问:
Investigate the stability of the equilibrium point (0,0) by constructing a suitable Lyapunov function.
I'm struggling to follow along with any online notes, as I understand there is no general method for constructing a Lyapunov function, any help would be appreciated.

Consider the following dynamical system

$\dot x = −x + 4y$ , $\dot y = −x − y^3$

Investigate the stability of the equilibrium point $(0,0)$ by constructing a suitable Lyapunov function. [Hint: try a function of the form $L(x, y) = x^2 + ay^2$]

答:
With

$\dot x = -x + 4y \tag 1$

and

$\dot y = -x - y^3 \tag 2$

and

$L(x, y) = x^2 + ay^2, \tag 3$

we have

$\dot L(x, y) = \dfrac{L(x, y)}{\partial x} \dot x + \dfrac{L(x, y)}{\partial y} \dot y$
$= (2x)(-x + 4y) + (2ay)(-x - y^3)$
$= -2x^2 + 8xy - 2axy - 2ay^4$
$= -2(x^2 + ay^4) + (8 - 2a)xy; \tag 4$

if we choose

$a = 4, \tag 5$

then

$8 - 2a = 0, \tag 6$

and hence (4) becomes

$\dot L(x, y) = -2(x^2 + 4y^4) < 0 \tag 7$

for all

$(x, y) \ne (0, 0); \tag 8$

furthermore

$L(0, 0) = 0, \tag 9$

and

$L(x, y) = x^2 + 4y^2 > 0 \tag{10}$

whenever (8) binds. Thus $L(x, y)$ as in (10) is a Lyapunov function for the system (1)-(2), and the origin $(0, 0)$ is a stable equilibrium point.

Note Added in Edit, Sunday 29 November 2020 9:46 AM PST:  $(0, 0)$ is in fact the unique equilibrium point of (1)-(2), for with

$\dot x = \dot y = 0, \tag{11}$

equations (1), (2) yield

$-x + 4y = 0, \tag{12}$

$-x - y^3 = 0; \tag{13}$

from (12),

$x = 4y, \tag{14}$

and substituting this into (13) we find

$-4y - y^3 = 0, \tag{15}$

or

$y(y^2 + 4) = 0; \tag{16}$

for every real $y$,

$y^2 + 4 > 0, \tag{17}$

which forces

$y = 0; \tag{18}$

then both (12) and (13) affirm that

$x = 0 \tag{19}$

as well, and we see that $(0, 0)$ is the sole equilibrium point of (1)-(2).  End of Note.


问:
About Wilf's inequality on complex number
I created the image below from various bits of the book "The Cauchy-Schwarz Master Class". In particular, it's from a question in chapter 6.

Could someone explain to me why $\theta_j$ is not more than $2\psi$ or how one arrive at the inequality just before (14.54) in the image.

(Edited on 29-11-2020)

I would like to draw your attention to an answer to this question:

GM-AM-QM inequalities for complex numbers?!

It seems the constant $2\psi$ also appeared in the answer but the person just said it works.

答:
The complex AGM inequality states that if $z_k, k=1,..,n$ are (nonzero) complex numbers for which the argument spread is at most $2\psi, \psi <\pi/2$, or in other words, choosing say the principal argument in $(-\pi, \pi]$ or $[0, 2\pi)$ but having a fixed choice for all $z_k$, we have $|\arg(z_k)-\arg (z_l)| \le 2\psi <\pi$,

then $(|z_1..z_n|)^{1/n}\cos \psi \leq |\frac{z_1+..z_n}{n}|$

and the proof follows from the usual AGM inequality for positive numbers, noting that by a rotation which changes neither LHS, nor RHS (so send the lowest most argument of the $z_k$ taken in $(-\pi, \pi]$ say to $-\psi$, then all the others are at most $\psi$ by the spread hypothesis), we can assume that the arguments are all in $[-\psi, \psi]$ so $\Re z_k \geq |z_k|\cos \psi >0$ hence:

$(|z_1..z_n|)^{1/n}\cos \psi \leq (\Re (z_1)..\Re (z_n))^{1/n} \leq \frac{\Re z_1+..\Re z_n} {n} = \frac{\Re (z_1+..z_n)}{n} \le |\frac{z_1+..z_n}{n}|$
(note that if we let $\psi=\pi/2$ we still can apply the argument above, but the inequality is vacuous)

Apply this here with $z_k=1/(z-r_k)$ noting that the hypothesis shows that the arguments of $z-r_k$ hence of $z_k$ have spread at most $2\psi(z)$ for some $\psi(z) < \pi/2$ - considering $w_k=z-r_k$ their arguments are contained in the angle drawn by definition, so their difference is at most $2\psi$ as the picture says  (the angle in Wilf's inequality depends on $z$ of course and can get very close to $\pi/2$ if $z$ gets very close to a side of the polygon) and get the bound in the line above (14.54).

