问:
The group of additive sequences of integers

Let $G$ be the additive group of the sequences $(a_n)_{n\ge 1} \subset \mathbb{Z}$ and let $f:G \to \mathbb{Z}$ be a group homomorphism. We denote by $e_i$ the sequence $(0,0,...,0,1,0,0...)$ (we just have a $1$ on the $i$-th position, everything else is zero). Show that the set $\{i\ge 1 | f(e_i) \ne 0\}$ is finite.

I thought that I should consider the set of sequences of finite support i.e. the subgroup generated by the $e_i$s. Let's denote it by $H$. We have that $f(H)$ is a subgroup of $\mathbb{Z}$, so $f(H)=m\mathbb{Z}$ for some integer $m$. We also have that $f(G)=n\mathbb{Z}$ for some positive integer $n$. But I don't know how to proceed any further. I think that we should aim for a contradiction. I can only think about something having to do with divisibility, but I don't know.

答:
Suppose $I = \{i \geq 1 \mid f(e_i) \neq 0\}$ is infinite.
We can suppose $I = \mathbb{N}$.

Let $H_k = \prod_{i=1}^k\{0\} \times \prod_{i \geq k+1}\mathbb{Z}$.

$f:G \to \mathbb{Z}$ induce an monomorphism $\bar{f} : G/\ker{f} \to \mathbb{Z}$

Now we consider the projective system
$$\varphi: G/M_{k+1} \to G/M_k, \text{ where } M_k = H_k + \ker{f}$$

Let $f_k: G/M_k \to \mathbb{Z}/n_k\mathbb{Z}$ where $f(M_k) = n_k \mathbb{Z}$

We have a continuous monomorphism of topoligical groups :
$$\varprojlim G/M_k \to \varprojlim\mathbb{Z}/n_k\mathbb{Z}$$

from $ H_k \subset M_k$ we deduce a conitnuous monomorphism
$$\varprojlim G/H_k \to \varprojlim G/M_k $$
By composition we have a continous monomorphism
$$\varprojlim G/H_k \to \varprojlim\mathbb{Z}/n_k\mathbb{Z}$$

But $\varprojlim G/H_k$ is isomorph to $G$, and $\varprojlim \mathbb{Z}/n_k\mathbb{Z}$
is somthing like $\prod_{k=1}^N\mathbb{Z}_{p_k}$ where $\mathbb{Z}_{p_k}$
is $p_k$-adic ring!


问:
Detect my mistake: $\int_{-\infty}^{\infty}\frac{dx}{(x^{2}+4)^{n}}$
Detect my mistake:
$\displaystyle\int_{-\infty}^{\infty}
\frac{{\rm d}x}{\left(x^{2} + 4\right)^{n}}\ $.

My try:

We shall take a half circle around the upper part of the plane.

The arc will tend to zero.

Now we should calculate residue:
$\displaystyle%
\frac{\left(x - 2{\rm i}\right)^{n}}
{\left(x^{2} + 4\right)^{n}} =
\frac{1}{\left(x + 2{\rm i}\right)^{n}}$.

We shall take $\left(n - 1\right)$ derivatives:
$$
\frac{\partial^{n - 1}}{\partial x^{n - 1}}
\left[\frac{1}{\left(x + 2{\rm i}\right)^{n}}\right] =
\frac{\left(2n - 2\right)!}{n!\left(x + 2{\rm i}\right)}
= \frac{\left(2n - 2\right)!}{n!\left(4{\rm i}\right)}.
$$
Now the residue is incorrect, which leads to the integral result to be incorrect.

Does anyone see my mistake $?$.

ContentLicense: CC BY-SA 4.0

Body: As @leoli noted,$$\frac{d^{n-1}}{dx^{n-1}}(x+2i)^{-n}=(-n)(-n-1)\cdots(-(2n-2))(x+2i)^{-(2n-1)},$$i.e. you shifted the exponent the wrong way. So the integral is$$\begin{align}\frac{2\pi i}{(n-1)!}\lim_{x+2i}\left.\frac{d^{n-1}}{dx^{n-1}}(x+2i)^{-n}\right|_{x=2i}&=\frac{2\pi i(-1)^{n-1}(2n-2)!}{(n-1)!^2}\left.(x+2i)^{1-2n}\right|_{x=2i}\\&=\frac{(2n-2)!2\pi}{(n-1)^24^{2n-1}}.\end{align}$$


问:
Correspondence Theorem for Lie algebras
I am trying to prove the following:

Let $L$ be a Lie algebra and $I$ an Ideal of $L$. There exists a bijection between the Ideals of the quotient algebra $L/I$ and the Ideals of $L$, that contain $I$.

Let $J$ be an Ideal of $L$, which contains $I$ $\Rightarrow$ $J/I$ is an Ideal of $L/I$

Let $K$ be an Ideal of $L/I$ $\Rightarrow$ $J:=\{z\in L \mid z+I \in K\}$ is an Ideal of $L$, which contains $I$

I already know that this holds true for vector spaces (as it is true for modules).
Knowing that, how can I show that this also holds true for Lie algebras?

答:
I think this well-know proof holds for Lie algebras.

Let $J$ be an ideal of $L$ that contain $I$. Consider the map
$$\begin{array}\ \phi:& L & \rightarrow & L/I \\ & J &\mapsto & J/I\end{array}$$
Note that $J/I$ is an ideal of $L/I$ since for $X+I \in J/I$ and $Y+I\in L/I$ we have $[X+I,Y+I]=[X,Y]+I$, where $[X,Y]$ is in $J$ because it is an ideal of $L$.

Now, let $K$ be an ideal of $L/I$ and consider the map
$$\begin{array}\ \psi:& L/I & \rightarrow & L \\ & K &\mapsto & \{ X\in L\ |\ X+I \in K\}\end{array}$$

We need to check that this set is in fact an ideal of $L$, but this is easy. For $X\in \psi(K)$ and $Y\in L$ we have $[X,Y]\in L$ and $[X,Y] +I=[X+I,Y+I] \in K$, because $K$ is an ideal.

$\psi(\phi(J))=\psi(J/I) = \{X\in L\ |\ X+I \in J/I\}=J$.

$\phi(\psi(K))=\phi(\{Y\in L\ |\ Y+I \in K\})=\{Y+I\in L/I \ |\ Y+I\in K\} = K$.

This show the bijection.

Title: Correspondence Theorem for Lie algebras


问:
Is there a rigorous definition for matrix derivatives?
I know that,

A function $f: \mathbb{R}^n \to \mathbb{R}$ is said to be differentiable at $x$ if there exists a vector $v$ such that,
$$
    \lim_{h \to 0} \frac{f(x+h) - f(x) - v^Th}
    {\|h\|} = 0.
$$
When $v$ exists, it is given by the "gradient" $\nabla f(x) = \left(\frac{\partial f}{\partial x_1}, ..., \frac{\partial f}{\partial x_n}\right)(x)$

Does there exist a similar definition for "matrix derivative"

https://en.wikipedia.org/wiki/Matrix_calculus#Derivatives_with_matrices

答:
Matrix derivation is just a particular case of Fréchet derivative between two Banach spaces. Which by the way is very similar in term of definition to the definition of the derivative of a function $f : \mathbb R^n \to \mathbb R$ provided in the question.

Applied to matrix derivatives, you just have to consider a map $f : V \to M$ where $M$ is a linear space of matrices endowed with the norm of your choice and $V$ a Banach space that can be (or not) of finite dimension.


问:
Given a traffic matrix $A$ what does it mean for the pair-wise dot products of rows to be non-zero, and can we enforce this somehow?
The traffic matrix $A$ encompasses the information regarding the paths of origin-destination nodes in a directed graph. Columns represent od-pairs and rows the directed edges of the underlying graph. The element $(i, j)$ in the matrix is one if the edge $i$ belongs to the (directed) path in the od-pair $j$, and otherwise it is zero. Paths for the od-pairs chosen are arbitrary. That is, we can and we will leave out some of the paths. For example in the directed $K_3$ the path between nodes $1$ and $2$ could be $(1, 2)$ or $(1, 3), (3, 2)$.

What I would like to know is that what do we know about such matrix $A$ that satisfy the following property: the pair-wise dot products of the rows are non-zero. In other words, each pair of rows has at least one common column full of ones.

So, how could we talk about this in the language of graph theory/linear algebra/lattices and, does this question result in a decision problem. Specifically, if we want to enforce the aforementioned property, do we have an algorithm to reformulate the matrix $A$?

答: The property means that every pair of edges has a common od-pair.  That is, for all $i_1$ and $i_2$ with $i_1<i_2$, there exists at least one $j$ with both edges $i_1$ and $i_2$ in the path for od-pair $j$.


问:
Proof $p(x),p'(x), p''(x),p'''(x)$ from a basis por polynomials of degree three
I need to proof that given a polynomial $p(x)$ of degree three, the set $\left\{p(x),p'(x), p''(x),p'''(x)\right\}$ from a basis. Let $p(x)=a_0+a_1x+a_2x^2+a_3x^3$, with $a_3\neq 0$. I calculated the derivatives

$p'(x)= a_1+2a_2x+3a_2x^2$,

$p''(x)=2a_2+6a_3x$, and

$p'''(x)=6a_3$.

This set will be a basis if this system has a solution

$\left( \begin{matrix}
a_0 & a_1 & 2a_2 & 6a_3 \\
a_1 & 2a_2 & 6a_3 & 0 \\
a_2 & 3a_3 & 0 & 0 \\
a_3 & 0 & 0 & 0 \\
 \end{matrix} \right|\left.  \begin{matrix}a_0\\a_1\\a_2\\ a_3 \end{matrix}\right)$

And that's true because the determinant of this matrix is $103a_3^4$, i.e, is different of zero since $a_3\neq 0$.

That's correct?

答:
You should use $(b_0,b_1,b_2,b_3)$ for the last column, since you want to show that $A$ is invertible; you need that $Ax=b$ has a unique solution for any $b$.

And to make the determinant easier, you may switch the rows.


问:
How do you solve a simple simultaneous equation to get unkowns for a sinusoid (with no phase shift) given a couple of points?
This is probably a very simple question, but I am struggling to find the answer.

Given:

\begin{equation}
y_1 = A \cos(\omega t_1)\\
y_2 = A \cos(\omega t_2)\\
y_3 = A \cos(\omega t_3)
\end{equation}

(I can go up to as much as $y_n = A \cos(\omega t_n)$. Using only $y_1$ and $y_2$ is also, probably, fine.)

So I know the $y's$ and $t's$. How do I find $A$ and $\omega$?

I, essentially, want to use that to compute the peak of a discrete signal I'm assuming is a cosine for a very short interval which includes that peak. So I thought if I can obtain the equation of the sinuoid then I can differentiate it, set it to $0$ and then solve for the $t$. That way I can obtain $t$ and $y$ at the maximum/peak.

答:
$$
\begin{cases}
y_1 = A\cos(\omega t_1)\\
y_2 = A\cos(\omega t_2)\\
\end{cases}\tag{1}
$$
If you have two unknowns, $A, \omega$ you need only two equations.

If there is another equation, the risk is that the third is not compatible with the other two, like in this example
$$
\begin{cases}
a+b=10\\
a-b=4\\
2a+3b=1\\
\end{cases}
$$
first two equations have $a=7,b=3$ as result, but this doesn't satisfy the third equation. The system is inconsistent.

Back to $(1)$.

You get $A$ from the first equation
$A=\frac{y_1}{\cos\omega t_1}$ and plug it in the second:

$y_2=\frac{y_1}{\cos\omega t_1}\cos \omega t_2$

$y_2\cos\omega t_1-y_1\cos\omega t_2=0$

which can be solved with numerical methods once you have the actual values.

There is no formula to solve something like

$1.2 \cos (3.6 \omega)-0.75 \cos (2.7 \omega)=0$

Once you get $\omega$ you substitute and find $A$.


