问:
Finding the anti-derivative of complex absolute value
I am wondering if there is an anti-derivative of the complex absolute value and if it exists, how to find it.

I am looking for

$
\int |z|^2 dz = ?
$

Any help is appreciated!

答:
No. One the complex plane, only analytic functions have derivatives or anti-derivatives. And neither $|z|$ nor $|z|^2$ is analytic anywhere. These functions are real-valued everywhere, and the only complex-analytic functions whose images are not 2-dimensional are constants.

Integration in $\Bbb C$ is along curves. So to have an integral, you must define the curve it is along. The only exception is when the funtion you are analyzing is conservative, so the integral along any closed curve is always $0$. In that case, it doesn't matter which path you take between two points - the value of the integral is dependent on only the end points. However in $\Bbb C$, the condition for being conservative happens to coincide with the condition for being analytic. Thus you cannot use integration to define an anti-derivative unless the integrand is analytic.


问:
Which one is the larger value of $\min(x,y)$ and $\max(0,x+y-1)$, for all $x,y\in [0,1]$?
Which one is the larger value of $\min(x,y)$ and $\max(0,x+y-1)$, for all $x,y\in [0,1]$?

I'm trying as follows.

If $x\leq y$, then $$\min(x,y)=x$$

Now we have $x+y-1\leq 2y-1$.

Now I confused to conclude the value of
$$\max(0,x+y-1).$$

If $x> y$, then $$\min(x,y)=y$$

Now we have $x+y-1> 2y-1$.

Now I also confused to conclude the value of
$$\max(0,x+y-1).$$

So, can we compare the value of $\min(x,y)$ and $\max(0,x+y-1)$, for all $x,y\in [0,1]$?

答: You have $x-1\leq 0$ and $y-1\leq 0$ so $x+y-1$ is less or equal to both $x$ and $y$. Therefore $\min{(x,y)}\geq\max{(0,x+y-1)}$

ContentLicense: CC BY-SA 4.0


问:
Direct Product Subgroups
My lecturer gave us a warning and I'm not sure I understand it:

"A subgroup of A x B need not be a direct product A1 x B1"

He gave us this example:

A = C4 =

B = C2

< (a,b) > <= A x B

= {(a,b), (a^2, 1), (a^3, b), (1,1)}

Which he says is not a direct product, but I don't understand this. Is this not just A1 x B1 where A1 = C4 and B1 = C2? Why is this not a direct product?

答:
The direct product of $C_4$ and $C_2$, in your notation, consists of the following elements:

$$ \{(1,1), (a,1), (a^2,1), (a^3,1), (1,b), (a,b), (a^2,b), (a^3,b)\}. $$

While the subgroup you mention has the following elements:

$$ \{(1,1), (a,b), (a^2,1), (a^3,b)\}. $$

Title: Direct Product Subgroups


问:
How to calculate the number of side of a polygon?

I stuck on this problem. Please suggest any hint on how to solve this problem. Thanks in advance.

答:
Hint: What is $ \angle BEA + \angle CAE $?

Express that in 2 different ways.

 $\angle BAE = \frac{360^\circ}{n}   $

 $\angle BEA + \angle CAE = 45^\circ $

 $\frac{1}{n} + \frac{ 2}{n} = \frac{ 180^\circ}{ 45^\circ }$


问:
stability and asymptotic stability: unstable but asymptotically convergent solution of nonlinear system
Consider nonlinear systems of the form $X(t)'=F(X(t))$, where $F$ is smooth (assume $C^\infty$). Is it possible to construct such a system (preferably planar system) so that $X_0$ is an unstable equilibrium, but all nearby solution curves tend to $X_0$ as $t \to \infty$? If so, how? A conceptual construction is enough. What will the phase portrait look like?</p>

答:
Definition of stability:

Suppose $X^∗$ ∈ $\mathbb{R}^n$ is an equilibrium point for the differential equation
$$X′ = F(X).$$
Then $X^∗$ is a stable equilibrium if for every neighborhood $\mathcal{O}$ of $X^∗$ in $\mathbb{R}^n$ there is a neighborhood $\mathcal{O}_1$ of $X^∗$ in $\mathcal{O}$ such that every solution $X(t)$ with $X(0) = X_0$ in $\mathcal{O}_1$ is defined and remains in $\mathcal{O}$ for all $t > 0$.

and

Definition of asymptotic stability:

If $\mathcal{O}_1$ can be chosen so that, in addition to the properties for stability, we have $\lim_{t \to \infty} X(t) = X^∗$, then we say that $X^∗$ is asymptotically stable.

The exact problem:

In the definition of an asymptotically stable equilibrium point, we required that the equilibrium point also be stable. This requirement is not vacuous. Give an example of a phase portrait (a sketch is sufficient) that has an equilibrium point toward which all nearby solution curves (eventually) tend, but which is not stable.

