问:
Find the Number of Group Homomorphisms from $\mathbb{Z}/10\mathbb{Z}$ to $A_4$.
Is this reasoning correct?

$\mathbb{Z}/10\mathbb{Z}$ is isomorphic to $\mathbb{Z}_{10}$ which is generated by $1$, hence we may look at the number of homomorphisms between $\mathbb{Z}_{10}$ and $A_4$ for the answer to the original question. A homomorphism $\phi$ must map the element $1$ of $\mathbb{Z}_{10}$ to an element of $A_4$ such that its order divides both $10$ (by properties of homomorphisms) and $12$ (by Lagrange's Theorem). Thus, the only possible candidates for the mapping of $1$ must have order $1$ or $2$. This in turn yields $4$ possible elements in $A_4$, namely, $()$, $(12)(34)$, $(13)(24)$, and $(14)(23)$. All of these mappings preserve the group operation, thus the answer to the original question is $4$.

答:
Your argument is correct. A slightly different way to see it is to note that choosing an element $\sigma\in A_4$ defines uniquely a homomorphism $f_\sigma\colon\mathbb{Z}\to A_4$ by $f_\sigma(n)=\sigma^n$ and this induces a homomorphism $\mathbb{Z}/10\mathbb{Z}\to A_4$ if and only if
$$
10\mathbb{Z}\subseteq\ker f_\sigma
$$
Conversely, a homomorphism $\mathbb{Z}/10\mathbb{Z}\to A_4$ induces a homomorphism $\mathbb{Z}\to A_4$ by composing with the canonical projection.

The kernel of $f_\sigma$ is easy to compute: it is $k\mathbb{Z}$ where $k$ is the order of $\sigma$, because the image of $f_\sigma$ is precisely $\langle\sigma\rangle$.

Now we know that $k\mathbb{Z}\supseteq 10\mathbb{Z}$ if and only if $k\mid 10$ and so the order of $\sigma$ can only be $1$ or $2$. This makes for just four choices of $\sigma$, namely $()$, $(12)(34)$, $(13)(24)$ and $(14)(23)$.


问:
Number of connected components quotient space
Be $=\{(,)∈ℝ^2:=0 \text{ or } =1\}$ with topology $T$ induced by the usual of $\mathbb R^2$. Let be $$ equivalent relation in $$ with equivalent classes are
$$[(0,0)]=[(0,1)]=\{(0,0),(0,1)\},\quad [(,)]=\{(,)\} \text{ if } ≠0.$$

Let be $(/,/)$ the quotient space and $:→/$ the projection. I want to know the number of connected components of $/∖(0,0)$.

I think they should be $3$, $1$, $4$ or $2$ but I can't see it.

Any help is appreciated!

答:
The four connected components of $Y = X/R\setminus\{[0,0]\}$ are
$$
\pi( \mathbb R_{<0} \times \{0\} ),\
\pi( \mathbb R_{>0} \times \{0\} ),\
\pi( \mathbb R_{<0} \times \{1\} ),\
\pi( \mathbb R_{>0} \times \{1\} ).
$$
Do you see how these are all open and closed in $Y$, connected and their union is disjoint and all of $Y$?

Here $\mathbb R_{<0}$ and $\mathbb R_{>0}$ denote the sets of negative and positive real numbers, respectively.

ContentLicense: CC BY-SA 4.0


问:
How to proof that in a tree there is always one vertex with max $\frac{v}{2}$ distance to all other vertices?
I found the statement that in every simple, undirected and connected graph with $v$ vertices, there is one vertex that has a maximum distance of $\frac{v}{2}$ to all other vertices.

I have limited the statement to trees because if it is true for trees it should be true for all simple, undirected and connected graphs as adding edges to trees creates circles which can only shorten the distance but not increase it.

I am fairly certain that the statement is correct but I am unsure how to proof it. I was thinking about showing it for a tree with only two leaves and thereby a maximum sum of distances between the vertices and showing that even in this case it holds, so it should hold for any other tree. Does that make sense?

答:
Given a tree with $v$ vertices, let $a$ and $b$ be two vertices such that the distance $D$ between them is maximal (i.e. there are no two vertices with a greater distance between them.) Clearly $D \le v-1 \,$.

Suppose $D$ is even, and let $c$ be the middle vertex on the path from $a$ to $b$. Clearly the distance from $c$ to $a$ and to $b$ is $\frac{D}{2} \,$. If there were any vertex (call it $f$) whose distance to $c$ was larger than $\frac{D}{2} \,$, then the distance from $a$ (or $b$) to $f$ would be greater than $D$, contradicting the original assumption that the distance between $a$ and $b$ was maximal. Thus vertex $c$ has a maximum distance of $\frac{D}{2} \le \frac{v-1}{2} \,$ to every other vertex.

The case when $D$ is odd can be proven in a similar way.


问:
Prove U = $\mathbb{R}^2$\ {(0,0)} is open in $\mathbb{R}^2$
I have to prove that  $U = \mathbb{R}^2\setminus \{(0,0)\}$ is open in $\mathbb{R}^2$. I know that $U$ is open if $U = \operatorname{inw}(U)$ (collection of all interior points of $U$) (this is the Dutch notation, not sure about the English one). I know the definition of a interior point too, but I don't know how I can solve this question. I think you can use the balls to cover $U$ including the point $(0,0)$?

答: Points in $\mathbb{R}^2 $ with the euclidean topology are closed sets (every $T_1$ space satisfies this claim). The complement of a closed set is open by definition.


问:
How change equation $y=a(1-e^{bx})$ as linear regression to approximate $a$ and $b$?
Assume data $(x_i, y_i),i=1,2,...,n $ have a form like $y=a(1-e^{-bx})$.
The question is how to approximate $a$ and $b$ by linear regression.
There is similar question asked here.
Here's what I tried:
$$\ln y = \ln a + \ln(1-e^{-bx})=\ln a + \ln(1-({e^{-x}})^{b})=\ln a + \ln(1-(X)^{b})$$
then
$$Y = A +  \ln(1-(X)^{b})$$
which is not linear relation between $y$ and $x$.
Or as another way:
$$\frac{y}{a}-1=-e^{-bx}\Longrightarrow \ln {(a-y)} - \ln a = \ln(e^{-bx})=-bx$$
which is not linear relationship between $y$ and $x$ again.
I also did'nt find any way to simplify $\ln(1-e^{-bx})$.
$$\ln(1-e^{-bx})=\ln(1-\frac{1}{e^{bx}})=\ln(\frac{e^{bx}-1}{e^{bx}})=\ln(e^{bx}-1)-bx$$
How get rid of $1$? and find linear relationship between $y$ and $x$?

答:
I am afraid that you can't linearize this model.

To solve your problem, two possible approaches:

if you have large enough $x$'s, the ordinate of the horizontal asymptote gives you $a$ (approximately  the $y$ of the largest $x$).

if you can't estimate the asymptote, hypothesize some $a$ and draw $b$ by

$$b=\frac{\ln\left(1-\dfrac ya\right)}x$$ for every $(x,y)$ pairs. Take the average $b$ and compute the fitting error. Now you have the fitting error as a function if $a$ and you can use a numerical minimizer.

Additional remark:

If you have three equidistant $x$'s, let $x_k=x_0+kh$, you can use the relation

$$\frac{y_2-y_1}{y_1-y_0}=e^{bh}$$ to obtain an approximation of $b$.


问:
Poisson Distribution - sum of RVs
Question:

$X$ balls are thrown to $n$ bins (each ball has an equal chance to get to each bin). Let $X_1,\dots, X_n$ be the amount of balls in each cell.
a. Show that if $X \sim \text{Poisson}(\lambda)$ then $\displaystyle X_i \sim \text{Poisson} \left(\frac \lambda n\right)$ and that the $X_i$'s are independent.

My thought:
I know how to prove it the other way around, but not from the sum to the parts of it...

Here's a chance to reverse the steps (I'm not sure how I can justify the last move, and how I should figure indepndence:
$P(X=k)=e^{-\lambda} \frac {\lambda^k}{k!}=
e^{-(\frac \lambda n)n} \frac {((\frac \lambda n)n)^k}{k!}=
\frac {e^{-(\frac \lambda n)n}}{k!} \sum_{m=0}^k \frac {k!}{m!(k-m!)}(\frac \lambda n)^m{((\frac \lambda n)(n-1))^{k-m}}=
 {e^{-(\frac \lambda n)n}} \sum_{m=0}^k \frac 1{m!(k-m)!}(\frac \lambda n)^m{((\frac \lambda n)(n-1))^{k-m}}=$
$ {e^{-(\frac \lambda n)n}} \sum_{m=0}^k \frac {(\frac \lambda n)^m}{m!}\frac {((\frac \lambda n)(n-1))^{k-m}}{(k-m)!}=$
$ \sum_{m=0}^k {e^{-(\frac \lambda n)}} \frac {(\frac \lambda n)^m}{m!} {e^{-(\frac \lambda n(n-1))}}\frac {((\frac \lambda n)(n-1))^{k-m}}{(k-m)!}=$
$\sum_{m=0}^k P(X_1=m)P(\sum_{i=2}^nX_i=k-m)$

答:
MGF's are easy and by far the most convenient way to solve your problem.

$$
\begin{align}
\psi_X(u) &= \mathbb{E} e^{uX}\\
&=\sum_{k=0}^{\infty} e^{uk} \frac{\lambda^k}{k !} e^{-\lambda}\\
&=e^{-\lambda} \sum_{k=0}^{\infty} \frac{(\lambda e^{u})^k}{k !}\\
&=\exp\{ \lambda(e^u -1) \}
\end{align}
$$

Now, since you know that $X_i$'s are independent, you know that $\psi_X(u) = \prod_{i=1}^n \psi_{X_i}(u) $. All you need to prove that the only way this is going to happen is that $X_i \sim  Poisson(\frac{\lambda}{n}) $.

You can start by computing an MGF of $X_i$ following the computation above.

ContentLicense: CC BY-SA 3.0

Title: Poisson Distribution - sum of RVs


问:
Can you suggest some challenging ring theory problems?
I would like to work on some wonderful excercises (not particularly hard) in ring theory, subjects PID, UFD or polynomial rings.
Can you introduce a resource that has newer exercises?
Or if you can remember a good one, please let us know about it too! :)

Thank you very much!

答:
There are lots of places to find exercises. Any general algebra textbook worth its salt will have plenty of exercises on ring theory (in particular the subjects you've named).

Off my bookshelf, there's (in approximately increasing order of difficulty):

Dummit and Foote's "Abstract Algebra" (chs 8 and 9)

Aluffi's "Algebra: Chapter 0" (ch V.2)

Lang's "Algebra" (ch II.5)

Again, though, almost any algebra book will do. All three of these are available (perhaps unscrupulously) in PDF form online, and (for better or for worse) complete or partial solutions to many problems (certainly from Dummit and Foote) are available on this website. You might also have luck looking at Number Theory textbooks (though I don't know much number theory, so I don't have any recommendations).

I hope this helps ^_^


问:
Two Multivariable Limits
I need to solve the following limit:$$\lim\limits_{(x,y)\to(0,0)}\frac{x+y}{\sqrt{x^2+y^2}\cdot(1+\ln(x^2+y^2))}$$. I expect for it to be $0$, but not sure how to prove it. I tought using a substitution $t=x^2+y^2$ but not sure what to do with part $x+y$. Also i tried comparing this to get that $0\le f(x,y)\le 0$.

答: $|x+y| \leq \sqrt 2 \sqrt {x^{2}+y^{2}}$. So $0 \leq |f(x,y)| \leq \frac {\sqrt 2 t} {t(1+2\ln t)}$ where $t =\sqrt {x^{2}+y^{2}}$.  Now use the fact that $\ln t \to -\infty$ as $t \to 0+$.

Title: Two Multivariable Limits


问:
Algorithm of $n = x^2 y$ to simplify a squrae root of n
we can simplify a square root $$\sqrt 8 = \sqrt {2^2 \cdot 2} = 2\sqrt 2$$

I need a good algorithm to simplify a big integer $n$ of thousands digits.

$$\text{for constant } n \in \mathbb N \\x, y \in \mathbb N \\ \text{ } \\ \text{find positive integer solution }(x,y) \text{ in}\\ n = x^2 y$$

答:
It's unlikely that there is a good algorithm because Wikipedia says

every known algorithm for computing a square-free factorization [of  integer] computes also the prime factorization.


问:
Functional equation $f(x+y)=f(x)f(y)-f(xy)+1$
I'm new to solving functional equations and found the following functional equation from a collection of functional equations.

Find all functions $f: \mathbb{Q}\to \mathbb{R}$ for $$f(x+y)=f(x)f(y)-f(xy)+1$$ for all $x,y\in \mathbb{Q}$.

By substituting $x,y\to 0$, we get
$$f(0)=f(0)^2-f(0)+1$$
$$f(0)^2-2f(0)+1=0$$
$$f(0)=1$$
And then doing the substitution $x\to 1,\ y\to -1$, we get
$$f(0)=f(1)f(-1)-f(-1)+1$$
$$f(1)f(-1)-f(-1)=0$$
$$f(-1)(f(1)-1)=0$$
$$f(-1)=0\vee f(1)=1$$

So, no we have three possible cases to consider and I have no idea on how to proceed on any of them.
I would appreciate any hints before giving the full solution as in the best case that's all I need right now.

答:
Our equation is. for all values of x and y belonging to the set of rational numbers there exists f(x+y) = f(x)f(y) - f(xy) + 1, such that f(x+y) is real.
or, f(xy) = f(x)f(y) - f(x+y) +1
if y = 0 then  f(x*0) = f(x)f(0) - f(x+0) +1
or, f(o) = f(x)f(0) - f(x) +1
or, f(0) -1 = f(x)f(0) - f(x) = f(x){f(0)-1}
or, f(x) = 1
therefore, f(x+y) = 1.
This is the only solution I guess. [I am not sure though. :) ]</p>


问:
Any commutative associative operation can be extended to a function on nonempty finite sets
This is a fact we use very frequently in general mathematics when we write such notations as $1+2+3+4$: since we know that $+$ is commutative and associative, we can just "drop the parentheses" and not worry about order of operations. Of course I believe this, but how does one prove this in full generality? Even stating it is giving me trouble. Here's my attempt:

Assume an operation $\oplus:S\times S\to S$ is provided satisfying $x\oplus y=y\oplus x$ and $x\oplus(y\oplus z)=(x\oplus y)\oplus z$ for all $x,y,z\in S$.