问:
If a random variable X takes negative numbers, do you still use the same formula to find the expected value and variance?
I'm revising probability questions after taking a break from that area for several years. I've put together a simplified version of my question and attached the probabilities as a table if that is easier to read. My question is, if you're given a distribution with negative values, do you still find the expected value and variance using the standard formula that I've put together below? I can't recall ever having seen an example with negative values before, so I was wasn't sure if it was a trick question. Also, does anyone know how to enter problems like this into Wolfram Alpha, so that I can check I got the right expected value and variance as I go through my revision problems? Expected value is usually straight forward enough, but variance calculations are so easy to mess up. Thanks!

```<code>E(X) = -1*0.4 + 1*0.35 + 3*0.25 = 0.7
</code>```

enter image description here

答:
When in doubt, go for the definition.

A random varible is a function $X:\Omega \rightarrow \mathbb R$.

A continous random varible has a density function $f_X(x)$ such that $P(a\leq X \leq b)=\int_{a}^{b}f_X(x)dx$. We demand $\int_{-\infty}^{\infty}f_X(x)dx=1$.

Expectation of $X$ is defined as $\Bbb E[X]=\int_{-\infty}^{\infty}xf_X(x)dx$.

Variance of $X$ is defined as $Var(X)=\Bbb E[(X-\Bbb E[X])^2]$.

If you meet these definitions, you can do whatever you want.


问:
Expected value in a Poisson process
The problem is from this lecture video. Fish are caught in a Poisson process with parameter $\lambda$. You fish for 2 hours, if in that time you catch at least one fish, then you stop at 2 hours time. Otherwise, you continue to fish until you catch one fish. The question is, what is the expected number of fish $E[N]$. Where $N$ is the random variable for the number of fish.

In the lecture, the professor states that the answer is simply:
$$E[N]=\lambda\tau + P_P(0,2)(1)$$

Here, $P_P$ is the probability distribution for the number of arrivals $k$ in some interval $\tau$:
$$P_P(k, \tau) = \frac{(\lambda\tau)^ke^{-\lambda\tau}}{k!}$$

I don't get why the first term does not have a factor of $1-P_P(0,2)$ which is based on the total expectation theorem:
$$E[N]=P(F)E[N|F] + P(F^c)E[N|F^c]$$

Here, let $F$ be the event that you fish for 2 hours and thus $F^c$ will be the event that you fish for more than 2 hours since you have not caught any fish in the first 2 hours.

答: You always fish for the first two hours.  The expected number of fish caught in this time is $2\lambda$.  Your only fish for the period after two hours if you have caught no fish in the first two hours, and the lecturer represents the probability of this event as $P_P(0,2)$.  The total expectation is $$2\lambda +P_P(0,2)\cdot1$$  (We have $\tau=2$.)  There is no factor of $P(F)$ on the first term, because you always fish for the first two hours.

Title: Expected value in a Poisson process


问:
Determine Jacobi Matrix of a unit simplex
Let $\Phi$ be a affine linear mapping with $\Phi(K)=\hat K$, where $\hat K$ is the unit simplex and $\Phi$ is of the form $$\Phi(\hat x)=A\hat x+b$$ with $A\in\mathbb R^{d\times d}$ and $b\in\mathbb R^d$.

How can I determine $\hat D\Phi(\hat x)$, the Jacobian matrix of $\Phi$ at the point $\hat x$?

答: The Jacobian matrix of an affine linear mapping is the matrix of this mapping, hence $\hat D = A$. It does not depend on the point $\hat x$.


问:
Showing $\int_0^\infty\frac {r} {x^2}e^{-x^2-\frac {r^2} {x^2}}dx=\int_0^\infty e^{-x^2-\frac {r^2} {x^2}}dx$
As part of a parameter integral problem, I need to show that given the function $$
\operatorname{F}\left(r\right) =
\int_{0}^{\infty}\exp\left(-x^{2} - \frac{r^{2}}{x^{2}}\right){\rm d}x\,,\qquad \operatorname{F}\hspace{0.1mm}'\!\left(r\right) =
-2\operatorname{F}\left(r\right)\,,
$$
which amounts to what the title demands. However I cannot find a way to do so, either by integration by parts, substitution or anything else, even though it should be simple.

What am I missing $?$.

答: Use a substitution $x\mapsto r/x$. It may be easier to follow if you write $y=r/x$ so the right-hand side transforms to $\int_0^\infty e^{-r^2/y^2-y^2}dy$.


问:
Prove the collection of isolated points of a set in $\mathbb{R}^n$ is countable.
Let $S \subset \mathbb{R}^n$ be a bounded set and let $B$ be the set of isolated points of $S$ ($x \in S$ is isolated if there exists an open neighbourhood $U$ of $x$ such that $U \cap S = \{x\}$. I claim $B$ is countable.

First I will assume there exists a countable basis $\beta$ for the topology on $\mathbb{R}^n$.  Then for each $x \in B$, there exists an open set, call it $U_{x}$, in $\beta$ such that $U_{x} \cap S = \{x\}$.  Then we can see that $B$ is in bijective correspondence with the collection of sets $\{U_{x} \}_{x \in B } $.  Since $\{U_{x} \}_{x \in B }$ is a subset of the countable set $\beta$, it is countable and hence $B$ is countable.

Questions about my proof:  Is my proof correct?
Also, did I use the axiom of choice in choosing the $U_{x}$?  If so, is there a proof without it?

答:
Your argument is fine, and you don’t need the axiom of choice.

If $\beta$ is countably infinite, by definition there is a bijection from $\Bbb N$ to $\beta$, so we can enumerate $\beta=\{B_k:k\in\Bbb N\}$, and for each isolated point $x$ of $S$ we can let

$$k(x)=\min\big\{k\in\Bbb N:B_k\cap S=\{x\}\big\}\,.$$

Your $U_x$ is then $B_{k(x)}$, and $k(x)$ is a well-defined natural number, not an arbitrary choice.


问:
Solving 3 simultaneous cubic equations
I have three equations of the form:

$$i_1^3L_1+i_1K+V_1+(i_2+i_3+C)Z_n=0$$
$$i_2^3L_2+i_2K+V_2+(i_1+i_3+C)Z_n=0$$
$$i_3^3L_3+i_3K+V_3+(i_1+i_2+C)Z_n=0$$

where $L_1,L_2,L_3,K,V_1,V_2,V_3,C$ and $Z_n$ are all known constants.

What methods can I use to obtain the values of $i_1,i_2$ and $i_3$ ?

答:
Any interactive method for solve this equations is difficult to calculate each interate manually. I recommend an already established method that has many theorems that guarantee convergence. Newton's method with guaranteed convergence theorems for Kantorovich ( see in ArXiV a expository Article on the
Kantorovich's Theorem on Newton's Method ) is easy to implement in many computational software for numerical or algebraic computing.</p>

ContentLicense: CC BY-SA 3.0

Title: Solving 3 simultaneous cubic equations


问:
Basis Element of the Product topology vs. Element of the Basis
Hi i am reading Topology by Munkres and have some doubts under theorem 20.3 regarding basis element of product topology. It is written that:

Let $B=(a_1,b_1)\times...\times(a_n,b_n)$ be a basis element for the product topology and let $\bar x =(x_1,...,x_n)$ be an element of $B$.

My doubts are as follows:

I am unable to understand the above quoted statement. What does the notation $(a_1,b_1)\times(a_2,b_2)$ stand for?

What is the difference between basis element $B$ and an element of $B$? I mean a basis element for the product topology on the cartesian product was defined by $U\times V$ previously in the book where U and V are open sets in X and Y respectively. Then why is it now $(a_1,b_1)\times(a_2,b_2)$ ? And how is it different from the tuple x =(x_1,...,x_n)?

For reference i am attaching the screenshot where i have highlighted the quoted text.

答:
$\Bbb R^n$ is a Cartesian product of $n$ copies of $\Bbb R$. The standard base for the topology of $\Bbb R$ is the set of all open intervals $(a,b)$ for $a < b$.

In theorem 15.1 and the example after it Munkres discusses the $\Bbb R^2$ case and notes that the set of products of intervals then is a (standard) base product of copies of $\Bbb R$, as we have here.

So this remark in the proof refers back to that discussion.


答:
Your idea is correct.

Another possible approach would be using the fact that polynomials of different degrees are always linearly independent.

Title: Proof $p(x),p'(x), p''(x),p'''(x)$ from a basis por polynomials of degree three

Body:
I need to proof that given a polynomial $p(x)$ of degree three, the set $\left\{p(x),p'(x), p''(x),p'''(x)\right\}$ from a basis. Let $p(x)=a_0+a_1x+a_2x^2+a_3x^3$, with $a_3\neq 0$. I calculated the derivatives

$p'(x)= a_1+2a_2x+3a_2x^2$,

$p''(x)=2a_2+6a_3x$, and

$p'''(x)=6a_3$.

This set will be a basis if this system has a solution

$\left( \begin{matrix}
a_0 & a_1 & 2a_2 & 6a_3 \\
a_1 & 2a_2 & 6a_3 & 0 \\
a_2 & 3a_3 & 0 & 0 \\
a_3 & 0 & 0 & 0 \\
 \end{matrix} \right|\left.  \begin{matrix}a_0\\a_1\\a_2\\ a_3 \end{matrix}\right)$

And that's true because the determinant of this matrix is $103a_3^4$, i.e, is different of zero since $a_3\neq 0$.

That's correct?


问:
Find all roots of a quintic polynomial given 1?
Let's say I have solved for one of the roots $a$ of a quintic equation using Newton's method. I read somewhere that you can "simply" divide the equation by $x - a$ to get a quartic and solve from there.

How would I get this quartic equation from my quintic euqation (using a formula), and after doing this, how would I go about solving it numerically?

答:
If $p(x)=x^5+Ax^4+Bx^3+Cx^2+Dx+E$ has root $a$, then

$$p(x)=(x-a)(x^4+A'x^3+B'x^2+C'x+D')$$

where

$$\begin{align}
A&=A'-a\\
B&=B'-aA'\\
C&=C'-aB'\\
D&=D'-aC'\\
E&=-aD'
\end{align}$$

We can easily unwind the first two equations to

$$\begin{align}
A'&=A+a\\
B'&=B+aA'=B+aA+a^2
\end{align}$$

and the last two (provided $a\not=0$) to

$$\begin{align}
D'&=-E/a\\
C'&=(D'-D)/a=-(E/a^2+D/a)
\end{align}$$

Alternatively you could keep going in the "forward" direction with

$$\begin{align}
C'&=C+aB'=C+aB+a^2A+a^3\\
D'&=D+aC'=D+aC+a^2B+a^3A+a^4
\end{align}$$

Either way, if you have a (good) numerical approximation to $a$, you can compute (approximately) the coefficients $A'$, $B'$, $C'$, and $D'$ of a quartic and then apply your favorite root-finding algorithm to that polynomial. If you succeed in getting a second root, you can repeat the basic idea outlined here, to factor that root out of the quartic leaving a cubic, etc. (Once you get things down to a quadratic, the quadratic formula will finish things off.)

Note, however, that a quintic may have only one real root, so some caution is called for if you're trying to make this mindless and mechanical; in particular, Newton-Raphson is not going to help with a quartic that has no real roots.


问:
Failing a basic integration exercise; where did I go wrong?
(This is a basic calculus exercise gone wrong where I need some feedback to get forward.)

I've attempted to calculate an integral by first integrating it by parts and then by substituting. The result I got is not correct though. Can I get a hint about where started to make mistakes?

Here are my steps:

$$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx$$

Applying integration by parts formula

$$\int uv' \, dx = uv - \int u'v \, dx$$

where $u=\sqrt{x}$, $v'=\sin\sqrt{x}$, $u'=\frac{1}{2\sqrt{x}}$ and $v=-\cos\sqrt{x}$, resulting in:

$$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx = \sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{x}} \cdot -\cos\sqrt{x} \, dx$$

Next, I tried substituting $x$ with $g(t)=t^2$ in the remaining integral, i. e. replacing each $x$ in the integral with $t^2$ and the ending $dx$ with $g'(t)=2t\,dt$. I would later bring back the $x$ by substituting $t=\sqrt{x}$ after integration. Continuing from where we left by substituting:

$$\sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{x}} \cdot -\cos\sqrt{x} \, dx$$
$$\Longrightarrow \sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{t^2}} \cdot -\cos\sqrt{t^2}\cdot2t \, dt$$

Then I pulled out the constant multipliers from the integral:

$$\sqrt{x} \cdot -\cos\sqrt{x} - \frac{1}{2} \cdot -1 \cdot 2 \int \frac{1}{\sqrt{t^2}} \cdot \cos\sqrt{t^2}\cdot t \, dt$$

which turned out to eliminate each other (resulting in just $\cdot1$), so we end up with:

$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \int \frac{1}{\sqrt{t^2}} \cdot \cos\sqrt{t^2}\cdot t \, dt$$

Reducing the integrand by reducing $\frac{1}{\sqrt{t^2}}\Rightarrow\frac{1}{t}$, which again is eliminated by multiplying with the integrand's $t$, and reducing $\cos\sqrt{t^2}\Rightarrow\cos t$. Therefore resulting in:

$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \int \cos t \, dt$$

where the integral can be solved as $\int \cos t \, dt = \sin t + C$. So now we are at:

$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \sin t + C$$

The correct answer however is
$$\int \sqrt{x} \sin\sqrt{x} \, dx = 4 \sqrt{x} \sin\sqrt{x} - 2 (x - 2) \cos\sqrt{x} + C$$

So something somewhere in my process went horribly wrong. What?

答:
The mistake is
$$v'=\sin\sqrt{x}\,dx \to  v=\int \sin\sqrt{x}\,dx\neq -\cos\sqrt{x}+k, \quad k\in \Bbb R$$

$$\int \sin\sqrt{x}\,dx=2\left(-\sqrt{x}\cos \left(\sqrt{x}\right)+\sin \left(\sqrt{x}\right)\right)+k',\quad k'\in \Bbb R$$


问:
Payoff in static Bayesian Games
Let $ A_i $ be the set of strategies of player i and $ \Theta_i $ the set of possible types of player i. The payoff function in a static Bayesian Game is defined as

$ u_i: A_i \times \Theta_i \rightarrow  \mathbb{R} $

I do understand that the cross product of the types and actions determine the payoff in a (static) Bayesian Game, but I wonder why it is actions cross player's types? So far, I thought that the player's types are decisive for the action as they fix what action is a best response for each type of any player. Hence, I assumed that the payoff function is determined as

$ u_i: \Theta_i \times A_i \rightarrow  \mathbb{R} $

where for every type of player i an action is choosen and mapped in the set of real numbers.

Can anybody tell me what the rationale is behind defining the payoff function as actions cross player's types and not the other way around?

答:
A convention. A type is chosen exogenously by "Nature" while an action is chosen by the player after observing the type. So often one would write the payoff function as $u(a_i;\theta_i)$ using semi-colon to emphasize that players do not control what type they are when they are choosing their actions.

In addition this convention is similar to the notation in multivariate calculus where a variable is typically written first followed by a parameter, say $f(x,t)=tx^2$, where $t$ is a parameters with$t\in\{1,2,3\}$.

Title: Payoff in static Bayesian Games


问:
The group of additive sequences of integers

Let $G$ be the additive group of the sequences $(a_n)_{n\ge 1} \subset \mathbb{Z}$ and let $f:G \to \mathbb{Z}$ be a group homomorphism. We denote by $e_i$ the sequence $(0,0,...,0,1,0,0...)$ (we just have a $1$ on the $i$-th position, everything else is zero). Show that the set $\{i\ge 1 | f(e_i) \ne 0\}$ is finite.

I thought that I should consider the set of sequences of finite support i.e. the subgroup generated by the $e_i$s. Let's denote it by $H$. We have that $f(H)$ is a subgroup of $\mathbb{Z}$, so $f(H)=m\mathbb{Z}$ for some integer $m$. We also have that $f(G)=n\mathbb{Z}$ for some positive integer $n$. But I don't know how to proceed any further. I think that we should aim for a contradiction. I can only think about something having to do with divisibility, but I don't know.

答:
This is a proof which is completely elementary (although not too simple). The idea is to prove that, for $n$ sufficiently big, $f(e_{n})=0$.

Consider a sequence of the kind $x=(2^{n_{1}},2^{n_{2}},...)$, where $(n_{i})$ is an increasing sequence of natural numbers (I'll explain later how I'm going to choose that sequence). For every $k$, we have

$$
f(x)=f(2^{n_{1}},...,2^{n_{k}},2^{n_{k+1}},...)=\sum_{i=1}^{k}2^{n_{i}}f(e_{i})+f(0,...,0,2^{n_{k+1}},2^{n_{k+2}},...)=\sum_{i=1}^{k}2^{n_{i}}f(e_{i})+2^{n_{k+1}}f(0,...,0,1,2^{n_{k+2}-n_{k+1}},2^{n_{k+3}-n_{k+1}},...).
$$

Call $b_{k}=f(0,...,0,1,2^{n_{k+2}-n_{k+1}},2^{n_{k+3}-n_{k+1}},...)$, so that

$$f(x)=\sum_{i=1}^{k}2^{n_{i}}f(e_{i})+2^{n_{k+1}}b_{k}.$$

Therefore,

$$b_{k}=\dfrac{1}{2^{n_{k+1}}}\left[f(x)-\sum_{i=1}^{k}2^{n_{i}}f(e_{i})\right].$$

Now, applying the Triangle Inequality,

$$|b_{k}|\leq \dfrac{1}{2^{n_{k+1}}}\left[|f(x)|+\sum_{i=1}^{k}2^{n_{i}}|f(e_{i})|\right].$$

By choosing $(n_{i})$ appropriately, we can make the right hand side of the above inequality tend to $1/2$. Simply choose by recursion the sequence so that for every $k$,

$$\sum_{i=1}^{k}2^{n_{i}}|f(e_{i})|<\dfrac{1}{2}2^{n_{k+1}},$$

and we have convergence to $1/2$. Since $b_{k}$ is an integer for every $k$, we have $b_{k}=0$ for $k\geq N$ ($N$ being a sufficiently large natural number).

Now, simply notice that $b_{k}=f(e_{k+1})+2^{n_{k+2}-n_{k+1}}b_{k+1}$ (this follows directly from the definition of $b_{k}$). Using this, we deduce immediately that $f(e_{k})=0$ for large $k$.

Hope this helps!


问:
On the integral $\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$
I came across this integral

$$\mathcal{J} = \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$$

According to W|A it equals $\frac{1}{2}$. However, I cannot find a way to crack it. It smells like a Beta integral , but I do not see any obvious subs. One could start by setting $u=x^2$ but there is no clear path after that.

A promising way might be the following

\begin{align*}
\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}} &= \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{\left ( 3x^2 +2 \right ) x^2 +3}}\\
 &=\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{3\left ( x^2+ \frac{1}{3}   \right )^2 + \frac{8}{3}}} \\
 &= \cdots
\end{align*}

A clever trigonometric sub might clear things but I don't see something. On the other hand , I don't the theory of elliptic integrals is needed here nor complex analysis ( would be interesting to see a solution using contours, though )

So, any ideas how to evaluate it?

P.S: Are there techniques available for these type of problems?

答:
According to Maple, the integral is
$$\frac{\sqrt{3}\, \boldsymbol{\mathit{K}}\! \left(\frac{\sqrt{3}}{3}\right)}{6}$$
Due to different conventions, this would be $Sqrt[3] EllipticK[1/3]/6$ in Mathematica or Wolfram Alpha.


问:
Calculate limit using Maclaurin series
$$\lim_{x \to 0}\frac{1-(\cos{x})^{\sin{x}}}{x^3}$$

So what I have tried is defining three functions, $f(x)=\cos{x}, g(x) = \sin{x}, h(x) = x^3$

What I did next was do Maclaurin expansion for three functions up until the third order:

$f(x) = 1 - \frac{x^2}{2} + o(x^3)$

$g(x) = x - \frac{x^3}{3} + o(x^3)$

$h(x) = x^3 + o(x^3)$

So when I plug in these values I get:

$$\lim_{x \to 0}\frac{1-(1-\frac{x^2}{2}+o(x^3))^{x-\frac{x^3}{3} + o(x^3)}}{x^3}$$

So after plugging in $x=0$ I get $\frac{0}{0}$ and when I try to apply L'Hôpital's rule I find myself in a situation where I need to calculate the derivative of the denominator several times (which would take too long). I'm positive that I'm correct until this point (I checked on wolframalpha, the result is 1/2).

Does anyone know how to proceed from here? Do I need to calculate the derivative of the denominator several times or is there a smarter way?

答:
We can derive the limit using series expansion as follows:

\begin{align*}
\color{blue}{\lim_{x\to 0}}&\color{blue}{\frac{1-\left(\cos x\right)^{\sin x}}{x^3}}\\
&=\lim_{x\to 0}\frac{1-\exp\left((\sin x)\ln\left(\cos x\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\exp\left(\left(x-\frac{x^3}{6}+\mathcal{O}\left(x^4\right)\right)