But now since by substitution LHS is $|a_n/P(z)|^{1/n}\cos \psi$ as $P(z)=a_n(z-r_1)..(z-r_n)$ and RHS is $|\frac{P'(z)}{nP(z)}|$ by the usual formula for $(P'/P)(z)$ as the sum of the reciprocals of $1/(z-r_k)$, (14.54) follows by rewritting and moving the $\cos \psi$ in the denominator of RHS

ContentLicense: CC BY-SA 4.0


问:
expected value of number of selections from $\{ 1,2, \dots, n\}$ with replacement such that the selection process stops when the sum exceeds $n$
I was going through my old notebooks and saw this problem:

Determine $\lim _{n \rightarrow \infty} \mathbb{E}(X)$ where $X$ denotes the random variable corresponding to the output of the following algorithm, where all random choices made are mutually independent and has an integer $n \geq 1$ as input:

```<code>INPUT: n

sumOfSelections = 0;
j = 0;

while sumOfSelections <= n
    do
       j = j + 1;
       x_j = uniformly random element in {1,2,...,n};
       sumOfSelections = sumOfSelections + x_j
endwhile;
return j
</code>```

I really have a weak intuition for these kinds of problems and am basically lost on how to proceed. My first instinct is that the average value of x_j is $\frac{n+1}{2}$ and so $X$ is 2 on average, but I think this line of attack is wrong.

How does one get $\mathbb{E}(X)$ not by averaging arguments but by using the probabilities of the outcomes of the algorithm?

答:
Essentially, we roll a fair $n$-sided die, with faces number from $1$ through $n$, until the sum of the numbers rolled equals $n$ or more.  The questions asks for the expected number of rolls.

Let $E_s$ be the expected number of rolls remaining, if the sum of the numbers rolled so far is $s$.  We have $$E_s=\begin{cases}
0,&s\geq n\\
1+\frac1n\sum_{j=1}^nE_{s+j},&s<n
\end{cases}$$
The second case comes from the fact that if the current sum $s<n$ we have to roll once more, and for each $j$, with probability $\frac1n$ , the next roll is $j$, and we have an average of $E_{s+j}$ additional rolls yet to come.

We want to compute $E_0$.  For a specific $n$, this is easily done by working backwards.  We know that $E_{n-1}=1$ since any roll will get us to $n$.  Now,$$\begin{align}
E_{n-2}&=1+\frac1nE_{n-1}=\frac{n+1}n\\
E_{n-3}&=1+\frac1n(E_{n-2}+E_{n-1})=\left(\frac{n+1}n\right)^2\\
E_{n-4}&=1+\frac1n(E_{n-3}+E_{n-2}+E_{n-1})=\left(\frac{n+1}n\right)^3
\end{align}
$$

and it's easy to guess the pattern.  I haven't proved that this pattern will continue. Can you do it?

If the pattern does indeed continue, then $$E_0=\left(\frac {n+1}n\right)^{n-1}\to e\text{ as }n\to\infty$$


问:
The path of complex structure.
$J = \left( {\begin{array}{*{20}{c}}
0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)$. If $I(t)$ is a path in $\rm M(2n, \mathbb R)$ ($2n \times 2n $ real matrix) such that $I(0)=0$ and $I(t)J+JI(t)=0$, then for sufficiently small $t$ can we find a path $J(t) \in \rm M(2n, \mathbb R)$ such that (1) $J(0)=J$, (2) $J^2(t)+I^2(t)=-I_{2n}$, (3) $I(t)J(t)+J(t)I(t)=0$?</p>

答:
This is not what was asked for exactly, but the path exists in the complex domain...

Let the dot denote matrix multiplication to avoid function dependence ambiguity.

Take:

$J(t)=J.\left(\mathbb{I}_{2n}+iI(t)\right)\in \rm M(2n, \mathbb C): I(0)=0,J(0)=J$.

$J,I(t)\in \rm M(2n, \mathbb R)$

We have that:

$\{J(t),I(t)\}=J(t).I(t)+I(t).J(t)$

$=J.\left(\mathbb{I}_{2n}+iI(t)\right).I(t)+I(t).J.\left(\mathbb{I}_{2n}+iI(t)\right)$

$=\{J,I(t)\}.\left(\mathbb{I}_{2n}+iI(t)\right)$,

and so, because $\{J,I(t)\}=0$,

$\{J(t),I(t)\}=0$.

We also have that:

$J(t)²=J.\left(\mathbb{I}_{2n}+iI(t)\right).J.\left(\mathbb{I}_{2n}+iI(t)\right)$,

$J(t)²=J.J.\left(\mathbb{I}_{2n}-iI(t)\right).\left(\mathbb{I}_{2n}+iI(t)\right)$,

$J(t)²=J².\left(\mathbb{I}_{2n}+I(t)²\right)$,

but:

$J² = \left( {\begin{array}{*{20}{c}}
0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right).\left( {\begin{array}{*{20}{c}}
0&{ - {I_n}}\\{{I_n}}&0\end{array}} \right)=-\mathbb{I}_{2n}$,

and so:

$J(t)²+I(t)²=-\mathbb{I}_{2n}$.

ContentLicense: CC BY-SA 3.0