Claim: Given any finite set $\emptyset\subset A\subseteq S$, there exists a unique $z\in S$ such that for any function $f:{\cal P}(A)\to{\cal P}(A)$ which satisfies $\emptyset \subset f(B)\subset B$ for all $B\subseteq A$ with $|B|\ge 2$ and any function $g:{\cal P}(A)\to S$ which satisfies $g(\{x\})=x$ for all $x\in S$ and $g(B)=g(f(B))\oplus g(B-f(B))$ for all $|B|\ge 2$, $g(A)=z$.

The operation $\oplus$ does not necessarily have an identity element, so we do not attempt to define an empty sum. Intuitively, this element $z$ represents the finite sum of the elements in $A$, so if $A=\{1,2,3\}$ and $f(\{1,2,3\})=\{1\}$ and $f(\{2,3\})=\{3\}$, then
$$z=g(\{1,2,3\})=g(\{1\})\oplus g(\{2,3\})=g(\{1\})\oplus (g(\{3\})\oplus g(\{2\}))=1\oplus(3\oplus 2).$$
There has got to be a better way to say that, but this is the only way I can think of to capture all the possibilities of parenthesization, and still be amenable to a formal proof. And now that I've stated it, how should I prove it? I suppose I should induct on something, but I've no idea what.

Edit: The goal here is to be able to define an operation $F$ such that $F(\{x_1,\dots,x_n\})=x_1\oplus\cdots\oplus x_n$ and be assured that the operation is well defined and satisfies $F(A\cup B)=F(A)\oplus F(B)$, when $A$ and $B$ are disjoint finite nonempty subsets of $S$.

答:
So you want to prove that if $S$ is a set and $+$ is a binary commutative associative law : $S \times S \to S$, then there is a unique function $\sum$ from non empty finite subsets of $S$ to $S$ satisfying :

- if $x \in S, \sum \{x\} = x$

- if $A \cap B = \emptyset, \sum A \cup B = \sum A + \sum B$.

Prove this by induction of the size of the subset : if $|A| = 1$ then we have no choice.

If $|A| \ge 2$, suppose we can define $\sum A$ in several ways : $A = B_1 \cup B_2 = C_1 \cup C_2$, where the pairs are disjoint. Using the induction hypothesis, the sums of the subsets $B_i$ and $C_j$ are well-defined and we need to show that $\sum B_1 + \sum B_2 = \sum C_1 + \sum C_2$.

Let $D_{ij} = B_i \cap C_j$. Suppose for now that none of them is empty. Then $\sum B_1 = \sum D_{11} + \sum D_{12}$ and so on, and the claim boils down to proving $\forall a,b,c,d \in S, (a+b)+(c+d) = (a+c)+(b+d)$.

And this is easy : $(a+b)+(c+d) = a+(b+(c+d)) = a+((c+d)+b) = a+(c+(d+b)) = (a+c)+(d+b) = (a+c)+(b+d)$.

The cases where one or more of the $D_{ij}$ is empty are done in a similar way (and are even easier). Or you could add an element $\star$ to $S$ and extend $+$ by defining $\star + x = x + \star = x$, prove that $+$ still is commutative and associative, and finally define $\sum \emptyset = \star$


答:
Perhaps this is on the easier side (not sure what difficulty you're looking for), but sections 34 through 41 of Durbin's "Modern Algebra: An Introduction" are dedicated almost entirely to the properties of polynomial rings and polynomial quotient rings with the intent of making a beeline to some intro Galois theory. It starts fairly slowly and approachable. It's full of all kinds of exercises on polynomial rings and theorems important to the subject. There seem to be solutions online to many problems, too.

It's my favorite algebra reference, and always one of my first go-to's. Can't recommend it enough.

Title: Can you suggest some challenging ring theory problems?

Body:
I would like to work on some wonderful excercises (not particularly hard) in ring theory, subjects PID, UFD or polynomial rings.
Can you introduce a resource that has newer exercises?
Or if you can remember a good one, please let us know about it too! :)

Thank you very much!


问:
sum of four squares is less then one
You are given four numbers $x_1, x_2, x_3$ and $x_4$ in $[0, 1]$. What is the probability that the sum of the squares of these four numbers is less then one.

答: Let us use some geometrical intuition (this is possible because we have uniform probability and no number has a particular privilege with respect to another in the extraction). Simplify the problem to the case of three numbers $x_1,x_2,x_3$: picking them with uniform probability distribution you are asking what is the probability that $x_1^2+x_2^2+x_3^2\leq 1$ with the condition $x_1,x_2,x_3\in[0,1]$. This is basically computing the volume of $1/8$ of a sphere of radius 1 (you are excluding all negative coordinates) over the volume of a cube of edge 1 (which is 1 in any dimension). In four dimensions, analogously, you can compute the volume of the 4D sphere and then exclude the negative coordinates combinations: namely you should divide the volume of the aforementioned sphere by 16. This leads to the final result which should be $\pi^2/32$.

Title: sum of four squares is less then one


问:
Prove that $C^1[0,1]$ is not Hilbert space
Problem: On $C^1[0,1]$, we define an inner product by

$$\langle x, y \rangle = x(0)y(0) + \int_0^1 x^\prime (t) y^\prime (t) dt.$$
Prove that $(C^1[0,1],\langle \cdot,\cdot \rangle)$ is not Hilbert space.

My attempt: Suppose that $(C^1[0,1],\langle \cdot,\cdot \rangle)$ is Hilbert space.

Next, we denote $\Vert \cdot \Vert$ be the norm induced by $\langle \cdot,\cdot \rangle$. Thus, for all $x\in C^1[0,1]$ we have
$$\Vert x \Vert= \sqrt{x^2(0)+\int_{0}^{1}[x'(t)]^2dt}.$$

Consider $(x_n)_n \subset C^1[0,1]$ defined by $x_n(t)=\dfrac{t^{n+1}}{n+1}$. I will prove that $(x_n)_n$ is Cauchy sequence as following
$$\lim_{n\to+\infty}\Vert x_{n+p}-x_n \Vert^2= \lim_{n\to+\infty}\int_{0}^{1}t^{2n}(t^p-1)^2dt=\int_{0}^{1}\lim_{n\to+\infty}t^{2n}(t^p-1)^2dt=0.$$
Since, $(C^1[0,1],\langle\cdot,\cdot\rangle)$ is Hilbert space, we have $(x_n)$ converges to $x_0 \in (C^1[0,1],\Vert\cdot\Vert)$. Now, I just have to show that $x_0 \notin C^1[0,1]$ then the problem will be solved but I have stucked.

Thanks for any help.

P/S: In addition, I also try to solve this problem by choosing a sequence that is the Cauchy sequence but not convergent in $(C^1[0,1],\Vert \cdot \Vert)$ but it does not work.

答:
Here is a different solution. Your norm is
$$\|f\| = \sqrt{|f(0)|^2+\|f'\|_2^2}$$
and we know that $(C[0,1],\|\cdot\|_2)$ is not complete with the non-convergent Cauchy sequence given with
$$f_n(x) := \begin{cases}
 0,   & \text{if $x \in \left[0,\frac1{2}\right]$}\\
 n\left(x-\frac{1}{2}\right), & \text{if $x \in \left[\frac1{2},\frac1{2}+\frac1n\right]$}\\
 1,              & \text{if $x \in \left[\frac1{2}+\frac1n,1\right]$}
\end{cases}$$
Therefore it makes sense to define $$g_n(x):=\int_0^x f_n(t)\,dt$$
which is a $C^1[0,1]$-function with $g_n' = f_n$. For $m \ge n$ we have
\begin{align}
\|g_m-g_n\|^2 &= |g_m(0)-g_n(0)|^2 + \int_0^1 |g_m(t)-g_n(t)|^2\,dt\\
&= \int_0^1 |f_m(t)-f_n(t)|^2\,dt\\
&= \int_{\frac12}^{\frac12+\frac1m} (m-n)^2\left(t-\frac12\right)^2\,dt + \int_{\frac12+\frac1m}^{\frac12+\frac1n}\left(1-n\left(x-\frac12\right)\right)^2\,dt\\
&= (m-n)^2\int_{0}^{\frac1m} t^2\,dt + \int_{\frac1m}^{\frac1n}(1-nt)^2\,dt\\
&= \frac{(m-n)^2}{3m^3}+\frac{\left(1-\frac{n}{m}\right)^3}{3n}\\
&\le \frac1{3m}+\frac1{3n} \xrightarrow{m,n\to\infty} 0
\end{align}

so $(g_n)_n$ is a Cauchy sequence in $(C^1[0,1],\|\cdot\|)$. Suppose that $g_n \xrightarrow{\|\cdot\|} g$ for some $g \in C^1[0,1]$.

In particular this means $g_n(0) \to g(0)$ and $g_n' \xrightarrow{L^2} g'$ which implies that $g'$ is the $L^2$-limit of $(f_n)_n$. We have
$$\left\|f_n-\chi_{\left\langle\frac12,1\right]}\right\|_2^2 = \int_{\frac12}^{\frac12+\frac1n} \left(1-n\left(x-\frac12\right)\right)^2\,dt = \int_0^{\frac1n} (1-nx)^2\,dt = \frac1{3n} \xrightarrow{n\to\infty} 0$$
so $g' = \chi_{\left\langle\frac12,1\right]}$ almost everywhere. Since $g(0)=0$, it follows that
$$g(x) = \int_0^x g'(t)\,dt = \left(x-\frac12\right)\chi_{\left\langle \frac12,1\right]}(x)$$
which is not differentiable at $x=\frac12$. This is a contradiction.


问:
A possible characterization for the median of a r.r.v.
Let $X$ be a real random variable. It is clear that any median $m\in\mathbb{R}$ of $X$ satisfies that
$$\text{E}[|X - m|] = \min_{x\in\mathbb{R}}\text{E}[|X - x|]$$.

My question is the following. If $b\in\mathbb{R}$ is a number such that $\text{E}[|X - b|] = \min_{x\in\mathbb{R}}\text{E}[|X - x|]$, is $x$ a median of $X$?

Thanks in advance.

答: Yes, if you define a median as a value $m$ where  $\mathbb P(X<m) \le \frac12$ and $\mathbb P(X\le m) \ge \frac12$. Your $b$ will have this property.


问:
Is the following differentiating under the integral sign correct?
Suppose $$\frac{\delta f[u]}{\delta u(x)}\equiv \frac{\partial f}{\partial u}-\frac{\partial }{\partial x}\frac{\partial f}{\partial u_x}+\left(\frac{\partial }{\partial x}\right)^2\frac{\partial f}{\partial u_{xx}}+...$$
and $$f[u]=\int_\mathbf{R} \phi(u,u_x)dx$$
What then is $$\frac{\partial }{\partial x}\frac{\delta f[u]}{\delta u(x)}$$?

My guess would be

$$\frac{\partial }{\partial x}\int_\mathbf{R} \frac{\partial \phi}{\partial u}-\frac{\partial }{\partial x}\frac{\partial \phi}{\partial u_x}dx$$

Then $$\left[\frac{\partial \phi}{\partial u}-\frac{\partial }{\partial x}\frac{\partial \phi}{\partial u_x}\right]_{-\infty}^{\infty}$$ Would that be right?

Anybody?

答: Have been informed that I misunderstood the notation in my notes. The problem is resolved now! Sorry about that!</p>


问:
Showing derivative operator is self-adjoint
Consider the Hilbert space $L^2(\mathbb{R})$, and unbounded operator $Au:=iu’$ with domain
$$D(A)= \{u \in L^2(\mathbb{R}) | u \text{ is absolutely continuous and } u’ \in L^2(\mathbb{R})\} $$
I’m showing that it’s self-adjoint. It’s not hard to show that it’s symmetric, and suffice to show that $D(A^*) \subseteq D(A)$, which I don’t know how to deal with.
Any help will be appreciated.

答:
Let $u\in D(A^\ast)$. For all $\phi\in C_c^\infty(\mathbb{R})$ we have
$$
\langle A^\ast u,\phi\rangle=\langle u,A\phi\rangle=-i\int u\phi'=i\langle u',\phi\rangle,
$$
where $u'$ is the distributional derivative of $u$.

Thus
$$
|\langle u',\phi\rangle|=|\langle A^\ast u,\phi\rangle|\leq \|A^\ast u\|_2\|\phi\|_2.
$$
Hence $u'\in L^2(\mathbb{R})$ by the Riesz representation theorem.

Now let
$$
\tilde u(x)=\int_0^x u'(y)\,dy.
$$
This function is absolutely continuous and has weak derivative $u'$. It follows that there exists a constant $C\in\mathbb{R}$ such that $u=\tilde u+C$ a.e. as explained here.


问:
Is the sequence defined by $a_{1}=\frac43, a_{n+1} = \sqrt{5a_{n}-6}$ monotonically decreasing?
I want to know where is my mistake in the proof by induction that $(a_n)$ is monotonically decreasing.

Basis of induction: $a_{2} - a_{1} = \sqrt{\dfrac{2}{3}} - \dfrac43 < 0$

Assume $a_{n+1} - a_{n} < 0 $

Then $a_{n+2} - a_{n+1} =  \sqrt{5a_{n+1}-6} -  \sqrt{5a_{n}-6}= ( \sqrt{5a_{n+1}-6}- \sqrt{5a_{n}-6})\dfrac{ \sqrt{5a_{n+1}-6}+ \sqrt{5a_{n}-6}}{ \sqrt{5a_{n+1}-6}+ \sqrt{5a_{n}-6}}=\dfrac{5a_{n+1}-6-5a_n+6}{ \sqrt{5a_{n+1}-6}+ \sqrt{5a_{n}-6}}=\dfrac{5(a_{n+1}-a_n)}{ \sqrt{5a_{n+1}-6}+ \sqrt{5a_{n}-6}}<0$

So $a_{n+1} - a_n < 0 \;\;\forall n \in \Bbb{N}$, hence this sequence is monotonically decreasing.

But how can this be true if $a_3 \notin \Bbb{R}$?

答: A sequence is a function from the natural numbers to the real line, that is you have $a_n: \mathbb{N} \rightarrow \mathbb{R}$, where the pedex $n$ is used as $x$ in the standard $f(x)$ notation. This sequence is not well defined if $a_n < 6/5$, but for the values with $a_n \ge 6/5$ you can apply your proof, that basically means your sequence is not well defined for every $n \ge 3$, since you proved inductively that increasing $n$, for sure you do not get a higher value for $a_n$. Since the definition of monotonically decreasing sequence is that $a_{n+1} \le a_n$ for all $n \in \mathbb{N}$, your sequence is not monotonically decreasing.