\ln\left(1-\frac{x^2}{2}+\mathcal{O}\left(x^4\right)\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\exp\left(\left(x-\frac{x^3}{6}+\mathcal{O}\left(x^4\right)\right)
\left(-\frac{x^2}{2}+\mathcal{O}\left(x^4\right)\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\exp\left(-\frac{x^3}{2}+\mathcal{O}\left(x^4\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\left(1-\frac{x^3}{2}+\mathcal{O}\left(x^4\right)\right)}{x^3}\\
&\,\,\color{blue}{=\frac{1}{2}}
\end{align*}

Title: Calculate limit using Maclaurin series


问:
Is $n = 0$ in proofs involving mathematical induction a rigorous expression?
I have had this confusion from high school. In proofs involving mathematical induction, we always say for $n = 0$, blablabla, so that a certain condition holds for $n = 0$. After I learned logic, I found "$n = 0$" may not be a rigorous way of expression. It is in fact a replacement of $n$ by $0$ in $Q\left(n\right)$, where $Q$ is an arbitrary predicate, and we want to prove $\left(\forall n \in \mathbb{N}\right)Q\left(n\right)$. Replacement does not indicate any equality at all. And assume that you would like to finish the proof by asserting $Q\left(n+1\right)$ from $Q\left(n\right)$, then everyone would agree that "for $n = n+1$" is a ridiculous expression. I am wondering why people are still insisting on using the type of expression "$n = 0$" in proofs without explanation in a book at an introductory level. Especially in classes, wouldn't it cause confusions to students who do not understand logic?

答:
"n=0" is a perfectly clear statement.

If using the Peano axioms for the natural numbers you found them on $0$ (some versions start at $1$) then $0$ is defined to be the unique natural number which is not a successor. Given that $n$ is a natural number, you then have a test for whether it is equal to zero or not.

The induction step is that $Q(Sn)$ (which is informally $Q(n+1)$) follows from $Q(n)$ and the Peano axioms imply that if $Q(0)$ is also true then $Q(n)$ is true for all natural numbers.

Now if you want to go back to the foundational arguments of Set Theory, successorship and induction are further abstracted and have to be realised in a different context and a rather different language.

But I learned induction as an "obvious" property of natural numbers before I was introduced to Peano.

I think that Peano, set theory and informal introduction are three different levels of abstraction/precision. As tools, mathematicians use them as necessary, choosing which is appropriate to their context. Later we prove that they are consistent with each other and that the arguments are valid whatever level they are expressed at (though some mathematicians and systems exclude induction, or forms of induction).

I think your issue probably results from mixing the levels of abstraction/precision. Statements like $n=n+1$ do sometimes get used. They are best avoided, because they are formally false and can be confusing. But in context they often represent clunky abbreviations for true statements.


问:
What are some examples of "visually beautiful" math texts?
Are there good examples of math textbooks that seamlessly combines art, cartoon drawing or computer graphics with the texts? I know that there are a few books on the market, which are not text-book per se, that are geared towards general audience, where there are plenty of artistic figures naturally generated by the phenomena under study, usually things of geometric nature, Escher tiling, or things like chaos.

Wolfram's a New Kind of Science comes to mind. What are some other examples of "visually beautifu" or "artistic" math textbooks or texts.

答:
How about Conway, Burgiel, and Goodman-Strauss's "The Symmetries of Things"?

It's broadly about group theory, but it takes the most stunning visual approach I've ever seen. It has lots of full page illustrations, and basically every page has a margin full of pretty pictures related to the current topic. Here's some examples:

And one more to show that this style goes all the way through the book:

It's a really breathtaking book, and the exposition is similarly well done. I'm excited to be able to share it!

I hope this helps ^_^


答:
$v'=\sin\sqrt{x}\ $ does not imply $\ v=-\cos\sqrt{x}$.

By the chain rule, $\frac{d}{dx}(-\cos(x^\frac{1}{2})) = \frac12x^{-\frac12} \times \sin\left(x^\frac12\right) \neq \sin\sqrt{x}$.

In fact, to find $v$, which equals $\int \sin\sqrt{x}\ dx$, you have to use a substitution like $u^2 = x.$ Then by integrating by parts, you should get that $v = 2\sin(\sqrt{x}) - 2\sqrt{x}\cos(\sqrt{x}).$

You can continue down this path, but it is a bit ugly, so let's see if there's a simpler overall approach. The original integral is $\int \sqrt{x}\cdot\sin\sqrt{x}\,dx$. It's best to just start with the substitution $u=\sqrt{x}\quad (^*)$.

Then the integral becomes $2\int u^2 \sin(u) du$, which is much nicer to work with. I think you solve this by integrating by parts twice.

$(^*)$ I originally used the substitution $\ x = u^2\ $ but I think this is inaccurate because $\sqrt{u^2} = |u|,\ $ so the integral would actually become $\ 2 \large{\int}$ $u\ |u| \sin(|u|)\ du,\ $ which I'm not sure is correct. Basically the substitution $\ x = u^2\ $ doesn't tell us if $u=\sqrt{x}\ $ or $\ u=-\sqrt{x}.\ $ Using the substitution $\ u=\sqrt{x}\ $ leaves no room for ambiguity.

Title: Failing a basic integration exercise; where did I go wrong?

Body:
(This is a basic calculus exercise gone wrong where I need some feedback to get forward.)

I've attempted to calculate an integral by first integrating it by parts and then by substituting. The result I got is not correct though. Can I get a hint about where started to make mistakes?

Here are my steps:

$$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx$$

Applying integration by parts formula

$$\int uv' \, dx = uv - \int u'v \, dx$$

where $u=\sqrt{x}$, $v'=\sin\sqrt{x}$, $u'=\frac{1}{2\sqrt{x}}$ and $v=-\cos\sqrt{x}$, resulting in:

$$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx = \sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{x}} \cdot -\cos\sqrt{x} \, dx$$

Next, I tried substituting $x$ with $g(t)=t^2$ in the remaining integral, i. e. replacing each $x$ in the integral with $t^2$ and the ending $dx$ with $g'(t)=2t\,dt$. I would later bring back the $x$ by substituting $t=\sqrt{x}$ after integration. Continuing from where we left by substituting:

$$\sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{x}} \cdot -\cos\sqrt{x} \, dx$$
$$\Longrightarrow \sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{t^2}} \cdot -\cos\sqrt{t^2}\cdot2t \, dt$$

Then I pulled out the constant multipliers from the integral:

$$\sqrt{x} \cdot -\cos\sqrt{x} - \frac{1}{2} \cdot -1 \cdot 2 \int \frac{1}{\sqrt{t^2}} \cdot \cos\sqrt{t^2}\cdot t \, dt$$

which turned out to eliminate each other (resulting in just $\cdot1$), so we end up with:

$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \int \frac{1}{\sqrt{t^2}} \cdot \cos\sqrt{t^2}\cdot t \, dt$$

Reducing the integrand by reducing $\frac{1}{\sqrt{t^2}}\Rightarrow\frac{1}{t}$, which again is eliminated by multiplying with the integrand's $t$, and reducing $\cos\sqrt{t^2}\Rightarrow\cos t$. Therefore resulting in:

$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \int \cos t \, dt$$

where the integral can be solved as $\int \cos t \, dt = \sin t + C$. So now we are at:

$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \sin t + C$$

The correct answer however is
$$\int \sqrt{x} \sin\sqrt{x} \, dx = 4 \sqrt{x} \sin\sqrt{x} - 2 (x - 2) \cos\sqrt{x} + C$$

So something somewhere in my process went horribly wrong. What?


问:
Show that the $C'$ is on the circumcircle of $A'B'C$
At the triangle $ABC$ with angles $\alpha, \beta, \gamma$ the points $A'$, $B'$ and $C'$ are between $B$ and $C$, $A$ and $C$, $A$ and $B$ respectively, such that the circumcircles of $AB'C'$ and $A'BC'$ have only the point $C'$ in common.

It holds that the two circumcircles have the same tangent in $C'$ and that the angle $\angle A'C'B'$ is equal to $\alpha+\beta$.

Show that the $C'$ is on the circumcircle of $A'B'C$.

What condition must be satisfied so that $C'$ is on the circumcircle of $A'B'C$ ? Do we have to Find the radius of the circle and the calculate the distance of the center to $C'$ ?

答:
You are almost there. Look at the $A'C'B'C$ quadrilateral. You have $\angle A'C'B'=\alpha+\beta$. In the $ABC$ triangle you have $$\angle ACB=\angle B'CA'=180^\circ-\alpha-\beta$$
Then $$\angle B'CA'+\angle A'C'B'=180^\circ$$
Therefore $A'C'B'C$ is a cyclic quadrilateral, meaning all vertices lie on a circle, which is the circumcircle of $A'B'C$. Three non-collinear points uniquely define a circle.


答:
Rather unfortunately, the notation that is used is not ideal.  A better way to conceptualize the problem is to write the model as follows.  As stated in the lecture, let $N_\tau$ be the number of fish caught in $\tau$ hours, if you were to simply fish for that amount of time, irrespective of the number of fish caught.  Then $$N_\tau \sim \operatorname{Poisson}(\lambda \tau).$$  Now let $S$ be the random number of fish caught under the conditions of the problem, in which if $N_2 = 0$, then you continue to fish until one more is caught.  The reason I use a different variable here is because $S$ is not Poisson and there is no definite stopping time specified; thus I do not want to use the name $N$ as I don't want to confuse it with $N_\tau$.  Now we can write $$S = \begin{cases}N_2, & N_2 \ge 1 \\ 1, & N_2 = 0. \end{cases}$$  So $$\operatorname{E}[S] = \operatorname{E}[N_2 \mid N_2 \ge 1]\Pr[N_2 \ge 1] + \operatorname{E}[1 \mid N_2 = 0]\Pr[N_2 = 0].$$  I think this is what you had in mind when applying the law of total expectation.  Rather than introducing a new event $F$, one can simply use the existing notation $N_2 \ge 1$ to represent the event that one fishes for only $2$ hours.  All that is left is to execute the calculation:

$$\Pr[N_2 \ge 1] = 1 - e^{-2\lambda}, \\ \operatorname{E}[N_2 \mid N_2 \ge 1] = \sum_{n=1}^\infty n \frac{e^{-2 \lambda}}{1 - e^{-2 \lambda}} \frac{(2 \lambda)^n}{n!} = \frac{2\lambda}{1-e^{-2\lambda}},$$ hence $$\operatorname{E}[S] = 2\lambda + e^{-2\lambda}$$ as claimed.  But why does this simplify so nicely?  Well, let's look at it another way now.  We write $$S = N_2 + \mathbb 1(N_2 = 0),$$ where $\mathbb 1$ is the indicator function that equals $1$ if $N_2 = 0$ and $0$ otherwise.  Equivalently, $\mathbb 1(N_2 = 0)$ is a Bernoulli random variable in which the parameter $p = \Pr[N_2 = 0]$.  Then by linearity of expectation, $$\operatorname{E}[S] = \operatorname{E}[N_2] + \operatorname{E}[\mathbb 1(N_2 = 0)] = 2\lambda + \Pr[N_2 = 0] = 2\lambda + e^{-2\lambda}.$$  This is what the solution in the lecture did.  Note we did not use the law of total expectation--we did not condition on the outcome of $N_2$--but we did use the linearity of expectation.  The reason this works is because we reframed the model for $S$ in a way that is not directly conditioned on $N_2$.

Title: Expected value in a Poisson process

Body:
The problem is from this lecture video. Fish are caught in a Poisson process with parameter $\lambda$. You fish for 2 hours, if in that time you catch at least one fish, then you stop at 2 hours time. Otherwise, you continue to fish until you catch one fish. The question is, what is the expected number of fish $E[N]$. Where $N$ is the random variable for the number of fish.

In the lecture, the professor states that the answer is simply:
$$E[N]=\lambda\tau + P_P(0,2)(1)$$

Here, $P_P$ is the probability distribution for the number of arrivals $k$ in some interval $\tau$:
$$P_P(k, \tau) = \frac{(\lambda\tau)^ke^{-\lambda\tau}}{k!}$$

I don't get why the first term does not have a factor of $1-P_P(0,2)$ which is based on the total expectation theorem:
$$E[N]=P(F)E[N|F] + P(F^c)E[N|F^c]$$

Here, let $F$ be the event that you fish for 2 hours and thus $F^c$ will be the event that you fish for more than 2 hours since you have not caught any fish in the first 2 hours.


问:
UMVUE for $e^{-\lambda}$
Let $X_1\dots,X_n$ be $Poisson(\lambda)$. Show that $$T= \left( \frac{n-1}{n} \right)^{\sum_{i=1}^{n}X_i}$$ is an UMVUE for $e^{-\lambda}$

I know that $\sum_{i=1}^{n}X_i$
it is a sufficient and complete statistic since the distribution belongs to the exponential family. But I don't know how to go on to show that $T$ is an UMVUE.

Any hint?

答:
Since $Y = \sum_{i=1}^nX_i \sim$ Poisson$(n\lambda)$, we can compute $E[T] = E\left[\left(\frac{n-1}{n}\right)^Y\right]$ directly,

\begin{eqnarray}
E[T] &=& E\left[\left(\frac{n-1}{n}\right)^Y\right]\\
&=& \sum\limits_{k=0}^\infty\left(\frac{n-1}{n}\right)^k\frac{e^{-n\lambda}(n\lambda)^k}{k!}\\
&=& e^{-n\lambda}\underbrace{\sum\limits_{k=0}^\infty\frac{\left(n\lambda\left(\frac{n-1}{n}\right)\right)^k}{k!}}_{e^{n\lambda\frac{n-1}{n}}}\\
&=& e^{-n\lambda}e^{n\lambda\frac{n-1}{n}} \\
&=& e^{-\lambda}
\end{eqnarray}

So $T$ is unbiased. By Lehmann-Scheffe, it is UMVUE.

Title: UMVUE for $e^{-\lambda}$


问:
$(x+2)\cos\frac1{x+2} - x\cos\frac1x > 2$ for $x\in[1,\infty)$
Let $$f(x) = x\cos\frac{1}{x}$$ for x in $[1,\infty)$

Now I need to prove or disprove the difference $$f(x+2) - f(x) > 2$$ for all $x$ in the domain.

I tried a lot but I don't seem to be getting anywhere. My approach was to try and use graphs, but somehow that doesn't seem to work out that well.

Any ideas or suggestions please?

答:
Note that cosine function is strictly concave on $[0,\frac{\pi}{2}]\supset[0,1]$. It implies that when $x\ge 1$,
$$\frac{2}{x+2}\cos 0 +\frac{x}{x+2}\cos\frac{1}{x} < \cos\left(\frac{2}{x+2}\cdot 0+\frac{x}{x+2}\cdot \frac{1}{x}\right)=\cos\frac{1}{x+2}.$$
The conclusion follows. </p>


答:
Let us make the limit more "friendly" by a change of variable, $t:=\sin x$. We now have

$$\lim_{t\to0}\frac{1-(1-t^2)^{t/2}}{\arcsin^3t}=\lim_{t\to0}\frac{1-(1-t^2)^{t/2}}{t^3}.$$

Then by the generalized binomial theorem,

$$\lim_{t\to0}\frac{1-1+\dfrac t2t^2-\dfrac t2\left(\dfrac t2-1\right)\dfrac{t^4}2+\cdots}{t^3}$$ and higher order terms. You can conclude.

Title: Calculate limit using Maclaurin series

Body:
$$\lim_{x \to 0}\frac{1-(\cos{x})^{\sin{x}}}{x^3}$$

So what I have tried is defining three functions, $f(x)=\cos{x}, g(x) = \sin{x}, h(x) = x^3$

What I did next was do Maclaurin expansion for three functions up until the third order:

$f(x) = 1 - \frac{x^2}{2} + o(x^3)$

$g(x) = x - \frac{x^3}{3} + o(x^3)$

$h(x) = x^3 + o(x^3)$

So when I plug in these values I get:

$$\lim_{x \to 0}\frac{1-(1-\frac{x^2}{2}+o(x^3))^{x-\frac{x^3}{3} + o(x^3)}}{x^3}$$

So after plugging in $x=0$ I get $\frac{0}{0}$ and when I try to apply L'Hôpital's rule I find myself in a situation where I need to calculate the derivative of the denominator several times (which would take too long). I'm positive that I'm correct until this point (I checked on wolframalpha, the result is 1/2).

Does anyone know how to proceed from here? Do I need to calculate the derivative of the denominator several times or is there a smarter way?


问:
Let $G$ be a group with $|G|=102 ( = 2 \cdot 3 \cdot 17)$, with $|Z(G)| = 2$. Prove the following:
Let $G$ be a group with $|G|=102 ( = 2 \cdot 3 \cdot 17)$, with $|Z(G)| = 2$. Prove the following:

$|Z(G/Z(G))| = 1$

$G/Z(G)$ has a subgroup of order 17

$G$ has a subgroup of order 34

I am preparing for my Algebra exam and I have been stuck on this question for a bit.

1.
So the only thing I know is that $|Z(G/Z(G))| = 1$, which means that $Z(G) \in G/Z(G)$ is the only element for which $A \cdot Z(G) = Z(G) \cdot A$ for every $A \in G/Z(G)$. But I don't know how to prove that there are no other subgroups for which that holds. At first I thought it might be a general property, but apparently it isn't (according to this example: $Z(G/ Z(G)) = 1$ Prove or disprove.).

So I know that it has to do something with the order of the groups, but I don't know how to use the information I have to get the conclusion. So far what I know is that with Lagrange I get $|G/Z(G)| = 102/2 = 51 = 3 \cdot 17$. I see that both are prime but I don't know how this is supposed to help me any further.

2.
I thought of maybe using the fact that $|Cl(A)|$, for every $A \in G/Z(G)$, has to be a divisor of $|G/Z(G)|$, but then I have that $|Cl(A)|$ for every $A$ has to be either 3 or 17, but how to prove that there exists one of 17 I don't know. Or maybe I can use the Class equation but I'm not sure.
I unfortunately don't know how to get any further with this question.

3.
I suppose that this will follow from (1) and (2), since if $G/Z(G)$ has a group of order 17, since $|Z(G)| = 2$, it follows with Lagrange that $G$ has a subgroup of order 34.

Can someone please help me by giving hints that will help me solve this problem? This is a rather introductory course, I've seen other people solve this with 'Sylow theory' but we haven't seen that yet in this course.

答:
Hints:

We have that $\;\left|G/Z(G)\right|=\frac{102}2=51\;$. Now, how many groups of order $\;51\;$ are there? This question is pretty symple if you know semidirect product, but if you don't it is also pretty simple knowing only Sylow Theorems

2)+ 3) are immediate consequences of (1) and Sylow Thms. and the given info (what happens when you take the product of two subgroups $\;HK\;$ ,with $\;K\lhd G\;$ ?)


问:
Is there any trick to calculate $\int\ln(x+\sqrt{x^2-1})dx$ faster?
$$\int\ln(x+\sqrt{x^2-1})dx$$
To evaluate the integral we can apply integral by parts (using $u=\ln(x+\sqrt{x^2-1})$ and $dv=dx$). but my question is: is there other approach to evaluate the integral faster? I think there might be something because we can write it as:

$$\ln\left(x+\sqrt{x^2-1}\times\dfrac{x-\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\right)=\ln\left(\cfrac1{x-\sqrt{x^2-1}}\right)=-\ln(x-\sqrt{x^2-1})$$
But I don't know how to use this.

答:
Use $x=\cosh t$

$$\int \sinh t \ln(\cosh t + \sinh t) dt = \int t \sinh t dt = t\cosh t - \sinh t$$

then substitute back.


问:
How to do a fast numerical computation of an oscillatory integral including $\operatorname{HeunC}$ function using Mathematica?
I am trying to numerically compute the following integral in Mathematica

$$\int_{1}^{1000} dx \,(x+2)e^{-2Ia(x+2)}\operatorname{HeunC}[-4Ib,-4Ib,1+4Ib,1,-4Ib,-x/2]$$

where $\operatorname{HeunC}$ is the confluent Heun function. Mathematica can numerically compute this integral for the known values of $a$ and $b$. There are two problems here: When I increase the value of $a$ and $b$, the integral becomes highly oscillatory and it takes a long time to be computed. Also, even for the small value of $a$ and $b$, Mathematica spends a long time to compute this integral when the upper bound is $1000$, but it is pretty fast when I decrease the upper bound. Could you please help me with this problem?

答:
It looks like HeunC was introduced in Mathematica only recently and may have implementation issues.

For shortness let's introduce
$$
F(x) = \operatorname{HeunC}(-4ib, -4ib, 1+4ib, 1,-4ib; -x/2)
$$

Evaluating it with MachinePrecision gives completely wrong results (obtained with https://www.wolframcloud.com/):