The $C^1$ example I found with the aid of Mathematica (I guess $C^\infty$ is not possible):

```<code>StreamPlot[{x^2 + (-(x^2 + 1/16) (y - 1/2)^3) Boole[y >= 1/2], -y + (y - 1)^2 y Boole[y >= 1] - 500 (y - 2)^4 Boole[y >= 2] + (x + 1)^3 Boole[x <= -1] + (x - 1)^3 Boole[x >= 1]}, {x, -4, 4}, {y, -4, 4}]
</code>```

The phase portrait is here (I don't have privilege to post image; you can always run Mathematica with the above code to get the phase portrait). The example is unprecedentedly ugly, but at least it works.

This is inspired by the system
$$x'=x^2,$$
$$y'=-y.$$
and in a small neighborhood of 0 (actually $[-1,1] \times [-\infty, 1/2]$), the system is exactly the above simple simple. All solutions nearby 0 goes back to 0, since nearby solutions in the left half plane naturally converge to 0, and nearby solutions in the right half plane first converge to the $x$-axis, then wind back to the left half plane.

ContentLicense: CC BY-SA 3.0

Body:
Definition of stability:

Suppose $X^∗$ ∈ $\mathbb{R}^n$ is an equilibrium point for the differential equation
$$X′ = F(X).$$
Then $X^∗$ is a stable equilibrium if for every neighborhood $\mathcal{O}$ of $X^∗$ in $\mathbb{R}^n$ there is a neighborhood $\mathcal{O}_1$ of $X^∗$ in $\mathcal{O}$ such that every solution $X(t)$ with $X(0) = X_0$ in $\mathcal{O}_1$ is defined and remains in $\mathcal{O}$ for all $t > 0$.

and

Definition of asymptotic stability:

If $\mathcal{O}_1$ can be chosen so that, in addition to the properties for stability, we have $\lim_{t \to \infty} X(t) = X^∗$, then we say that $X^∗$ is asymptotically stable.

The exact problem:

In the definition of an asymptotically stable equilibrium point, we required that the equilibrium point also be stable. This requirement is not vacuous. Give an example of a phase portrait (a sketch is sufficient) that has an equilibrium point toward which all nearby solution curves (eventually) tend, but which is not stable.

The $C^1$ example I found with the aid of Mathematica (I guess $C^\infty$ is not possible):

StreamPlot[{x^2 + (-(x^2 + 1/16) (y - 1/2)^3) Boole[y >= 1/2], -y + (y - 1)^2 y Boole[y >= 1] - 500 (y - 2)^4 Boole[y >= 2] + (x + 1)^3 Boole[x <= -1] + (x - 1)^3 Boole[x >= 1]}, {x, -4, 4}, {y, -4, 4}]

The phase portrait is here (I don't have privilege to post image; you can always run Mathematica with the above code to get the phase portrait). The example is unprecedentedly ugly, but at least it works.

This is inspired by the system
$$x'=x^2,$$
$$y'=-y.$$
and in a small neighborhood of 0 (actually $[-1,1] \times [-\infty, 1/2]$), the system is exactly the above simple simple. All solutions nearby 0 goes back to 0, since nearby solutions in the left half plane naturally converge to 0, and nearby solutions in the right half plane first converge to the $x$-axis, then wind back to the left half plane.


问:
A Regular Space that isn't Tychonoff
A known example of a regular space that isn't Tychonoff is the "Mysior Plane" $X=X_0\cup\{p\}$, where

$$X_0=\{(x,y)\in\mathbb{R}^2 \ |\ y\geq 0\}$$

and $p\notin X_0$. The topology here is defined by declaring every point $(x,y)$ with $y>0$ to be isolated, an open set around a point $(x,0)$ be on the form $U_x\setminus F$, where
$$U_x=\{(x,y)\ |\ 0\leq y\leq 2\}\cup\{(x+y,y)\ |\ 0\leq y\leq 2\}$$
and $F$ is a finite set, and a neighborhood of $p$ to be on the form $U_n=\{p\}\cup\{(x,0)\ |\ x\geq n\}$ for $n\in\mathbb{N}$.

However, I'm having some trouble showing this space in not Tychonoff:

Let $H_j=\{(x,0)\ |\ j-1\leq x \leq j\}$, $H$ be the $x$-axis and let $f:X\rightarrow\mathbb{R}$ be a continuous function such that $f(H_1)=0$. It can be shown that the sets $K_j=f^{-1}(0)\cap H_j$ are infinite for each $j\in\mathbb{R}$. Why does that implies that $f(p)=0$?

答:
You have a typo in the definition of $K_j$: $K_j=f^{-1}[\{0\}]\cap H_j$. The proof that $K_j$ is infinite for each $j\in\Bbb Z^+$ is by induction on $j$. Clearly $K_1=H_1$ to get the induction started. For the induction step it will be convenient to let

$$U_x^-=\{\langle x,y\rangle:0\le y\le 2\}$$

and

$$U_x^+=\{\langle x+y,y\rangle:0\le y\le 2\}$$

for each $x\in\Bbb R$.

Now assume that $K_n$ is infinite. Then there is a countably infinite $C\subseteq K_n$. Let $\langle c,0\rangle\in C$. Suppose that $G$ is a $G_\delta$-set containing $\langle c,0\rangle$, i.e., $G=\bigcap_{n\in\Bbb Z^+}G_n$ for some open nbhds $G_n$ of $\langle c,0\rangle$. Then each $G_n$ contains all but finitely many points of $U_c^+$, so $G$ contains all but countably many points of $U_c^+$. Now

$$\begin{align*}
U_c^+\setminus f^{-1}[\{0\}]&=\bigcup_{n\in\Bbb Z^+}\left(U_c^+\cap f^{-1}\left[\left[\frac1n,\to\right)\right]\right)\\
&\qquad\cup\bigcup_{n\in\Bbb Z^+}\left(U_c^+\cap f^{-1}\left[\left(\leftarrow,-\frac1n\right]\right]\right)
\end{align*}$$

is a subset of $U_c^+$ that is a countable union of closed sets and does not contain $\langle c,0\rangle$, so its complement is a $G_\delta$-set containing $\langle c,0\rangle$. We’ve just seen that this complement contains all but countably many points of $U_c^+$, so $U_c^+\setminus f^{-1}[\{0\}]$ must be countable. $C$ is countable, so $B=\bigcup\limits_{\langle c,0\rangle\in C}\big(U_c^+\setminus f^{-1}[\{0\}]\big)$ is countable, and so is

$$P=\{\langle x,0\rangle:\langle x,y\rangle\in B\}\,.$$

Let $H=H_{n+1}\setminus P$; clearly $H$ is infinite. (In fact $|H|=|\Bbb R|$.)

Now let $\langle x,0\rangle\in H$. Then

$$U_x\cap U_c^+\cap f^{-1}[\{0\}]\supseteq U_x^-\cap U_c^+\cap f^{-1}[\{0\}]\ne\varnothing$$

for each $\langle c,0\rangle\in C$. That is, every open nbhd of $\langle x,0\rangle$ meets $f^{-1}[\{0\}]$, and $f^{-1}[\{0\}]$ is closed, so $\langle x,0\rangle\in f^{-1}[\{0\}]$. Thus, $H\subseteq f^{-1}[\{0\}]$, and therefore $K_{n+1}=H_{n+1}\cap f^{-1}[\{0\}]$ is infinite. This completes the induction step.

We now know that $K_n$ is infinite for each $n\in\Bbb Z^+$. In particular,

$$U_n\cap f^{-1}[\{0\}]\ne\varnothing$$

for each $n\in\Bbb Z^+$, so $p\in\operatorname{cl}f^{-1}[\{0\}]=f^{-1}[\{0\}]$, i.e., $f(p)=0$.

You may also find the answers to this question of some interest.

Title: A Regular Space that isn't Tychonoff


问:
Induction on a contractive sequence
By definition I know that a contractive sequence is called contracting if there exists $l\in (0,1)$ such that for $n>N$:

$$|x_{n+2}-x_{n+1}|\le l |x_{n+1}-x_n|;\forall n \in\mathbb{N}$$

There is an inequality $|x_{n+1}-x_n| \le  l^{n-1} |x_2-x_1|$ which seems related to the contractive sequences that is written in my lecture notes but I am not sure how this holds true. Can anyone guide me on some steps or show me how this inequality holds true? It will be really helpful.

Title: Induction on a contractive sequence

Body: $$|x_{n+1}-x_n| \le l|x_n-x_{n-1}| \le l^2|x_{n-1}-x_{n-2}| \le \cdots \le l^{n-1} |x_2-x_1|.$$


问:
Need to calculate $\displaystyle{\lim_{n \to \infty}} \sqrt[n]{(\ln(n))^{2009}-1}$
Calculate:

$$\displaystyle{\lim_{n \to \infty}} \sqrt[n]{(\ln(n))^{2009}-1}$$

I thought about some trick like this one:

$$\displaystyle{\lim_{n \to \infty}} \sqrt[n]{(\ln(n))^{2009}-1} = \displaystyle{\lim_{n \to \infty}} \sqrt[n]{\left(\frac{e^{2009*\ln[\ln(n)]}-1}{2009*\ln[\ln(n)]}\right) * 2009*\ln[\ln(n)]}$$

But unfortunately it doesn't seem to lead anywhere.

It is important also to add that I can not use L-Hospitals rule.

Body:
$\ln n=t$

take log on both side
$$\log L=\lim_{t\to \infty}  \frac{{(t)}^{2009}-1}{e^t}$$

By taylor series$$=\lim \frac{1-\frac{1}{t^{2009}}}{\frac{1}{t^{2009}}+\frac{t}{t^{2009}}...+\frac{t^{2010}}{2010!t^{2009}}+....}=0$$

$$L=e^0=1$$


问:
$x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$
I'm doing some exercise to prepare for my multivariable analysis exam. I didn't understand the second part of this question.

Given the function

$$f(x,y)=(x^2+y^2+1)^2 - 2(x^2+y^2) +4\cos(xy)$$

Prove that the taylor polynomial of degree $4$ of $f$ is equal to
  $5+x^4+y^4$.

First, $4\cos(xy) = 4 - 2(xy)^2 + 4R_3 $

$(x^2+y^2+1)^2=x^4+2 x^2 y^2+2 x^2+y^4+2 y^2+1$

Therefore: $(x^2+y^2+1)^2 - 2(x^2+y^2)=x^4+2 x^2 y^2+y^4+1$

Therefore: $f(x,y)=x^4+y^4+5+4R_3$

I don't know exactly why I can now conclude that Taylor Polynomial of degree 4 must be $5+x^4+y^4$, but I don't know exactly why.

Now the second question is: $x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$

New edit

I understand this now thanks to hint of Hagen von Eitzen, thanks !

The third question is:

Determine what kind of stationary point you have in $(0,0)$.

答:
$$x^4 + y^4 \ge \frac{(x^2+y^2)^2}{2} \\
2x^4+2y^4\ge x^4+2x^2y^2+y^4 \\
x^4-2x^2y^2+y^4 \ge0 \\ (x^2-y^2)^2\ge0$$

Notice that $x^2-y^2\in R$, so all real number with an even exponent are always greater or iqual to zero.

Title: $x^4+y^4 \geq \frac{(x^2+y^2)^2}{2}$


问:
Find the limit of trigonometric function
$$
\lim_{x\to+\infty}(\frac{2}{\pi}\text{arctan}\;x)^{x}
$$

Even though I've expanded via Taylor series, I can't figure out

答:
If you compose Taylor series (that is to say working one piece at the time), not only you will get the limit but also the asymptotics of the function.
$$y=\Big[\frac{2}{\pi } \tan ^{-1}(x)\Big]^x\implies \log(y)=x \log\Big[\frac{2}{\pi } \tan ^{-1}(x)\Big]$$
$$\tan ^{-1}(x)=\frac{\pi }{2}-\frac{1}{x}+\frac{1}{3 x^3}-\frac{1}{5
   x^5}+O\left(\frac{1}{x^7}\right)$$
$$\frac{2}{\pi } \tan ^{-1}(x)=1-\frac{2}{\pi  x}+\frac{2}{3 \pi  x^3}-\frac{2}{5 \pi
   x^5}+O\left(\frac{1}{x^7}\right)$$
$$\log\Big[\frac{2}{\pi } \tan ^{-1}(x)\Big]=-\frac{2}{\pi  x}-\frac{2}{\pi ^2 x^2}+\frac{2 \left(\pi ^2-4\right)}{3 \pi ^3
   x^3}+O\left(\frac{1}{x^4}\right)$$
$$\log(y)=-\frac{2}{\pi  }-\frac{2}{\pi ^2 x}+\frac{2 \left(\pi ^2-4\right)}{3 \pi ^3
   x^2}+O\left(\frac{1}{x^3}\right)$$
$$y=e^{\log(y)}=e^{-\frac 2\pi } \left(1-\frac{2}{\pi ^2 x}+\frac{2 \left(3-4 \pi +\pi ^3\right)}{3 \pi ^4
   x^2}\right)+O\left(\frac{1}{x^3}\right)$$

Make $x=123.456$; the "exact" value is $0.5282144850$ while the above expansion gives $0.5282144663$.


问:
Proving Inner Product
Suppose $A$ is an $n \times n$ matrix.

How do I show that the inner product;

<A$\vec{u}$,$\vec{v}$> = <$\vec{u}$,A$^T$$\vec{v}$>

答:
Using the fact that the inner product $\langle u, v \rangle$ can be written as $u^Tv$, we have

$$\langle Au, v \rangle = (Au)^Tv = (u^TA^T)v = u^T(A^Tv) = \langle u, A^Tv \rangle$$.

Title: Proving Inner Product


问:
Compute principal invariants for an extension of amount $\lambda$ in direction $\boldsymbol{e}$
I'm trying to solve an exercise taken from Gurtin's book (Ex. 3 pg 52)

Let $f$ be an extension of amount $\lambda$ in direction $e$: $$f(p)=f(q)+U(p-q)$$
where $$U = I + (\lambda  -1) e \otimes e$$ with $\lambda >0$ and $p,q$ are material points.

Compute $C$ and the list of principal invariants $i_1(C),i_2(C),i_3(C)$, where $C$ is the right Cauchy Green strain tensor defined as $C=U^2$

First I compute $U^2$ explicitely:

$$U^2 =  (I + (\lambda  -1) e \otimes e ) (I + (\lambda  -1) e \otimes e ) = I + 2(\lambda -1) e \otimes e + (\lambda -1)^2 |e|^2 e \otimes e$$
I don't think I can simplify more, right?

To compute the invariants, I should use the fact that $C=U^2$ is symmetric and positive definite and hence the principal invariants are easy to compute because we only need the eigenvalues, but here I don't really know how I could move, because $e$ may be any direction, so it seems a really weird question.

EDIT
$$i_1(C)=\text{tr}(C)=\lambda^2$$

$$i_2(C)=\frac{1}{2}[\text{tr}(C)^2 - \text{tr(C^2)}]$$

Now $C^2 = U^2 U^2 = I+(\lambda^4 -1) e \otimes e$ and hence $$\text{tr}(C^2)=\text{tr}(U^4) = \lambda^4$$ so $$i_2(C)=\frac{1}{2}\lambda^2 (1 - \lambda^2 )$$

Finally, $$i_3(C)=\det(C)=\det(I + (\lambda -1) e \otimes e)$$

Now, $\det$ is the product of the eigenvalues, and we can see immediately that

$\lambda$ is an eigenvalue: $$(I-(\lambda -1) e \otimes e)e = \lambda e$$

all the other eigenvalues are $1$: take $v$ eigenvector, which is orthogonal to $e$ because $U$ is symmetric $$(I-(\lambda -1)e \otimes e)v = v - (\lambda-1) (e\otimes e) v=v$$ because $(e \otimes e )v = 0$

Therefore $$i_3(C) = \lambda 1= \lambda$$

答: We have $U^2 = I + (\lambda^2-1)e\otimes e$. So the characteristic polynomial is $$\begin{align} \det(tI - U^2) &= \det((t-1)I - (\lambda^2-1)e\otimes e) \\ &= (t-1)^3 \det\left(I - \frac{\lambda^2-1}{t-1}e\otimes e\right) \\ &= (t-1)^3\left(1 - {\rm tr}\left(\frac{\lambda^2-1}{t-1}e\otimes e\right)\right) \\ &= (t-1)^3 \left(1 - \frac{(\lambda^2-1)}{t-1} \right) \\ &= (t-1)^3 - (\lambda^2-1)(t-1)^2 \\ &{\color{red}{= (t-1)^2(t-\lambda^2)}} \\ &= (t^2-2t+1)(t-\lambda^2) \\ &= t^3 - (\lambda^2+2)t^2 + (2\lambda^2+1)t - (\lambda^2). \end{align}$$It seems that the invariants are $\lambda^2+2$, $2\lambda^2+1$ and $\lambda^2$.


答:
For $n\geq 9$, we have $\ln(n)>2$. Also, $\ln(n)<n$. Hence,
$$
1<\left(\ln(n)\right)^{2009}-1<\left(\ln(n)\right)^{2009}<n^{2009}\\
\Rightarrow1<\sqrt[n]{\left(\ln(n)\right)^{2009}-1}<n^{2009/n}=\left(n^{1/n}\right)^{2009}
$$
Using $\lim_{n\to\infty}n^{1/n}=1$ and the squeeze theorem, it follows that your limit is $1$. See here, for instance, about proofs of this limit.