问:
'Contravariance has more mathematical structure in it whereas covariance is more geometrical and easy to understand.' Is it a serious claim?
In Shastri's Basic Algebraic Topology Remark 1.8.8, the author wrotes that

The difference between covariance and contravariance is simply in the fact that covariance preserves the direction of the arrow whereas contravariance reverses it. However, in practice, it turns out that contravariance has more mathematical structure in it whereas covariance is more geometrical and easy to understand.

I am really curious as to why the author thinks that way. Unfortunately, he doesn't provide further justification of that. My question is:

Is this a serious claim? (i.e. Is there a sense in which it is true?)

If the answer to 1 is 'yes', what's the underlying reason for this phenomenon? (maybe there is a categorical justification?)

Though I am just a beginner in algebraic topology, answers from all levels are welcomed.

As Connor Malin comments below, every contravariant functor is a covariant functor from $\mathsf{C^{op}}$ to $\mathsf D$, so the two concepts are duals of each other, which further mystifies the claim.

答:
Is it a serious claim? I claim this doesn't have a serious answer: It is opinion-based.

Remark 1.1.8 follows the definition of covariant and contravariant functors, thus it is certainly a statement about functors used in practice. There are other contexts where the words "covariant" and "contravariant" are used, for example in the context of (multi-)linear algebra; see Lutz Lehmann's comment and this question. I guess that most people (including physicists and engineers) who have heard these word associate it with tensors. But this context is irrelevant here.

Have a look at the contents of the book. It has 13 chapters, but one cannot say that contravariant functors (essentially cohomology functors) are overrepresented and provide more mathematical structure than covariant functors.

There are some nice things which you can do with cohomology, for example you can introduce the cup product and get the cohomology ring or define cohomology operations. But all that can be dualized, for example you can give homology a coring-structure. Many people have never heard about this concept, thus it may appear somewhat strange.

Contravariant functors are often representable ("Brown's representability theorem") which led to the fruitful concept of a spectrum. Another example is $K$-theory which arises via classifying vector bundles on spaces $X$; this gives a very important (generalized) cohomology theory. The construction is geometrical and easy to understand although it is contravariant.

This shows that many important concepts of algebraic topology are based (in a natural way) on contravariance. However, other things like homotopy groups are covariant.

I think we need on equal terms both co- and contravariance. The whole complex of duality, for example duality in manifolds, relies on both types.


问:
mathematical statistic question
Y~Erlang(n,1) , how to solve E(|Y-n|))?

I think that we have to divide integral by 2 parts, one with y>n, other one with y<n. But I couldn't progress anymore. someone know about this?

答:
Consider the distribution of $Z_j=Y_j-1$, where $Y_j \sim Exponential(1)$, and $Y \stackrel{iid}{=} \sum_j Y_j$, (iid)  then $Y-n = \sum_{j=1}^{n}Z_j$.

Derive the distribution of $Z_j, \ f_{Z_j}(z_j) = e^{-z+1}, z \geq -1$, and find its expectation. Then $\mathbf{E}[Y-n]=\sum_j\mathbf{E}[Z_j]$, because $Z_j$ are also iid.

Title: mathematical statistic question


问:
Determine the remainder, when $77^{200}$ is divided by $54$.
Is there any fast way to get the remainders of the division of large numbers?

Like$-$

$$\frac{77^{200}}{54}$$

Body:
As $77\equiv23\pmod{54}, 77^n\equiv23^n$

As  Carmichael Function $\lambda(54)=18$

and as $(77,54)=1$

and as  $200\equiv2\pmod{18}$

$$23^{200}\equiv23^2{\pmod{54}}\equiv-11\equiv-11+54$$


问:
Is this cancellation argument used to find the sum of a geometric series correct?
Here is a common argument used to prove that the sum of an infinite geometric series is $\frac{a}{1-r}$ (where $a$ is the first term and $r$ is the common ratio):
\begin{align}
S &= a+ar+ar^2+ar^3+\cdots \\
rS &= ar+ar^2+ar^3+ar^4+\cdots \\
S-rS &= a \\
S(1-r) &= a \\
S &= \frac{a}{1-r} \, .
\end{align}
I am sceptical about the validity of this argument. It feels like there is something amiss about a proof involving infinite series that makes no mention of the fact that they are typically defined as the limit of their partial sums. The third line involves 'cancelling' all of the terms other than $a$. This makes it seem like an infinite series is actually an infinite string of symbols, rather than a limiting expression. If $a+ar+ar^2+ar^3+\cdots$ is simply a shorthand for
$$
\lim_{n \to \infty}\sum_{k=0}^{n}ar^k \, ,
$$
and $ar+ar^2+ar^3+ar^4+\cdots$ a shorthand for
$$
\lim_{n \to \infty}\sum_{k=1}^{n}ar^k \, ,
$$
then I am struggling to see how the terms actually cancel. Perhaps the third line can be written more formally as
\begin{align}
S-rS &= (a+ar+ar^2+ar^3+\cdots)-(ar+ar^2+ar^3+ar^4+\cdots) \\
&= \lim_{n \to \infty}\sum_{k=0}^{n}ar^k - \lim_{n \to \infty}\sum_{k=1}^{n}ar^k \\
&= \lim_{n \to \infty}\left(\sum_{k=0}^{n}ar^k - \sum_{k=1}^{n}ar^k\right) \\
&= \lim_{n \to \infty}\left(a +\sum_{k=1}^nar^k - \sum_{k=1}^{n}ar^k\right) \\
&= \lim_{n \to \infty}a \\
&= a \, .
\end{align}
There is another concern I have. Geometric series only converge when $|r|<1$. However, the argument used above seems to apply regardless of whether $|r|<1$, which would yield nonsensical results such as
$$
1+2+4+8+\cdots = -1 \, .
$$
So is the argument rigorous, and if so, why are my fears misplaced?

答: Your argument is rigorous if you know that your series converges. Namely you first should know that the series converges and afterwards you may use that method to compute its limit. Prior to know that the series converges the method is wrong since, if the series diverges, in the first step you are subtracting two infinities ($S-rS = \infty - \infty$). Your fears are indeed justified.


问:
How to find a Lyapunov function
I have the following system: $$\dot{x_1} = x_2(x_3-2) \\ \dot{x_2} = x_1(x_3-2) \\ \dot{x_3}=-x_3^3 $$ and I want to determine its equilibrium points together with their stability. To find the equilibrium points, I wrote $$x_2=0 \text{ or } x_3=2 \\ x_1=0 \text{ or } x_3=2  \\ x_3 =0$$
Thus, only equilibrium point $x^*$ for the system is the origin $0$. I found $\det{Df(0)} = 0$ so I expect that the origin is a non-hyperbolic equilibrium point, probably asymptotically stable. To determine its type, I want to find a Liapunov function. Now, I assumed I can take $V(x)=c_1x_1^2 +c_2x_2^2+c_3x_3^3$ for some positive constants $c_i$. We have $V(0)=0$ and $V(x)>0$ for any $x \neq 0$. Then, $$\dot{V}(x)=\nabla V\cdot f(x) = 2c_1x_1x_2(x_3-2)+2c_2x_1x_2(x_3-2)-2c_3x_3^4.$$

$(\star )$ Now, let $c_1=c_2=1$ and $c_3 =2$ so that we get $$\dot{V}(x)=4x_1x_2(x_3-2)-4x_3^4.$$

However, I do not know how to determine the sign of this expression to conclude the result. I want to ask

Is it enough to find a Lyapunov function which is always positive or negative to say that $x^*$ is stable/unstable/asymptotically stable?

Similar to the first question but am I allowed to set arbitary values for $c_i$'s? (The $\star$ part)

How can I deduce the result?

答:
One can precisely describe all trajectories of the system with  (almost) no computation. As polfosol said, the dynamics of $x_3$ is independent from $x_1$, $x_2$. Without explicitly computing the solution, we can immediately say that $x_3(t)\to0$ as $t\to+\infty$.

Next, we call $u:=x_1+x_2$ and $v:=x_1-x_2$ and find that
$$\begin{cases}
\dot u &= (x_3-2)u\\
\dot v &= (2-x_3)v.
\end{cases}$$
Hence, for $t$ large, $\dot u\sim -2u$ and $\dot v\sim 2v$
and therefore $u\to0$
and $v$ (which cannot change sign) satisfies
$v\to+\infty$ if $v(0)>0$, $v\to-\infty$ if $v(0)<0$
and $v\equiv0$ if $v(0)=0$  (these can easily be proved rigorously).

Reverting to the original variables, we conclude that any trajectory tends to $\infty$ approaching the line
$$x_1+x_2=0\\
x_3=0$$
except for the ones starting from the plane $x_1-x_2=0$ (stable manifold), which approach the equilibrium $(0,0,0)$.

Title: How to find a Lyapunov function


问:
The multivariate normal distribution and its calculation
Let a random vector $X = (X_1, X_2,\ldots, X_p)^{\mathrm{T}}$ has the multivariate normal distribution with mean vector $\mu_1$ and covariance matrix $\Sigma_1 > 0$, i.e. $X \sim \mathcal{N}_p(\mu_1, \Sigma_1)$, and has the density $f_1(x)$, $x \in \mathbb{R}^p$.
Let a random vector $Y = (Y_1, Y_2,\ldots, Y_p)^{\mathrm{T}}$ has the multivariate normal distribution with mean vector $\mu_2$ and covariance matrix $\Sigma_2 > 0$, i.e. $Y \sim \mathcal{N}_p(\mu_2, \Sigma_2)$, and has the density $f_2(y)$, $y \in \mathbb{R}^p$.

I encountered
$$
D_p = \int_{\mathbb{R}^p}\int_{\mathbb{R}^p} (x - y)^{\mathrm{T}}(x - y) \, f_1(x)f_2(y) \, dy dx.
$$

Mathematica 12 showed me:
\begin{align*}
D_1
& = \int_{-\infty}^\infty\int_{-\infty}^\infty (x - y)^2 f_1(x) f_2(y) \, dydx \\
& = (\mu_1 - \mu_2)^2 + (\sigma_1^2 + \sigma_2^2),
\end{align*}
where
$$
f_1(x) = \frac{1}{\sqrt{2\pi}\,\sigma_1} \exp\biggl[-\frac{1}{2}\left(\frac{x - \mu_1}{\sigma_1}\right)^2 \, \biggr],
\quad
f_2(y) = \frac{1}{\sqrt{2\pi}\,\sigma_2} \exp\biggl[-\frac{1}{2}\left(\frac{y - \mu_2}{\sigma_2}\right)^2 \, \biggr],\\
$$

\begin{align*}
D_2
& = \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty
 [(x_1 - x_2)^2 + (y_1 - y_2)^2 ] f_1(x_1, y_1) f_2(x_2, y_2) \, dy_2dy_1 dx_2dx_1 \\
& = (\mu_{11} - \mu_{21})^2 + (\mu_{12} - \mu_{22})^2
    + (\sigma_{11}^2 + \sigma_{12}^2) + (\sigma_{21}^2 + \sigma_{22}^2)\\
& = (\mu_1 - \mu_2)^{\mathrm{T}}(\mu_1 - \mu_2) + \mathrm{tr}(\Sigma_1) + \mathrm{tr}(\Sigma_2)
\end{align*}
where
$$
\mu_1 = (\mu_{11}, \mu_{12})^{\mathrm{T}}, \quad
\mu_2 = (\mu_{21}, \mu_{22})^{\mathrm{T}},
$$
$$
\Sigma_1 =
\begin{pmatrix}
\sigma_{11}^2 & \rho_1\sigma_{11}\sigma_{12} \\
\rho_1\sigma_{11}\sigma_{12} & \sigma_{12}^2\\
\end{pmatrix}, \quad
\Sigma_2 =
\begin{pmatrix}
\sigma_{21}^2 & \rho_2\sigma_{21}\sigma_{22} \\
\rho_2\sigma_{21}\sigma_{22} & \sigma_{22}^2\\
\end{pmatrix},
$$
$$
|\rho_1| \leq 1, \quad
|\rho_2| \leq 1.
$$

I am guessing
\begin{equation}
D_p = (\mu_1 - \mu_2)^{\mathrm{T}}(\mu_1 - \mu_2) + \mathrm{tr}(\Sigma_1) + \mathrm{tr}(\Sigma_2).
\tag1\label1
\end{equation}

How can I prove equation \eqref{1}?

答:
Suppose that $X$ and $Y$ are independent. Put $c = \mu_1 - \mu_2$, $A = X-EX$, $B = Y-EY$. Hence
$$D_p = E(X-Y)^{T}(X-Y) =E(X-EX - (Y- EY) + c)^{T}(X-EX - (Y-EY)  +c) = E(A-B+c)^{T}(A-B+c).$$

We have $A \sim N_p(0, \Sigma_1)$ and $B \sim N_p(0, \Sigma_2)$. Hence $EA=EB=0$, $EA^{T}=EB^{T}=0$, $DA = \Sigma_1$, $EA^{T}A = tr(\Sigma_1), EB^{T}B = tr(\Sigma_2)$. Thus

$$D_p = E(A-B)^{T}(A-B) + E(A-B)^{T}c + Ec^{T}(A-B)+Ec^{T}c$$
and

$$D_p - c^{T}c = E(A-B)^{T}(A-B) = EA^{T}A - EB^{T}A - EA^{T}B + E B^{T}B = EA^{T}A +E B^{T}B $$
because $EB^{T}A = EB^{T} EA = 0$ and $EA^{T}B = EA^{T}EB = 0$ by independence.

It follows that

$$D_p  = c^{T}c + tr(\Sigma_1) + tr(\Sigma_2) = (\mu_1 - \mu_2)^{T}(\mu_1 - \mu_2) +  tr(\Sigma_1) + tr(\Sigma_2)$$
q.e.d.


答:
I think the proofs generally start (or at least they should) with something like "Suppose an infinite geometric series converges to the sum $S \,$", after which all the steps would be justified.

The 'cancelling' you mentioned can be avoided by including an extra step:

\begin{align}
S &= a+ar+ar^2+ar^3+\cdots \\
rS &= ar+ar^2+ar^3+ar^4+\cdots \\
rS + a &= a+ar+ar^2+ar^3+\cdots \\
&= S\\
S-rS &= a \\
S(1-r) &= a \\
S &= \frac{a}{1-r} \, .
\end{align}

Also, since the proof assumes the existence of $S$, it clearly will not apply if $|r| \ge 1 \,$.