```<code>F[b_,x_] = HeunC[-4I b,-4I b,1+4I b,1,-4I b,-x/2];
F[10,500.]
-2.333438484634705*^4219+4.27023838819976*^4218 I
F[10,500.`1000]
0.00398375247818748....-0.0000498007848027014.... I
F[1,500.]
-1.84935*^270-1.79429*^266 I
F[1,500.`1000]
0.0039527529500429...-0.00049881082215... I
</code>```

Plotting the function for small $x$ suggests that its absolute value monotonicaly decreases and never exceeds 1, so values with $10^{4219}$ magnitude are obviously wrong. I guess this leads to slow (and simply wrong) numerical integration in your case.

One of the possible solutions is to implement HeunC by yourself. We know that HeunC is a solution to
$$
y''(z) + \left(\frac{\gamma}{z} + \frac{\delta}{z-1} + \epsilon\right) y'(z) + \frac{\alpha  z - q}{z(z-1)} y(z) = 0\\
y(0) = 1
$$
In your case
$$
q = \alpha = \epsilon = -4ib\\
\gamma = 1 + 4ib\\
\delta = 1\\
z = -\frac{x}{2}
$$
Rewriting the equation in terms of $F$, $b$ and $x$ gives:
$$
F''(x) + \left(
\frac{1+4ib}{x} + \frac{1}{x+2} + 2ib
\right) F'(x) + \frac{2ib}{x} F(x) = 0\\
F'(0) = 1
$$

The problem is that you cannot just take this equation and ask Mathematica to solve it numerically. This is because

this second order equation has only one initial condition,

the point $x = 0$ is a singularity.

The former is a consequence of the latter. The usual solution of the problem is to step a bit away from the $x=0$ and to start numerical integration from $x=x_0 > 0$. Anyway, to start integration from $x = x_0$ one needs initial conditions, both for $F(x_0)$ and $F'(x_0)$.

One possibility is to ask Mathematica for them, since we know that $F(x)$ may be expressed in terms of HeunC. Another approach is to use series expansion. The series expansion can be derived from the expansion for HeunC, or may be obtained quite easily from the equation itself (Frobenius method).

Assume that $F(x) = \sum_{n=0}^\infty c_n x^n$. We already know that $c_0 = 1$. Multiplying the equation with $x$ and taking limit when $x \to 0$ yields
$$
(1 + 4 ib) F'(0) + 2 i b F(0) = 0 \implies c_1 = F'(0) = -\frac{2ib}{1 + 4ib}
$$
Now let's multiply the equation with $x(x+2)$:
$$
(x^2+2x) F''(x) + \left(
2ib x^2 + (2 + 8ib) x + 2 + 8 ib
\right) F'(x) + (2ib x + 4) F(x) = 0
$$
Let's now plug $F(x) = \sum_n c_n x^n$ and collect the coefficients before $x^n$ term:
$$
n(n-1) c_n + 2(n+1) n c_{n+1} +
2ib (n-1) c_{n-1} + (2 + 8ib) n c_n + (2 + 8ib) (n+1) c_{n+1}
+ 2ib c_{n-1} + 4ib c_n = 0
$$
$$
\left[2(n+1) n + (2 + 8ib) (n+1)\right]c_{n+1} +
\left[n(n-1) + (2 + 8ib) n + 4ib\right] c_n +
\left[2ib(n-1) + 2ib\right] c_{n-1} = 0,\\
2 (n+1) (1 + 4ib + n) c_{n+1} +
\left[n(n+1) + 4ib(1+2n)\right] c_n +
2ib n c_{n-1} = 0.
$$
This is a three-terms recurrence for $c_n$ from which we may obtain as many $c_n$ as we need.
$$
c_{n+1} = -\frac{\left[n(n+1) + 4ib(1+2n)\right] c_n +
2ib n c_{n-1}}{2 (n+1) (1 + 4ib + n)}
$$
From the theory of the confluent Heun's function it is known that the radius of convergence of its series at $z=0$ is $R = 1$. Since we use $x/2$ as its argument, the radius of convergence of $\sum_{n=0}^\infty c_n x^n$ should be equal to $2$. Numerical experiment agrees with it:

Here's a plot of $\frac{|c_n|}{|c_{n+1}|}$ which illustrate that the limit for $n \to \infty$ is 2.
We may take $x_0 = 1$ and expect that series $\sum_{n=0}^\infty c_n x_0^n$ converge fast enough for plactical usage. Estimating truncation error with the first neglected term gives us that $M=50$ terms are enough for double precision (16 digits).