Title: Need to calculate $\displaystyle{\lim_{n \to \infty}} \sqrt[n]{(\ln(n))^{2009}-1}$

Body:
Calculate:

$$\displaystyle{\lim_{n \to \infty}} \sqrt[n]{(\ln(n))^{2009}-1}$$

I thought about some trick like this one:

$$\displaystyle{\lim_{n \to \infty}} \sqrt[n]{(\ln(n))^{2009}-1} = \displaystyle{\lim_{n \to \infty}} \sqrt[n]{\left(\frac{e^{2009*\ln[\ln(n)]}-1}{2009*\ln[\ln(n)]}\right) * 2009*\ln[\ln(n)]}$$

But unfortunately it doesn't seem to lead anywhere.

It is important also to add that I can not use L-Hospitals rule.


答:
Here it is a proof based on probability and copulas theory for the sake of curiosity.

Let $C:[0,1]^{2}\to[0,1]$ be a copula.

Since it is nondecreasing in each coordinate and it has uniform margins, one has that
\begin{align*}
\begin{cases}
C(u,v) \leq C(u,1) = u\\\\
C(u,v) \leq C(1,v) = v
\end{cases} \Rightarrow C(u,v)\leq\min\{u,v\}
\end{align*}
Moreover, we do also know that $C$ is 2-increasing. Particularly, $V_{C}([u,1]\times[v,1])\geq 0$, that is to say,
\begin{align*}
C(1,1) - C(u,1) - C(1,v) + C(u,v) \geq 0 \Rightarrow C(u,v)\geq u + v - 1
\end{align*}
Once $C(u,v)\geq 0$, we conclude that $C(u,v)\geq\max\{u+v-1,0\}$.

Gathering both results, one concludes that
\begin{align*}
\max\{x + y - 1,0\} \leq C(x,y) \leq \min\{x,y\} \Rightarrow \max\{x + y - 1,0\} \leq \min\{x,y\}
\end{align*}

More generally, if $(X,Y)\sim H$, $X\sim F$ and $Y\sim G$, where $H(x,y) = C(F(x),G(y))$, due to the Sklar's theorem, we can conclude that
\begin{align*}
\max\{F(x) + G(y) - 1,0\} \leq H(x,y) \leq \min\{F(x),G(y)\}
\end{align*}
for some copula $C$, and we are done.

Such result is known as the Fréchet-Hoeffding inequality.

Hopefully this helps!

Body:
Which one is the larger value of $\min(x,y)$ and $\max(0,x+y-1)$, for all $x,y\in [0,1]$?

I'm trying as follows.

If $x\leq y$, then $$\min(x,y)=x$$

Now we have $x+y-1\leq 2y-1$.

Now I confused to conclude the value of
$$\max(0,x+y-1).$$

If $x> y$, then $$\min(x,y)=y$$

Now we have $x+y-1> 2y-1$.

Now I also confused to conclude the value of
$$\max(0,x+y-1).$$

So, can we compare the value of $\min(x,y)$ and $\max(0,x+y-1)$, for all $x,y\in [0,1]$?


问:
Finding the largest negative integral value of a in this inequality.
The inequality is :$$\sin^2x+a\cos x+a^2>1+\cos x$$

I have simplified it to:$$\cos^2x+(1-a)\cos x-a^2<0$$

My approach was that since this is in the form of a parabola the minimum value for x must be $\frac{-b}{2a}$ form so after applying conditions I get :
$\frac{a-1}{2}$ which should be equal to -1 since that is the minimum value of cos.

But I get answer -1 which is wrong, where have I flawed?

答:
Hint $$\cos x = 1 \implies 1+(1-a)-a^2<0 \implies 0 < a^2+a-2=(a-1)(a+2) \implies a<-2 \text{ or } a> 1.$$
Therefore $a$ can't be $-1$.


问:
Is $\varphi:x \mapsto A\cdot x$ an orthogonal projection for M
I got the transformation
$\varphi:x \mapsto A\cdot x$
and the matrix
$M = \begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix}$.

I have to check whether $\varphi$ is the orthogonal projection for $M$.

To check if it is a projection, I check if the Eigenvalues of $M$ are either $0$ or $1$.

I calculated the Eigenvalues and the corresponding Eigenvectors:

$\lambda_1 = 1, \enspace v_1= \begin{pmatrix} -1 \\ 1 \end{pmatrix},$

$\lambda_2 = 0, \enspace v_2= \begin{pmatrix} 0 \\ 1 \end{pmatrix},$

So it is a projection. But how do I check whether it is an orthogonal projection?

I received the hint to look for the Eigenvectors and the dot-product, but I don't get the connection. Can someone please enlighten me?

答:
If this linear map is really an orthogonal projection onto the line spanned by $(-1,1)$, then it should map vectors orthogonal to $(-1,1)$ to $0$. That is, the $0$-eigenspace should be spanned by a vector orthogonal to $(-1,1)$. Is your vector $v_2$ orthogonal to $v_1$?

You may not yet know about the spectral theorem, but if the matrix were representing an orthogonal projection, it would have to satisfy $A^2=A$ and $A=A^T$. Your matrix is not symmetric.


答:
(Fill in the gaps. If you're stuck, show your work and explain why you're stuck.)

Hint:

Your quadratic is in $ \cos x$, which is restricted to $ -1 \leq \cos x \leq 1$.

If you want to prove that a upward opening quadratic $f(x) $ on a restricted domain of $ [p, q]$ is $< 0$, then we just need to show that $f(p) < 0$ and $ f(q) < 0$.

In particular, why do we not need to check the turning point?

Hence, conclude that the largest negative integral values of $a$ is -3.

Title: Finding the largest negative integral value of a in this inequality.

Body:
The inequality is :$$\sin^2x+a\cos x+a^2>1+\cos x$$

I have simplified it to:$$\cos^2x+(1-a)\cos x-a^2<0$$

My approach was that since this is in the form of a parabola the minimum value for x must be $\frac{-b}{2a}$ form so after applying conditions I get :
$\frac{a-1}{2}$ which should be equal to -1 since that is the minimum value of cos.

But I get answer -1 which is wrong, where have I flawed?


问:
2nd order linear differential equation with non-constant coefficients
Considering the equation $2y''+(x+1)y'+3y=0$ where $X_0=2$. Find the general term in each solution. That is, the general term for Y1,Y2 where $y=A_0(Y_1)+A_1(Y_2)$

I've solved this as

$$y=A_0[1-3/4(x-2)+3/8(x-2)^3+1/64(x-2)^4+...] +A_1[(x-2)-3/4(x-2)^2+1/24(x-2)^3+9/64(x-2)^4+...]$$

I did this using the recurrence relationship:

$$A_{n+2}= -3A_{n+1}/2(n+2) - (n+3)A_n/2(n+1)(n+2)$$  where $n=1,2,3,...$

Furthermore, I confirmed Y1,Y2 as a fundamental set of solutions by verifying that $W(Y_1,Y_2)(X_0)$ does not equal zero.

I'm not sure how to find the general term of Y1 and Y2. I've read my book again and didn't find any information to clarify. I appreciate any help!

答:
Similar to Help on solving an apparently simple differential equation,

Let $y=\int_Ce^{xs}K(s)~ds$ ,

Then $2(\int_Ce^{xs}K(s)~ds)''+(x+1)(\int_Ce^{xs}K(s)~ds)'+3\int_Ce^{xs}K(s)~ds=0$

$2\int_Cs^2e^{xs}K(s)~ds+x\int_Cse^{xs}K(s)~ds+\int_Cse^{xs}K(s)~ds+\int_Ce^{xs}K(s)~ds=0$

$\int_C(2s^2+s+3)e^{xs}K(s)~ds+\int_C2se^{xs}K(s)~d(xs)=0$

$\int_C(2s^2+s+3)e^{xs}K(s)~ds+\int_C2sK(s)~d(e^{xs})=0$

$\int_C(2s^2+s+3)e^{xs}K(s)~ds+[se^{xs}K(s)]_C-\int_Ce^{xs}~d(sK(s))=0$

$\int_C(2s^2+s+3)e^{xs}K(s)~ds+[se^{xs}K(s)]_C-\int_Ce^{xs}(sK'(s)+K(s))~ds=0$

$[se^{xs}K(s)]_C-\int_Ce^{xs}(sK'(s)-(2s^2+s+2)K(s))~ds=0$

$\therefore sK'(s)-(2s^2+s+2)K(s)=0$

$sK'(s)=(2s^2+s+2)K(s)$

$\dfrac{K'(s)}{K(s)}=2s+1+\dfrac{2}{s}$

$\int\dfrac{K'(s)}{K(s)}ds=\int\left(2s+1+\dfrac{2}{s}\right)~ds$

$\ln K(s)=s^2+s+2\ln s+c_1$

$K(s)=cs^2e^{s^2+s}$

$\therefore y_c=\int_Ccs^2e^{s^2+(x+1)s}~ds$

But since the above procedure in fact suitable for any complex number $s$ ,

$\therefore y_n=\int_{a_n}^{b_n}c_n(m_nt)^2e^{(m_nt)^2+(x+1)m_nt}~d(m_nt)=m_n^3c_n\int_{a_n}^{b_n}t^2e^{m_n^2t^2+m_n(x+1)t}~dt$

For some $x$-independent real number choices of $a_n$ and $b_n$ and $x$-independent complex number choices of $m_n$ such that:

$\lim\limits_{t\to a_n}t^3e^{m_n^2t^2+m_n(x+1)t}=\lim\limits_{t\to b_n}t^3e^{m_n^2t^2+m_n(x+1)t}$

$\int_{a_n}^{b_n}t^2e^{m_n^2t^2+m_n(x+1)t}~dt$ converges

For $n=1$, the best choice is $a_1=0$ , $b_1=\infty$ , $m_1=\pm i$

$\therefore y=C_1\int_0^\infty t^2e^{-t^2}\cos((x+1)t)~dt$ or $C_1\int_0^\infty t^2e^{-t^2}\sin((x+1)t)~dt$

Hence $y=C_1\int_0^\infty t^2e^{-t^2}\sin((x+1)t)~dt+C_2\int_0^\infty t^2e^{-t^2}\cos((x+1)t)~dt$


问:
I don't think W/O replacement events are dependent.

Let say there're cards numbered 1 to 10. P(A) is the probability you pick a card whose number is the multiple of 2. P(B) is the probability you pick a card whose number is the multiple of 5. So now you draw two cards without replacement. You don't place the first card you just drew back in the original deck. Then, you draw again from that 9 remaining cards.

In the picture, P(A) is shown as the ordered pairs colored red, and P(B) is shown as the ordered pairs shaded with light blue.

Then I counted them all.

P(A) = (9+9+9+9+9) / (100-10) = 45/90 = 1/2

P(B) = (9+9) / (100-10) = 18/90 = 1/5

P(A∩B) = 9 / 90 = 1/10

So here, I found out that P(A∩B) is actually the multiple of P(A) and P(B), which shouldn't be if they are dependent, meaning they are independent. I've also found out, in a different model, that the probabilities could be dependent each other in the case of W/O replacement such as this one.

I know events with replacements are independent. But it feels like events without replacements doesn't mean dependency. It's more like the case of contraposition of the first one. Events with replacements mean independent, which means, by the contraposition rule, dependent events are the ones done without replacements.

But all the other internet pages say like "if they are done without replacement, then they are dependent". This claim is so prevalent. But one of us, they or I, must be wrong. What am I missing here?

答:
A=the first card is even;

B=the second card is 5 or 10.

$P(A)=\frac{1}{2}$ as you said, but $P(B)=P(B\cap A)+P(B\cap A^C)=P(A)P(B|A)+P(b\cap A^C)=\frac{1}{2}\frac{1}{5}+\frac{4}{5} \frac{2}{9}+\frac{1}{5} \frac{1}{9} =\frac{1}{10}+\frac{1}{5}=\frac{3}{10}$

I hope no errors were made.


问:
Find a limit $\lim_{x\rightarrow0}\frac{\int_{\cos(\sqrt{x})}^{1}\frac{\cos(t)}{1-t}dt}{\sqrt{x}}$
Find
$$\lim_{x\rightarrow0}\frac{\int_{\cos(\sqrt{x})}^{1}\frac{\cos(t)}{1-t}dt}{\sqrt{x}}$$
Substitute $y = \sqrt{x}$, by Mean Value Theorem for integrals, set $L = \lim_{y\rightarrow0}\frac{\frac{\cos(c)}{1-c}(1-\cos(y))}{y} = [\cos(y) \leq c \leq 1 \rightarrow 1- c \leq1-\cos(y)] = \lim_{y\rightarrow0}\frac{(\frac{\cos(c)\dot(1-c)}{1-c})}{y} \leq L$. We know that $0 < \cos(c) \leq \cos(1)$, thus by bounds $L \geq \lim_{y\rightarrow0}\frac{\cos(c)}{y} = +\infty$.

To me, the interval of integration is shrinking yet the denominator inside tends to approach $0$. How to solve it properly?

答:
If we use the fundamental theorem of calculus, we have that
$$\frac d {dx} \left(\int_{\cos(\sqrt{x})}^{1}\frac{\cos(t)}{1-t}\,dt\right)=\frac{\sin \left(\sqrt{x}\right) \cos \left(\cos \left(\sqrt{x}\right)\right)}{2
   \sqrt{x} \left(1-\cos \left(\sqrt{x}\right)\right)}$$ Now, using L'Hospital rule, we are looking for the limit of
$$\frac{\sin \left(\sqrt{x}\right) \cos \left(\cos
   \left(\sqrt{x}\right)\right)}{1-\cos \left(\sqrt{x}\right)}$$ Compsing Taylor series around $x=0$, we have
$$\frac{\sin \left(\sqrt{x}\right) \cos \left(\cos
   \left(\sqrt{x}\right)\right)}{1-\cos \left(\sqrt{x}\right)}=\frac{2 \cos (1)}{\sqrt{x}}+O\left(x^{1/2}\right)$$ which tends to $+\infty$.


问:
A question about $\sigma$-Algebra determine whether the diagonal it is measurable.
Let $X = Y$ be uncountable sets. $S = T$ the $\sigma$- algebra of all
sets $A$ such that $A$ or $A^c$

Is it countable. Let W be the $\sigma$-algebra defined on

$X × X$ where $E \in W$ if $E$ or $E^c$ is countable.

Determine what relationship is
meets between $S \times T$ and $W$. Who is larger? Determine whether the diagonal
it is measurable.

答:
I understand that by $S \times T$ you mean the product $\sigma$-algebra of $S$ with $T$.

Let us prove that $W \subsetneq S\times S$.

Let $E$ be any countable subset of $X \times X$. Then $E=\{(x_i,y_i) \in X \times X: i\in \mathbb{N}\}$. Note that, for all $i \in  \mathbb{N}$, $\{(x_i,y_i)\} = \{x_i\}\times \{y_i\}$. So $\{(x_i,y_i)\} \in S\times S$. So
$$ E=\bigcup_{i \in \mathbb{N}} \{(x_i,y_i)\} \in S\times S $$

Let $E$ be any set such that $E^c$ is countable. Then, by what we have just proved, $E^c \in S\times S $. So, $E \in S\times S $.

So if $E$ is any subset of $X \times X$ such that $E$ or $E^c$ is countable, then  $E \in S\times S $.

In other words, $W \subseteq S\times S$.

Now let us prove that $W \neq S\times S$. Let $a$ be any element of $X$. It is immediate that $$\{a\} \times (X\setminus \{a\}) \in  S\times S $$ Since $X$ is uncountable, we have that $\{a\} \times (X\setminus \{a\})$ is uncountable and $(\{a\} \times (X\setminus \{a\}))^c$ is also uncountable. So $$\{a\} \times (X\setminus \{a\}) \notin W $$

The diagonal:

Let $\Delta=\{(x,x) : x \in X\}$. Let us prove that $\Delta \notin S\times S$.

The simplest way to do it is to note that we can define a measure $\mu$ in $(X, S)$ by $\mu(E)=0$ if $E$ is countable and $\mu(E)=1$ if $E$ is co-countable. It is easy to check that $\mu$ is in fact a measure.

Now consider the product measure $\mu \times \mu$ and its outer measure $(\mu \times \mu)^*$. It is easy to see that $(\mu \times \mu)^*(\Delta)=1$ and $(\mu \times \mu)^*(\Delta^c)=1$. So we have
$$(\mu \times \mu)^*(X \times X)=1 < 2 = (\mu \times \mu)^*(\Delta) +  (\mu \times \mu)^*(\Delta^c)$$

So $\Delta$ is not measurable in the Carathéodory's extension sense. So $\Delta \notin S\times S$.

Remark: To see that $(\mu \times \mu)^*(\Delta)=1$, note that, if $\Delta$ is covered by a countable union of rectangles $A \times B$ (where $A, B \in S$), at least for one of such rectangles $A$ and $B$ must be co-countable.


问:
Showing that a transformation maps the exterior of a disk onto the exterior of an ellipse
I'm given a transformation T:

$u(x,y)=2x+\frac{2x}{x^2+y^2}$,  $v(x,y)=2y-\frac{2y}{x^2+y^2}$