Title: Is this cancellation argument used to find the sum of a geometric series correct?

Body:
Here is a common argument used to prove that the sum of an infinite geometric series is $\frac{a}{1-r}$ (where $a$ is the first term and $r$ is the common ratio):
\begin{align}
S &= a+ar+ar^2+ar^3+\cdots \\
rS &= ar+ar^2+ar^3+ar^4+\cdots \\
S-rS &= a \\
S(1-r) &= a \\
S &= \frac{a}{1-r} \, .
\end{align}
I am sceptical about the validity of this argument. It feels like there is something amiss about a proof involving infinite series that makes no mention of the fact that they are typically defined as the limit of their partial sums. The third line involves 'cancelling' all of the terms other than $a$. This makes it seem like an infinite series is actually an infinite string of symbols, rather than a limiting expression. If $a+ar+ar^2+ar^3+\cdots$ is simply a shorthand for
$$
\lim_{n \to \infty}\sum_{k=0}^{n}ar^k \, ,
$$
and $ar+ar^2+ar^3+ar^4+\cdots$ a shorthand for
$$
\lim_{n \to \infty}\sum_{k=1}^{n}ar^k \, ,
$$
then I am struggling to see how the terms actually cancel. Perhaps the third line can be written more formally as
\begin{align}
S-rS &= (a+ar+ar^2+ar^3+\cdots)-(ar+ar^2+ar^3+ar^4+\cdots) \\
&= \lim_{n \to \infty}\sum_{k=0}^{n}ar^k - \lim_{n \to \infty}\sum_{k=1}^{n}ar^k \\
&= \lim_{n \to \infty}\left(\sum_{k=0}^{n}ar^k - \sum_{k=1}^{n}ar^k\right) \\
&= \lim_{n \to \infty}\left(a +\sum_{k=1}^nar^k - \sum_{k=1}^{n}ar^k\right) \\
&= \lim_{n \to \infty}a \\
&= a \, .
\end{align}
There is another concern I have. Geometric series only converge when $|r|<1$. However, the argument used above seems to apply regardless of whether $|r|<1$, which would yield nonsensical results such as
$$
1+2+4+8+\cdots = -1 \, .
$$
So is the argument rigorous, and if so, why are my fears misplaced?


问:
Converse to extreme value theorem?
It's well known that continuous functions achieve minima on compact sets. An even stronger result is that lower semicontinuous functions achieve minima on compact sets.

Question. If an extended-real function $f$ attains a minimum on every compact subset of a topological space  $X$, does it follow that $f$ is lower semicontinuous?

If it helps, I'm happy to assume that $X$ is a convex and compact subset of a locally convex TVS and that $f$ is convex.

答:
What about the following function $f\colon\mathbb{R}\to\mathbb{R}$?
$$
f(x) =
\begin{cases}
0, & \text{if $x\neq 0$},\\
1, & \text{if $x = 0$}.
\end{cases}
$$
It attains a minimum on every (compact) set, but it is not l.s.c.

Title: Converse to extreme value theorem?


问:
How to calculate the arc length of a cardioid-like (the innerloop)?
I'm trying to calculate the arc length of the innerloop see picture of a cardioid-like $r=4a\cos^3\frac\theta3$ but don't know where to start. I know you have formulas like $L=\int_{a}^{b} \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2}} d \theta$, but don't know how to do this for the innerloop.

thanks in advance

答:
We can take out constants and multiply them back in at the end, so the curve is $r=\cos^3\frac{\theta}3$.

The required length is twice the length of the curve from $\theta=\pi$ to $\theta=\frac{3\pi}2$. Thus
$$\frac L2=\int_\pi^{3\pi/2}\sqrt{\cos^6\frac\theta3+\cos^4\frac\theta3\sin^2\frac\theta3}\,d\theta$$
$$=\int_\pi^{3\pi/2}\sqrt{\cos^4\frac\theta3}\,d\theta=\int_\pi^{3\pi/2}\cos^2\frac\theta3\,d\theta=\frac34\left(\frac\pi3-\frac{\sqrt3}2\right)$$
Therefore the length of the inner loop for $r=4a\cos^3\frac\theta3$ is $a(2\pi-3\sqrt3)$.


问:
When can i get away with approximating the expected value of a ratio as the ratio of expected values
I'm actually an engineering student so I'm not too good with probability and was hoping someone may be able to help with the following:

So I have a ratio of discrete random variables. I want to be able to know when I can get away with approximating its expected value as a ratio of expected values. For example, for the continuous case I came across the following formulation from a Taylor series expansion

$\text{E}[x/y]=\text{E}[x]/ \text{E}[y]- \text{Cov}[x ,y]/ \text{E}[y]^2 + \text{E}[x] \text{Var}[y]/ \text{E}[y]^3$

Does this apply for discrete random variables? if not, what is its analogous expression?

thank you!
hadi

答: First of all, $X$ and $Y$ should be independent. Now the problem is that in general $E[1/y] \neq 1/E[y]$. However, this is a reasonable approximation when $y$ is concentrated around some "large" value.</p>


问:
Integrate $\frac{1}{x\sqrt{a - bx^2}}$
I've been toiling for hours trying to answer this one:

Find $\int \frac{1}{x\sqrt{a - bx^2}}\,dx$

It's from Silvanus Thompson's 'Calculus Made Easy' (apparently not easy enough for me!).

The answer given is $\frac {1}{\sqrt{a}}\log\left(\frac{\sqrt{a}-\sqrt{a-bx^2}}{x\sqrt{a}}\right)$

I've attempted the question by letting $x=\sqrt{\frac ab}\sin(u)$, which I think is on the right track as I end up with the $\frac1{\sqrt a}$ term in my answer, but nothing else falls into place in terms of the given answer.

Thank you in advance for any help!

Andrew

答:
Hint Your method seems to work, provided anyway that $a, b > 0$: The substitution $x = \sqrt{\frac{a}{b}} \sin \theta$, $dx = \sqrt{\frac{a}{b}} \cos \theta \,d\theta$ transforms the integral to
$$\frac{1}{\sqrt{a}} \int \csc \theta \,d\theta .$$

Another option avoids trigonometric functions altogether: Rewrite the integral as
$$\int \frac{x \,dx}{x^2 \sqrt{a - b x^2}} .$$

What substitution does this form of the integrand suggest?

Additional hint Consider the substitution $u^2 = a - b x^2$, $du = -2 b x \,dx$.

Title: Integrate $\frac{1}{x\sqrt{a - bx^2}}$


问:
Complexity of transforming an indirect proof into a direct one
Suppose that ZFC is consistent, and let ZFC'=ZFC+Con(ZFC). We can define a reasonable notion of
the length of a proof inside ZFC', such that for any $n$ the set $P_n$ of all proofs of length $\leq n$
is finite. For any provable statement $\phi$ (I mean provable in $ZFC'$), denote by $c(\phi)$ the shortest possible
length of a proof of $\phi$.

If we denote by $L_n$ the (finite) set of all provable statements written with at most $n$ characters, then

the map $f_1$ defined by $f_1(n)={\sf max}_{\phi \in L_n} (c(\phi))$ is obviously not computable (otherwise provability in ZFC' would become decidable : we would only need to browse all proofs of length $\leq f_1(n)$).

On the other hand, the map $f_2$ defined by $f_2(n)={\sf max}_{\phi \in P_n} (c(\phi))$ is certainly computable (see fgp’s answer below).

Now consider a slightly more complicated case : denote by $Q_n$ the set of statements such that there is a proof (in ZFC') of $ZFC \vdash \phi$, whose length is $\leq n$. Is the map $f_3$ defined by $f_3(n)={\sf max}_{\phi \in Q_n} (c(\phi))$ computable ?

答:
It certainly seems so. If you pick an $\phi \in P_n$, then by definition it has a proof of length $\leq n$. Thus, $f_2(n) \leq n$, so to compute $f_2(n)$ you only have to find the smallest $m \leq n$ such that $P_m$ proves the same statements as $P_n$.

In other words, if $\Phi_i$ is the (finite!) set of statements proven by $P_i$, then $f_2$ can be defined as $$
  f_2(n) = \min \left\{ m \leq n\,:\, \Phi_m = \Phi_n \right\} \text{,}
$$
which certainly is computable.


问:
How to go about finding how many years between 1 and 2000 were leap years?
Question:

Assuming x represents the value of the year. If x is divisible by 4, then the year is determined as being leap. But, if x is divisible by 100, then, the year would be determined as not being leap, unless it would be divisible 400, which would then imply it being leap. How many years between 1 and 2000 would be leap?

Here is what I got:

Number of years divisible by 4 = $\frac{1996-4}{4}+1$ = 499

Number of years divisible by 100 = $\frac{1900-100}{100}+1$ = 19

Number of years divisible by 400 = $\frac{1600-400}{400}+1$ = 4

Number of years between 1 and 2000 would be leap = 499 - 19 + 4 = 484

Could anyone assist me into reformatting/ presenting my answers into a more mathematically proper presentation?

答:
Prior to the Gregorian calendar, a leap year would occur once every four years. In this case we have
$$j(x)=\left\lfloor\frac{x}{4}\right\rfloor$$
Once the Gregorian calendar was adopted, which occurred in 1582, we have that every 400 years contains 97 leap years. Hence
$$g(x)=\left\lfloor\frac{97x}{400}\right\rfloor$$
In this case we also have to account for the 12 leap years that occurred prior to 1582. Kudos to egreg for pointing this out.
$$g(x)=\left\lfloor\frac{97x}{400}\right\rfloor+12$$
Since 4 AD was not a leap year, we must substract one from the end result in both scenarios
$$j(x)=\left\lfloor\frac{x}{4}\right\rfloor-1$$
$$g(x)=\left\lfloor\frac{97x}{400}\right\rfloor+11$$
Ultimately what you're looking for is this
$$f(x)=\left\{\begin{array}{ll}
0, & x\lt 4 \\
\left\lfloor\frac x4\right\rfloor-1, & 4\le x\lt 1582 \\
\left\lfloor\frac{97x}{400}\right\rfloor+11, & x\ge 1582
\end{array}\right.$$
Where $x$ is an integer that represents a given year in the common era. Since we seek the total number of leap years between 1 and 2000, we have
$$f(1999)=495$$
And we're done. I hope this helps you understand.


答:
Let

$$S_n=\sum_{k=0}^n ar^n.$$

Then

$$(1-r)S_n=S_n-rS_n=a-ar^{n+1}$$ and

$$S_n=a\frac{1-r^{n+1}}{1-r}.$$

This is absolutely rigorous for any $r\ne1$.

And from this, for $|r|<1$,

$$S=\lim_{n\to\infty}S_n=\frac a{1-r}.$$

Now, for $|r|\ge1$, the infinite sum diverges as does the limit of $r^{n+1}$, so there is no paradox.


问:
Irreducible polynomial in integer polynomial ring with complex root is a reciprocal polynomial
I am working on this problem:

Let $p(x) \in \mathbb{Z}[x]$ be an irreducible polynomial. Show that $p(x)$ is a reciprocal polynomial (i.e., that its coefficients equidistant from either end are equal) if one of its roots is a complex number $z$ with $|z|= 1$.

My thinking so far is:

Let $z  = a+bi$. Then $z = a-bi$ is also a root of $p(x)$, so both $(x-(a+bi))$ and $(x-(a-bi))$ are factors of $p(x)$ $\Rightarrow$ $(x-(a+bi))(x-(a-bi)) = x^2 - 2ax + a^2 + b^2 = x^2-2ax + 1$ is a factor of $p(x)$ (using that $|z| = 1$, and thus, $a^2 + b^2 = 1$).

I suppose that, at this stage, I'm not sure how to use the fact that $p(x)$ is irreducible over $\mathbb{Z}[x]$ to obtain the result. It seems to me that $p(x)$ cannot have any other nontrivial polynomials over $\mathbb{Z}$ as factors, since $p(x) \in \mathbb{Z}[x]$. But then, how can I know about other factors of $p(x)$ ?

Thanks!

答:
The following proof is due to a classmate of mine:

Let $K$ be the splitting field of $p(x)$ over $\Bbb Z$ and $\sigma\in Aut(K/\Bbb Z)$ such that for every root $\alpha$ of $p(x), \sigma(\alpha)=\dfrac1{\alpha}$. But $\sigma$ sends any root of $p$ to some root of $p$, which means that the set of all roots of $p$ is the same as the set of reciprocals of all roots of $p$. Hence for every root of $p$, its reciprocal is also a root of $p$, and thus $p(x)$ is reciprocal.


问:
Theory of probability.
Theory of probability.
The random vector (x, y) has a density:
$ f(x,y)=Cxy ;  0\leqslant x \leqslant 2, 1 \leqslant y \leqslant 3 $
Find the constant C.
How can I do it?

答:
Massive hint:
$$
C\int_{X}\int_{Y}f_{XY}(x,y) = 1
$$

Title: Theory of probability.


问:
Use the sum $\sum_{k=1}^{n} k^4 = \frac15 n^5 + \frac12 n^4 + \frac13 n^3 - \frac{1}{30}n$ to find the value of sup$\{L( f ,Pn) : n\in \mathbb{N}\}.$
Suppose that $f : [0,1]\to \mathbb{R}$ is given by $f (x) := 8−x^4 $ for all $x \in [0,1].$ For each $n\in \mathbb{N}$, let
$P_n$ denote the partition of $[0,1]$into $n$ subintervals of equal width. Use the sum $\sum_{k=1}^{n} k^4 = \frac15 n^5 + \frac12 n^4 + \frac13 n^3 - \frac{1}{30}n$ to find the value of sup$\{L( f ,Pn) : n\in \mathbb{N}\}.$

So far I have let $P_n =\{0, \frac1n , \frac2n , ... , 1\}$ so that we get,

$m_i$ = inf$\{8-x^4: x\in [\frac{i-1}{n} , \frac{i}{n}]\} = 8-(\frac{i-1}{n})^4.$

Using this to obtain that,

$L(F,P_n)=\sum_{i=1}^{n} m_i (x_i-x_{i-1})$
$=\sum_{i=1}^{n} 8-(\frac{i-1}{n})^4\cdot \frac1n.$

But I dont know how to use this to then sub into the sum that I have been given

答:
Wait: the $m_i$s that you compute are not correct. Indeed
$$m_i = 8-\biggl(\frac{i}{n}\biggr)^4$$
What you are asked to do is to find
$$\lim_{n\to+\infty}\sum_{i=1}^n\biggl(8-\frac{i^4}{n^4}\biggr)\frac{1}{n},$$
since the sup turns out to be a limit due to the fact that the the terms of the sum are positive $\forall n\geq 1$. Note that you did not use the parenthesis in the correct way because $1/n$ multiplies the whole $m_i$ and not just the term $(i/n)^4$. Now split your summands to get (applying the formula for $\sum_{k=1}^n k^4$)
$$\frac{1}{n}\sum_{i=1}^n8 - \frac{1}{n^5}\sum_{i=1}^ni^4 = 8 - \frac{1}{5}- \frac{1}{2n}- \frac{1}{3n^2} + \frac{1}{30n^4}.$$
An easy limit computation leads to the final result, which is $8-1/5 = 39/5$.