Now we can obtain $F(x)$ numerically by solving the problem
$$
F''(x) + \left(
\frac{1+4ib}{x} + \frac{1}{x+2} + 2ib
\right) F'(x) + \frac{2ib}{x} F(x) = 0, \quad x \in [1, 1000]\\
F(1) = \sum_{n=0}^{50} c_n\\
F'(1) = \sum_{n=1}^{50} n c_n
$$

To minimize computational problems arising from numerical integration of an (piecewise) numerical solution I suggest evaluating integral in the same ODE problem. This is achieved by expanding the ODE problem with an auxiliary equation
$$
J'(x) = (x+2)e^{-2Ia(x+2)} F(x)\\
J(1) = 0.
$$
It is clear that
$$
J(x) = \int_1^x (t+2)e^{-2Ia(t+2)} F(t) dt
$$
and we're interested in $J(1000)$ only. Combining the problems into single ODE guarantees that ODE integration grid will be the same as quadrature integration grid. This cannot be achieved by separate NDSolve + NIntegrate calls.

Here's an implementation of this method in a notebook. But be warned - WorkingPrecision stongly affects the result. Using MachinePrecision is very fast, but the relative error is about $10^{-5}$. Using MachinePrecision=20 improves the result by an order of magnitude or two, but timing is much worse.

Body:
It looks like HeunC was introduced in Mathematica only recently and may have implementation issues.

For shortness let's introduce
$$
F(x) = \operatorname{HeunC}(-4ib, -4ib, 1+4ib, 1,-4ib; -x/2)
$$

Evaluating it with MachinePrecision gives completely wrong results (obtained with https://www.wolframcloud.com/):

F[b_,x_] = HeunC[-4I b,-4I b,1+4I b,1,-4I b,-x/2];
F[10,500.]
-2.333438484634705*^4219+4.27023838819976*^4218 I
F[10,500.`1000]
0.00398375247818748....-0.0000498007848027014.... I
F[1,500.]
-1.84935*^270-1.79429*^266 I
F[1,500.`1000]
0.0039527529500429...-0.00049881082215... I

Plotting the function for small $x$ suggests that its absolute value monotonicaly decreases and never exceeds 1, so values with $10^{4219}$ magnitude are obviously wrong. I guess this leads to slow (and simply wrong) numerical integration in your case.

One of the possible solutions is to implement HeunC by yourself. We know that HeunC is a solution to
$$
y''(z) + \left(\frac{\gamma}{z} + \frac{\delta}{z-1} + \epsilon\right) y'(z) + \frac{\alpha  z - q}{z(z-1)} y(z) = 0\\
y(0) = 1
$$
In your case
$$
q = \alpha = \epsilon = -4ib\\
\gamma = 1 + 4ib\\
\delta = 1\\
z = -\frac{x}{2}
$$
Rewriting the equation in terms of $F$, $b$ and $x$ gives:
$$
F''(x) + \left(
\frac{1+4ib}{x} + \frac{1}{x+2} + 2ib
\right) F'(x) + \frac{2ib}{x} F(x) = 0\\
F'(0) = 1
$$

The problem is that you cannot just take this equation and ask Mathematica to solve it numerically. This is because

this second order equation has only one initial condition,

the point $x = 0$ is a singularity.

The former is a consequence of the latter. The usual solution of the problem is to step a bit away from the $x=0$ and to start numerical integration from $x=x_0 > 0$. Anyway, to start integration from $x = x_0$ one needs initial conditions, both for $F(x_0)$ and $F'(x_0)$.

One possibility is to ask Mathematica for them, since we know that $F(x)$ may be expressed in terms of HeunC. Another approach is to use series expansion. The series expansion can be derived from the expansion for HeunC, or may be obtained quite easily from the equation itself (Frobenius method).

Assume that $F(x) = \sum_{n=0}^\infty c_n x^n$. We already know that $c_0 = 1$. Multiplying the equation with $x$ and taking limit when $x \to 0$ yields
$$
(1 + 4 ib) F'(0) + 2 i b F(0) = 0 \implies c_1 = F'(0) = -\frac{2ib}{1 + 4ib}
$$
Now let's multiply the equation with $x(x+2)$:
$$
(x^2+2x) F''(x) + \left(
2ib x^2 + (2 + 8ib) x + 2 + 8 ib
\right) F'(x) + (2ib x + 4) F(x) = 0
$$
Let's now plug $F(x) = \sum_n c_n x^n$ and collect the coefficients before $x^n$ term:
$$
n(n-1) c_n + 2(n+1) n c_{n+1} +
2ib (n-1) c_{n-1} + (2 + 8ib) n c_n + (2 + 8ib) (n+1) c_{n+1}
+ 2ib c_{n-1} + 4ib c_n = 0
$$
$$
\left[2(n+1) n + (2 + 8ib) (n+1)\right]c_{n+1} +
\left[n(n-1) + (2 + 8ib) n + 4ib\right] c_n +
\left[2ib(n-1) + 2ib\right] c_{n-1} = 0,\\
2 (n+1) (1 + 4ib + n) c_{n+1} +
\left[n(n+1) + 4ib(1+2n)\right] c_n +
2ib n c_{n-1} = 0.
$$
This is a three-terms recurrence for $c_n$ from which we may obtain as many $c_n$ as we need.
$$
c_{n+1} = -\frac{\left[n(n+1) + 4ib(1+2n)\right] c_n +
2ib n c_{n-1}}{2 (n+1) (1 + 4ib + n)}
$$
From the theory of the confluent Heun's function it is known that the radius of convergence of its series at $z=0$ is $R = 1$. Since we use $x/2$ as its argument, the radius of convergence of $\sum_{n=0}^\infty c_n x^n$ should be equal to $2$. Numerical experiment agrees with it:

Here's a plot of $\frac{|c_n|}{|c_{n+1}|}$ which illustrate that the limit for $n \to \infty$ is 2.
We may take $x_0 = 1$ and expect that series $\sum_{n=0}^\infty c_n x_0^n$ converge fast enough for plactical usage. Estimating truncation error with the first neglected term gives us that $M=50$ terms are enough for double precision (16 digits).

Now we can obtain $F(x)$ numerically by solving the problem
$$
F''(x) + \left(
\frac{1+4ib}{x} + \frac{1}{x+2} + 2ib
\right) F'(x) + \frac{2ib}{x} F(x) = 0, \quad x \in [1, 1000]\\
F(1) = \sum_{n=0}^{50} c_n\\
F'(1) = \sum_{n=1}^{50} n c_n
$$

To minimize computational problems arising from numerical integration of an (piecewise) numerical solution I suggest evaluating integral in the same ODE problem. This is achieved by expanding the ODE problem with an auxiliary equation
$$
J'(x) = (x+2)e^{-2Ia(x+2)} F(x)\\
J(1) = 0.
$$
It is clear that
$$
J(x) = \int_1^x (t+2)e^{-2Ia(t+2)} F(t) dt
$$
and we're interested in $J(1000)$ only. Combining the problems into single ODE guarantees that ODE integration grid will be the same as quadrature integration grid. This cannot be achieved by separate NDSolve + NIntegrate calls.

Here's an implementation of this method in a notebook. But be warned - WorkingPrecision stongly affects the result. Using MachinePrecision is very fast, but the relative error is about $10^{-5}$. Using MachinePrecision=20 improves the result by an order of magnitude or two, but timing is much worse.


问:
Prove that the map $f:[0,1]\to l_{\infty}([0,1];\Bbb{R})$ defined by $t\mapsto 1_{[t,1]}$ is Riemann integrable.

A map $f:[0,1]\to F$ (where $F$ is a banach space) is said to be Riemann
integrable if $\exists I\in F$ such that for every $\epsilon>0$ there
is $\delta>0$ satisfying $$\left\lVert I-\sum\limits_{i=1}^n
 f(\xi_i)(t_i-t_{i-1})\right\rVert_F<\epsilon$$ for all partions
$P=\{0=t_0<t_1<\cdots<t_n=1\}$ of $[0,1]$ with $\lVert
 P\rVert:=\underset{i}{\text{max}}(t_i-t_{i-1})<\delta$ and for every
choice of $\xi_i\in(t_{i-1},t_i)$

Here $l_{\infty}([0,1];\Bbb{R}):=\left\{f:[0,1]\to\Bbb{R}|\ \lVert f\rVert_\infty=\underset{t}{\text{sup}}| f(t)|<\infty\right\}$ is banach space with the sup norm $\lVert\cdot\rVert_\infty$.

The map $f:[0,1]\to l_{\infty}([0,1];\Bbb{R})$ is defined as $\displaystyle{f(t)=1_{[t,1]}}\ \forall t\in[0,1]$. Recall that-
$$
1_{[t,1]}(x)=\begin{cases}
1&\text{if }x\in [t,1]\\
0&\text{if }x\in [0,t)
\end{cases}$$
I have to find $I\in l_{\infty}([0,1];\Bbb{R})$ such that the expression above in the definition of riemann integrebility is satisfied.

I don't have any idea how to think about this $I$. Initially, I have tried with $I=1_{[0,1]}$ (which is basically the constant function $1$), but it will not work because for any partion $P=\{0=t_0<t_1<\cdots<t_n=1\}$ we have $\left| 1_{[0,1]}(0)-\sum\limits_{i=1}^n
 f(\xi_i)(0)(t_i-t_{i-1})\right|=\left| 1-1_{[\xi_1,1]}(0)t_1\right|=1$ (as $\xi_1>0$).

This implies $\left\lVert 1_{[0,1]}-\sum\limits_{i=1}^n
 f(\xi_i)(t_i-t_{i-1})\right\rVert_\infty\ge 1$. So, $I=1_{[0,1]}$ fails.

Then I tried with constant function $0$, but it fails in the similar fashion (by evaluating at $1$).

Can anyone help me in this regard? Thanks for help in advance.

答:
For partition $P$, the Riemann sum is the function $S:x \mapsto S(x)$ given by

$$S(x) = \sum_{j=1}^n \mathbf{1}_{[\xi_j,1]}(x)(t_j - t_{j-1}) = \sum_{j=1}^{n-1} t_j\mathbf{1}_{[\xi_j,\xi_{j+1})}(x) + \mathbf{1}_{[\xi_n,1]}(x)$$

As the partition norm tends to $0$, it appears that this sum -- which itself is a step function with respect to $x$ -- will converge to the integral $I:x \mapsto I(x) = x$.

To verify, note that if $x \in [0,1]$ then one of three cases holds:

(1) We have $x < \xi_1$ and $|S(x) - x| = x \leqslant t_1-0 \leqslant \|P\|$.

(2) We have $x \geqslant \xi_n$ and $|S(x) - x| = 1-x\leqslant 1-t_{n-1} \leqslant \|P\|$.

(3) We have $\xi_k \leqslant x < \xi_{k+1}$ for some $k \in \{1,2,\ldots,n-1\}$ and $$|S(x) - x|= |x - t_k| \leqslant \max(|t_{k+1}-t_k|, |t_k-t_{k-1}|)\leqslant \|P\|$$

Thus, $S(x) \to I(x) = x$ as $\|P\| \to 0$.


问:
What is the maximum number of digits a product has?
If I have $X \times Y = Z$ where $X,Y \neq 0$ and

$n$ is the number of digits in $X$

$m$ is the number of digits in $Y$

$k$ is the number of digits in $Z$

What is $k$ related to $m$ and $n$?

Thanks

答: Well if $X$ has $n$ digits and $Y$ has $m$ digits, then $X\leq 10^n$ and $Y\leq 10^m$. Therefore $X\cdot Y \leq 10^{m+n}$. It seems that the answer is is $m+n$. It think it is optimal because $3\cdot 4=12$.


问:
Condition for a finite sequence to be a basis of a subspace of the dual space
Let $E$ be a vector space and $(x_i^*)_{1\leq i\leq p}$ a family of elements of $E^*$. Write $G'$  for the subspace generated by the $x^*_i$ and let $F$ be the orthogonal of $G'$ in $E$. Assume the $x^*_i$ are linearly independent. We want to show the existence of a family $(x_i)_{1\leq i\leq p}$ of elements of $E$ such that $\langle x_i,x^*_j\rangle=\delta_{ij}$ for all $i,j$. There author says that

...$E/F$ and $G'$ can each be canonically identified with the dual of
the other; if the family $(x^*_i)$ is free, there is in $E/F$ a basis
$(\dot{x}_i)$ the dual of $(x^*_i)$ and every representative system of
the classes $\dot{x}_i$ has the required properties.

Let $\pi:E\rightarrow E/F$ be the canonical linear surjection. Then $^t\pi:(E/F)^*\rightarrow G'$ is an isomorphism (the transpose of $\pi$). This shows that we can identity $(E/F)^*$ and $G'$. But how should we identify $G'^*$ and $E/F$? Taking the transpose again gives another isomorphism $^{tt}\pi:(G')^*\rightarrow(E/F)^{**}$. This means that $(E/F)^{**}$ is finite dimensional and so the canonical mapping $c_{E/F}:E/F\rightarrow (E/F)^{**}$ is an isomorphism. Does the author mean that
$$\dot{x}_i:=c_{E/F}^{-1}(^{tt}\pi(x^{**}_i)),$$
where $(x^{**}_i)$ is the dual basis of $(x^{*}_i)$?

答:
The inclusion $i : G' \hookrightarrow E^*$ yields $^ti : E^{**} \to (G')^*$, and then $^ti \circ c_E : E \to (G')^*$ where $c_E : E \to E^{**}$ is the canonical injection. Now note that
\begin{align}
\ker({}^ti \circ c_E) &= \{x \in E : c_E(x) \circ i = 0\} \\
&= \{x \in E : (\forall z \in G') \ c_E(x)(z)=0\} = F,
\end{align}
so that $^ti \circ c_E$ induces an isomorphism $E/F \to (G')^*$ that sends the class of $x \in E$ to $c_E(x) \circ i$, the restriction of $c_E(x)$ on $G'$.