And I'm asked to show that the transformation maps the exterior of the disk of radius 2 $D:=\{(x,y)|x^2+y^2\geq4\}$ onto the exterior of the ellipse $\{(u,v)|u^2/25+v^2/9\geq1\}$.

My first attempt (which I thought was all I needed to do) was to simply put $u^2$ and $v^2$ onto the ellipse to get the disk, but that did not work.

My question is: what am I missing? because I though the problem was pretty straightforward.

After that I need to compute the Jacobian and use it to calculate the area between two ellipses, but again, the problem seems a little odd since the transformation should make the integral easier to compute but I computed the Jacobian and it does not seems like it is helping.

I'm not asking for the solution, I just want to know why my attempt did not work and I would appreciate any hints or explanations about how to solve it, or any insights about transformations that maybe I'm not taking into account!

答:
Under the assumption that $x^2+y^2=4$, you should have been able to prove that $$\frac{1}{25}\left(2x+\frac{2x}{x^2+y^2}\right)^2 + \frac{1}{9}\left(2y-\frac{2y}{x^2+y^2} \right)^2 = 1,$$so possibly you made an algebra mistake. But this would have only showed that $T$ takes the circle into the ellipse. Note that $$(\Bbb R^2 \setminus \{(0,0)\}) \setminus \{(x,y): x^2+y^2 = 4\} \quad \mbox{and} \quad (\Bbb R^2\setminus \{(0,0)\}) \setminus \{ (u,v) : u^2/25 + v^2/9 =1\}$$each have two connected components, so $T$ has to map connected components to connected components, by continuity. Then to check that $D$ gets mapped to the exterior of the ellipse (as opposed to the punctured interior), it would suffice to test one point: for example, $T(3,0) = (20/3,0)$.

In any case, Wolfram Alpha gives that the expression expands to $$\frac{8 x^2}{25 (x^2 + y^2)} + \frac{4 x^2}{25 (x^2 + y^2)^2} - \frac{8 y^2}{9 (x^2 + y^2)} + \frac{4 y^2}{9 (x^2 + y^2)^2} + \frac{4 x^2}{25} + \frac{4 y^2}{9},$$so with $x^2+y^2=4$ this indeed gives $1$.

I'd like to offer another point of view, though, simpler and more enlightening: it might be interesting to set $\Bbb R^2 \cong \Bbb C$, so that $$T(z) = 2\left(z + \frac{1}{z}\right).$$If $z(t) = 2{\rm e}^{{\rm i}t}$ parametrizes the circle, then we have $$T(z(t)) = 2\left(2{\rm e}^{{\rm i}t} + \frac{1}{2}{\rm e}^{-{\rm i}t}\right) = 4{\rm e}^{{\rm i}t} + {\rm e}^{-{\rm i}t} = 5\cos t + 3{\rm i} \sin t.$$This is explictly a parametrization of the ellipse, in view of ${\rm e}^{{\rm i}t} = \cos t + {\rm i}\sin t$.


问:
Basic Reference material about ODEs such as saparability with calculations and examples?
I am trying to show this kind of non-linear $y''''=y'y''/(1+x)$ in normal form. For example here if $y=e^{x}\rightarrow y^{(n)}=e^{x}\rightarrow x=-1$, where $y^{(n)}$ means $n$th differential, then $x=-1$, too weak idea. When I google with differential or anything like that, most of the material does not look the material that I need. I need to solve different type of problems such as this homework

$$\begin{cases} u'=(u+v)^{2} \\ v''=x+u'v'  \end{cases} $$

I am not requesting you to solve them but I am requesting some material because I find my book quite hard-reading in this section. The earlier chapter begun that something is something, without much further ado really why?, and now the next advanced chapters are referring to the past chapters. The idea is a rush introduction to this topic in an engineering course so I think it explains quite a bit about the pedagogy.

答:
Hint:

Let $r=x+1$ ,

Then $\dfrac{d^4y}{dr^4}=\dfrac{1}{r}\dfrac{dy}{dr}\dfrac{d^2y}{dr^2}$

$r\dfrac{d^4y}{dr^4}=\dfrac{dy}{dr}\dfrac{d^2y}{dr^2}$

Let $u=\dfrac{dy}{dr}$ ,

Then $r\dfrac{d^3u}{dr^3}=u\dfrac{du}{dr}$

$\int r\dfrac{d^3u}{dr^3}~dr=\int u\dfrac{du}{dr}~dr$

$\int r~d\left(\dfrac{d^2u}{dr^2}\right)=\int u~du$

$r\dfrac{d^2u}{dr^2}-\int \dfrac{d^2y}{dr^2}~dr=\int u~du$

$r\dfrac{d^2u}{dr^2}-\dfrac{du}{dr}=\dfrac{u^2}{2}+C_1$


问:
Evaluate $\oint_{|z-1/2|=3/2} \frac{\tan(z)}{z}$ with residue theorem
Here is my attempt at the problem:

$\oint_{|z-1/2|=3/2} \frac{\tan(z)}{z}$

We have that $z=0$ is a removable singularity, so we may consider that $\tan(z)=\frac{\sin(z)}{\cos(z)}$. From here we have that $\frac{\tan(z)}{z} = \frac{\sin(z)}{z\cos(z)}$. This have a pole of order 1 at $z=\frac{\pi}{2}$. We only consider this residue because its the only residue in the circle we are integrating over. So by the residue theorem we have  $\oint_{|z-1/2|=3/2} \frac{\tan(z)}{z} = 2\pi i \operatorname{Res}[\frac{\sin(z)}{z\cos(z)}, \frac{\pi}{2}]$. We can evaluate this residue by the following $\operatorname{Res}[\frac{\sin(z)}{z\cos(z)}, \frac{\pi}{2}] = \lim_{z \to \pi/2}(z-\frac{\pi}{2})\frac{\sin(z)}{z\cos(z)}= \lim_{z \to \pi/2}(z-\frac{\pi}{2})\frac{\sin(z)}{z\cos(z)-z\cos(\frac{\pi}{2})+z\cos(\frac{\pi}{2})}$. If we pull out a $\frac{1}{z-\frac{\pi}{2}}$ on the bottom we have $\operatorname{Res}[\frac{\sin(z)}{z\cos(z)}, \frac{\pi}{2}]$ = $\lim_{z \to \pi/2}\frac{\sin(z)}{z\cos'(\frac{\pi}{2})} = \lim_{z \to \pi/2}\frac{\sin(z)}{-z\sin(\frac{\pi}{2})} = \frac{1}{-\frac{\pi}{2}} = -\frac{2}{\pi}.$ So by multiplying the $2\pi i$ by this we end up with a solution of $-4i$ but the integral shouldn't be negative, so where did I go wrong on here?

答:
$\text{ For simple calculations, you need this :}$

$\cos z=-(z-\pi/2)+(z-\pi/2)^3/3!-(z-\pi/2)^5/5!+......$

$\text{ so that }$

$Res(f(z), z=\pi/2)=\lim_{z\rightarrow \pi/2}(z-\pi/2)f(z)=\lim_{z\rightarrow\pi/2}(-1)\times\frac{\sin z}{z}=-\frac{2}{\pi}$


问:
The norm of convex function - an example.
Given the fuction $f: R^n \times R^n \longrightarrow R$ defined by $$f(x, y): = \frac{1}{2} \| x - y \|^2_2$$
Is $f$ a convex function?

My take:

I know norm functions are convex and so I will show that the Hessian is positive semi definite.

Let $\frac{1}{2} \| x - y \|^2_2 = \frac{1}{2} \| 1^\top x - 1^\top y \|^2_2$ so that

$\nabla h(x) = (1^\top x - 1^\top y)$ and

$\nabla^2 h(x) = 1^\top \succeq 0$

I am not sure whether my steps are legit. Any help will be apprciated.

Body:
I don't really know what you did there, but it should be convex, in fact:

for all $t\in [0,1]$ and since the real plane is convex,
$$f(t(x_1,y_1)+(1-t)(x_2,y_2))=f(tx_1+(1-t)x_2,ty_1+(1-t)y_2)=$$
$$= \frac{1}{2}||t(x_1-y_1)+(1-t)(x_2-y_2)||_2^2 \le$$
$$\le \frac{1}{2}||t(x_1-y_1)||_2^2 + ||(1-t)(x_2-y_2)||_2^2=$$
$$= \frac{1}{2}t^2||x_1-y_1||_2^2 + \frac{1}{2}(1-t)^2||x_2-y_2||_2^2\le$$
and since $0\le t,(1-t)\le 1$ we obtain the definition of convexity of a function:
$$\le \frac{1}{2}t||x_1-y_1||_2^2 + \frac{1}{2}(1-t)||x_2-y_2||_2^2=$$
$$= tf(x_1,y_1) + (1-t)f(x_2,y_2)$$
Hope there are no mistakes.


问:
Application of "Riesz Interpolation" for $L^p$-Norm of a Fourier-Transformed Function
I am asking this question to try to decipher notes given by my professor.

We have a function, $f: \mathbb{R}^2 \rightarrow \mathbb{R} $, and we are trying to bound its $L^p$-Norm, where $p>2$. This function has a nice Fourier Transform, so we would like to bound the norm of $f$ by the norm of $\mathcal{F}[f] = \hat{f}$ somehow.

According to my professor's scrawlings, we can say that, for $p>2$,

$ ||f||_{p} \leq C ||\hat{f}||_{\frac{p}{p-1}} $, by "Riesz Interpolation".

I am unfortunately not able to figure what he means by this, or how this can be proven. He writes that it somehow comes from the fact that the $L^\infty$- and $L^2$-norms can be bounded in the following ways:

$||f||_{\infty} \leq C ||\hat{f}||_{1}, \ \ ||f||_{2} = ||\hat{f}||_{2}$.

I understand these two facts well enough, but am not seeing how we get the more general result above.

答:
Your professor is referring to the Riesz-Thorin interpolation theorem. It basically says that if a linear operator $T$ is bounded as a map $L^{p_1}\to L^{q_1}$ and also as a map $L^{p_2}\to L^{q_2}$, then it is also bounded as a map $L^{p_\theta}\to L^{q_\theta}$ for
$$
\frac{1}{p_\theta} = \frac{\theta}{p_1} + \frac{1-\theta}{p_2}, \frac{1}{q_\theta} = \frac{\theta}{q_1} + \frac{1-\theta}{q_2}, ~\theta\in[0,1].
$$
The full statement also gives you a bound on the operator norm. The idea is to apply this in the specific case that $T$ is the Fourier transform, using the two endpoint bounds you have already stated.

The resulting estimate is called the Hausdorff-Young inequality.


问:
Calculus Number question

I got this question from a friend who was trying to solve this, and I am currently having trouble with the proof. I know that $t_n$ is the fractional part of the sum of the sequence and the sequence diverges. Now, I'm trying to find a proof that there are infinitely many sums $t_i$ which when $l$ subtracted from $t_i$ gives $\pm\epsilon$. So how can I proceed in this proof?

答: All of the points between 0 and 1 will be accumulation points. Instead of writing out a formal proof I'll just note that because the series diverges one keeps "re-starting" in the modulo 1 arithmetic, and because the terms converge to 0 one moves up in smaller and smaller steps so that all points between 0 and 1 will be revisited arbitrarily closely arbitrarily many times.</p>

Title: Calculus Number question


问:
Skolem Functions vs Elementary Equivalence
I am just curious about something. I think this is trivially true, but I wanted to be sure.

Suppose we have two elementary equivalent $\mathcal{L}$-structures $\mathfrak{A}$ and $\mathfrak{B}$. Also, suppose $\text{Th}(\mathfrak{A})$ has built-in Skolem functions, namely, for all $\mathcal{L}$-formulas $\phi(v,w_1,\cdots,w_n)$ there is a function symbol $f$ such that $\text{Th}(\mathfrak{A})\vDash \forall \bar{w} ((\exists v \phi(v,\bar{w}))\rightarrow \phi(f(\bar{w}),\bar{w}))$. Can we say the same for $\mathfrak{B}$, in this case?

答: Yes. Specifically, by elementary equivalence the same function symbols serve as Skolem functions in $\mathfrak{B}$: "$\forall\overline{w}((\exists v\phi(v,\overline{w}))\rightarrow\phi(f(\overline{w}),\overline{w}))$" is a first-order sentence, so if it's true in $\mathfrak{A}$ it's true in $\mathfrak{B}$.


问:
Does a non-fraction number have a reciprocal?
As far as I understand from this Wikipedia page, $\frac 73$ is a reciprocal of $\frac 37$;

But can a term "reciprocal" be applied to a non-fraction number?

For example, it would be correct to say:

reciprocal of $\frac 21$ is $\frac 12$

But would it also be correct to say:

reciprocal of 2 is $\frac 12$

?

答: Yes.  For instance, $\dfrac1{\pi}$ is the reciprocal of $\pi$.  $\dfrac1r$ is the reciprocal of $r$ for any nonzero $r\in \Bbb R$.


问:
An attempt at proving that $A=(0,1)$ is not compact on the real line with the usual topology.
I am supposed to show that the open interval $(0,1)$ on the real line $\mathbb{R}$ (with the usual topology) is not compact. I know that one of the examples of an open cover without a finite subcover, in this case, would be $A=\left\{\left(\dfrac{1}{n},1\right)\bigg\rvert n\in\mathbb{N}\right\}$, thus proving that $A$ is not compact.

However, I am trying to prove the same using a different example of an open subcover, $$\mathscr{F}=\left\{\left(\dfrac{1}{2^n},\dfrac{3}{2^n}\right)\bigg\rvert n\in\mathbb{N}\right\}.$$ If it can be shown that the family of sets $\mathscr{F}$ forms an open cover for $(0,1)$, i.e., $\bigcup\limits_{n\in\mathbb{N}}G_n$, where $G_n=\left(\dfrac{1}{2^n},\dfrac{3}{2^n}\right)$, it is straightforward to check that for any $x=\dfrac{1}{2^n}$, where $n\in\mathbb{N}$, it only lies in one of these $G_n$'s. So, removing any one of these open sets from our open cover, in an attempt to find a finite subcover, the family of sets won't form a finite subcover of $(0,1)$. Thus, $(0,1)$ is not compact.

Where I am getting stuck is in actually formally proving that any $x\in(0,1)$ lies in one of these $G_n$'s. My intuition here is that by the use of Archimedean principle, for any $x\in(0,1)$, there exists some $k\in\mathbb{N}$ such that $x>\dfrac{1}{k}$. So, let us choose $k$, for a given $x$, such that $\dfrac{1}{k}<x<\dfrac{1}{k-1}$. Again, since this $k>1$, there exists some $n_0\in\mathbb{N}$ such that $2^{n_0-1}<k<2^{n_0}\implies \dfrac{1}{2^{n_0}}<\dfrac{1}{k}<\dfrac{1}{2^{n_0-1}}$. My claim is that $x\in \left(\dfrac{1}{2^{n_0}},\dfrac{3}{2^{n_0}}\right)$, and approach was to try and prove that $\left(\dfrac{1}{k},\dfrac{1}{k-1}\right)\subseteq \left(\dfrac{1}{2^{n_0}},\dfrac{3}{2^{n_0}}\right)$, but I am unable to proceed ahead from this point.

Any help would be highly appreciated, in either proving or disproving the claim that I make in the last sentence.

答:
I would note that $G_1=(1/2,3/2)$ which covers $(1/2,1)$, and then noting that
$$\frac{3}{2^{n+1}} \ge \frac{1}{2^n}  \iff \frac{3}{2} \ge 1  \mbox{, for all }n\in \mathbb{N}$$
which is true.
This gives us that two consecutive intervals have non-empty intersection so that
$$\bigcup_{n\le N}G_{n}=\left(\frac{1}{2^N},\frac{3}{2}\right) \mbox{, for all } N\in \mathbb{N}$$
Now if we have $x\in (0,1)$ this means $x >0$ and $\frac{1}{2^n}$ converges to $0$ so by definition if we get $\epsilon =x$, there exists $N_0$ such that $\forall n\ge N_0$, $\frac{1}{2^n}<x$.

So we have that $x\in \bigcup_{n\le N_0}G_{n}$ (finite union) $\Longrightarrow$ there exists an interval $G_k$ with $k\le N_0$.