问:
Bounded solution ODE
I have the system

$$x'(t) = y(t)$$
$$y'(t) = -y(t)^3 - p(x(t))$$

with initial condition $(x(0),y(0)) = (x_0,y_0)$ and $p(x(t))$ is a polynomial of odd degree with positive leading coefficient. I have to show that the solution $(x(t),y(t))$ is bounded for all $t \geq 0$. I am not sure how to start. I am guessing I need to look for a monotone quantity. Any help would be appreciated.

答:
Take some integral $P$ of $p$. The leading term then has even degree, so all sets $\{x:P(x)<c\}$ are bounded. Assume that the integration constant is chosen so that $P$ only takes non-negative values.

Then $L(x,y)=\frac12y^2+P(x)$ is a Lyapunov function of the system,
$$
\frac{d}{dt}L(x(t),y(t))=-y(t)^4
$$
means that if $c=L(x(0),y(0))$, then $L(x(t),y(t))\le c$ for all $t\ge 0$. The level sets of $L$ are likewise bounded, as also $|y|\le \sqrt{2c}$. Moreover, the solution can only become stationary in $y=0$ if also $0=p(x)=P'(x)$, that is, in one of the equilibrium points of the system. Else the solution descends towards one of the minimum points of $P$.

Title: Bounded solution ODE


答:
$$I=\int \frac{dx}{x\sqrt{a-bx^2}}$$
Let $x=\sqrt\frac{a}{b}\sin t$, then
$$I=\int \frac{\sqrt{a/b} \cos t!dt}{\sqrt{a/b} \sin t \sqrt{a} \cos t}.$$
$$\implies I=\frac{1}{\sqrt{a}} \int \csc t dt=\frac{1}{\sqrt{a}}\int \csc t \frac{\csc t+\cot t}{\csc t+ \cot t}dt.$$
Let $\csc t+ \cot t=u$
$$I=\frac{1}{\sqrt{a}} \int \frac{-du}{u}=\frac{-1}{\sqrt{a}}\ln(\csc t+\cot t)=-\frac{1}{\sqrt{a}}ln\frac{1-\sqrt{1-c^2x^2}}{cx}+C,c=\sqrt{b/a}$$

Title: Integrate $\frac{1}{x\sqrt{a - bx^2}}$

Body:
I've been toiling for hours trying to answer this one:

Find $\int \frac{1}{x\sqrt{a - bx^2}}\,dx$

It's from Silvanus Thompson's 'Calculus Made Easy' (apparently not easy enough for me!).

The answer given is $\frac {1}{\sqrt{a}}\log\left(\frac{\sqrt{a}-\sqrt{a-bx^2}}{x\sqrt{a}}\right)$

I've attempted the question by letting $x=\sqrt{\frac ab}\sin(u)$, which I think is on the right track as I end up with the $\frac1{\sqrt a}$ term in my answer, but nothing else falls into place in terms of the given answer.

Thank you in advance for any help!

Andrew


答:
For a moment, let's gloss over the finer points of how leap years are determined, and find the number of multiple of $4$'s there are between $4$ and $1996$:
\begin{align}
1996-4 &= 1992 \\
1992 \div 4 &= 498 \, .
\end{align}
$498$ is not quite right because of what's called the Fence Post Problem. How many multiple of $4$'s are there between $4$ and $8?$ Not one but two! So there are actually $499$ multiple of $4$'s between $4$ and $1996$. Now let's try to think about the multiples of $100$ that are not multiples of $400$. There are $19$ multiples of $100$ in the range: $100,200,300,\ldots,1900$. Of those $19$, four of them are multiples of $400$. So there are $15$ exceptions that we have to deal with. We are left with
$$
499-15=484
$$
leap years.

This answer is technically incorrect because $4$ AD was not a leap year, as k170 has pointed out. So we are left with $483$ leap years (unless someone points out some other idiosyncrasy about the calendar system!).

Title: How to go about finding how many years between 1 and 2000 were leap years?

Body:
Question:

Assuming x represents the value of the year. If x is divisible by 4, then the year is determined as being leap. But, if x is divisible by 100, then, the year would be determined as not being leap, unless it would be divisible 400, which would then imply it being leap. How many years between 1 and 2000 would be leap?

Here is what I got:

Number of years divisible by 4 = $\frac{1996-4}{4}+1$ = 499

Number of years divisible by 100 = $\frac{1900-100}{100}+1$ = 19

Number of years divisible by 400 = $\frac{1600-400}{400}+1$ = 4

Number of years between 1 and 2000 would be leap = 499 - 19 + 4 = 484

Could anyone assist me into reformatting/ presenting my answers into a more mathematically proper presentation?


问:
How long will it take Marie to saw another board into 3 pieces?
So this is supposed to be really simple, and it's taken from the following picture:

Text-only:

It took Marie $10$ minutes to saw a board into $2$ pieces. If she works just as fast, how long will it take for her to saw another board into
  $3$ pieces?

I don't understand what's wrong with this question.  I think the student answered the question wrong, yet my friend insists the student got the question right.

I feel like I'm missing something critical here. What am I getting wrong here?

答:
The student is absolutely correct (as Twiceler has correctly shown).

The time taken to cut a board into $2$ pieces (that is $1$ cut) : $10$ minutes

Therefore, The time taken to cut a board into $3$ pieces (that is $2$ cuts) : $20$ minutes

The question may have different weird interpretations as I am happy commented:<br

Time taken to cut it into one piece = $0$ minutes

So Time taken to cut it into $3$ pieces = $0 \times 3$ minutes = $0$ minutes.

So $0$ can be an answer. but it is illogical just like the teacher's answer

and as Keltari said

Another correct answer would be 10 minutes. One could infer, "If she works just as fast," that "work" is the complete amount of time to do the job.  -Keltari

This is logical but you can be sure that this is not what the question meant, but the student has chosen the most relevant one. The teacher's interpretation is mathematically incorrect.

The teacher may have put the question for the students to have an idea of Arithmetic Progression and may have thought that the students will just answer the question without thinking hard. In many a schools, at low grades children are thought that real numbers consists of all the numbers. Only later in higher grades do they learn that complex numbers also exist. (I learned just like that.) So the question was put as a question on A.P. thinking that the students may not be capable of solving the answer the correct way.

Or as Jared rightly commented:

This is simultaneously wonderful and sad. Wonderful for the student who was level-headed enough to answer this question correctly, and sad that this teacher's mistake could be representative of the quality of elementary school math education. – Jared

Whatever may be the reason, there is no doubt that the student has been accurate in answering the question properly and that the teacher's answer is illogical.


答:
We have
$$C\int_{X}\int_{Y}f_{XY}(x,y) dydx= 1.$$ Then

$$C\int_0^2\int_1^3 xy dy dx = 1,$$ where

$$\int_0^2\int_1^3 xy dy dx = \int_0^2 4x dx = 8.$$

Finally, we have that $$C = \frac{1}{8}.$$

Title: Theory of probability.

Body:
Theory of probability.
The random vector (x, y) has a density:
$ f(x,y)=Cxy ;  0\leqslant x \leqslant 2, 1 \leqslant y \leqslant 3 $
Find the constant C.
How can I do it?


问:
Is there a "superposition" principle for non-linear PDEs?
I have following sets of PDES:

$\frac{\partial \omega_1'}{\partial t}+u_2'\frac{\partial \omega_1'}{\partial x_2}+u_3'\frac{\partial \omega_1'}{\partial x_3}= \nu(\frac{\partial^2 \omega_1'}{\partial x_2^2}+\frac{\partial^2 \omega_1'}{\partial x_3^2})$

I want to assume a Beltrami flow, which basically neglects the non-linear terms in this PDE to zero, hence the PDE simplifies to:

$\frac{\partial \omega_1'}{\partial t}=\nu(\frac{\partial^2 \omega_1'}{\partial x_2^2}+\frac{\partial^2 \omega_1'}{\partial x_3^2})$

The functions $u_2'$ and $u_3$' can be obtained as:

$\frac{\partial \omega_1'}{\partial x_2}=\frac{\partial^2 u_3'}{ \partial x_2^2}+\frac{\partial^2 u_3'}{\partial x_3^2} \\
    \frac{\partial \omega_1'}{\partial x_3}=-\frac{\partial^2 u_2'}{ \partial x_2^2}-\frac{\partial^2 u_2'}{\partial x_3^2}$.

I found the solution for $\omega_1'$ using the separation of variables as:

$\omega_1'=\sum_{j=0}^{p}\sum_{k=0}^{n}(C_{1,k} cos(\sqrt{a_k}x_2)+C_{2,k} sin(\sqrt{a_k}x_2))(C_{3,j} cos(\sqrt{b_j}x_3)+C_{4,j} sin(\sqrt{b_j}x_3))e^{-\nu(a_k+b_j)t}$ Thus $u_2'$ and $u_3'$ can be easily found using the above-mentioned relation. However in order to fulfill the boundary conditions as

$u_2'(x_2=0)=0 \\
    u_2'(x_2=L_2)=0\\
    u_3'(x_2=0)=0\\
    u_3'(x_2=L_2)=0$
I need to look at the whole sum of the solution of $\omega_1'$, therefore adding each soltuin for each $a_k$, thus superpositioning the solutions. Each solution individually solves

$u_2'\frac{\partial \omega_1'}{\partial x_2}+u_3'\frac{\partial \omega_1'}{\partial x_3}=0$.

However once I take a sum of each solution this does not work any more.

The question therefore is, weather there is a technique for THIS non linear PDE to combine single solutions correctly?

I would highly appreciate any kind of help.

答:
I know there are attempts to project superposition principles for non-linear PDE systems, although the cases in which these work are very special. I am no expert in this field, but I may refer you to the following article, which deals with non-linear superposition in terms of certain equations (Navier-Stokes is included!):
https://ieeexplore.ieee.org/document/4342629

In any case, I hope experts can help you out, your approach surely is very exciting and it's nice to see people deal with special solutions of Navier-Stokes!


问:
Reading the primal solution from dual simplex tableau
I was following ptrickJMT's video on how to solve a minimization linear programming problem and he did something that I did not understand why it works. He starts with the following minimization problem:

$$
\begin{matrix}\min &  C = 3x_1 + 9x_2 \\
s.t: \\
& 2x_1 + x_2 \ge 8 \\
& x_1 + 2x_2 \ge 8 \\
&x_1, x_2 \ge 0
\end{matrix}
$$

He proceeds to find the dual form of this example:

$$
\begin{matrix}\max &  P = 8y_1 + 8y_2 \\
s.t: \\
& 2y_1 + y_2 \le 3 \\
& y_1 + 2y_2 \le 9
\end{matrix}
$$

He proceeds to solve it using the regular simplex algorithm and arrives in the following matrix:

$$
\left[\begin{array}{ccccc|c}
2 & 1 & 1 & 0 & 0 & 3 \\
-3 & 0 & -2 & 1 & 0 & 3 \\
\hline
8 & 0 & 8 & 0 & 1 & 24
\end{array}\right]
$$

Up to here, it's fine, I can see that the optimal solution is $C = P = 24$, and that $y_1 = 0$ and $y_2 = 3$. He also claimed that the optimal solution of the primal problem is $x_1 = 8$ and $x_2 = 0$, he did so by reading coefficients of the slack variables as seen below. I did not understand why this is true, can anyone clarify this to me?

$$
\left[\begin{array}{ccccc|c}
2 & 1 & 1 & 0 & 0 & 3 \\
-3 & 0 & -2 & 1 & 0 & 3 \\
\hline
8 & 0 & \color{red}{8} & \color{red}{0} & 1 & 24
\end{array}\right]
$$

Thank you.

答:
He probably used the complementary slackness theorem. For this purpose I write the condition constraints with slack variables and therefore equalities.

$$
\begin{matrix}\min &  C = 3x_1 + 9x_2 \\
s.t: \\
& 2x_1 + x_2 -s_1= 8 \\
& x_1 + 2x_2 -s_2= 8 \\
&x_1, x_2 \ge 0
\end{matrix}$$

$$\begin{matrix}\max &  P = 8y_1 + 8y_2 \\
s.t: \\
& 2y_1 + y_2 +z_1= 3 \\
& y_1 + 2y_2 +z_2= 9
\end{matrix}
$$

The complementary slackness theorem states:

$x_j^*\cdot z_j=0 \ \forall \ \ j=1,2, \ldots , n$

$y_i^*\cdot s_i=0 \ \forall \ \ i=1,2, \ldots , m$

$s_i \text{ are the slack variables of the primal problem.}$

$z_j \text{ are the slack variabales of the dual problem.}$

From the dual optimal solution we know that $y_2^*=3$. That means that $3\cdot s_2=0\Rightarrow s_2=0$. That again means the second constraint of the primal is

$$x_1+2x_2=8\quad (1)$$

And last but not least we know that the optimal primal value is equal to the optimal dual value (strong duality theorem).  That means that $8\cdot y_1^*+8\cdot 0=y_1^*+8\cdot 3=24$. Therefore the equation for the optimal value of the primal is

$$3x_1 + 9x_2=24 \quad (2)$$

$$x_1 + 3x_2=8\quad (2a)$$

The solution follows straightforward if you subtract $(1)$ from $(2a)$. If you are more familiar with this theorems it is not very time-consuming. In this case more or less two minutes, I guess.


问:
PDF and CDF of $R=\frac {X}{X+Y}$
Let $X$ and $Y$ be two exponentially distributed random variables with $\lambda$ and $\mu$ parameters respectively. Find the CDF and PDF of $R=\frac {X}{X+Y}$

I'm able to find the PDF of $S=X+Y$. But I'm unsure about the bounds of the integral when I want to calculate the PDF of $R$.