答:
Note
$$\int\ln\left(x+\sqrt{x^2-1}\right)dx=\int \cosh^{-1}xdx \overset{IBP}=x \cosh^{-1}x -\sqrt{x^2-1}$$

Title: Is there any trick to calculate $\int\ln(x+\sqrt{x^2-1})dx$ faster?

Body:
$$\int\ln(x+\sqrt{x^2-1})dx$$
To evaluate the integral we can apply integral by parts (using $u=\ln(x+\sqrt{x^2-1})$ and $dv=dx$). but my question is: is there other approach to evaluate the integral faster? I think there might be something because we can write it as:

$$\ln\left(x+\sqrt{x^2-1}\times\dfrac{x-\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\right)=\ln\left(\cfrac1{x-\sqrt{x^2-1}}\right)=-\ln(x-\sqrt{x^2-1})$$
But I don't know how to use this.


问:
Detect my mistake: $\int_{-\infty}^{\infty}\frac{dx}{(x^{2}+4)^{n}}$
Detect my mistake:
$\displaystyle\int_{-\infty}^{\infty}
\frac{{\rm d}x}{\left(x^{2} + 4\right)^{n}}\ $.

My try:

We shall take a half circle around the upper part of the plane.

The arc will tend to zero.

Now we should calculate residue:
$\displaystyle%
\frac{\left(x - 2{\rm i}\right)^{n}}
{\left(x^{2} + 4\right)^{n}} =
\frac{1}{\left(x + 2{\rm i}\right)^{n}}$.

We shall take $\left(n - 1\right)$ derivatives:
$$
\frac{\partial^{n - 1}}{\partial x^{n - 1}}
\left[\frac{1}{\left(x + 2{\rm i}\right)^{n}}\right] =
\frac{\left(2n - 2\right)!}{n!\left(x + 2{\rm i}\right)}
= \frac{\left(2n - 2\right)!}{n!\left(4{\rm i}\right)}.
$$
Now the residue is incorrect, which leads to the integral result to be incorrect.

Does anyone see my mistake $?$.

答:
Real Approach Using Beta Functions (as a verification)
$$
\begin{align}
\int_{-\infty}^\infty\frac{\mathrm{d}x}{(x^2+4)^n}
&=\frac2{4^n}\int_{-\infty}^\infty\frac{\mathrm{d}x}{(x^2+1)^n}\tag1\\
&=\frac1{2^{2n-1}}\int_0^\infty\frac{\mathrm{d}x}{x^{1/2}(x+1)^n}\tag2\\
&=\frac1{2^{2n-1}}\frac{\color{#C00}{\Gamma\!\left(\frac12\right)}\color{#090}{\Gamma\!\left(n-\frac12\right)}}{\color{#00F}{\Gamma(n)}}\tag3\\
&=\frac1{2^{2n-1}}\frac{\color{#C00}{\sqrt\pi}\color{#090}{(2n-2)!\sqrt\pi}}{\color{#00F}{(n-1)!}\color{#090}{2^{2n-2}(n-1)!}}\tag4\\
&=\frac\pi{2^{4n-3}}\binom{2n-2}{n-1}\tag5
\end{align}
$$
Explanation:

$(1)$: substitute $x\mapsto2x$

$(2)$: substitute $x\mapsto\sqrt{x}$

$(3)$: apply the Beta function integral

$(4)$: apply $\Gamma\!\left(n+\frac12\right)=\frac{(2n)!\sqrt\pi}{4^nn!}$

$(5)$: collect factorials into a binomial coefficient

Contour Approach Using Residues (computes the residue)
$$\newcommand{\Res}{\operatorname*{Res}}
\begin{align}
\Res_{x=2i}\left(\frac1{(x-2i)^n}\frac1{(x+2i)^n}\right)
&=\color{#C00}{\left[(x-2i)^{-1}\right]\frac1{(x-2i)^n}}\color{#090}{\frac1{(x-2i+4i)^n}}\tag6\\
&=\color{#C00}{\left[(x-2i)^{n-1}\right]}\color{#090}{\left(1+\frac{x-2i}{4i}\right)^{-n}\frac1{(4i)^n}}\tag7\\
&=\binom{-n}{n-1}\frac1{(4i)^{n-1}}\frac1{(4i)^n}\tag8\\
&=(-1)^{n-1}\binom{2n-2}{n-1}\frac1{(4i)^{2n-1}}\tag9\\
&=\frac{-i}{4^{2n-1}}\binom{2n-2}{n-1}\tag{10}
\end{align}
$$
Thus, the integral around the contour $[-R,R]\cup Re^{i[0,\pi]}$ for $R\gt2$ is
$$
2\pi i\frac{-i}{4^{2n-1}}\binom{2n-2}{n-1}=\frac{\pi}{2^{4n-3}}\binom{2n-2}{n-1}\tag{11}
$$
which is the integral we want because the integral around the semi-circle tends to $0$.

Note that $(6)$ and $(7)$ show that the residue is the coefficient of $(x-2i)^{n-1}$ in $\frac1{(x+2i)^n}$. This is
$$
\frac1{(n-1)!}\left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^{n-1}\frac1{(x+2i)^n}\tag{12}
$$
which is the same quantity that is being computed in the question. Using the generalized binomial theorem, which can be proven by taking derivatives of $(x+2i)^{-n}$, is just a simplification.


问:
If $4$ distinct integers $a, b, c, d$ are randomly selected from $n$ distinct integers, what is the probability that $a \lt b \lt c \lt d$?

a. If $i$ distinct integers $r_1, r_2, \ldots,r_i$, i $\leq n$, are randomly selected (one after other) from n distinct integers, what is the probability that $r_1 < r_2 < \cdots < r_i$?

b. Does the probability you got in part a of the question change if < is replaced with $\leq$?

I know that a question similar to part 'a' of this  question can be solved using the concept of Decision tree that orders the elements of the list, an application of which is to analyse the complexity of comparison-based sorting algorithm.

I came across a solution to part 'a' of the question that says the required probability is $C(n, i)/P(n,i)$, which is nothing but equal to the solution of the Decision tree based approach, which is $1/i!$.

What I don't understand is this - how $C(n,i)$ is equal to favourable number of outcomes and $P(n,i)$ is equal to total number of outcomes?

Figure: A decision tree for sorting three distinct elements.

Image taken from Discreet Mathematics and its Applications by Kenneth Rosen

答:
The probability does change if you replace $<$ with $\leq$ and you're sampling from your set of $n$ distinct integers with replacement.

In this situation, the number of unordered samples of size $i$ becomes ${n+i-1 \choose i}$ while the number of ordered samples of size $i$ is $n^i$.

So the probability that $r_1\leq \ldots \leq r_i$ equals $\frac{{n+i-1 \choose i}}{n^i}$


问:
How this expansion correlate with the normal formula of Taylor expansion?
I usually see this kind of approximations in physics and I don't understand how it is correlate the taylor expansion formula. Consider the function $f(x)$ and now for $a<<1$ they write

$$f(x+a)= f(x) + f'(x)\cdot a + \frac{1}{2!}f''(x)\cdot a^2 + O(a^2)$$ now I don't understand how it is correlate to the familliar formula?
$$f(x+x_0)= f(x_0)+f'(x)\cdot (x-x_0)+ \frac{1}{2!}f''(x)\cdot(x-x_0)^2 + O(x^2)$$

why the first term is not $f(a)$? who is my $x_0$ in this case?

答:
Consider the function $h(t) = f(x+t)$ and write its Taylor expansion around $t_0 = 0$:
$$
h(t) = h(0) + h'(0)t + \frac{1}{2}h''(0)t^2 + o(t^2).
$$

Evaluate at $a$ and you get
$$
f(x+a) = f(x) + af'(x) + \frac{1}{2}a^2f''(x) + o(a^2).
$$


问:
The group of additive sequences of integers

Let $G$ be the additive group of the sequences $(a_n)_{n\ge 1} \subset \mathbb{Z}$ and let $f:G \to \mathbb{Z}$ be a group homomorphism. We denote by $e_i$ the sequence $(0,0,...,0,1,0,0...)$ (we just have a $1$ on the $i$-th position, everything else is zero). Show that the set $\{i\ge 1 | f(e_i) \ne 0\}$ is finite.

I thought that I should consider the set of sequences of finite support i.e. the subgroup generated by the $e_i$s. Let's denote it by $H$. We have that $f(H)$ is a subgroup of $\mathbb{Z}$, so $f(H)=m\mathbb{Z}$ for some integer $m$. We also have that $f(G)=n\mathbb{Z}$ for some positive integer $n$. But I don't know how to proceed any further. I think that we should aim for a contradiction. I can only think about something having to do with divisibility, but I don't know.

答:
So this all appears in Section 94 of Fuchs's Infinite Abelian Groups II, and this is due to Sasiada. The general result is that this property holds whenever the image $G$ is countable and reduced (i.e., has no non-trivial divisible subgroup, where a divisible group is one where, given any $x\in G$ and $n\in \mathbb N$, there exists $y\in G$ such that $x=n\cdot y$). I will write the proof in the case of $G=\mathbb Z$, but it is a simple matter to generalize this.

Let $P$ be the product of countably many copies of $\mathbb{Z}$ and $G=\mathbb{Z}$. Let $\eta:P\to G$ be a homomorphism such that $\eta(e_n)\neq 0$ for infinitely many $n$. Of course, by removing those for which $\eta(e_n)=0$, we may assume that $\eta(e_n)$ is non-zero for all $n$. Since $\bigcap_m mG=0$, there exists a sequence of integers
$$ 1=k_1<k_2<\cdots$$
such that $\eta(2k_n!e_n)\not\in k_{n+1}G$ for each $n=1,2,\dots$. The set of elements $(g_1,g_2,\dots)$ in $P$ with each $g_i$ equal to either $0$ or $k_i!$ has cardinality $2^{\aleph_0}$, and $G$ has strictly smaller cardinality. Thus there there are two different elements of this set that have the same image under $\eta$. Their difference is an element $x=(h_1,h_2,\dots)$ with each $h_i$ either $0$, $\pm k_i!$ or $\pm 2k_i!$.

But now we obtain a contradiction. If $j$ is the first index for which $h_j\neq 0$ (and there must be such an index as $x\neq 0$) then $\eta(h_j)\not\in k_{j+1}G$, and
$$ \eta(h_j)=\eta(x)-\eta(0,\dots,0,h_{j+1},h_{j+2},\dots)=-\eta(0,\dots,0,h_{j+1},\dots)\in k_{j+1}G.$$

(There appears to be a typo in the proof of 94.2 in Fuchs. He only requires that $\eta(k_i!e_i)\not\in k_{i+1}G$ for all $i$, whereas it seems to me that you need the factor of $2$ because it looks like he gets the possibilities for the entries of $x$ wrong, and omits the potential $2k_i!$, and only has $0$ or $\pm k_i!$ as options for $h_i$.)


答:
Note that this is an inverse function and hence IBP is standard (see below). As soon as you recognize
$$
\ln(x+\sqrt{x^2-1})=\mbox{arcosh}(x)
$$
you can simplify the IBP computations, since no computation is really needed:

The following formula holds for any continuous bijective function:
$$
\int f^{-1}(x) dx= x f^{-1}(x)- \int x (f^{-1})'(x) dx = x  f^{-1}(x)- \int f( f^{-1}(x)) (f^{-1})'(x) dx \\
 = x  f^{-1}(x)- \int f(u) du $$

[Side note: there is a picture proof of this formula in the Book "Proofs without Words".
The proof is basically just observing that
$$
\int_{a}^x f(x) dx + \int_{f(a)}^{f(x)} f^{-1}(u) du
$$
graphically the signed between the rectangle with corners $(0,0), (0,f(x)), (x,0), (x,f(x))$ and the rectangle with corners $(0,0), (0,f(a)), (a,0), (a,f(a))$
The picture proof can be seen in this answer.
]

Now set
$$f(x)= \cosh(x) \\
f^{-1}(x)=\mbox{arcosh}(x)$$
and integrate $\int \cosh(u) du$.

If you accept the above formula as a "trick", then that's it.

Note Whenever when you know the integral of a bijection $f(x)$, the above formula gives you the integral of $f^{-1}$.

Since the formula is just IBP, it explains why $\int \ln(x) dx, \int \arcsin(x) ds$.... are always easy to calculate by parts.


问:
Question about angles in a triangle having a median line

If PS = SR and x = y, is it true that angle QSR must be 90 degrees?

More generally, if PS = SR, is it true that x+y must equal 90?

I have tried drawing circumcircles, expanding the triangle into a parallelogram, and drawing an altitude from S onto QR.  I can't seem to prove QSR is 90.  Which makes me suspect it doesn't have to be.  On the other hand, I can't seem to visualize a triangle (or draw one) where it is clear that PS=SR and x=y, but angle QSR is not 90.  I also feel like using the laws of sines/cosines doesn't prove helpful.

If angle QSR is 90 and PS=SR, x need not equal y.  If angle PQR is 90 and PS=PR, x need not equal y.  If angle PQR is 90 and PS=SR and x=y, then, yes, angle QSR is 90.  There's a lot of combinations of givens and unknowns we can consider.  But for the particular 2 givens, PS=SR and x=y, and the unknown, angle QSR, I seem to be at a loss.  Any thoughts?

EDIT: I have solved this.  The answer to both questions is "no."  For a fixed value of y and a fixed length of QR, x can be written as a function of the length of PR, call it b.  Then taking the example of the fixed values y=30 and QR=2/sqrt(3), as b increases from 2, x strictly decreases from 60 (to 0), and angle QSR decreases from 90 degrees.  Then at some value of b>2 (approximately b=3.1), x will be equal to 30, while QSR will be less than 90.  Thus, there exists a triangle where PS=SR and x=y, but angle QSR is less than 90 degrees.  The same example shows that x and y can both be equal to 30, and thus x+y need not equal 90.

答:
The statement $PS=SR$ means that $S$ is at the midpoint of $PR$, and so $QS$ is a median.

Easy answer first: considering your second "more general" question, observe that $y$ can be obtuse, so the answer is "no", $x+y$ is clearly not always $90^\circ$.

Now the hard one: Taking $x=y$ we immediately have the similarity $\triangle PQR \sim \triangle PQS$ (with the shared $\angle SPQ$). We thus know that $\frac{|PR|}{|PQ|}=\frac{|PQ|}{|PS|}$ and so $|PQ|^2 = |PR||PS| = 2|PS|^2$, so $|PQ| = \sqrt{2\,}|PS|$.