Hope I made no mistakes.


答: Let $0<x<1$. Consider the interval $(\frac {\ln (\frac 1  x)} {\ln 2},\frac {\ln (\frac 3  x)} {\ln 2})$. The length of this interval is easily seen to be $\frac {\ln 3} {\ln 2}$. Since any interval of length greater than $1$ contains an integer there exist a positive integer $n$ in this interval. You can now verify that $x \in (\frac 1 {2^{n}},\frac 3 {2^{n}})$. Note that since $0 <x<1$ and $x \in (\frac 1 {2^{n}},\frac 3 {2^{n}})$ the integer $n$ is necessarily a positive integer.

Title: An attempt at proving that $A=(0,1)$ is not compact on the real line with the usual topology.

Body:
I am supposed to show that the open interval $(0,1)$ on the real line $\mathbb{R}$ (with the usual topology) is not compact. I know that one of the examples of an open cover without a finite subcover, in this case, would be $A=\left\{\left(\dfrac{1}{n},1\right)\bigg\rvert n\in\mathbb{N}\right\}$, thus proving that $A$ is not compact.

However, I am trying to prove the same using a different example of an open subcover, $$\mathscr{F}=\left\{\left(\dfrac{1}{2^n},\dfrac{3}{2^n}\right)\bigg\rvert n\in\mathbb{N}\right\}.$$ If it can be shown that the family of sets $\mathscr{F}$ forms an open cover for $(0,1)$, i.e., $\bigcup\limits_{n\in\mathbb{N}}G_n$, where $G_n=\left(\dfrac{1}{2^n},\dfrac{3}{2^n}\right)$, it is straightforward to check that for any $x=\dfrac{1}{2^n}$, where $n\in\mathbb{N}$, it only lies in one of these $G_n$'s. So, removing any one of these open sets from our open cover, in an attempt to find a finite subcover, the family of sets won't form a finite subcover of $(0,1)$. Thus, $(0,1)$ is not compact.

Where I am getting stuck is in actually formally proving that any $x\in(0,1)$ lies in one of these $G_n$'s. My intuition here is that by the use of Archimedean principle, for any $x\in(0,1)$, there exists some $k\in\mathbb{N}$ such that $x>\dfrac{1}{k}$. So, let us choose $k$, for a given $x$, such that $\dfrac{1}{k}<x<\dfrac{1}{k-1}$. Again, since this $k>1$, there exists some $n_0\in\mathbb{N}$ such that $2^{n_0-1}<k<2^{n_0}\implies \dfrac{1}{2^{n_0}}<\dfrac{1}{k}<\dfrac{1}{2^{n_0-1}}$. My claim is that $x\in \left(\dfrac{1}{2^{n_0}},\dfrac{3}{2^{n_0}}\right)$, and approach was to try and prove that $\left(\dfrac{1}{k},\dfrac{1}{k-1}\right)\subseteq \left(\dfrac{1}{2^{n_0}},\dfrac{3}{2^{n_0}}\right)$, but I am unable to proceed ahead from this point.

Any help would be highly appreciated, in either proving or disproving the claim that I make in the last sentence.


答:
You need to show those intervals "continuously overlap".  That is for any $k \in \mathbb N$ we have  $\frac 1{2^{k+1}} < \frac 1{2^k} < \frac 3{2^{k+1}} <  \frac 3{2^k}$.  That follows as $\frac 12 < 1 < \frac 32 < 3$.

This means $\cup (\frac 1{2^n},\frac 3{2^n}) = (\lim_{n\to \infty} \frac 1{2^n}, \frac 3{2^1}) = (0, \frac 32)$ and so any $x \in (0,1) \in \cup (\frac 1{2^n},\frac 3{2^n})$

However if you can't use limits in that manner, consider:

Let $x \in (0,1)$.  Let $K = \{n\in \mathbb N| \frac 1{2^n} < x\}$.  $K$ is non-empty. (Proof:  Let $n > \log_2 \frac 1x$, then $n \in K$.)

Now the well ordering principle of natural numbers says $K$, being a subset of $\mathbb{N}$, must have a least element $k = \min K; k \in K$ and $k-1 \not \in K$.  If $k= 1$ are done $\frac 12 < x < 1 < \frac 32$ so $x \in (\frac 12, \frac 32)$.  If $k > 1$ then $k\in K\implies \frac 1{2^k} < x$ and $k-1 \not \in K\implies x \le \frac 1{2^{k-1}} < \frac 3{2^k}$.  So $x \in (\frac 1{2^k}, \frac 3{2^k})$.


问:
Distribution of $\tau/x^2$ where $\tau$ is first hitting time
I am considering the hitting time $\tau=\inf\{t:|B_t|\geq x\}$. I am told to consider the martingale $\cosh(\lambda B_t)\textrm{e}^{-\frac12\lambda^2t}$. I know that by letting $\lambda=\sqrt{2\theta}$, we can prove that the moment generating function of $\tau$ is $\mathbb{E}(\textrm{e}^{-\theta\tau})=\frac1{\cosh\sqrt{2\theta}x}$. The follow-up question from this is to show that $\tau/x^2$ therefore has a distribution independent of $x$, but I am not sure how to go about this. Am I supposed to be differentiating the MGF? All help is appreciated, thanks!

答: By your equation, $$E(e^{-\theta(\tau/x^2)}) = E(e^{-(\theta/x^2)\tau})= \frac{1}{\cosh(\sqrt{2\theta/x^2}x)}=\frac{1}{\cosh(\sqrt{2\theta})},$$ which is independent of $x.$


问:
Find infimum of set A: $A=\{\frac{m^4 +2n^2}{2m^2 -m^2n + n^2}: m, n \in \mathbb{N}\}$
Find infimum of set A:

$$A=\{\frac{m^4 +2n^2}{2m^2 -m^2n + n^2}: m, n \in \mathbb{N}\}$$

So, I know that supremum is in $\infty$ for $ \displaystyle  \lim_{m \to +\infty}$ and $n=1$.

I know that infimum $=\frac{3}{2}$ and is obtained when $n=1$, $m=1$, but I don't know how to actually calculate that. It seams that the expression is not monotonous, so I wouldn't seek try to express it using one variable. On the other hand, there are only 2 variables no it is unlikely that it is inequality of means

答:
if $n=3$ we obtain the sequence
$$\biggl{\{}\frac{m^4+18}{-m^2+9}\biggr{\}}_{m\in \mathbb{N}}$$
and this diverges at $-\infty$ for $m\rightarrow  +\infty$


问:
Writing trace of a linear operator in terms of inner products
This fact may be obvious, but say that I have a linear operator $T: V \to V$, where $V$ is finite-dimensional, and an orthonormal basis $\{e_1, \ldots, e_n\}$ for $V$. I want to write an expression for the trace of $T$, which is supposed to take the form
\begin{align*}
\text{trace}(T) = \langle Te_1, e_1 \rangle + \ldots + \langle Te_n, e_n\}.
\end{align*}
I am trying to understand why this holds in general. If this were the standard basis for $\mathbb{R}^n$ and we were using the standard dot product, $\langle Te_i, e_i \rangle$ would give the $i$th entry of the $i$th column, and summing would therefore sum the diagonal of the matrix. But that doesn't generalize to any basis and any inner product.

答:
Orthonormal expansion says that $$x = \sum_{k=1}^n \langle x,e_k\rangle e_k$$for all $x \in V$, provided the basis $\mathcal{B} = (e_k)_{k=1}^n$ is orthonormal (proof: write $x = \sum_{i=1} x_ie_i$ for some coefficients $x_i$ and hit both sides with $\langle \cdot, e_k\rangle$ to obtain $x_k = \langle x,e_k\rangle$). That said, you should recall how the trace is defined:

pick any basis $\mathcal{B}$ for $V$.

write each $T(e_j)$ as a combination $\sum_{i=1}^k a_{ij} e_i$, so that you have a matrix $[T]_{\mathcal{B}} = (a_{ij})_{i,j=1}^n$.

define ${\rm tr}(T) = {\rm tr}[T]_{\mathcal{B}}$, where in the right we have a matrix trace.

show that the definition above does not depend on the choice of $\mathcal{B}$, by using that the matrix trace is invariant under conjugation.

Now you just want to exploit the fact that the chosen $\mathcal{B}$ is orthonormal. Orthonormal expansion gives $$T(e_j) = \sum_{i=1}^n \langle T(e_j),e_i\rangle e_i.$$So $[T]_{\mathcal{B}} = (\langle T(e_j),e_i\rangle)_{i,j=1}^n$. Then $${\rm tr}(T) = \sum_{i=1}^n \langle T(e_i),e_i\rangle.$$

If the basis is not orthonormal, one needs to write $g_{ij} = \langle e_i,e_j\rangle$ (and $(g_{ij})_{i,j=1}^n$ is no longer necessarily the identity matrix). Then $$x = \sum_{i=1}^n x_i e_i \implies \langle x,e_j\rangle = \sum_{i=1}^n x_i g_{ij} \implies x_i = \sum_{j=1}^n g^{ij}\langle x,e_j\rangle,$$where $(g^{ij})_{i,j=1}^n$ is the inverse matrix of $(g_{ij})_{i,j=1}^n$. So in general we'll have $${\rm tr}(T) = \sum_{i,j=1}^n g^{ij}\langle T(e_i),e_j\rangle.$$This is not a mistake: the trace is given in terms of the inner product by a double sum, which now accounts for the fact that the basis may be non-orthonormal.


问:
Maximum-sum path down a triangle of numbers
We have a triangle of $n$ positive and integer number. we start from top or (head) of this triangle and in each step we are going to adjacent number in next row. goal is finding path of maximum  consecutive number.

example: selected number in following figure has length at most $23$.

the best algorithm for finding this path has order of $O(n^2)$.

It's so not very sensible for me how we can find the path (what is the fastest logical way of solving this puzzle)?
$$\begin{array}
0&&&(3)\\
&&(7)&&4\\
&2&&(4)&&6\\
8&&5&&(9)&&3\end{array}$$

答:
The algorithm to find the largest-sum path in $O(n^2)$ time is as follows:

for each number in the second-last row $c$, look at the two numbers below it, $l$ and $r$. Replace $c$ with $c+l$ or $c+r$, whichever is larger, and note the direction that must be followed to achieve the larger value ($L$ or $R$ correspondingly)

recurse in the smaller triangle, prepending to the strings of $L$ and $R$ from previous steps

eventually you are left with one number, which is the largest possible sum, and the path to achieve it

$$\begin{array}
0&&&3\\
&&7&&4\\
&2&&4&&6\\
8&&5&&9&&3\end{array}\to
\begin{array}
0&&3\\
&7&&4\\
10L&&13R&&15L\end{array}$$
$$\begin{array}
0&&3\\
&7&&4\\
10L&&13R&&15L\end{array}\to
\begin{array}
0&3\\
20RR&&19RL\end{array}\to23LRR$$
This works because we are going to pass through the second-last row regardless of what path we take, and in that row the choice we want to make (to maximise the sum) is clear. It is a kind of dynamic programming.


问:
Proof by Lewis Carroll that an urn cannot contain two balls
I am self-learning probability theory and solving some problems from the book Probability and Random processes by Grimmett and Stirzaker. Exercise 1.4.4 poses a nice problem. I would like someone to verify, if my solution is correct and thought process is correct.

What do you think of the following 'proof' by Lewis Carroll that an urn cannot contain two balls of the same color? Suppose that the urn contains two balls, each of which is either black or white. In the obvious notation $P(BB)=P(BW)=P(WB)=P(WW)=\frac{1}{4}$. We add a black ball, so that $P(BBB)=P(BBW)=P(BWB)=P(BWW)=\frac{1}{4}$. Next, we pick a ball at random, the chance that the ball is black is (using conditional probabilities)

$$1 \cdot \frac{1}{4}+\frac{2}{3}\cdot \frac{1}{4} + \frac{2}{3}\cdot \frac{1}{4} + \frac{1}{3}\cdot \frac{1}{4}=\frac{2}{3}$$

However, if there is a probability $\frac{2}{3}$ that a ball, chosen randomly from three, is black, then there must be two black and one white, which is to say that originally there was one black and one white ball in the urn.

Solution.

I think, the fallacy in this proof has to do with mistaking the unconditional probability $P(BBB)$ for the conditional probabilities $P(BBB|\text{atleast one ball is black})=\frac{1}{4}$.

\begin{align*}
P(\text{Ball drawn at random is black}) = 1\cdot\frac{1}{8} + \frac{3}{8}\cdot\frac{2}{3} + \frac{3}{8}\cdot \frac{1}{3} + \frac{1}{8}\cdot 0 = \frac{1}{2}
\end{align*}

This makes perfect intuitive sense, because without any knowledge of what ball was added to the urn, the unconditional probability that a ball drawn at random is black is simply $\frac{1}{2}$; the ball is equally likely to be black or white.

Surely, the urn can have two balls of the same color! I'm not sure, how to argue this. But, what I can tell for sure is,

\begin{align*}
P(BBB|\text{atleast one ball is black}) = P(BWW |\text{atleast one ball is black})= \frac{1}{4}
\end{align*}

答:
This is equivalent to this problem.  Suppose there are two families each with two children one has two girls, the other has two boys.  If you pick child at random the probability is $\frac 12$ it's a girl.  But the probability is $\frac 12$ then that means that family had to have a boy and a girl.  Do you see the error there?

There is no "that family".  There are two possible families with different probabilities.  We can't assume a global probability for any possible family will determine anything about the family we eventually pick.  That's bait and switch.

So the urn problem is exactly.  Of the many possible urns we get the probability of drawing a black ball is $\frac 23$.  But that doesn't say anything about the condition of one particular urn arrangement which we had no guarentee of being "the" urn.


问:
How to determine the order of the poles of a complex function using its Laurent series?
I'm trying to find the poles and determine the orders of the complex function
$$f(z)=\frac{1}{(z+2i)^2}-\frac{z}{z-i}+\frac{z+1}{(z-i)^2}$$
I think that $z=-2i$ is a pole of order 2 and $z=i$ is also a pole of order 2, but I'm not sure how to show this. I know that if $f(z)$ has a pole at $z=z_0$ of order $m$ then the principal part of its Laurent series is finite: $$\frac{b_m}{(z-z_0)^m}+...+\frac{b_1}{z-z_0}$$ So, I tried to expand $f(z)$ in two Laurent series, one centered at $z_0=-2i$ and other at $z_0=i$.

For that I wrote $$-\frac{z}{z-i}+\frac{z+1}{(z-i)^2}=-1-\frac{i}{z-i}+\frac{z-i+i+1}{(z-i)^2}=-1-\frac{i}{z-i}+\frac{1}{z-i}+\frac{i+1}{(z-i)^2}$$ $$=-1+\frac{1-i}{z-i}+\frac{i+1}{(z-i)^2}$$

For the Laurent series of $f(z)$ centered at $z=-2i$, the first term is already part of this series, right? So I just need to express the sum of the second and third terms as the series centered at $-2i$. Those terms are analytic at $-2i$, right? So they can be expressed as a Taylor series only, so $\frac{1}{(z+2i)^2}$ would be the principal part of the Laurent series, making $z=-2i$ a pole of order 2.