答: For $0<r<1$ we have $P(R \leq r)=\int_0^{\infty}\int_{(1-r)x/r}^{\infty} \lambda e^{-\lambda x} \mu e^{-\mu y} dydx$ since $R \leq r$ iff $X \leq r(X+Y)$ iff $Y\geq (1-r)x/r$. The value is $\frac {\lambda r} {\lambda r+\mu (1-r)}$. PDF is its derivtive, namely, $\frac {\lambda \mu} {(\lambda r+\mu (1-r))^{2}}$.

Title: PDF and CDF of $R=\frac {X}{X+Y}$


问:
$\sum_{k=1}^{n} \binom{n}{k}k^{r}$

Find:$$\sum_{k=1}^{n} \binom{n}{k}k^{r}$$

For r=0 the sum is obviously $2^{n}$.

For r=1 the sum is $n2^{n-1}$.

For r=2 the sum is $n(n+1)2^{n-2}$.

Here's what I've tried:

$$\frac{d(1+x)^{n}}{dx}=n(1+x)^{n-1}=\binom{n}{1}+2\binom{n}{2}x+3\binom{n}{3}x^{2}+\cdots+n\binom{n}{n}x^{n-1}$$

$$\frac{d[x\cdot n(1+x)^{n-1}]}{dx}=\binom{n}{1}+2^{2}\binom{n}{2}x+3^{2}\binom{n}{3}x^{3}+\cdots+n^{2}\binom{n}{n}x^{n-1}$$

So if we continue like this and put x=1 we will get our result.However I cannot generalize this.

I think the answer has something to do with the Stirling Numbers of the Second kind.

Another way to present this question:
    Find $A_{1},A_{2},\cdots ,A_{r}$ such that $$k^{r}=A_{1}k+A_{2}k(k-1)+A_{3}k(k-1)(k-2)+\cdots$$

If someone can answer the second question I can solve the first one.

答:
You're right that it has to do with Stirling numbers. Wikipedia's article on Stirling numbers of the second kind answers your second question:

If we let $$(x)_n=x(x-1)(x-2)\cdots(x-n+1) \,$$ (in particular, $(x)_0 = 1$ because it is an empty product) be the falling factorial, we can characterize the Stirling numbers of the second kind by $$ \sum_{k=0}^n \left\{\begin{matrix} n \\ k \end{matrix}\right\}(x)_k=x^n$$

Title: $\sum_{k=1}^{n} \binom{n}{k}k^{r}$


问:
Linear differential equation system and general variation of parameters
I need some feedback to complete my solution for the following exercise:

Given a matrix $ A\in\mathbb{R}^{d\times d}$ and a vector $ b\in\mathbb{R}^{d}$, take a look at the initial value problem (IVP)

$$ y'=Ay+b\quad\land\quad y(0)=y_{0}\in\mathbb{R}^{d}$$

Determine the solution for the IVP by using variation of parameters. Calculate the
occurring integral explicitly in case of $ b\in\operatorname{img}(A)$.

Well, one can find $y(x)=\mathrm{e}^{xA}y_{0}$ as solution for the homogeneous differential equation. Now using the method of variation of parameters by inserting $ y(x)=\mathrm{e}^{xA}y_{0}(x)$ in the system and knowing that the matrix exponential is invertible

\begin{align}
 &&A\mathrm{e}^{xA}y_{0}(x)+\mathrm{e}^{xA}y_{0}'(x) &= A\mathrm{e}^{xA}y_{0}(x)+b \\
\Leftrightarrow && \mathrm{e}^{xA}y_{0}'(x) &= b\\
\Leftrightarrow && y_{0}'(x) &=\mathrm{e}^{-xA}b \\
\Leftrightarrow && y_{0}(x) &= y_{0} + \left(\int_{0}^{x}\mathrm{e}^{-tA}b\mathrm{d}t\right)
\end{align}
How to go on from that point onwards? Can I do something like that or?
\begin{align}
\overset{?}{\Rightarrow} && y_{0}(x) &= y_{0} + \left(\int_{0}^{x}\mathrm{e}^{-tA}\mathrm{d}t\right)b \\
\Leftrightarrow && y_{0}(x) &= y_{0}+A^{-1}b\mathrm{e}^{-xA}
\end{align}

Thanks in advance.

答:
You stumbled evaluating the integral in the last step.
$$\eqalign{
d\left(e^{-tA}\right) &= -Ae^{-tA}dt \\
e^{-xA}-I &= \int_{0}^{x}-Ae^{-tA}dt \\
A^{-1}(I-e^{-xA}) &= \int_{0}^{x}e^{-tA}dt \\
}$$
Using the function
$$\phi(x) = \frac{e^x-1}{x}$$
this result can be written as
$$\eqalign{
\int_{0}^{x}e^{-tA}dt
&= \phi(-xA)\;x \\
}$$
And your variational approach yields
$$\eqalign{
y_0(x) &= y_0 + \phi(-xA)\left(xb\right) \\
y(x) &= e^{xA}y_0  + \phi(xA)\left(xb\right) \\
}$$


问:
Proving statement about the ODE $y'=y+y^4$

Is there a solution to the problem $$\left\{\begin{matrix}
y'=y+y^4\\
y(x_0)=y_0
\end{matrix}\right.$$
  which is defined on $\mathbb{R}$? ($x_0,y_0$ might be any real numbers)

It's easy to prove that for all $(x,y)\in\mathbb{R}^2$ there exists an open interval $I$ (with $x_0\in I$) where the problem has a unique solution. However is the maximal interval always $(-\infty,\infty)$? I know that the answer is no, but that's just because I found a solution for particular values of $x_0$ and $y_0$ and checked its domain. But is there a way to prove that $I$ need not be $I=(-\infty,\infty)$ without actually solving the problem for a certain initial condition? In other words, how can I prove that the unique solution need not be defined on $\mathbb{R}$?

答:
This is a Bernoulli equation, and can thus be solved directly. Setting $u(x)=y(x)^{-3}+1$ results in the equation
$$
u'(x)=-3y(x)^{-4}y'(x)=-3u(x)\implies u(x)=u(x_0)e^{-3(x-x_0)}
$$
so that
$$
y(x)=\frac{y(0)e^{x-x_0}}{\sqrt[3]{1+y_0^3(1-e^{3(x-x_0)})}}
$$
where the cube root is extended as odd function to negative values.

The solution has a singularity in finite time if the denominator has a root, which is the case if $1+y_0^{-3}>0$, that is, $y_0>0$ or $y_0<-1$.


问:
(Uniform) continuity of a nonlinear function on a locally convex topological vector space
Let $E_1$ be a (nontrivial) vector space, $P$ be a family of seminorms on $E_1$, $\tau_1$ denote the topology generated by $P$, $(E_2,\tau_2)$ be a topological space, $f:E_1\to E_2$ at $x\in E_1$.

Can we show that $f$ is $(\tau_1,\tau_2)$-continuous at $x$ if and only if $f$ is $(p,\tau_2)$-continuous for all $p\in P$?

EDIT: If not, is the proof in the reference below wrong or is there anything special in that situation?

The following instance of this setting (found in Linde's Probability in Banach Spaces) is confusing me:

In his proof, the author has found a single $K$ for which $\hat\mu$ is (uniformly) $p_K$-continuous.

Translated to the general setting in this question, this seems to indicate that it's sufficient to have $(p,\tau_2)$-continuity for a single $p\in P$. Is this really true?

On the other hand, if I got it right, he has even shown uniform continuity. Is this the crucial ingredient?

答:
Only one direction is true.

If $f$ is $(p,\tau_2)$-continuous for any $p \in P$, then $f$ is $(\tau_1,\tau_2)$-continuous. This follows simply from the fact that $\tau_1$ is finer (larger) than the topology of $p$.

The other direction is not true in general. Let $p,q$ be two nonequivalent norms on a vector space $E$. W.l.o.g. assume that there is no constant $M>0$ such that $p \le Mq$. Let $\tau_1$ be the topology generated by $P=\{p,q\}$ and $\tau_2$ the standard topology on $\Bbb{R}$. Then $p : E \to \Bbb{R}$ is $(\tau_1,\tau_2)$-continuous but not $(q,\tau_2)$ continuous.


问:
proof verification $f(\overline{A})\subseteq \overline{f(A)}$
let $X$ be compact $f $ continuous and $A\subseteq X$ prove that $f(\overline{A})\subseteq \overline{f(A)}$

solution:

take $f(x) \in f(\overline{A})\Rightarrow x \in \overline{A}\Rightarrow$

$\exists \: x_n\subseteq A : x_n\rightarrow x$ now since $x_n\subseteq A$ and
$f$ is cont. $f(x_n)\subseteq f(A)$ and $f(x_n)\rightarrow f(x)\Rightarrow$

$f(x) \in  \overline{f(A)}$

答:
Your reasoning is correct, but you need to connect the ideas in a better way. Here is a better way to prove it

Let $x \in f(\bar{A}),$ thus there exists $x_{0} \in \bar{A}$ such that $f\left(x_{0}\right)=x .$ Thus, we have that $x_{n} \in A$ with $x_{n} \rightarrow x_{0} .$ since $f$ is continuous, it follows that $\lim _{n \rightarrow+\infty} f\left(x_{n}\right)=f\left(\lim _{n \rightarrow+\infty} x_{n}\right)=f\left(x_{0}\right)=x .$ Therefore, since $f\left(x_{n}\right) \in f(A), \forall n$
and $f\left(x_{n}\right) \longrightarrow x,$ then $x \in \overline{f(A)} .$ We conclude that :
$$f(\bar{A}) \subseteq \overline{f(A)}$$


问:
Show that $\{ f^{-1}(E) : E \in σ(S)\} = \sigma \{f^{-1}(E) : E ∈ S\}$.
Given a collection of subsets of a set $X$ let $\sigma$(S) denote the $\sigma$-algebra generated by $S$. Let $X, Y$ be sets, $f : X → Y$ and $S$ be a collection of
subsets of $Y$ . Show that $\{f^{-1}(E) : E \in\sigma(S)\} = \sigma\{f^{-1}(E) : E \in S\}$.

I have already shown that both of these sets are $\sigma$-algebras, so this is not an issue. I have also shown that  $B = \{E \subset Y : f^{-1}(E) \in X1\}$ is a $\sigma$-algebra on $Y$ (where $X_1$ is the first set). I wanted to use this to say that is $S \subset B$ and $B$ is a $\sigma$-algebra implies that $\sigma$(S) $\subset$ B and thus show that A $\in$ X$_1$ implies A $\in$ X$_2$. I don't really know if this is correct or how to go about finishing it so any help would be appreciated.

答: Let $\mathcal A=\{f^{-1} (E):E \in \sigma (S)\}$ and $\mathcal B=\sigma \{f^{-1} (E):E \in S\}$. Consider $\{E \in \sigma (S): f^{-1}(E) \in \mathcal B\}$. This is a sigma algebra which contains $S$. Hence it contains $\sigma (S)$. This proves that $\mathcal A \subseteq \mathcal B$. The other inclusion is simple: $\{f^{-1} (E):E \in S\} \subseteq \mathcal A$ and $\mathcal A$ is a sigma algbra. Hence $\mathcal B \subseteq \mathcal A$.


问:
Linear Combination of Normally Distributed Random Variables
$\xi_{1}, \ldots, \xi_{5}$ and independent random variables with $N_{1,4}$ distribution. Find distribution of

$\eta = \xi_{1}-\frac{\xi_{2}+\xi_{3}+\xi_{4}+\xi_{5}}{2}$

and its CDF.

My solution

We know that if $X \in N_{a,\alpha^2}$ and $Y \in N_{b,\beta^2}$ then $X+Y=Z \in N_{a+b,\alpha^2 + \beta^2}$

Also, $X=cZ \in N_{a,(c\alpha)^2}$, where $c =const$

So,

$\eta \in N_{1-\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right), 4-(1+1+1+1)} = N_{0,0}$

Basically, I get Dirac delta function which is weird (or not?). Is it correct?

答: If $X$ and $Y$ are independent random variables then $var (X-Y)=var (X)+var (Y)$, not $var (X-Y)=var (X)-var (Y)$.


答:
Another way to get the answer summarizes as

$${2000\over4}-{2000\over100}+{2000\over400}-1$$

That is, start by including year 2000 and assume for the moment that every fourth year is a leap year. That gives the $2000/4$. Next assume you throw away the leap day from every century mark. That gives the $-2000/100$. Then remember that there is a leap year in every fourth century mark. That gives the $2000/400$. And finally, remember that you don't want to include the year 2000. That gives the $-1$.

Remark: As k170 points out, the Julian calendar omitted leap day in the year 4, so there's an argument for subtracting another $1$ from the answer. However, the Julian calendar otherwise included leap days every four years, not skipping century marks, so there is an argument for adding back another dozen to the answer, since the Gregorian calendar, whose main innovation was the skipping of leap day in three out of every four century marks, wasn't invented until the sixteenth century. The reason for saying "a dozen or so" instead of something more precise is that the exact answer is geopolitically dependent: Different countries adopted the Gregorian calendar in different centuries. (See egreg's comment below k170's answer, which I only just now noticed.)


问:
Domain of the Total Derivative
Suppose I have some map $F: \mathbb{R^{m}} \rightarrow \mathbb{R^{n}}$. I'm struggling to get an intuitive understanding of the meaning of the total derivative $dF_p: \mathbb{R^{m}} \rightarrow \mathbb{R^{n}}$ at a point $p \in \mathbb{R^{m}}$. This will obviously be represented some $n \times m$ matrix. Now, each entry in this matrix will be of the form $\frac{\partial F_i}{\partial x_j}$, and thus will be a function of $(x_1,...,x_m)$.

As I understand, we input the values of $p \in \mathbb{R^{m}}$ into this matrix, thus giving an $n \times m$ matrix of real numbers, and then this defines the map $dF_p: \mathbb{R^{m}} \rightarrow \mathbb{R^{n}}$.

However, the Regular Value Theorem states that we require this map to be surjective. This implies that we will be able to apply $dF_p$ to any point $q \in \mathbb{R^{m}}$ (otherwise how are we able to $hit$ all of the values in $\mathbb{R^{n}}$?) But this map is the differential at $p$, and so how does this make sense to apply it anywhere other than $p$?

