much better/shorter representation. Applying the principle of my scheme to them, for instance for $N=11$ I can make this route:

and it indicates, that likely we need the look to the right of the table far more than possible in the image of that small segment of the first table.

Body:
What I'm trying is this.

We go the inverse way; starting on value $w_0=N$ and then $w_1=w_0/2^{\nu_2(w_0) }$.

After that, possibly one of $\{3w_1,3w_1+1,3w_1+2\}$ is a perfect power of $2$. If it is, we've solved the problem.

If not, possibly one of $ 3^2 \cdot w_1 +\{0,1,2,3,4,5,6,7,8\}$ is a perfect power of $2$ . If yes, then we've solved the problem.

If not, possibly one of $ 3^3 \cdot w_1 +\{0,1,\ldots,3^3-1\}$ is a perfect power of $2$.

We can proceed the obvious way.

For $N=w_0=100$ I got
$2^{A_1}=4$ so $A_1=2$ and $w_1=25$

Then in the interval $25\cdot 3 \to 75+\{0,1,2\}$ there is no perfect power of $2$

Then in the interval $25 \cdot 3^2 \to 225+\{0,1,...,8\}$ there is no perfect power of $2$

Then in the interval $25 \cdot 3^3 \to 675+\{0,1,...,26\}$ there is no perfect power of $2$

Then in the interval $25 \cdot 3^4 \to 2025+\{0,1,...,80\}$ there is a perfect power of $2$ (namely $2^{11}$and we have the solution (which was already given by the OP):

$ \qquad \qquad 1 \cdot 2^{11} \backslash 3 \backslash 3\backslash 3\backslash 3 \cdot 2^2 = N (= 100)$

where "$\backslash$" means integer division and we proceed from left to right without compacting the integer-divisions.

However, I didn't attempt yet to find out, whether this is an optimal solution in any way...

For instance, using $w_1=25$ gives $4$ steps of multiplication by $3$ (inlcusive the residue correction) but it we use $w_1=w_0=100$ then I needed $57$ multiplications by $3$ (inclusive the residue corrections). So there are clearly optimality criteria to define/satisfy.

update1

I found it interesting, that a simple restriction is then equivalent with the Collatz-transformation. If we throw away the requirement that in the listed example we do not increase the exponent at 3 until a perfect power of 2 occurs in the interval $25*3^k + \{0,1,2,...,3^k-1\}$ but instead at a occuring even number (at odd residue) we remove the even factor and then proceed taking that value for iteration, we simply have the $(3a+1)/2^A$-transformation, and we could use the orbit of that transformation as upper limit for the optimal length, giving a basic normalization of the representation of a number.

As a perhaps more interesting option I looked at another restriction:

In the example check the intervals using $N=25$ again:

$$ 25\cdot 3 \to w_{1,k}=75+\{1\}_k \\
 25 \cdot 3^2 \to w_{2,k}=225+\{1,3,...,7\}_k \\
  25 \cdot 3^3 \to w_{3,k}=675+\{1,3,...,25\}_k \\
 \vdots \\
 25 \cdot 3^e \to w_{e,k}=25 \cdot 3^e +\{1,3,...,3^e-2\}_k
$$
Test the even numbers $w_{e,k}=2^j \cdot r_k$ (= remove the even prime factor $2^j$) along increasing $e$ and $k$ until first time $r_{k}<25$. And iterate, until we arrive at $r_k=1$.

This gives a trajectory of strictly decreasing intermediate values, being another candidate for the normalization.
Unfortunately this way needs sometimes even more steps of the basic operation than using the $3a+1$-rule, so again this method does not always produce an optimal (shortest) representation of some $N$.
However, it seems an interesting question to prove, that such a rule is always applicable at all, leading towards $1$.

To get a better overview in the possibilities of finding an optimal representation the following picture :

Here we may start at the left column, with some power of 2, and walk right for the $ /3$ steps, until $2$ or $1$ occurs. If we assume, that this process covers all natural numbers, it might be the best strategy for getting a shortest trajectory for any number $N$. If we count the steps down and then steps right towards some number $N$  as length of the trajectory, it seems  that  lengthes towards some duplicates (for some small $N$ visible in the more rightpart and colored orange) are always longer. To help intuition I placed  duplicate indicator lines for the values $N=8$ and $N=42$  in the picture.

For the question of whether all numbers are covered by this scheme I brute force sorted all occuring intermediate values smaller than $1000$ and using the first $120$ rows of the table I got the following list of occuring numbers $N$ (the smaller part of them occur multiple times):

    0    1    2    3    4    5    6    7    8    9
   10   11   12   13   14   15   16   17   18   19
   20   21   22   23   24   25   26   27   28   29
   30   31   32   33   34   35   36   37   38   39
   40   41   42   43   44   45   46   47   48   49
   50   51   52   53   54   55   56   57   58   59
   60   61   62   63   64   65   66   67   68   69
   70   71   72   73   74   75   76   77   78   79
   80   81   83   84   85   86   87   88   89   90
   92   93   94   95   97   98   99  100  101  102
  103  105  106  107  109  110  112  113  115  116
  118  119  121  122  124  126  127  128  129  131
  132  133  134  136  138  140  141  142  143  145
  147  149  151  153  155  157  159  161  163  166
  168  170  172  174  177  179  181  184  186  189
  191  194  196  199  201  202  204  207  210  212

But I think it is not too difficult to prove that the infinite table indeed covers all natural numbers.

Looking at the solutions in @HandeBruijns's answer, for instance $N=11$, he has a much better/shorter representation. Applying the principle of my scheme to them, for instance for $N=11$ I can make this route:

and it indicates, that likely we need the look to the right of the table far more than possible in the image of that small segment of the first table.