For the Laurent series of $f(z)$ centered at $z=i$ the sum of the second and third terms are already part of this series. So I need to write the first term as a series centered at $i$, but this term is analytic at that point so it would be a Taylor series, making the sum of the last two terms the principal part of the Laurent series, making $z=i$ a pole of order 2.

Am I on the right path? How do I express the sum of the last two terms a series centered at $z=-2i$ and the first term as a series centered at $z=i$?

答: It is not necessary to find the exact Lauernt expansion: To find then order of  pole at $i$, for example, you just observe that $\frac  1 {(z+2i)^{2}}$ is analytic in some disk around  $i$ so it has  a power series expansion of the type $\sum\limits_{n=0}^{\infty} a_n (z-i)^{n}$ in that  disk. This gives the nature of the Laurent series expansion around $i$ and you can can infer that the oder of the pole is $2$.


问:
Find all pairs of primes $p, q$ such that 3$p^2q+16pq^2$ equals to square of natural number
Find all pairs of prime numbers $p, q$ such that 3$p^2q+16pq^2$ equals to square of natural number

My attempt: I've been trying to calculate equation through square root but now I'm stuck, please help

答:
After factoring as $pq(3p+16q)=n^2$
We claim that p divides n and q divides n since both p and q divide $n^2$. Thus $n=k.pq$ for some kind belongs to natural numbers.

$3p+16q=k^2 pq 3/q+16/p=k^2$

from here we obtain p=2,q=3 and p and q are primes there are no more solutions.


答:
It is a bit of work to grind through, but one can show
$\operatorname{tr} ( ABC) = \operatorname{tr} ( CAB)$ (cyclic property).
Hence for any invertible $V$ we have
$\operatorname{tr} (V^{-1} TV ) = \operatorname{tr} (V V^{-1} T ) = \operatorname{tr} T$.

If $e_k$ is the standard basis, we have $\operatorname{tr} T = \sum_k \langle T e_k, e_k \rangle$. Let $v_k$ be an orthonormal basis and $V = [v_1 \cdots v_n ]$.

Then $\operatorname{tr} T = \operatorname{tr} (V^T TV) = \sum_k \langle V^TTV e_k, e_k \rangle = \sum_k \langle TV e_k,Ve_k \rangle = \sum_k \langle T v_k, v_k \rangle$.

Title: Writing trace of a linear operator in terms of inner products

Body:
This fact may be obvious, but say that I have a linear operator $T: V \to V$, where $V$ is finite-dimensional, and an orthonormal basis $\{e_1, \ldots, e_n\}$ for $V$. I want to write an expression for the trace of $T$, which is supposed to take the form
\begin{align*}
\text{trace}(T) = \langle Te_1, e_1 \rangle + \ldots + \langle Te_n, e_n\}.
\end{align*}
I am trying to understand why this holds in general. If this were the standard basis for $\mathbb{R}^n$ and we were using the standard dot product, $\langle Te_i, e_i \rangle$ would give the $i$th entry of the $i$th column, and summing would therefore sum the diagonal of the matrix. But that doesn't generalize to any basis and any inner product.


问:
Evaluating Complex Line Integrals
Calculate $\int_{\gamma}\frac{\Re(z)}{z-\frac{1}{2}}dz$ and $\int_{\gamma}\frac{\Im(z)}{z-\frac{1}{2}}dz$ when $\gamma$: $|z|=1$ is positively oriented.

This is what I have tried to do, starting with the first line integral.

Since $\frac{\Re(z)}{z-\frac{1}{2}}$ is not analytical/holomorphic at any point in the plane - it does not satisfy the Cauchy-Riemann equations - we cannot use Cauchy's integral formulae immediately. However, $|z|=1$ implies that the conjugate to $z$, $z*=\frac{1}{z}$, because $1=|z|^{2}=zz*$. Thus, we may write:
 \begin{align*} \int_{\gamma}\frac{\Re(z)}{z-\frac{1}{2}}dz &=\frac{1}{2}\int_{\gamma}\frac{z+\frac{1}{z}}{z-\frac{1}{2}}dz \\ &=\frac{1}{2}\int_{\gamma}\frac{z^{2}+1}{z(z-\frac{1}{2})}dz \\
 &=\frac{1}{2}\int_{\gamma}(1-\frac{\frac{z}{2}+1}{z(z-\frac{1}{2})}dz \\ &=\frac{1}{2}\int_{\gamma}dz-\frac{1}{2}\int_{\gamma}\frac{\frac{z}{2}+1}{z(z-\frac{1}{2})}dz  \\ &=-\frac{1}{2}\int_{\gamma}\frac{-2}{z}+\frac{\frac{5z}{2}}{z-\frac{1}{2}}dz  \\ &= \int_{\gamma}\frac{1}{z}dz-\frac{5}{4}\int_{\gamma}\frac{z}{z-\frac{1}{2}}dz \\ &=2\pi{i}(1-\frac{5}{8}) \\ &=\frac{3\pi{i}}{4} \end{align*}

Note that I rewrote the right hand side two times, using partial fractions. Similarly,I have attempted to the second line integral, however, this time a only used partial fractions once.

\begin{align*} \int_{\gamma}\frac{\Im(z)}{z-\frac{1}{2}}dz &= \frac{1}{2i}\int_{\gamma}\frac{z-\frac{1}{z}}{z-\frac{1}{2}}dz \\ &=\frac{1}{2i}\int_{\gamma}\frac{z^{2}-1}{z(z-\frac{1}{2})}dz \\ &=\frac{1}{2i}\int_{\gamma}(1-\frac{2}{z}+\frac{3}{z-\frac{1}{2}})dz \\ &=\frac{1}{2i}\int_{\gamma}dz-\frac{1}{2i}\int_{\gamma}\frac{-2}{z}dz-\frac{1}{2i}\int_{\gamma}\frac{3}{z-\frac{1}{2}}dz \\ &= \frac{1}{i}\int_{\gamma}\frac{1}{z}dz-\frac{3}{4i}\int_{\gamma}\frac{1}{z-\frac{1}{2}}dz \\ &=2\pi{i}(\frac{1}{i}-\frac{3}{4i}) \\ &=\frac{5\pi}{4}\end{align*}

As I have not studied complex analysis, I am unsure whether any of these two answers are correct. I would very much appreciate if someone more experienced person could help me check, most likely correct, my calculations.

答:
You have some problem on partial fraction $\displaystyle \frac{z^2 + 1}{z(z-1/2)} \neq  1 - \frac{\frac z 2 + 1}{z(z-1/2)}$.

This should be $\displaystyle \frac{z^2 + 1}{z(z-1/2)} = 1 + \frac{\frac z 2 + 1}{z(z-1/2)} = 1+\frac{5/2}{z - 1/2} - \frac 2 z$.

And this should give you $\displaystyle \frac{2 \pi i}{2} ( \frac 52 - 2) = \pi/2$

On the other,
$\displaystyle \frac{z^2 - 1}{z(z-1/2)} = 1 + \frac 2 z - \frac{3/2}{z - 1/2}$ while you have written $\displaystyle \frac{z^2 - 1}{z(z-1/2)} = (1-\frac{2}{z}+\frac{3}{z-\frac{1}{2}})dz $

Also that on here
$$=\frac{1}{2i}\int_{\gamma}(1-\frac{2}{z}+\frac{3}{z-\frac{1}{2}})dz \\
=\frac{1}{2i}\int_{\gamma}dz-\underbrace{\frac{1}{2i}\int_{\gamma}\frac{-2}{z}dz-\frac{1}{2i}\int_{\gamma}\frac{3}{z-\frac{1}{2}}dz}_{\text{why - here ??}} \\
 $$
This is wrong step.

Title: Evaluating Complex Line Integrals


问:
Path connected sets (How to intutively find which set is path connected before attempting a formal proof )
I have been preparing for an objective masters exam and  general topology is in the syllabus. The instructor in undergrad course was not interested in teaching and so I self studied and managed to do average level understanding of topology.

Problem arises in path connected set . Proving a set  path connected by definition is not easy and questions are often asked in exam whether a set is path connected or not? But rigorious proof is not asked as I have to just mark the correct options. So, I am asking for if there is some intution .  Consider that  I am given a set and  I have to see whether it's path connected or not , then I can use that intutive method which is less rigorious and time saving also.

If yes , can you tell me by using it on this question : Let A be the following subset of $\mathbb{R}^2$ : A={$(x,y)$:${(x+1)}^2+y^2 \leq  1$} $\bigcup $ {$(x,y): y = x sin(1/x), x>0$}. (This was 1 of the question asked in that exam).

I will apply your method/explanation/intution to other asked question for practice.

I shall be really thankful for any help received!

答:
Here's a very vague intuition for it: Being connected means that you can't pull apart the space into several pieces. Being path-connected means that you can actually travel from any one point to any other.

You can see this quite easily with the topologist's sine curve; the two path-connected components can't be pulled apart without changing the topology, but there's no way to move continuously from one to the other. Similarly, with the one-point compactification of the long line, the point at the end can't be pulled away, but it's "too far away" to actually reach from anywhere else on the long line.

It's worth mentioning that manifolds are connected if and only if they are path connected. So any distinguishing example is going to be somewhat not intuitive.


问:
what is the meaning of this symbol $ ∨?$
what is the meaning of this symbol $ ∨?$

My attempt : last time  i was asked this  same problem  with different pattern Here

Earlier  i was thinking that   $ ∨$ similar to union symbolics
But union symbolics is $\cup$

So here im confuse

答:
The most common use of this symbol is as logical operator "or", which connects two statements.

So for two statements $A$ and $B$ the expression $A\vee B$ would read "A or B".

As many other symbols this has other uses too, so it depends on the context. You linked a set-theory related topic.

The other symbol "$\wedge$" is the logical "and".

Edit: Also note that $\cup$ is a set-theoretic operation, while $\vee$ is used for statements. So in two different settings.

You define the set theoretic $\cup$ of two sets $X$ and $Y$ as the set

$X\cup Y:=\{x: x\in X\vee x\in Y\}$

It is a preference of style if you use those symbols, or just write "or", "and" instead.
Outside of the field of mathematical logic, most people avoid these symbols, as they are somewhat unpleasant to read.


问:
Finding a bijection that is not its inverse
Let $S$ be the set of binary strings of length $n$ i.e. each $x\in S$ is a string of zeroes and ones. Let $W(x)$ for $x\in S$ be the Hamming weight of $x$ i.e. number of ones in $x$. I am looking for a bijection $f:S\to S$ which

is NOT its inverse

"mixes" strings of a given weight to different weight classes. So this bijection doesn't work: Shift all strings of a weight class by a constant number because this would map strings of a given weight to strings of the same weight.

答:
Think of a binary string of length n as a (row) vector of n components. That is, regard "110101" as the vector $(1,1,0,1,0,1)$. Now take an $n\times n$ matrix $A$ consisting of 0's and 1's whose determinant is an odd number.

Using this we can construct a bijection $f:S\to S$ by the following method:
$ f(x) = xA\pmod 2$ here $xA$ means multiplying the row vector $x$ by the matrix $A$ getting a new row vector, and then mod 2 means replacing all ODD numbers in the result by 1 and evens by 0, getting a binary string. This will be a bijection on $S$.
(In algebra jargon, this is an invertible linear transformation on a finite dimensional vector space over the field of 2 elements).

The choice of matrices $A$ gives you many bijections


问:
When is an element of a free module over a principal ideal domain contained in a basis?
I'm trying to show the following:

Let $R$ be a principal ideal domain and let $M$ be free $R$-module of rank $n$. Let $Y=\{y_1,\ldots,y_n\}$ be a basis of $M$ and $x\in M$ with $x=y_1a_1+\cdots+y_na_n$. Then $x$ is contained in a base of $M$ if and only if $\gcd(a_1,\ldots,a_n)=1$.

I have shown that if $x$ belongs to a basis of $M$, then $\gcd(a_1,\ldots,a_n)=1$.

This is my idea for the other direction:

If $n=1$, then $a_1\in R^*$, and we are done. Now suppose $n\geq 2$. Since $\gcd(a_1,\ldots,a_n)=1$, there are $s_1,\ldots,s_n\in R$ such that $a_1s_1+\cdots+a_ns_n=1$, then using this equality I tried to get some non-zero $y\in M$ such that $\langle x\rangle\cap \langle y\rangle=0$. Supposing the existence of such $y$,  using Zorn's lemma you can get a maximal nonzero submodule $N$ such that $\langle x\rangle\cap N=0$, then I showed that $\langle x\rangle\oplus N=M$, and since $M$ is free and $R$ a principal ideal domain $N$ is free and has a basis $\{x_1,\ldots,x_m\}$, and thus $\{x,x_1,\ldots,x_m\}$ would be a basis of $M$.

My question is: is there such an $y$? Also, I would like to know whether this idea is too complicated, and if so, I want to know alternative approaches to it.

Thanks

答:
Let $R$ be a PID $M$ a free $R$-module of finite rank $n$. Let $\{x_1,\dots,x_k\}$ be $k\leq n$ elements in $M$, and denote $N$ their $R$-span in $M$. Then the claim is that the $k$-ple $\{x_1,\dots,x_k\}$ can be completed to a basis of $M$ over $R$ if and only if
$$
rk(N)=k\qquad\text{and}\qquad\text{$M/N$ is torsion-free.}
$$
Indeed, if $\{x_1,\dots,x_k,x_{k+1},\dots,x_n\}$ is an $R$-basis for $M$, then $M=N\oplus N^\prime$ where $N^\prime$ is the submodule of $M$ generated by$\{x_{k+1},\dots,x_n\}$ and $M/N\simeq N^\prime$. Thus $rk(N)=k$ because $rk(N^\prime)\leq n-k$ and $rk(N)+rk(N^\prime)=n$, and $M/N$ is free because it is isomorphic to a submodule of a free module.

Viceversa, consider the quotient map
$$
\pi:M\longrightarrow M/N.
$$
Let $\{z_1,\dots,z_r\}$ be a basis of $M/N$. Choose $y_i\in M$ such that $\pi(y_i)=z_i$ and let $N^\prime\subset M$ the submodule generated by the $y_i$'s. Since the $z_i$ are $R$-linearly independent so are the $y_i$, i.e. they form a basis of $N^\prime$.
For any $m\in M$ write
$$
\pi(m)=a_1z_1+\cdots+a_rz_r,\qquad a_i\in R.
$$
Then $m-\sum_{i=1}^ra_iy_i\in\ker(\pi)=N$, from which follows readily that $M=N\oplus N^\prime$ and thus $\{x_1,\dots,x_k,y_1,\dots,y_r\}$ is a basis for $M$ containing $\{x_1,\dots,x_k\}$. The claim is proved.

Now, let $x\in M$, $x\neq0$, and let $\{y_1,..,y_n\}$ be a basis of $M$. Then
$x=a_1y_1+\dots+a_ny_n$ for some $a_i\in R$. Let $d=\operatorname{gcd}(a_1,\dots,a_n)$.

If $d\neq1$, the element $y=\frac1dx\in M$ gives a torsion element in $M/Rx$.
On the other hand, suppose that $d=1$ and $z\in M/Rx$ is such that $rz=0$ for some non-zero $r\in R$. Choose $y\in M$ such that $\pi(y)=z$ and write $y=\sum_{i=1}^nb_iy_i$. Then $ry=sx$ for some $s\in R$, i.e.