However this doesn't appear to get anywhere in terms of demonstrating a right angle at $Q$ so we will examine for other possibilities.

Consider the line $PR$ with $S$ at its midpoint. Now select a point between $P$ and $S$, say $T$. Now imagine point $Q$ sliding on the line produced from $T$ perpendicular to $PR$. With $Q$ close to $PR$, angle $x$ will be large and angle $y$ will be small. When $Q$ is far from $PR$, angle $x$ will be small and angle $y$ will be large. So somewhere between these limits is a point where $x=y$ for any $T$ between $P$ and $S$. For $T$ at $P$, the same process produces an equal $x$ and $y$. Combined with the relationship above between $|PQ|$ and $|PS|$, we know where this is, so there is a continuum of different triangles that can have this property; not just a right triangle.

Tidying this up in retrospect, it's actually clear that with $S$ at the midpoint of $PR$,
$$ \text{angle }x = \text{angle }y \;\iff \; \triangle PQS \sim \triangle PRQ \;\iff\; |PQ| = \sqrt{2\,}|PS| $$
so the range of possible triangles fulfilling these conditions is illustrated in my gif above.


问:
An attempt to revisit the criticality of the Navier-Stokes Equation
I am looking for feedback on this attempt to revisit the criticality of the Navier-Stokes equation (NSE) for incompressible fluids in 3 dimensions. It has been said, that the NSE is supercritical (see  T. Tao, Why proving global regularity conjecture for Navier-Stokes is hard (terrytao.wordpress.com, March 18, 2007)), because the rescaling that preserves the equation is

$$\vec{u} \rightarrow \frac{1}{k}\vec{u}$$

while the energy is an integral of $|\vec{u}|^2$ and thus the rescaling that preserves the energy is given by

$$\vec{u} \rightarrow \frac{1}{k^{3/2}} \vec{u}$$

thereby showing that for the NSE in 3 dimensions energy is being rescaled more severely; this amplifies the rescaling at small scales $k\ll1$. Hence, the NSE appears to be energy-supercritical.

The attempt here is to show that in this reasoning, the assumption is only true in part; becaise the rescaling that preserves the NSE is not $\vec{u} \rightarrow \frac{1}{k}\vec{u}$, but it is actually

$$\vec{u}^\prime = k^{\alpha_x-\alpha_t}\vec{u},$$

where

$$
\begin{split}
(x,y,z)^\prime &= k^{\alpha_x}(x,y,z)\\
t^\prime & = k^{\alpha_t}t
\end{split}
$$

and these are identical with the above rescaling laws only when $\alpha_x=1$, $\alpha_t=2$ but not in general. This fact was discovered only recently, see A. Ercan and M. L. Kavvas, Chaos 25, 123126 (2015).

But this changes everything, because to keep the NSE invariant under scaling transformation, the velocity and energy transform as
$$
\begin{split}
\vec{u}^\prime & = k^{\alpha_x-\alpha_t}\vec{u} \\
E^\prime & = k^{5\alpha_x-2\alpha_t} E
\end{split}
$$
and the energy $E$ at a given moment $t$, would be scale-invariant if $\vec{u}$ transformed according to the law

$$\vec{u}^\prime = k^{-\frac{3}{2}\alpha_x}\vec{u}.$$

Then, appllying T. Tao's definitions of critical, subcritical and supercritical (see T. Tao, Current developments in mathematics 2006 , 255(2008))  we find that the NSE is

energy-subcritical, when $\frac{\alpha_t}{\alpha_x}>\frac{5}{2}$,

energy-critical, when $\frac{\alpha_t}{\alpha_x}=\frac{5}{2}$,

energy-supercritical, when $\frac{\alpha_t}{\alpha_x}<\frac{5}{2}$.

Do you find this reasoning correct/credible? NSE criticality is quite important and relates to the smoothness of solutions

Your kind thoughts and suggestions are welcome. It is better to have them put together as an answer, pointing mistakes or showing an alternative explanations, rather than having extended discussion/comments

Thank you

答:
You are only scaling the magnitude of the velocity field, whereas the scaling that is actually preserved by Navier-Stokes equation is
$$
u^{(\lambda)}(x,t) = \frac{1}{\lambda}u\left(\frac{t}{\lambda^2},\frac{x}{\lambda}\right), \quad p^{(\lambda)}(x,t) = \frac{1}{\lambda^2}u\left(\frac{t}{\lambda^2},\frac{x}{\lambda}\right).
$$
One can then find the critical Sobolev spaces of the Navier-Stokes equation in $\mathbb{R}^3$ by computing how Sobolev norms of different orders $H^\alpha$ scale with $\lambda$. (Using the Fourier version of the Sobolev norm makes this much easier). Then sub/super/criticality can be established by seeing which norms decrease/increase/stay the same as $\lambda\to 0$. This computation is carried out at the end of these notes and a more qualitative treatment is given in this post.


问:
sufficient conditions for integrability of a bivariate function
Consider the bivariate function

$$f(x,y):=\frac{1}{x^a y^b |x-y|^{c}}$$

on $\left[0,\infty\right[\times [0,\infty[$.

I am trying to find sufficient conditions on the positive real exponents $a,b,c$ such that the integral
$\;\int_0^\infty\int_0^\infty f(x,y) \,dx \,dy\;$
exists.

Necessarily we have the following set of conditions:

$$a<1\\b<1\\c<1\\a+c>1\\b+c>1$$

How figure these necessary conditions ?
I split the double integral

$$\int_0^\infty\int_0^\infty=
\int_0^1\int_0^1+\int_0^1\int_1^\infty+\int_1^\infty \int_0^1 + \int_1^\infty \int_1^\infty.$$

Are the conditions above also sufficient for $f$ to be in $L^1$?

Thanks a lot

答:
Substituting $ (x,y) = (u \lambda, u(1-\lambda)) $ for $u > 0$ and $0 < \lambda < 1$, we have

$$ \left| \frac{\partial(x, y)}{\partial(u, \lambda)} \right| = u, $$

and so, we get

\begin{align*}
\int_{(0,\infty)^2} f(x, y) \, \mathrm{d}x\mathrm{d}y
&= \int_{0}^{1} \int_{0}^{\infty} u f(u\lambda, u(1-\lambda)) \, \mathrm{d}u\mathrm{d}\lambda \\
&= \int_{0}^{1} \int_{0}^{\infty} \frac{u}{u^{a+b+c} \lambda^{a}(1-\lambda)^{b}\left| 2\lambda - 1 \right|^{c}} \, \mathrm{d}u\mathrm{d}\lambda.
\end{align*}

Since

$$ \int_{0}^{\infty} \frac{\mathrm{d}u}{u^{a+b+c-1}} = \infty $$

regardless of the values of $a, b, c$, it follows that the integral always diverges.


问:
Alternate proof for Vieta's formula (formula for the summing the roots of a polynomial)
I just saw Vieta's formula for the first time, where it was stated that given some polynomial
$$p(x)=a_nx^n+\cdots+a_0,$$

let $x_1,\ldots,x_n$ denote its roots. Then $$\sum_{i=1}^n x_i=-\frac{a_{n-1}}{a_{n}}.$$

I initially tried to use an inductive argument, starting with a linear base case, and supposing that the formula held for all polynomials of degree $m < n + 1$, and trying to move from here, by factoring out a term of $x$. This failed for what are perhaps obvious reasons.

My second attempt was successful, and I think I've found what is the standard proof, namely using the fundamental theorem of algebra to get linear factors, expanding the right side of the equation, setting the terms equal, and deriving the formula from there. This basic idea  can be found in detail at The art of problem solving.

I'm interested in an alternate proof. To clarify, I'm not looking for a proof that doesn't depend on the fundamental theorem of algebra in any way at all, but simply a different method of proving the result.

答:
Let's appeal to combinatorics and calculus! Let's GO!

Groundwork (After this section it's smooth sailin, I promise!):

Consider a polynomial of degree $n$ factored out as so:

$$ P(x) = a (x-x_1)(x-x_2)...(x-x_n)$$

Now we introduce a notation:

$$P_q(x) =  \frac{P}{(x-x_q) }$$

Basically the notation denotes to deleting a root out of the zero set, so for example:

$$P_1 (x) = \frac{P}{x-x_1}$$

For repeated deletion, just stack on the numbers:

$$P_{123} = \frac{P}{(x-x_1)(x-x_2)(x-x_3)}$$

Note that we can shuffle the indices and it's the quantity under speaking is invariant:

$$P_{123} = P_{321} = P_{132}$$

Are all the same thing!

And one more property:

$$P_{ii} = 0$$

That is repeated number in the string makes the expression zero totally

OK! Hope that was simple enough, now it's show time, we start by observing pattern's with the derivatives of $P$:

$$ \frac{d}{dx} P = \sum_{q=1}^n P_q$$

And,

$$ \frac{d^2}{dx^2} P = \sum_{q,w=1}^n P_{qw}$$

In the above example, take for example in the summation $P_{12}$ , this is same as $P_{21}$ due to the order which we remove roots not mattering. Now we can remove the repeats by taking out the ways to permute two numbers and inducing an order on the indices (note this [1]):

$$ \frac{d^2}{dx^2} P = 2! \sum_{q \leq w , (q, w) \geq 1}^n P_{qw}$$

Now consider taking the first kth derivatives,

$$ \frac{d^k}{dx^k} P = \sum_{ \text{k indices} , ( \text{k indices} ) \geq 1}^n P_{ \text{k indices}}$$

Now by applying the arguement in [1],  we can induce an order in the summation index and factor out repeats:

$$ \frac{d^k}{dx^k} P = k! \sum_{ \text{ ordered k indices}, \text{ k indices} \geq 1 }^n P_{k indices}$$

Recall that ordered k indices greater with each index having a domain from 1 to n , is basically stating the unique ways to pick out $k$ numbers from the numbers $ \leq n$
Evaluating at $x=0$

$$ \frac{d^k}{dx^k} P = k! (-1)^{n-k} \sum_{ \text{ordered k indices } , \text{k indices} \geq 1}^n \text{ product of n-k roots} $$

By maclaurain's theorem, the polynomial expansion of $P$ has coefficent for $x^k $ as:

$$ a_k  =  \frac{1}{k!} \frac{d^k}{dx^k} P(0) = (-1)^{n-k} \sum_{ \text{ordered k indices} , \text{k indices} \geq 1}^{ n  }  \text{product of n-k roots} $$

Which is exactly the statement of Vieta's Formula


问:
Trying to prove that the poset $U(\mathcal{A})$ is a lattice
Let $\mathcal{A}$ be a finite family of sets. For $\mathcal{A}' \subseteq \mathcal{A}$, define $\cup \mathcal{A}'=\bigcup_{A \in \mathcal{A}'}A$. Let $U(\mathcal{A})=\{\cup \mathcal{A}' : \mathcal{A}' \subseteq \mathcal{A}\}$, considered as a poset ordered by inclusion.

I'm trying to show that $U(\mathcal{A})$ is a lattice.

$U(\mathcal{A})$ is a lattice if every pair $x,y \in L$ has a unique largest common lower bound, called their $\textbf{meet}$, written $x \wedge y$, and also has a unique smallest common upper bound, called their $\textbf{join}$ and written $x \vee y$.  That is to say, $\forall z \in U(\mathcal{A})$

$$z \leq x \text{ and } z \leq y \rightarrow z \leq x \wedge y$$

$$z \geq x \text{ and } z \geq y \rightarrow z \geq x \vee y$$

My thoughts:

At first I just wanted to say that the join of two elements would be their union, but then I realized that their union is not necessarily a well defined element.

Eventually I realized that this question wasn't asking to give an explicit construction of the meet and join, but rather just show that they exist. Furthermore, since $\mathcal{A}$ is a finite family of sets, the poset $U(\mathcal{A})$ is bounded, and therefore meets and joins will indeed exist, and I just need to show that they are unique.

If anyone could help me out with this that'd be great, I've been stuck on this for longer than I'd like to admit haha.

答:
If $x_i\in U(\mathcal{A})$ for $i\in\{0,1\}$, then

$$x_0\lor x_1=\bigcup\{A\in\mathcal{A}:A\subseteq x_0\cup x_1\}\in U(\mathcal{A})\,,\tag{1}$$

and

$$x_0\land x_1=\bigcup\{A\in\mathcal{A}:A\subseteq x_0\cap x_1\}\in U(\mathcal{A})\,.$$

$(1)$ can actually be simplified. There are $\mathcal{A}_i\subseteq\mathcal{A}$ for $i\in\{0,1\}$ such that each $x_i=\bigcup\mathcal{A}_i$, and $(1)$ reduces to

$$x_0\lor x_1=x_0\cup x_1=\bigcup(\mathcal{A}_0\cup\mathcal{A}_1)\in U(\mathcal{A})\,.$$


问:
Can you raise $\pi$ to a real power to make it rational?
We're all familair with this beautiful proof whether or not an irrational number to an irrational power can be rational. It goes something like this:

Take $(\sqrt{2})^{\sqrt{2}}$

If it's rational, then you proved it, if it's irrational, take $((\sqrt{2})^{\sqrt{2}} ){^\sqrt{2}} = 2$ and you've proved it.

I'm wondering if you can raise $\pi$ or $e$ to a certain non-trivial real power to make it rational? And if not, where is the proof that it can't be done?

p.s. - I almost left out the real part, but then I realized that $e^{i\pi} = -1$.

答: Of course. Pick any positive rational $p$ and let $x=\log_\pi p$, then $\pi^x=p$.</p>


问:
How to solve ODE with these values given?
I'm a ninth grader and my math teacher gave me this task:

$$
y'(x) = \frac{3}{(x+1)^2} + 2 - 0.02x
$$

$$
y(0) = 0
$$

$$
y(5) = 10
$$

And he said I need to get answer with analytical method. Can you help me solve this?

edit

I just found out that my classmate got same task but different conditions.

$$
y'(x) = \frac{3}{(x+1)^2} + 2 - 0.02x
$$

$$
y(0) = 0
$$

$$
y(5) = 3
$$

He got the right answer:
$$
y(x) = 31.525 - 31.525  \exp(-0.02x)
$$

答: You can integrate directly to obtain $$y(x)= C - 0.01 x^2 + 2 x - \frac{3}{(x + 1)} $$ Now you can use the hypothesis $y(0)=0$ and $y(5)=10$, to find the value of $C$. Is it what you mean by analytical method?  Doing this, you will have to solve a linear equation system: $$C-3 = 0\ \ \&  \ C-0.01(5)^2+2(5)-\frac{3}{5+1}=10 $$  which I assume you can solve, but which can be seen as inconsistent, since $C=3$ & $C=0.75$. This is weird.