答:
You don't apply the matrix to $p$. Informally, you apply it to a vector pointing away from $p$. The idea is this: for each point $p\in\mathbb R^n$ at which $F$ is differentiable,you may find a linear map $\mathrm dF_p:\mathbb R^n\to\mathbb R^m$ such that
$$F(p+v)\approx F(p)+\mathrm dF_p(v).$$
Here, $\approx$ means that a certain limit is $0$ as $v$ goes to $0$, but that's not important for my answer. So if you want to approximate $F(p+v)$, you only need to apply the linear map to $v$. You don't apply it to points in the domain, but to vectors pointing between points. This may not be obvious in the setting of vector spaces as domains, but it's a bit more natural in the setting of affine spaces: an affine space is a set of points $A$ together with a vector space $V$ which describes vectors pointing between the points. If we have a differentiable map between the point sets of affine spaces, the differential is a linear map between the vector spaces associated with the affine spaces. So the differential of a function at a point has a genuinely different domain and range than the function itself. It's just that in the setting of vector spaces, the set of points and the set of vectors pointing between the points is the same, so the difference is less apparent.

Title: Domain of the Total Derivative


问:
Integration factor differential equations?
I have the equation: $$2y\:\mathrm{d}x- (\ln{y}+2x-1)\:\mathrm{d}y=0$$

I have to solve this.So,I noticed that $\frac{\partial P}{\partial y}$ is different from $\frac{\partial Q}{\partial x}$ so I have to apply the integrating factor. $P$ here is $2y\:\mathrm{d}x$ and $Q$ is $\ln{y}+ 2x-1$

So: $$\frac{1}{Q}\left(\frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x}\right)= \left[\frac{1}{\ln{y}+2x-1}\right] \times(2-2) =0$$

This cant be zero because I have to find the integrating factor using this so where is my mistake?

答: Note that if we set $M(x,y)=2y,~~N(x,y)=-\ln x-2x+1$ then $M_y-N_x=4$. So $$\mu(y)=\exp\left(\int\frac{4}{2y}dy\right)$$</p>


问:
How to integrate $ x = \int \frac{ky^2}{\sqrt{1-k^2y^4}} dy $?
The problem is to make the following integral stationary:
$$ \int_{x_1}^{x_2} \frac{\sqrt{1+y'^2}}{y^2}dx $$
to simplify the Euler equation, I tried to change the independent variable:
$$ \int_{y_1}^{y_2} \frac{\sqrt{1+x'^2}}{y^2}dy, \: \: \: \: \: \: \: \: y=y\left( x \right),\: y'=\frac{dy}{dx} $$
with the correspondent Euler equation:
$$ \frac{d}{dy}\frac{\partial F}{\partial x'}-\frac{\partial F}{\partial x}=0 $$
thus
$$
\begin{aligned}
\frac{\partial F}{\partial x}=0 \Rightarrow \frac{\partial F}{\partial x'} &= k\\
                                               \frac{x'}{y^2 \sqrt{1+x^2}} &= k\\
                                                        \frac{dx}{dy} = x' &= \frac{ky^2}{\sqrt{1-k^2y^4}}
\end{aligned}
$$
and I get:
$$ x = \int \frac{ky^2}{\sqrt{1-k^2y^4}} dy $$
Now, can I change $ ky^2 $ to a new single arbitrary variabel to simplify the integrand? Or are there a more effective method?

答:
As said in comments and answer, using $\frac{u}{\sqrt{k}}$ you will end with
$$\int \frac{ky^2}{\sqrt{1-k^2y^4}}\, dy=\frac{E\left(\left.\sin ^{-1}(u)\right|-1\right)-F\left(\left.\sin
   ^{-1}(u)\right|-1\right)}{\sqrt{k}}$$

If you need to solve for $u$ the equation
$$f(u)=E\left(\left.\sin ^{-1}(u)\right|-1\right)-F\left(\left.\sin
   ^{-1}(u)\right|-1\right)=a$$ you could use the series expansion
$$f(u)=\sum_{n=0}^\infty b_n \,u^{4n+3}$$ where the $b_n$'s form the sequence
$$\left\{\frac{1}{3},\frac{1}{14},\frac{3}{88},\frac{1}{48},\frac{35}{2432},\frac{63}{5
   888},\frac{77}{9216},\frac{429}{63488},\frac{1287}{229376},\frac{935}{196608},\cdots\right\}$$ Using the above coefficients, the fit is quite good : the absolute error is

$<0.1$% if $u \lt 0.921$

$<0.01$% if $u \lt 0.874$

$<0.001$% if $u \lt 0.829$

Using series reversion
$$u=t \left(1-\sum_{n=1}^\infty c_n \,t^{4n} \right)\qquad \text{where} \qquad t=\sqrt[3]{3a}$$ the first $c_n$'s forming the sequence
$$\left\{\frac{1}{14},\frac{15}{4312},\frac{61}{181104},\frac{85325}{2119641216},\frac{
   1214019}{227508157184}\right\}$$

For a test example, using $a=0.5$ the above would give as estimate
$u=0.99010375$ while the "exact" solution is
$u=0.99010360$


答: For $r$ in $R$'s support $[0,\,1]$,$$P(R\le r)=P(\tfrac{X}{Y}\le\tfrac{r}{1-r})=\int_{0}^{\infty}(1-e^{-\lambda yr/(1-r)})\mu e^{-\mu y}dy=\tfrac{\lambda r}{\lambda r+\mu\left(1-r\right)},$$so the PDF is $\tfrac{\lambda\mu}{\left(\lambda r+\mu\left(1-r\right)\right)^{2}}$. In the special case $\lambda=\mu$, $R\sim U(0,\,1)$.

Title: PDF and CDF of $R=\frac {X}{X+Y}$

Body:
Let $X$ and $Y$ be two exponentially distributed random variables with $\lambda$ and $\mu$ parameters respectively. Find the CDF and PDF of $R=\frac {X}{X+Y}$

I'm able to find the PDF of $S=X+Y$. But I'm unsure about the bounds of the integral when I want to calculate the PDF of $R$.


问:
Solving the intersection of two functions with fixed point iteration
I have to find the intersection(s) of two functions $\frac{1}{\sqrt{\ln(x)}}$ and $5-x^2+\sqrt[3]{x}$ with the fixed point iteration method. I suppose that I have to find the roots of the equation f1-f2 = 0, but when I try to iterate this equation is repels and goes to inf or -inf. I also tried to rearrange both equations to x, but I can't see how can I rearrange the equation with the logarithmic function. Could you please point me in the right direction, how these equations should be solved and how should I write the fixed point function? Thank you!

Sorry if this is a very newbie question but I just had to get back to mathematics and I didn't use it for a long time.

答:
i dont know about fixed point iteration. Didnt hear about it.
But maybe you can just try to rewrite in terms of "x".
i.e.
$$\begin{cases} y=\frac{1}{\sqrt{lnx}}\\y=5-x^2+x^{1/3}\end {cases}$$ $\implies\begin{cases} x=e^{\frac{1}{y^2}}\\y=5-x^2+x^{1/3}\end {cases} \\ $
then try to plug x into second equation.


问:
A problem of a function, a composite function and inverse function
Let $f:\mathbb{R}\rightarrow\mathbb{R}$, $g:\mathbb{R}\rightarrow\mathbb{R}$ and $h:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable functions such that $f(x) = x^3+3x+2$, $g(f(x))=x$ and $h(g(g(x)))=x$, $\forall$ $x\in\mathbb{R}$. Then

A) $g'(2) =\frac{1}{15}$

B) $h'(1)=666$

C) $h(0)=16$

D) $h(g(3)) = 36$

Note: The answer is B,C.

Body:
Here $g(x)=f^{-1}(x) \implies g^{-1}(x)=f(x)$, $f(g(x))=x$. Next $$h(g(g(x)))=x \implies h^{-1}(x)=g(g(x)) \implies h(x)= g^{-1}(g^{-1}(x))=f(f(x)).$$
Eventually, as $f(x)=x^3+3x+2$, we get $$h(x)=f(x^3+3x+2)=(x^3+3x+2)^3+3(x^3+3x+2)+2$$
So from here it follows that $h(0)=16$, and $h(g(3))=f(f(g(3))=f(3)=38,$ $h'(1)=666, g'(3)=\frac{1}{3}$
So options (B) and (C) are correct.


问:
log of maximum likelihood function.
Given 100 observations and density function : $ f(x;\theta) = \frac{1}{\pi \cdot \left(1+\left(x-\theta \right)^2\right)}$

find $\hat{\theta}_{MLE}$.

My Attempt:

I have found the joint likelihood $L(\theta;x_1,x_2\dots x_{100}) = \prod _{i=1}^{100}\left(\frac{1}{\pi \cdot \left(1+\left(x_i-\theta \right)^2\right)}\right)\:$

I have tried to find $l=\log(L)$ but I failed.

答:
There's no good form of an answer. In cases $n=1$ and $n=2$ there's a simple way to write an answer, but in case of big $n$ we may write only
$$\theta = argmax_{\theta \in \Theta} L(\theta, x_1, \ldots, x_{n}).$$
The problem is that we can not find the minimum of polynomial of degree $2n= 200$ explicitly.

Title: log of maximum likelihood function.


问:
Find correct to 6 decimal places, the x-coordinate of the point on the curve y = ln x which is closest to the origin.
I must use the Newton Raphson method.

Can someone please explain the steps I must take in order to reach the answer. For alot of this im having to self study the material to catch up and without instruction, im quite lost on how to approach the problem. I wasnt given any initial values, so am i to assume that i must use the distance formula or..

答:
The closest point of the curve has an $x$-coordinate that is the solution of the equation
$$f(x)=x^2+\log x =0$$
as $f\left(\frac12\right)=\frac14-\log 2<0$ and $f(1)=1>0$, for the intermediate values theorem the root is in the interval $(0.5,1)$

So we take $x_0=0.5$ and $x_n=x_{n-1}-\frac{f(x_{n-1})}{f'(x_{n-1})}$

$$
\begin{array}{l|r|r}
 n & x_n & error\\
\hline
 0 & 0.5 &\\
 1 & 0.647716 & 0.147716 \\
 2 & 0.652917 & 0.00520115 \\
 3 & 0.652919 & 1.77\times 10^{-6} \\
 4 & 0.6529186404 & 1.9\times 10^{-13} \\
\end{array}
$$

The solution with $6$ exact decimal places is $x^*=0.652919$


问:
Determining the height of an mountain from two points of view.
From viewpoint A west of the mountain you can see the peak of the mountain from a angle of elevation 17.4 degrees. From viewpoint B which is 3000 meters to the north from viewpoint A you can see the peak of the mountain from the a angle of elevation of 14.5 degrees. What is the height of the mountain from sea level, when view points are 200 meter over the sea level.

答:
If the base of the mountain is $L$ and its top is $T$ and we have viewpoints $A$ and $B$ where $B$ is north of $A$,

we have $AL = h \cot 17.4^0, BL = h \cot 14.5^0 \,$ (where $LT = h = H - 200$ is its relative altitude to viewpoints $A$ and $B$, $H$ is altitude of its peak above the sea level).

Now using Pythagoras, $BL^2 - AL^2 = 3000^2$. Write $BL, AL$ in terms of $h$ and solve for $h$ and then add $200$ to get $H$.


问:
Mean and variance of $N=\min\{{n\geq1|X_1+X_2+ \cdots +X_n > x}\}$
Let $X_r (r \geq 1)$ be an independent random variable that is uniformly distributed over $[0,1]$ and $x$ be a number between $0$ and $1$. $N$ is defined as follows:

$$N= \min\{{n \geq 1 | X_1 + X_2 + \cdots + X_n > x}\}$$

prove that for any $x$, $P( N > n ) = \frac {x^n}{n!}$

And calculate the mean and the variance of $N$.

I am self-studying and I found this problem in my book. I have tried various methods but have gotten nowhere so far. I would really appreciate some help.

答:
Wow, this is a hard question! Apologies in advance if I make any careless mistakes.

Note that $P(N > k)$ is asking: "after $k$ tries, what is the probability that the sum of the r.v.'s $X_1, X_2, \cdots X_k$ is still less than $x$?" This way you need more than $k$ r.v's to go beyond $x$.  So $P(N > k) = P(X_1 + X_2 + \cdots X_k < x)$.

I'm going to cheat and knock it down to the case of small variables first. When $k=1, P(N > k) = P(X_1 < x) = x$, so the formula works. When $k=2, P(N > k) = P(X_1 + X_2 < x)$. We condition on the first variable. That is,
\begin{align}
P(X_1 + X_2 < x) &= \int_0^x P(X_2 <x -x_1 \mid X_1 = x_1) f_{X_1}(x_1) dx_1 \\
\end{align}
where $f_{X_1}(x_1)$ is the PDF of the random variable $X_1$.
However, since $f_{X_1}$ is uniformly distributed, $f_{X_1}(x_1) = 1$ for all $x_1 \in [0,1]$. Hence,
\begin{align}
P(X_1 + X_2 < x) &= \int_0^x P(X_2 <x -x_1 \mid X_1 = x_1) (1) dx_1 \\
\end{align}
By independence, $P(X_2 <x -x_1 \mid X_1 = x_1) = P(X_2 < x -x_1)$, so,
\begin{align}
P(X_1 + X_2 < x) &= \int_0^x P(X_2 <x -x_1) dx_1 \\
&= \int_0^x \int_0^{x-x_1} 1 dx_2 dx_1.
\end{align}
Now, $\int_0^{x-x_1} 1 dx_2$ gives $(x-x_1)$, and $\int_0^x (x-x_1) dx_1 = \frac{x^2}{2}$.

For $k=3$, you can verify by a similar procedure that the integral you need is,
\begin{align}
P(X_1 + X_2 + X_3 < x) = \int^x_0 \int^{x-x_1}_0 \int^{x-x_1-x_2}_0 1 dx_3 dx_2 dx_1.
\end{align}
This gets hard to evaluate real quick, but luckily for us, we have a secret weapon! Notice that $x-x_1$ does not participate in the inner two integrals (they are integrating with respect to $x_2$ and $x_3$, hence you may let $x-x_1 = u$, and get,
\begin{align}
P(X_1 + X_2 + X_3 < x) = \int^x_0 \int^{u}_0 \int^{u-x_2}_0 1 dx_3 dx_2 dx_1.
\end{align}
But by the case of $k=2$, you know the inner two integrals works out to be,
\begin{align}
\int^{u}_0 \int^{u-x_2}_0 1 dx_3 dx_2 = \frac{u^2}{2!} = \frac{(x-x_1)^2}{2!}
\end{align}
Unsurprisingly,
$$
\int^x_0 \frac{(x-x_1)^2}{2!} dx_1 = \frac{x^3}{3!}.
$$
By now you should see a pattern. For $k=4$, you will eventually end up with
$$
\int^x_0 \frac{(x-x_1)^3}{3!} dx_1,
$$
for which you want to show this integral works out to be $\frac{x^4}{4!}$. For $k=5$, you will end up with
$$
\int^x_0 \frac{(x-x_1)^4}{4!} dx_1,
$$
for which you want to show this integral works out to be $\frac{x^5}{5!}$. What you need to prove now is that in general, the following preposition,

Proposition: $$\int^x_0 \frac{(x-x_1)^n}{n!} dx_1 = \frac{x^{n+1}}{(n+1)!}.$$

There are many ways to do this, (brute forcing binomial expansion of $(x-x_1)^n$ is perhaps the easiest), so I will not prove it here.