$$
rb_i=sa_i,\qquad\text{for all $i=1,\dots,n$}
$$
Thus $r\operatorname{gcd}(b_i)=s$, i.e. $r$ divides $s$ in $R$. Therefore $y=\frac sr x\in Rx$ and $\pi(y)=z=0$, thus proving that $M/Rx$ is torsion-free.


问:
How i can find the fourier transform of $\frac{\sinh(ax)}{\sinh(\pi x)}$ where,$ |a| < \pi$
Using a rectangular contour in the complex plane, bypassing the poles at $z=0$ and $z=i$, i got $$\int_{-\infty}^{+\infty}\frac{\sinh(ax)e^{ikx}}{\sinh(\pi x)}dx - e^{-k}\int_{-\infty}^{+\infty}\frac{\sinh[a(x+i)]e^{ikx}}{\sinh(\pi x)}dx = \sin(a)e^{-k}$$

and now?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{-\infty}^{\infty}
{\sinh\pars{ax}\expo{\ic kx} \over \sinh\pars{\pi x}}\,\dd x
\,\right\vert_{\,a\ \in\ \pars{-\pi,\pi}}} =
2\,\Re\int_{0}^{\infty}
{\sinh\pars{ax}\expo{\ic kx} \over \sinh\pars{\pi x}}\,\dd x
\\[5mm] \stackrel{2\pi x\ \mapsto\ x}{=}\,\,\, &\
{1 \over \pi}\,\Re\int_{0}^{\infty}
{\sinh\pars{\alpha x}\expo{\ic \kappa x} \over \sinh\pars{x/2}}\,\dd x\quad\mbox{where}\quad
\left\{\begin{array}{rcl}
\ds{\alpha} & \ds{\equiv} & \ds{a \over 2\pi}
\\[1mm]
\ds{\kappa} & \ds{\equiv} & \ds{k \over 2\pi}
\end{array}\right.
\end{align}
Then,
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{-\infty}^{\infty}
{\sinh\pars{ax}\expo{\ic kx} \over \sinh\pars{\pi x}}\,\dd x
\,\right\vert_{\,a\ \in\ \pars{-\pi,\pi}}}
\\[5mm] = &\
{1 \over \pi}\,\Re\int_{0}^{\infty}
{\expo{-\pars{-\alpha - \ic\kappa}x} -
\expo{-\pars{\alpha - \ic\kappa}x}
\phantom{A}\over \expo{x/2} - \expo{-x/2}}\,\dd x
\\[5mm] = &\
{1 \over \pi}\,\Re\int_{0}^{\infty}
{\expo{-\pars{-\alpha - \ic\kappa + 1/2}x}\,\,\, -
\expo{-\pars{\alpha - \ic\kappa + 1/2}x}
\phantom{A}\over 1 - \expo{-x}}\,\dd x
\\[5mm] = &\
{1 \over \pi}\,\Re\int_{0}^{\infty}
{\expo{-x} - \expo{-\pars{\alpha - \ic\kappa + 1/2}x}
\phantom{A}\over 1 - \expo{-x}}\,\dd x
\\[2mm] - &\
{1 \over \pi}\,\Re\int_{0}^{\infty}
{\expo{-x} - \expo{-\pars{-\alpha - \ic\kappa + 1/2}x}
\phantom{A}\over 1 - \expo{-x}}\,\dd x
\\[5mm] = &\
{1 \over \pi}\,\Re\bracks{%
\Psi\pars{{1 \over 2} + \alpha - \ic\kappa} -
\Psi\pars{{1 \over 2} - \alpha - \ic\kappa}}
\end{align}
See $\ds{\color{black}{\bf 6.3.22}}$ in A & S Table

In addition,
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{-\infty}^{\infty}
{\sinh\pars{ax}\expo{\ic kx} \over \sinh\pars{\pi x}}\,\dd x
\,\right\vert_{\,a\ \in\ \pars{-\pi,\pi}}}
\\[5mm] = &\
{1 \over 2\pi}\bracks{%
\Psi\pars{{1 \over 2} + \alpha - \ic\kappa} -
\Psi\pars{{1 \over 2} - \alpha + \ic\kappa}}
\\[2mm] - &\
{1 \over 2\pi}\bracks{%
\Psi\pars{{1 \over 2} - \alpha - \ic\kappa} -
\Psi\pars{{1 \over 2} + \alpha + \ic\kappa}}
\end{align}

With Euler Reflection Formula
( see ${\color{black}{\bf 6.3.7}}$ in A & S Table ):
\begin{align}
&\bbox[5px,#ffd]{\left.\int_{-\infty}^{\infty}
{\sinh\pars{ax}\expo{\ic kx} \over \sinh\pars{\pi x}}\,\dd x
\,\right\vert_{\,a\ \in\ \pars{-\pi,\pi}}}
\\[5mm] = &\
{1 \over 2}\braces{%
\cot\pars{\pi\bracks{{1 \over 2} - \alpha + \ic\kappa}} -
\cot\pars{\pi\bracks{{1 \over 2} + \alpha + \ic\kappa}}}
\\[5mm] = &\
{1 \over 2}\bracks{%
\tan\pars{{a \over 2} - \ic\,{k \over 2}} +
\tan\pars{{a \over 2} + \ic\,{k \over 2}}}
\\[5mm] = &\
\bbx{\sin\pars{a} \over \cosh\pars{k} + \cos\pars{a}}
\\ &
\end{align}


问:
Some problem in proving a martingale
This is the original question:

Let $U$ be uniform on $[0, 1],$ and consider a sequence of random variables $X_1, X_2, \ldots$ which are i.i.d. with Bernoulli($U$) distribution given $U.$ Let $S_n = X_1+\cdots+X_n$. Show that $M_n = \frac{S_n}{n}$ is a martingale with respect to the filtration $\mathcal{F}_n = \sigma(X_1, \ldots , X_n)$.

I am experiencing calculating problem in my proof, which goes as follows:

First we have:
$$E\left[\frac{S_{n+1}}{n+1} \mid \mathcal{F}_n \right] = E\left[\frac{S_n+X_{n+1}}{n+1} \mid \mathcal{F}_n\right] = \frac{S_n}{n+1}+\frac{1}{n+1}E[X_{n+1}\mid\mathcal{F}_n]$$

Then since $\mathcal{F}_n$ is the sigma field generated by a set of 0-1 random variable, then $$E[X_{n+1}\mid\mathcal{F}_n](\omega)=\sum_{(x_1,\ldots,x_n)\in\{0,1\}^n}c_{(x_1,\ldots,x_n)}1(X_1=x_1,\ldots,X_n=n)(\omega)$$
where $c_{(x_1,\ldots,x_n)}=E[X_{n+1}\mid X_n =x_n, \ldots, X_1=x_1]$.

Then
\begin{equation}
\begin{aligned}
& E[X_{n+1}\mid X_n = x_n,\ldots,X_1=x_1] \\[8pt]
= {} & P(X_{n+1}=1\mid X_n=x_n,\ldots,X_1=x_1)\\[8pt]
= {} & \frac{P(X_{n+1}=1,X_n=x_n,\ldots,X_1=x_1)}{P(X_n=x_n,\ldots,X_1=x_1)}\\[8pt]
= {} & \frac{\int_0^1 P(X_{n+1}=1,X_n=x_n,\ldots,X_1=x_1\mid u)\,du}{\int_0^1 P(X_n=x_n,\ldots,X_1=x_1\mid u)\,du}\\[8pt]
= {} & \frac{\int_0^1 P(X_{n+1}=1\mid u)P(X_n=x_n\mid u) \cdots P(X_1=x_1\mid u)\,du}{\int_0^1 P(X_n=x_n\mid u) \cdots P(X_1=x_1\mid u) \, du}\\[8pt]
= {} &\frac{1}{n+2}\cdot\frac{(x_1+\cdots+x_n+1)!(n-(x_1 + \cdots + x_n))!}{(x_1+\cdots+x_n)!(n-(x_1+\cdots+x_n))!}\\[8pt]
= {} & \frac{x_1+\cdots+x_n+1}{n+2}
\end{aligned}
\end{equation}
So $$E[X_{n+1}\mid\mathcal{F}_n](\omega) = \sum_{k=0}^n c_k1(S_n=k)(\omega) = \sum_{k=0}^n \frac{k+1}{n+2}1(S_n=k)(\omega) = \frac{S_n+1}{n+2}(\omega)$$
So $$E[M_{n+1}\mid\mathcal{F}_n] = S_n\left(\frac{n+3}{(n+1)(n+2)}\right) + \frac{1}{(n+1)(n+2)}$$ which is apparently not $S_n$ on the right hand side.

I just wonder what is wrong with my calculation.

答:
If $\left( \frac{S_n} n \right)_{n=1}^\infty$ is a martingale then $\Pr\left( \frac{S_2} 2 =1 \mid \frac{S_1} 1 = 1 \right) = 1.$

But clearly $\Pr\left( \frac{S_2} 2 = \frac 1 2 \mid \frac{S_1} 1 = 1 \right) >0.$

Title: Some problem in proving a martingale


问:
Being the first to pick 1 of 2 cards out of a deck of 52 that will win you 1 million dollars.
Contestants in a game show are asked to from a line.
One by one, each of them will be given one card (face up) from a deck of 52 until someone gets either an Ace of Spades or an Ace of Clubs.
The first person who gets one of those cards will receive a sum of 1 million dollars.

Before the game begins, you have the chance of choosing where in the line to position yourself in.
How will you find out where to which position will be the one most likely to be the first to receive one of the 2 black Ace cards.

My first thought was to do a mock version of this and count how many cards were laid out before the first black Ace appeared. I repeated this for 25 times while recording the results for each trial to hopefully get an estimate to the average number of cards distributed before one of the black Aces appeared.

答:
The first-in-line is the winner with probability $\frac2{52}$

The second-in-line is the winner with probability $\frac{50}{52}×\frac2{51}$

In general, the $n$th-in-line is the winner with probability
$$f(n)=\frac2{53-n}\prod_{k=1}^{n-1}\frac{51-k}{53-k}$$

However, $f(n)$ is strictly decreasing because (as can be shown by manipulating the definition)
$$f(n+1)=\frac{51-n}{52-n}f(n)$$
Hence you should ask to be first-in-line to have the highest probability of winning.


问:
$\min(a,b+c)\leq \min(a,b)+ \min(a,c)$ for $a,b,c\geq 0$
Given $a,b,c\geq 0$, then this following is true
$$\min(a,b+c)\leq \min(a,b)+ \min(a,c)$$

I tried to prove it by using some cases, i.e.

Case 1: if $\,\,a\leq b \leq c\,\,$ and $\,\,a\leq c \leq b,\,\,$ it's clear that
\begin{align*}
\min(a,b+c)&\leq \min(a,b)+\min(a,c)\\
a &\leq a+a=2a
\end{align*}
Case 2: if $\,\,b\leq a\leq c\,\,$ and $\,\,c\leq a \leq b,\,\,$ we have

\begin{align*}
\min(a,b+c)&\leq \min(a,b)+\min(a,c)\\
a &\leq b+a
\end{align*}
and
\begin{align*}
\min(a,b+c)&\leq \min(a,b)+\min(a,c)\\
a &\leq a+c
\end{align*}
Case 3: $\,\,c\leq b\leq a\,\,$ and $\,\,b\leq c \leq a,\,\,$

I am not sure about this case, because although $a$ is the biggest from $b$ and $c$, it can satisfies $a\leq b+c$ or $b+c\leq a$. Is the third case must be done with two cases again? i.e for $a\leq b+c$ and $b+c\leq a$?

or there is another way to prove $\min(a,b+c)\leq \min(a,b)+ \min(a,c)$ for $a,b,c\geq 0$?

Any help will be appreciated. Thanks.

答: For your case 3: $\min\{a,b\} + \min\{a,c\}  = b+c$. Since $\min\{a,b+c\}\leq b+c$ (min is always less than both numbers) you get the inequality.


问:
Continued fraction expansion related to exponential generating function
A recent SciComp.SE Question motivates us to ask for a nice continued fraction expansion of the following Maclaurin series:

$$ f(x) = \sum_{n=0}^\infty \frac{B_n\; x^{n+3}}{n! (n+3)} = \int_0^x \frac{t^3}{e^t - 1} dt $$

where $B_n$ are the (first) Bernoulli numbers.

Although the positive power of $t$ in the numerator of the integrand removes the singularity at the origin, the exponential function has complex period $2\pi i$.  Therefore the radius of convergence for the above series is $2\pi$, the distance from the origin to the nearest singularities.

Where power series convergence is always limited by the proximity of singularities, continued fraction expansions typically have larger domains of convergence, perhaps including the whole real line.  Such a circumstance weighs in favor of rational over polynomial approximations over a widened real interval, and nice continued fractions often give apt starting values for a minimax rational approximation.

答:
Background:

It is evident from the integral form that the function $f(x)$ will have a triple root at $x=0$, that it is monotone increasing, and (due to the integrand's denominator growing exponentially) finitely bounded above (by exactly $\pi^4/15$ as it turns out).

Here are a few fairly precise function values that illustrate this trend:

$$\begin{array}{|r|c|}
\hline \text{x} & \text{f(x) (shown rounded)} \\ \hline
1.0 & 0.22480518802594 \\ \hline
2.0 & 1.17634259660700 \\ \hline
3.0 & 2.55221845329080 \\ \hline
4.0 & 3.87705416153119 \\ \hline
5.0 & 4.89989215833058 \\ \hline
6.0 & 5.58585538083094 \\ \hline
7.0 & 6.00316896121307 \\ \hline
8.0 & 6.23962379489192 \\ \hline
9.0 & 6.36657389887547 \\ \hline
10.0 & 6.43192189678183 \\ \hline
\end{array}$$

It was also evident in from the Question itself that the function has a power series expansion about the origin, but that the radius of convergence is only $2\pi$ owing to the nearest singularities (poles).  Since these singularities "accumulate" at infinity in the complex plane, it cannot be expected that a convergent power series representation can be centered "at infinity", i.e. a Laurent series.

Nevertheless it may prove desirable to investigate that in more detail, after first exploring the possibility of converting the Maclaurin series into a continued fraction.  Below the third-order Debye function being considered is graphed alongside its truncated Maclaurin series polynomials of order 7, 15, and 23 for visual consistency:

Fig. 1 Debye function of 3rd order and its Maclaurin polynomials

Notice the diminishing returns near the threshold radius of convergence, that the improvement in approximation going from order 15 to order 23 is slight in comparison with the earlier improvement from order 7 to order 15.  As nonconstant polynomials inevitably veer off to plus or minus infinity, it is natural to hope for better convergence from rational expressions (which allow for the horizontal asymptote seen here).

Graphic produced using gnuplot interface from Maxima; I'll post the Maxima code as a GitHub gist once I've polished it...


问:
Probability of coin heads
Four fair coins are flipped. If the outcomes are assumed independent, what is the probability that two heads and two tails are obtained?

I calculated the answer as $\frac{11}{16}$ via using brute-force computations; is there any logical approach for this?

答: The correct answer is $\frac6{16}$. In general, for $n$ fair coins and obtaining $k$ heads and $n-k$ tails the probability is $\binom nk/2^n$.

Title: Probability of coin heads


问:
Prove that the prime $p$ can only be $13$
Given that $p$ is a prime such that both $\frac{p-1}{4}$ and $\frac{p+1}{2}$ are also primes.Then prove that $p=13$.
My try:
Let $p_1,p_2$ be primes such that
$$\frac{p-1}{4}=p_1$$ and
$$\frac{p+1}{2}=p_2$$
So we get,
$$p=4p_1+1=2p_2-1$$
Now if i start keeping values ofcourse i am getting $p_1=3,p_2=7,p=13$ as the only prime triplets. But is there a formal way to prove $13$ is the only value of $p$.

答:
You have

$$\begin{equation}\begin{aligned}
4p_1 + 1 & = 2p_2 - 1 \\
4p_1 & = 2p_2 - 2 \\
2p_1 & = p_2 - 1 \\
p_2 & = 2p_1 + 1
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$

With $p_1$, consider its possible values modulo $3$. If it's $p_1 \equiv 1 \pmod{3}$, then $p_2 \equiv 0 \pmod{3}$, which is not allowed since $p_2 \gt 3$. Alternately, if $p_1 \equiv 2 \pmod{3}$, then $p_2 \equiv 2 \pmod{3}$ so $p = 2p_2 - 1 \implies p \equiv 0 \pmod{3}$. The only case where this may be possible is where $p_2 = 2$ giving $p = 3$, but then $p = 4p_1 + 1$ can't hold. This leaves the only possible case where $p_1$, $p_2$ and $p$ are all prime is where $p_1 = 3$, leading to your one case where $p = 13$.


答:
The number $X$ of Heads in four independent tosses of fair coins has $X \sim \mathsf{Binom}(n=4, p=1/2).$ You seek
$P(X = 2) = {4 \choose 2}(1/2)^4 = 6/16 =  0.375
 \ne 11/16.$

In R statistical software, where dbinom is a binomial PDF:

```<code>dbinom(2, 4, .5)
[1] 0.375
</code>```

By 'brute force' listing:

```<code>TTTT
TTTH, TTHT, THTT, HTTT
TTHH, THTH, HTTH, HHTT, HTHT, THHT
HHHT, HHTH, HTHH, THHH
HHHH
</code>```

You want the third row.

Title: Probability of coin heads

Body:
Four fair coins are flipped. If the outcomes are assumed independent, what is the probability that two heads and two tails are obtained?

I calculated the answer as $\frac{11}{16}$ via using brute-force computations; is there any logical approach for this?

Body:
The number $X$ of Heads in four independent tosses of fair coins has $X \sim \mathsf{Binom}(n=4, p=1/2).$ You seek
$P(X = 2) = {4 \choose 2}(1/2)^4 = 6/16 =  0.375
 \ne 11/16.$

In R statistical software, where dbinom is a binomial PDF:

dbinom(2, 4, .5)
[1] 0.375

By 'brute force' listing:

TTTT
TTTH, TTHT, THTT, HTTT
TTHH, THTH, HTTH, HHTT, HTHT, THHT
HHHT, HHTH, HTHH, THHH
HHHH

You want the third row.


问:
Scaling of objective function in optimization problems.
In a lot of optimization algorithms I have seen that they use a scaling factor to scale the objective function. I dont understand what is the reason behind this. Why do we need scaling of objective function in optimization algorithms? Does it work without scaling. Logically it should work but I am a litte bit confused now.

I hope you can answer me and thank you indeed

答:
Beyond the numerical efficiency perspective there is also an impact on the model.

If your feature variables are scaled then your objective function needs to be scaled, otherwise you would need disproportionately large coefficients to get to the same target value.

Scaling on power law distributed objective functions also give skewed results.
These objective functions require sublinear transformations.

From Skiena's Data Science Design Manual p276:

"Small-scale variables need small-scale targets, in order to be realized using small-scale
coefficients. Trying to predict GDP from Z-scored variables will require
enormously large coefficients. How else could you get to trillions of dollars from
a linear combination of variables ranging from -3 to +3?"


答:
Suppose $p\equiv1\bmod3$, then it is easy to verify that $\frac{p-1}4\equiv0\bmod3$, so $\frac{p-1}4=3$ and $p=13$.

Suppose $p\equiv2\bmod3$, then by similar logic $\frac{p+1}2\equiv0\bmod3$ and $p=7$, but then $\frac{p-1}4$ is non-integral.

Since $p>3$ by $\frac{p-1}4$ being prime, $p=13$.

Title: Prove that the prime $p$ can only be $13$

Body:
Given that $p$ is a prime such that both $\frac{p-1}{4}$ and $\frac{p+1}{2}$ are also primes.Then prove that $p=13$.
My try:
Let $p_1,p_2$ be primes such that
$$\frac{p-1}{4}=p_1$$ and
$$\frac{p+1}{2}=p_2$$
So we get,
$$p=4p_1+1=2p_2-1$$
Now if i start keeping values ofcourse i am getting $p_1=3,p_2=7,p=13$ as the only prime triplets. But is there a formal way to prove $13$ is the only value of $p$.


问:
$\lim_{x \to \infty} \Gamma(x+1)/\Gamma(x+1+1/x^2)$
Can someone please help me with the calculation of this limit?

$$\lim_{x \to \infty} \frac{\Gamma(x+1)}{\Gamma(x+1+1/x^2)}$$

I tried wolframalpha and seems to be 1, yet there are no "detailed steps" as to how it reaches the conclusion (I understand there wouldn't be even if I pay, as in other cases it shows me the first ones).

Thanks in advance,
Sergio

Body:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\lim_{x \to \infty}\,\,{\Gamma\pars{x + 1} \over \Gamma\pars{x + 1 + 1/x^{2}}}} =
\lim_{x \to \infty}\,\,{x! \over \pars{x + 1/x^{2}}!}
\\[5mm] = &\
\lim_{x \to \infty}\,\,{\root{2\pi}x^{x + 1/2}\,\,\expo{-x} \over \root{2\pi}\pars{x + 1/x^{2}}^{x\ +\ 1/x^{2}\ + 1/2}
\,\,\,\,\expo{-x - 1/x^{2}}\,\,}
\\[5mm] = &\
\lim_{x \to \infty}\,\,{x^{x + 1/2} \over
x^{x\ +\ 1/x^{2}\ + 1/2}\,\,\,
\bracks{\pars{1 + 1/x^{3}}^{x^{3}}}^{1/x^{2}\ +\ 1/x^{5}\ + 1/\pars{2x^{3}}}
\,\,\,\,}\,\expo{1/x^{2}}
\\[5mm] = &
\lim_{x \to \infty}x^{-1/x^{2}} =
\exp\pars{-\lim_{x \to \infty}{\ln\pars{x} \over x^{2}}} =
\exp\pars{-\lim_{x \to \infty}{1/x \over 2x}}
\\[5mm] = &
\bbx{1} \\ &
\end{align}


答:
You would have
$$P=\frac{\mbox{good cases}}{\mbox{total cases}}=\frac{\frac{4!}{2!2!}}{\sum_{i=0}^4\frac{4!}{i!(4-i)!}}=\frac{6}{16}$$

Title: Probability of coin heads


答: Think of it this way.  If there were only one winning card, it would have equal probability to be in each place in the deck, and (equivalently) each person in line would be equally likely to receive the winning card.  With multiple winning cards, it’s still the case that each person in line is equally likely to receive a winning card; but to win, you need to receive a winning card and be the first to do so.  This is most likely if you’re first in line, since the second condition is then guaranteed.

Title: Being the first to pick 1 of 2 cards out of a deck of 52 that will win you 1 million dollars.

Body:
Contestants in a game show are asked to from a line.
One by one, each of them will be given one card (face up) from a deck of 52 until someone gets either an Ace of Spades or an Ace of Clubs.
The first person who gets one of those cards will receive a sum of 1 million dollars.

Before the game begins, you have the chance of choosing where in the line to position yourself in.
How will you find out where to which position will be the one most likely to be the first to receive one of the 2 black Ace cards.

My first thought was to do a mock version of this and count how many cards were laid out before the first black Ace appeared. I repeated this for 25 times while recording the results for each trial to hopefully get an estimate to the average number of cards distributed before one of the black Aces appeared.


问:
Proving two functions are equivalent given equal integrals without fundamental theorem of calculus
I want to prove that given two functions $f$ and $g$ that are continuous on the interval $[a,b]$, $\int_a^x f(t) \, dt=\int_a^x g(t) \, dt \; \forall x \in [a,b]$ implies that $f=g$ on the interval without using the fundamental theorem of calculus or Lebesgue integration. The conclusion seems relatively intuitive, but I don't think claiming intuition and "matching integrands" is rigorous enough. I initially tried to use Cauchy sums to prove this result, but I got stuck after realizing that Cauchy sums are primarily used to prove that functions are Riemann integrable, whereas in this question, we want to prove that the two functions $f$ and $g$ are equivalent. Any guidance would be appreciated!

答: Note that $\int_c^{d} f(x)dx=\int_c^{d} g(x)dx$ if $a \leq c <d \leq b$. Suppose $f(s) >g(s)$ for some $s$. Let $r=f(s)-g(s)$. Then, by continuity,  we can find an interval $(c,d)$ such that $f(t)>g(t)+\frac  r 2$ for all $t$ in $(c,d)$. This implies that $\int_c^{d} f(x)dx> \int_c^{d} g(x)dx +\frac {r(d-c)} 2$, a contradiction.


问:
Finding a matrix of an odd determinant
I am trying to find a square matrix of size $n$ consisting of zeros and ones whose determinant is an odd number. Any ideas?

答:
All identity matrices have determinant $1$.

More generally, all permutation matrices have determinant $\pm1$, and remember that adding a multiple of any row to any other row does not change the determinant. This should already give you many, many examples.

Another example (not in the instances described above) of a zero-one matrix with odd determinant is a $2n×2n$ matrix with zeros on the diagonal and ones elswehere – this has determinant $1-2n$.

Finally, the most general way of constructing a zero-one matrix with odd determinant is to take any integral matrix with odd determinant and reduce modulo $2$ (odds to $1$, evens to $0$), since $\mathbb Z_2$ is a field.

Title: Finding a matrix of an odd determinant


问:
Using Triple Integrals to Compute Total Mass
The question I have is as follows:
A rectangular swimming pool with vertical sides that measures 50ft by 150ft has a depth that increases in linearly from the short end (3 feet deep) to the deep end (15 feet deep).  This is assuming that water weights 62.4 pounds per cubic foot. I'm trying to set up an integral to compute this and I'm just getting stuck on the z coordinate aspect of the triple integral.  would it just be bounded by 3<=z<=12?</p>

答:
Suppose that the $x$-axis runs parallel to the long side, with the origin at a corner where the depth is $3$. Then the depth $z$ at the point with coordinates $(x,y)$ is given by $z=3+\frac{12x}{150}$ (here $0\le x\le 150$, $0\le y\le 50$).

If we want to find the volume using a triple integral, then we are integrating $dz \,dx\,dy$, where  $z$ runs from $0$ to $3+\frac{12x}{150}$, then $x$ runs from $0$ to $150$, and finally  $y$ runs from $0$ to $50$. For the weight, integrate $62.4\,dz\,dx\,dy$ over the same region.

Remark: It is perhaps simpler to set up the problem using a double integral, or a one variable integration.  Most simply, we can  use basic geometry. A cross-section perpendicular to the $y$-axis is a trapezoid whose area is easy to calculate. Then we multiply by $50$.


问:
Solution to $y′=x+y,\, y(0)=1$
I am trying to learn numerical method for differentials. It is key for me to understand this stuff right now, but I've come across an obstacle.

The books I read about these all seem to often say 'exact solution is easily found to be' in more or less the same way. I clearly have an issue with this, because I do not know any practical method, or even the name of the method I am suppose to learn, to find this sort of solution 'easily' or sometimes at all.

We shall illustrate our methods by applying them to the simple problem
$y′=x+y, y(0)=1, (3)$ which we call our benchmark problem. This
differential equation in $(3)$ is clearly linear, and the exact solution
is easily found to be $y=2e^x − x − 1$.

I look at this, and I don't understand. I guess I can say I covered one of these ideas which is that $y'=y, y(0) = 1$ because it is glearingly obvious that the only thing that differentiates to itself is $e^x$.

Now, $y' = x+y$, how would I approach that? What is the method behind the solution to get there? I tried to look at the solution instead, but $y'$ of the solution for me is $2e^x - 1$.

This must be something really trivial, but I just don't get it.

答:
There is another way to solve $$y'=y+x$$ Let $y=z-x$ to make
$$z'-1=z\implies z'-z=1$$ Now, the homogeneous solution is $z=c e^{x}$. Use now the variation of parameter to get
$$c' \,e^{x}=1 \implies c'=e^{-x}\implies c=-e^{-x}+k\implies z=-1+ke^{x} $$ Back to $y$
$$y=-x-1+ke^{x}$$ and using the condition $k=2$
$$y=2e^{x}-x-1$$

Title: Solution to $y′=x+y,\, y(0)=1$


问:
Proof by Lewis Carroll that an urn cannot contain two balls
I am self-learning probability theory and solving some problems from the book Probability and Random processes by Grimmett and Stirzaker. Exercise 1.4.4 poses a nice problem. I would like someone to verify, if my solution is correct and thought process is correct.

What do you think of the following 'proof' by Lewis Carroll that an urn cannot contain two balls of the same color? Suppose that the urn contains two balls, each of which is either black or white. In the obvious notation $P(BB)=P(BW)=P(WB)=P(WW)=\frac{1}{4}$. We add a black ball, so that $P(BBB)=P(BBW)=P(BWB)=P(BWW)=\frac{1}{4}$. Next, we pick a ball at random, the chance that the ball is black is (using conditional probabilities)

$$1 \cdot \frac{1}{4}+\frac{2}{3}\cdot \frac{1}{4} + \frac{2}{3}\cdot \frac{1}{4} + \frac{1}{3}\cdot \frac{1}{4}=\frac{2}{3}$$

However, if there is a probability $\frac{2}{3}$ that a ball, chosen randomly from three, is black, then there must be two black and one white, which is to say that originally there was one black and one white ball in the urn.

Solution.

I think, the fallacy in this proof has to do with mistaking the unconditional probability $P(BBB)$ for the conditional probabilities $P(BBB|\text{atleast one ball is black})=\frac{1}{4}$.

\begin{align*}
P(\text{Ball drawn at random is black}) = 1\cdot\frac{1}{8} + \frac{3}{8}\cdot\frac{2}{3} + \frac{3}{8}\cdot \frac{1}{3} + \frac{1}{8}\cdot 0 = \frac{1}{2}
\end{align*}

This makes perfect intuitive sense, because without any knowledge of what ball was added to the urn, the unconditional probability that a ball drawn at random is black is simply $\frac{1}{2}$; the ball is equally likely to be black or white.

Surely, the urn can have two balls of the same color! I'm not sure, how to argue this. But, what I can tell for sure is,

\begin{align*}
P(BBB|\text{atleast one ball is black}) = P(BWW |\text{atleast one ball is black})= \frac{1}{4}
\end{align*}

答:
Probability of drawing a black ball, after adding a black ball, is $\frac{1}{3} p + \frac{2}{3} q + 1 (1-p-q) = \frac{3-2p-q}{3}$, where $p$ is the probability the urn had no black ball and q the probability it had one black ball at the start. This is equal to $\frac{2}{3}$ for $p = \frac{1-q}{2}$ and $p=0, q=1$ is one of the solutions.

The “proof” in the question has this embedded as an assumption.

The famous Monty Hall problem is in some ways related to this.


答:
Also any diagonal matrix with odd elements on the diagonal, in fact:

$$\det(A)=a_{1,1}  \cdots a_{n,n}$$

Title: Finding a matrix of an odd determinant

Body: I am trying to find a square matrix of size $n$ consisting of zeros and ones whose determinant is an odd number. Any ideas?


问:
Do axioms in knowledge base joined by 'AND'?
Do axioms in knowledge base joined by 'AND'?
For example, if 'A and B' is one of the axioms, can I just put {A,B} in my knowledge base, as they will be implied with 'AND' relation among them in the knowledge base anyway? Thanks.</p>

答: Note that a knowledge base is an ordered pair containing a TBox and an ABox. If you'd like to add the axiom $A\land B$ to the knowledge base, then simply add the axiom to the finite set of assertion components or the ABox, for the respective knowledge base.


答:
If you have a matrix $A$ such that $A^TA=I$ then

$$1=\det(I) = \det(A^TA) = \det(A)^2$$

Title: Finding a matrix of an odd determinant


Body:
Çlearly, $(p-1)/4\ne2,(p-1)/4\ge3\iff p\ge13$

So, if $(p-1)/4>3,$

Either $(p-1)/4=6k+1,k\ge1$

$(p+1)/2=12k+3=3(4k+1)$

Or $(p-1)/4=6k-1,k\ge1,p=?$


问:
$\iiint _{V} \frac{dV}{r^2}= \iint_{S} \frac{\vec{r}\cdot \hat{n}}{r^2}\,dS$
I have a question about a problem. It is triple integrals on surfaces. It is show
$$\iiint _{V} \frac{dV}{r^2}= \iint_{S} \frac{\vec{r}\cdot \hat{n}}{r^2}\,dS$$
Some help?

答:
As per Divergence theorem,
$\, \displaystyle \iint_{S} \vec{F} \cdot \hat{n}\,d$S $\displaystyle = \iiint_V (div \,\vec{F}) \,dV$

We have, $\, \displaystyle \vec{F} = \frac{\vec{r}}{r^2}$

$\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$

$r^2 = x^2 + y^2 + z^2$

$\displaystyle div \vec{F} =  \nabla \cdot \vec{F} = \frac{\partial}{\partial x} (\frac{x}{r^2})+ \frac{\partial}{\partial y} (\frac{y}{r^2}) + \frac{\partial}{\partial z} (\frac{z}{r^2})$

$\displaystyle \frac{\partial}{\partial x} ( \frac{x}{r^2}) = \frac{1}{r^2} - \frac{2x^2}{r^4}$

Can you similarly find $F_y, F_z$ and add them? You will see $\displaystyle div\vec{F} = \frac{1}{r^2}$