问:
Numerical integration: Solve upper bound given integral
Given the answer L to the integral of a function f(x), how can I solve for the upper bound b using numerical integration?

$$\displaystyle \int_0^{b} f(x) dx = L$$

The anti-derivative of f(x) is of course not trivially solvable (unless I've missed something - in my case it is the square root of a quartic function)

答:
Another method that does not require you to recompute all the work if your guess for $b$ was not accurate enough is following. Consider some grid $x_n = n h$ with stepsize $h$. Do the following steps
$$
J_0 = 0\\
J_{n+1} = J_n + \int_{x_n}^{x_{n+1}} f(x) dx.
$$
Here $J(x)$ is simply $\int_0^x f(t) dt$ and $J_n = J(x_n)$.

You may use any single-interval quadrature that you like, no need to split interval $[x_n, x_{n+1}]$ into smaller pieces.

After each step check that $J_{n+1} < L$. If it is not true, then you've localized $b$ in the $[x_n, x_{n+1}]$ interval. Indeed, $J(x_n) < L$ (otherwise you would stop at the previous step) and $J(x_{n+1}) \geq L$. The solution of the
$$
J(b) = L
$$
should be somewhere between the $x_n$ and $x_{n+1}$.

Now you need to solve
$$
\int_{x_n}^b f(x) dx = L - J_n.
$$
The interval $b - x_n$ is small (less than $h$) and you have a decent initial guess, that can be derived from a linear approximation:
$$
\frac{b_0 - x_n}{x_{n+1} - x_n} = \frac{L - J_n}{J_{n+1} - J_{n}}
$$

Let's approximate $\int_{x_n}^b f(x) dx$ with some quadrature formula with weights $w_i$ and abscissae $\xi_i \in [0, 1]$:
$$
\int_{x_n}^b f(x) dx \approx (b-x_n) \sum_{i=0}^M w_i f(x_n + \xi_i (b - x_n))
$$
Rewriting the equation in $F(b) = 0$ form we have:
$$
F(b) = J_n + (b-x_n)\sum_{i=0}^M w_i f(x_n + \xi_i (b - x_n)) - L = 0\\
F'(b) = \sum_{i=0}^M w_i f(x_n + \xi_i (b - x_n)) + (b - x_n)
\sum_{i=0}^M w_i \xi_i f'(x_n + \xi_i (b - x_n))
$$
and the Newton iterations for this equation will be
$$
b_{k+1} = b_{k} - \frac{F(b_k)}{F'(b_k)}.
$$
If exact value for $F'(b)$ is hard to compute, $F'(b) \approx f(b)$ can be used instead.


问:
Every planar graph can be colored with 4 colors max
Ok, that's a theorem. As it says in this question.

I just cannot understand how mathematics can prove such a thing; what does exactly this theorem say? How to prove it mathematically?

Thanks!

答:
The first proof was due to Appel and Haken in 1977. However, as pointed out by Willie Wong, it is not really readable. Very roughly, it first uses a classification of almost 1500 "unavoidable configurations" of the triangulation of a plane. Next, using computers, it is shown that every of these configuration "leads" to a 4-coloration.

Appel and Haken published an algorithmic approach to this problem in K. Appel and W. Haken, Every Planar Map is Four-Colorable, American Mathematical Society 1989.

In 1997, another proof was published but still use the computer in a similar way. This was due to N. Robertson, D. Sanders, P.D. Seymour and R. Thomas.

To answer to your question How can math prove such a thing?, we see that computer plays a fundamental role in this particuliar problem. In fact, computer science plays a very fundamental part in today's research in finite structures. For instance, the Classification of the Finite Simple Groups is another very famous example. Computers were needed in several parts of the proof as they still are.

Now it does not reduce these results to "uninteresting" simply because computers are needed to prove them. To the contrary, the efforts needed for the conception and the implementation of such algorithms is in itself a very remarkable work.

In conclusion, mathematics can prove such things through algorithmic approaches and clever implementations in computers.

As a final remark, note also the following result due to Grötzsch, 1959:

Every planar graph not containing a triangle is 3-colorable.


问:
Trying to find the value of $h$ such that the set of vectors spans $\mathbb{R}^3$
Find the value of $h$ so that the vectors $$a_1= (-1,h,7)$$ $$a_2=(4,-2,5)$$ $$a_3=(1,-6,2)$$ span $\mathbb{R}^3$.

So, my attempt involved setting $a_1,a_2,a_3$ to be the column vectors of a matrix and then reducing this matrix to echelon form. Then there was a theorem we learned in class that the vectors will span $\mathbb{R}^3$ exactly when every row of the echelon form has a pivot position. However, I got really confused reducing the matrix to echelon form with an $h$, I've always did it with actual numbers, and with this the algebra got really messy. Could somebody help me out? Thank you.

答:
The better question is, for what $h$ values does the set of vectors not span $\mathbb R^3$? And to find the answer to this question, note that the vectors don't span $\mathbb R^3$ iff they are linearly dependent.

If they are linearly dependent, there exists some $x,y$ such that $a_1=xa_2+ya_3$. Hence, we have the following relations: $$4x+y=-1,\;-2x-6y=h,\;5x+2y=7$$So, $x=-3$, $y=11$, and $h=-60$.

So, the set of vectors spans $\mathbb R^3$ iff $h\neq-60$.


问:
Unbiased estimator of a uniform distribution
For a random sample $X_1,X_2,\ldots,X_n$ from a $\operatorname{Uniform}[0,\theta]$ distribution, with probability density function
$$ f(x;\theta) =
\begin{cases}
1/\theta,  & 0 \le x \le \theta \\
0, & \text{otherwise}
\end{cases} $$

Let $X_{\max} = \max(X_1,X_2,\ldots,X_n).$ What is the value of k such that $\hat \theta  = kX_{\max}$ is an unbiased estimator of $\theta$ ?

I'm not sure if there is more to this question, because my intuitive answer answer is just  $k=1$. This is because if you order the sample like $$x_{(1)} \le x_{(2)} \le \cdots \le x_{(n)}$$
such that $x_{(n)} = E[X_{\max}]$.  and the fact that the distribution is uniform, the estimator of $\theta$ should just be $X_{\max}$.

Unbiased estimator -> $E\left[\widehat{\theta\,}\right] = kE[X_{\max}] = \theta$

Is my logic wrong here?

答:
Others have already provided excellent answers showing how $ X_{\mathrm{max}} = \mathrm{max}\{X_1,...,X_n\}$ is a biased estimator and $\frac{n+1}{n}X_{\mathrm{max}}$ is an unbiased estimator. But $\frac{n+1}{n}X_{\mathrm{max}}$ is not the only unbiased estimator.

Let $\hat{\theta} = 2\overline{X}_n$. Then $\hat{\theta}$ is also an unbiased estimator.

$$
 \begin{align*}
  \mathrm{bias}(\hat{\theta}) &= \mathbb{E}(\hat{\theta}) - \theta\\
  &= 2\mathbb{E}(\overline{X}_n) - \theta\\
  &= \frac{2}{n}\sum_{i=1}^n \mathbb{E} (X_i) - \theta\\
  &= \frac{2}{n}n\frac{\theta}{2} - \theta\\
  &= 0
 \end{align*}
$$


问:
$x_n$ doesn't converge to zero and is always smaller than zero and a properly divergent sequence $y_n$
There's a sequence $y_n\rightarrow\infty$ for $n\rightarrow\infty$.
Proof that if $x_n$ doesn't converge to zero and for all $n\in\mathbb{N}$ : $x_n<0$
then there's a subsequence $x_{n_j}$ of $x_n$ such that $x_{n_j}y_{n_j}\rightarrow-\infty$ for $j\rightarrow\infty$.

I have no clue how to proof this. The subsequence thing confuses me so much. If all values of $x_n$ are smaller than zero and are multiplied with $y_n$ shouldn't it automatically converge to $-\infty$?

Thanks for your help in advance!!

答:
Since $(x_n)_n$ doesn't converge to zero, there exists $\varepsilon > 0$ and a subsequence $(x_{n_j})_j$ such that $|x_{n_j}| \ge \varepsilon$ for all $j \in \Bbb{N}$.

Since $x_{n_j} < 0$, it has to be $x_{n_j} \le -\varepsilon$. For large enough $j$ we have $y_{n_j} \ge 0$ so $$x_{n_j}y_{n_j} \le -\varepsilon y_{n_j} \xrightarrow{j\to\infty} -\infty.$$


问:
Stability Analysis Finite Difference Methods Black-Scholes PDE
$\mathbf{Background}$

Let $u = u(x,t)$ be the solution of the following forward parabolic initial
boundary-value problem in one space dimension,
$$u_t - a u_{xx} = f(x,t) \text { for } 0 < x < L \text { and } 0 < t <T,$$

with the boundary conditions
\begin{align*}
u &= \gamma_0(t) \text { at } x = 0, \text { for } 0 < t < T,\\
u &= \gamma_L(t) \text { at } x = L, \text { for } 0 < t < T,
\end{align*}
and the initial condition
$$u = u_0(x) \text { for } \text { at } t = 0, \text { for } 0 < x < L.$$

The problem is a model of heat conduction in one space dimension and $a > 0$ is the thermal conductivity.

Define a finite difference grid on the rectangle $[0,L] \times [0,T],$
$$(x_p,t_n) = (p\Delta x,n\Delta t) \text { for } 0 \leq p \leq P \text { and } 0 \leq n \leq N , \text { with } \Delta x = \frac{L}{P}, \text { and } \Delta t = \frac{T}{N}.$$

Suppose $U_p^{n} \approx u(x_p,t_n),\, f_p^{n} = f(x_p,t_n),\, \gamma_0^n = \gamma_0(t_n),\, \gamma_L^n = \gamma_L(t_n) \text { and } u_{0p} = u_0(x_p).$

Put $\rho = a\Delta t / \Delta x^2.$
The explicit Euler method for the problem is:
\begin{equation}
U_p^{n+1} = f_p^n\Delta t + \rho U_{p-1}^n + (1-2\rho)U_p^n + \rho U_{p+1}^n \text { for } 1 \leq p \leq P-1 \text { and } 0 \leq n \leq N-1.
\end{equation}

If $\rho \leq 1/2,$ or equivalently if $2a\Delta t \leq \Delta x^2,$ then the explicit Euler method is stable:
\begin{multline}
\label{stability theorem explicit}
|| \mathbf{U}_{0:P}^n ||_{\infty} \leq
\max \big\{||(\mathbf{u_0})_{1:P-1}||_{\infty}, ||\gamma_0^{0:n}||_\infty, ||\gamma_L^{0:n} ||_\infty \big\} + \sum_{j=0}^{n-1}|| \mathbf{f}_{1:P-1}^{j}  ||_\infty \Delta t \,\,
\text { for } 0 \leq n \leq N.
\end{multline}

$\mathbf{My\,\,Question}$

I'm computing numerical solution of Black-Scholes PDE using explicit Euler method. I want to undertake stability analysis similar to what I have indicated for the numerical solution of heat equation using explicit Euler. We know that explicit Euler is conditionally stable. However, unlike heat equation, for Black-Scholes PDE, I can't find relevant literature that derives bounds on the discretization steps. I have been reading some textbooks that simply say that $\Delta t /\Delta x^2$ should be below a certain threshold but do not indicate that threshold explicitly.

Could someone suggest useful references (journal articles/books/research papers) that undertake stability analysis of this kind for computing numerical solution to Black-Scholes PDE using finite difference methods (explicit/implicit/Crank-Nicolson)?

答:
The recommendation above is to make the change of variable $y := \log(S/K)$ and the function redefinition referenced in the notes. Those two let you transform
the original problem
$$V_t(t, S) + (\sigma^2/2)S^2 V_{SS}(t, S) + rSV_S(t, S) - rV(t, S) = 0,$$
with appropriate initial condition, e.g. $V(0, S) = \phi(S)$, into the equivalent system
$$P_t(t, y) + (\sigma^2/2)P_{yy}(t, y) + (r - \sigma^2/2)P_y(t, y) - rP(t, y) = 0$$
with initial condition $P(0, y) = \gamma(y)$.

I point this out because you already have a stability analysis for the latter, and if you have stably generated its solution you'll likewise have the solution for the former. You know from your stability analysis that the explicit scheme needs $(\sigma^2/2)\Delta t / (\Delta y)^2 \leq 1/2$, that is
$$ \sigma^2 \Delta t / (\Delta y)^2 \leq 1.$$
With $y = \log(S/K)$ I simply computed the differential $\Delta y = (1/S)\Delta S$ and substituted to get
$$ \sigma^2 S^2 \Delta t / (\Delta S)^2 \leq 1.$$
In practice this might mean the explicit scheme for the standard Black-Scholes system requires dynamic step-sizing.


答:
Consider $y_n=n, x_n=-\frac1n$. We easily see that $x_ny_n$ doesn't automatically diverge to $-\infty$ just because $x_n$ are negative. This is the reason we can't have $x_n\to0$.

Next, consider $y_n=n$ and
$$
x_n=\cases{\frac1n& if $n$ is even\\-1& otherwise}
$$
Take a look at the sequence $x_ny_n$ to see why we need to work with a subsequence.

As for the proof itself, consider this: the fact that $x_n$ does not converge to $0$ (in conjunction with $x_n<0$) means

There is an $\varepsilon < 0$ such that there are infinitely many terms that fulfill $x_n<\varepsilon$

Can you guess, just from reading the above paragraph (and the fact that this is the hint I give) what a suitable subsequence might be?

Title: $x_n$ doesn't converge to zero and is always smaller than zero and a properly divergent sequence $y_n$

Body:
There's a sequence $y_n\rightarrow\infty$ for $n\rightarrow\infty$.
Proof that if $x_n$ doesn't converge to zero and for all $n\in\mathbb{N}$ : $x_n<0$
then there's a subsequence $x_{n_j}$ of $x_n$ such that $x_{n_j}y_{n_j}\rightarrow-\infty$ for $j\rightarrow\infty$.

I have no clue how to proof this. The subsequence thing confuses me so much. If all values of $x_n$ are smaller than zero and are multiplied with $y_n$ shouldn't it automatically converge to $-\infty$?

Thanks for your help in advance!!