Once you prove this, the fact that $P(N > k) = \frac{x^k}{k!}$ should be clear.

Another way to compute expectation (for discrete variables only) is by using this formula:
$$
E(X) = \sum_{n = 1}^{\infty} P(X \geq n)
$$
(Why is this true?) Then,
\begin{align}
E(N) &= \sum_{n = 1}^{\infty} P(N \geq n) \\
&= \sum_{n-1= 0}^{\infty} P(N > n-1) \\
&= \sum_{u = 0}^{\infty} P(N > u) \\
&= \sum_{u = 0}^{\infty} \frac{x^u}{u!} \\
&= e^x.
\end{align}

Variance is a little tricker. We know $E(N)$ already, so the $E(N)^2$ term isn't the problem. What is problematic is the $E(N^2)$ term!
\begin{align}
 E(N^2) &= \sum_{n=1}^{\infty} P(N^2 \geq n) \\
&=P(N^2 \geq 1) + P(N^2 \geq 2) + P(N^2 \geq 3) + P(N^2 \geq 4) + ...
\end{align}
Note,
\begin{align}
& P(N^2 \geq 1) = P(N > 0) \\
&  P(N^2 \geq 2) = P(N > 1) \\
& P(N^2 \geq 3) = P(N > 1) \\
& P(N^2 \geq 4) = P(N > 1)
\end{align}
So 1 contributes 1 $P(N > 0)$ term, 2 to 4 contributes 3 $P(N > 1)$ terms, and extrapolating, 5-9 contributes 5 $P(N > 2)$ terms, 9-16 gives 7 $P(N > 3)$ terms, and so on. We hence have the following sum,
\begin{align}
E(N^2) &= P(N > 0) + 3 P(N > 1) + 5P(N > 2) + 7P(N > 3) + \cdots \\
&= 1 + 3x + 5\frac{x^2}{2!} + 7\frac{x^3}{3!} + \cdots \\
&= e^x + 2x + 4\frac{x^2}{2!} + 6\frac{x^3}{3!} + 8\frac{x^4}{4!} + \cdots \\
&= e^x + 2x(1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \cdots)\\
&= e^x + 2xe^x.
\end{align}
Thus, $\text{Var}(N) = e^x + 2xe^x - e^{2x}$, which is somehow (magically) positive for $x \in [0,1]$.


答:
Write the equation as
$$x^2=\sqrt[3]{x}-\frac{1}{\sqrt{\log (x)}}+5$$ that is to say
$$x_{n+1}=\sqrt{\sqrt[3]{x_n}-\frac{1}{\sqrt{\log (x_n)}}+5 } $$ For sure, because of the square root and the logarithm, you must start with $x_0>1$

Title: Solving the intersection of two functions with fixed point iteration

Body:
I have to find the intersection(s) of two functions $\frac{1}{\sqrt{\ln(x)}}$ and $5-x^2+\sqrt[3]{x}$ with the fixed point iteration method. I suppose that I have to find the roots of the equation f1-f2 = 0, but when I try to iterate this equation is repels and goes to inf or -inf. I also tried to rearrange both equations to x, but I can't see how can I rearrange the equation with the logarithmic function. Could you please point me in the right direction, how these equations should be solved and how should I write the fixed point function? Thank you!

Sorry if this is a very newbie question but I just had to get back to mathematics and I didn't use it for a long time.


答:
Start by sketching the two functions.  It should be clear that the first (red) only gives real values when $x >0$, and that there are two intersections

You can certainly find the first of these by fixed point iteration: $f_1(x)= \frac{1}{\sqrt{\ln(x)}}$ has an inverse $g_1(y)=\exp\left(\frac{1}{y^2}\right)$  so if you try $x_{n+1} = g_1(f_2(x_n))$ iteratively then you will find you get convergence to about $1.042037$ from almost any starting point: for example starting with $x_0=2$ you get about $1.216284, 1.048651, 1.042242, 1.042044, 1.042037, \ldots$

That does not give you the second intersection as it is not easy to invert $f_2(x)=5-x^2+x^{1/3}$.  So we may need a different approach such as Newton's method with $x_{n+1}=x_n - \dfrac{f_1(x_n)-f_2(x_n)}{f_1'(x_n)-f_2'(x_n)}$.  It turns out this does converge on the second intersection of around  $2.283985$ if you start at $1.337$ or greater: for example starting at $x_0=2$ you get about $2.315419, 2.284270, 2.283985, \ldots$


问:
how to generate all possible equations with a set of number and operators?
i got a maths problem, for given that a set of character {1,2,3,4,5,6,7,8,9,+,-,*,/}. and then by using the set of characters to randomly generate 10(or let say N) characters in an array, i.e. $[1,+,1,*,3,5,7,8,1]$. after that I need to find all the possibility of equations by using this array of characters plus 1 character '='.

Therefore in this case, $[1,+,1,*,3,5,7,8,1]$
it would be like the following equations:
1*1=1 ,
3+5=8 ,
7+1=3+5 ,
7+1=8 ,
8-1=7 ,
15=8+7 ,
15+3=18
(maybe more)

it can form multiple digit operation, i.e 15+3=18

so my question is I am trying using a program to generate the equations all. Could you guys give me some ideas what kind of algorithms can do so? or any methods do it? thanks a lot!

have a nice day!

答:
Thinking out aloud ;)

If you arranged the numbers in all possible ways noting you must have at least 2 numbers and incriminating to the maximum selected.

Then reproduce each number set with the = in each valid location. This would be your starting point.

Where you can then place the selected modifiers in each valid place.

First one at a time.

Then combining them one at a time.

Then for each iteration check the formulas validity saving or rejecting as required.

This seems quite lengthy but is a place to start. I think there will be some constraints or further optimisations I've not YET thought of you can apply to decrease the number of interactions you'll need to do.

I say yet as I'll not stop thinking about this. ;)


问:
Integrate $\frac{1}{x\sqrt{a - bx^2}}$
I've been toiling for hours trying to answer this one:

Find $\int \frac{1}{x\sqrt{a - bx^2}}\,dx$

It's from Silvanus Thompson's 'Calculus Made Easy' (apparently not easy enough for me!).

The answer given is $\frac {1}{\sqrt{a}}\log\left(\frac{\sqrt{a}-\sqrt{a-bx^2}}{x\sqrt{a}}\right)$

I've attempted the question by letting $x=\sqrt{\frac ab}\sin(u)$, which I think is on the right track as I end up with the $\frac1{\sqrt a}$ term in my answer, but nothing else falls into place in terms of the given answer.

Thank you in advance for any help!

Andrew

答:
Hint:

$$\int\dfrac{dx}{x\sqrt{a-bx^2}}=\int\dfrac{bx\ dx}{bx^2\sqrt{a-bx^2}}$$

Let $\sqrt{a-bx^2}=u\implies du=\dfrac{{-bx}}{\sqrt{a-bx^2}}$  and $bx^2=a-u^2$

More generally for $$\int\dfrac{dx}{x\sqrt{a-bx^n}}=\int\dfrac{bx^{n-1}\ dx}{bx^n\sqrt{a-bx^n}}$$

set $\sqrt{a-bx^n}=y$

Title: Integrate $\frac{1}{x\sqrt{a - bx^2}}$


问:
Find length of adjacent line in triange?
I have the following setup:

(We know every point, length and angle illustrated in the figure, expect for $z$)

I am then trying to figure out the length $z$ going from the point $X$ to the middle of the baseline.

Here is what I have managed to think myself:

The angle at $C$ and $C'$ in the small triangles are the same as the angle at $X$ in the triangle formed by $\overrightarrow{CXF}$ and $\overrightarrow{C'XF}$ where $F$ is a point in the middle of the baseline. That is, $\theta_1 = \theta_2$. I then thought that applying the definitions of sine, cosine or tangent might work but I would only know the opposite line of the triangle so wasn't a success. I got stuck here :(

Can anybody help me out?

As a matter of fact, I know that the answer should be $z = - \frac{f B}{x' - x}$ but do not know how it is derived, which is what I am interested in.

答:
Drop a perpendicular from $X$ to the baseline. Call the base of the altitude $Y$.

We form two similar triangles, hence we have the equal ratios:

$$\frac fx = \frac z{CY}, \quad \frac f{x'} = \frac z{C'Y}$$

We also have $B = CY + C'Y = \dfrac {xz}f + \dfrac {x'z}f = z\left(\dfrac{x+x'}f\right)$.

Hence $z = \dfrac {fB}{x+ x'}$.


问:
Why is the following matrix multiplication correct?
$$
A=\left[\begin{array}{ll}
Q_{1} & Q_{2}
\end{array}\right]\left[\begin{array}{cc}
\Lambda_{1} & 0 \\
0 & 0
\end{array}\right]\left[\begin{array}{l}
Q_{1}^{T} \\
Q_{2}^{T}
\end{array}\right]=Q_{1} \Lambda_{1} Q_{1}^{T}
$$

Shouldn't the result be $\Lambda _{1}$ ?

答:
No, its correct, since
$[Q_1,Q_2]\cdot \begin{pmatrix}\Lambda_1 & 0\\ 0 & 0\end{pmatrix} = [Q_1\Lambda_1, 0].$


问:
How many different ways can the ministers rearrange themselves fulfilling the below order?
The  King  of  Zyorg  has  a  cabinet  with N ministers  and  they  have  assembled  for  a meeting. The King enters the meeting chamber to find that the ministers have arranged themselves so that friends sit with friends.

The seats are numbered $1,... N$ and we refer to the minister sitting on the $i$th seat in this initial arrangement as $i$.  So, the initial arrangement is of the ministers is $(1,2,3,... N)$.

The King orders the ministers to rearrange themselves and finds that some of the friends have just exchanged seats.  The enraged King then orders that they must rearrange themselves  so  that  for  every $k < N$ the  set  of  ministers  sitting  in  the  first $k$ chairs DOES NOT consist only of the ministers $1,2,...k$.

He then wonders if this is possible at all and quickly convinces himself that this is so.Then he wonders how many different ways can they rearrange themselves fulfilling his order.

For  example,  suppose $N=  3$.   The  initial  arrangement  is  of  course  $(1,2,3)$.   If  they are arrange themselves as (2,1,3) this will violate the King’s order because for $k= 2$ the ministers in positions 1 up to 2 are{1,2}.  On the other hand the arrangement $(3,1,2)$ fulfills the requirement.  For $k= 1$ we have $\{3\}$,  for $k= 2$ we have $\{3,1\}$.  You can check that there are exactly 3 arrangements that comply with the King’s order.  They are :  $(2,3,1),(3,1,2)$ and $(3,2,1)$.  So, for $N = 3$ the answer to this problem is 3.

Your task is to write out the answers for 3 values of N.

(a) N= 5 (b) N= 6 (c) N= 7

My Approach:

Well I tried to solve for all the Ks, making all of the combinations for all the k(s), where for every after 1 this formula must be followed $((k+1)!)*(k+1)!))$ and then every value evaluated for k must be added. This worked perfectly for $N = 3, 4$ but not for any of the sub-questions asked.

答:
Consider an arbitrary permutation of $n$ ministers and let $k^*$ be the least number $k$ such that the rule:

the set of ministers sitting in the first  chairs DOES NOT consist
only of the ministers 1,2,....

breaks. Obviously the sequence of the first $k^*$ ministers is an allowed permutation of $k^*$ ministers.

Let the number of allowed permutations of $n$ ministers be $c_n$.

Then due to the above-mentioned reason the overall number of permutations will be:
$$
n!=\sum_{k=1}^n c_{k}(n-k)! ,
$$
where a single term in the r.h.s. counts the combinations which start with an allowed permutation of $k$ ministers followed by an arbitrary permutation of the remaining $n-k$ ones. From this we obtain the recurrence formula:
$$
c_1=1;\quad c_n=n!-\sum_{k=1}^{n-1}c_k(n-k)! ,
$$
which gives rise to the following sequence of $c_n$ starting with $n=1$:
$$
1,1,3,13,71,461,3447,\dots
$$


问:
Proving set inclusion for a continuous map $f$.
Proving that for a continuous map $f: \overline A \rightarrow F $, where $F$ is a metric space, it holds that $f[ \overline A ] \subset \overline{ f[A] }$.

Proof:

$\overline{ f[A] }$ is closed, since $f$ is continuous then the inverse of the set must be closed. We have:

$f^{-1}[\overline{ f[A] }]$ is closed and it holds that $A \subset f^{-1}[\overline{ f[A] }]$.

So it must hold that $\overline A \subset f^{-1}[\overline{ f[A] }]$ because $\overline A$ is the smallest closed set that contains $A$.

So, therefore it actually holds that $\overline A = f^{-1}[\overline{ f[A] }]$ because $\overline A$ is the largest set in the domain of $f$.

Now we have $f[\overline A] = f \circ f^{-1}[\overline{ f[A] }] \subset \overline{ f[A] } $.

So it holds that $f[\overline A] \subset \overline{ f[A] }$.

$\blacksquare$

Is the proof correct? If so, is there a way to somehow solve it in a different way using sequences? Also, my proof seems not to utilise the fact that $F$ is a metric space. If it is right, then the condition must be more general then. The proof also must imply that if $f$ is surjective then an equality will hold.

答:
The proof is fine and you are right, you don't need $F$ metric space. It holds in general that if $X, Y$ are topological spaces, then $$f: X \rightarrow Y \text{  is continuous }$$ if and only if for every subset $A \in X$ we have $$f(\overline{A}) \subset \overline{f(A)}$$

For the proof with sequences, then you may want the domain is a metric space (first countability would be enough) so that you can use the characterization of the closure of sets by means of sequences. The codomain needs not to be assumed metrizable for your purposes.