Title: The path of complex structure.


问:
Image of a function containing a $\heartsuit$
Find the image of
$$f:\mathbb Z\to\mathbb N\cup\{\heartsuit\},m\mapsto\begin{cases}
2m+1&m>0\\
\heartsuit&m=0\\
-(2m-1)&m<0
\end{cases}$$
How do I approach this? I couldn't attend the lecture and cannot ask peers.

答:
Hint:

The image is the set of values within $\mathbb{N}\cup\heartsuit$ that the function reaches.

The function takes in values from $\mathbb{Z},$ the integers. So plug in $1, 2, 3,\dots$, and plug in $0$, and plug in $-1, -2, -3,\dots$.

Then make a guess for which values you could reach based on the patterns from the $1,2,3,\dots$ and the $-1,-2,-3,\dots$.


答:
From the declaration $f:\mathbb Z\to\mathbb N\cup\{\heartsuit\}$ you know that the image $f(\mathbb Z)$ must be a subset of the set $\mathbb N\cup\{\heartsuit\}$.

I don't know what kind of mathematical object $\heartsuit$ is, but apparently it is something that can be a member of a set.

You also have a definition of $f$.

Try a few examples of $f(m)$ and see what the results are.
What are the values of $f(-2),$  $f(-1),$  $f(0),$  $f(1),$ and $f(2)$?
All of those values belong to the image $f(\mathbb Z)$.

Now you just need to cover all the other values of $f(m)$:
$f(3)$, $f(4),$ $f(5),$ and so forth,
as well as $f(-3)$, $f(-4),$ $f(-5),$ and so forth.

Title: Image of a function containing a $\heartsuit$

Body:
Find the image of
$$f:\mathbb Z\to\mathbb N\cup\{\heartsuit\},m\mapsto\begin{cases}
2m+1&m>0\\
\heartsuit&m=0\\
-(2m-1)&m<0
\end{cases}$$
How do I approach this? I couldn't attend the lecture and cannot ask peers.


问:
What is big O notation of the following functions?
So I have few simple functions I wrote and I would like to find out how to calculate their big O notation.

I assume that the big O notaion for the first function is O(log b log a). And for the second one is O(2n). But I believe it can not be that easy and deep analysis has to be involved.

答:
You need to exactly choose operation with respect to which you count complexity. In first if you count divisions, then it is only one, so complexity is $O(1)$.

For second if you choose, for example, multiplication and length of list is $n$ then you can use that $O(2n)=O(n)=O(Cn)$ for any fixed $C>0$.


问:
$y(x)=\lambda \int_{-1}^1 (1-|x-t|)y(t)dt$ to differential equation

Convert $$y(x)=\lambda \int_{-1}^1 (1-|x-t|)y(t)dt$$ to differential equation with boundary conditions.

I know how to deal with this kind of problems but not when the kernel is a modulus function. I started solving it in the typical manner:

$$y'(x)=\lambda \int_{-1}^1  \frac{t-x}{|t-x|} y(t) dt.$$
It sort of rings the bell that $\frac{t-x}{|t-x|}$ could be $-1$ or $1$ depending on the values $x$ and $t$ on $[-1,1]$. Is there a way to determine those values? If I am right, it would be easy to solve this equation.

I also found another similar question (Transforming the integral equation $u(x) + \frac{\lambda}{2}\int_{0}^{1}|x - s|u(s)ds = ax + b$ into its equivalent differential equation) so I tried one of the methods in the answers:

$$\begin{align} \int_{-1}^1  (1-|x-t|) y(t) dt = \int_{-1}^{-x}  (1-|x-t|) y(t) dt + \int_{-x}^0  (1-|x-t|) y(t) dt \\\\+ \int_{0}^x  (1-|x-t|) y(t) dt + \int_{x}^1  (1-|x-t|) y(t) dt
\end{align}$$

Then I'm really not sure what happens with the signs:

$$\begin{align}
\int_{-1}^1  (1-|x-t|) y(t) dt =
\int_{-1}^{-x}  (1+x-t) y(t) dt +
\int_{-x}^0  (1-x+t) y(t) dt \\\\+
\int_{0}^x  (1-x+t) y(t) dt +
\int_{x}^1  (1+x-t) y(t) dt
\end{align}$$

Then by Leibnitz's Rule it should (?) follow that $$\frac{d^2}{dx^2} \int_{-1}^1 (1-|x-t|)u(t)dt=-4u(x)?$$ I feel like I just messed up with the signs...

答:
If $x\le-1$, then
$$y(x)=\lambda \int_{-1}^1 (1-|x-t|)y(t)dt=\lambda \int_{-1}^1 (1-(t-x))y(t)dt=\lambda \int_{-1}^1 (1-t+x)y(t)dt$$
and hence
$$ y'(x)=\lambda \int_{-1}^1 y(t)dt. $$
If $|x|<1$, then
$$y(x)=\lambda \int_{-1}^x (1-|x-t|)y(t)dt+\lambda \int_x^1 (1-(t-x))y(t)dt=\lambda \int_{-1}^x(1-x+t)y(t)dt+\lambda \int_x^1 (1-t+x)y(t)dt $$
and hence
$$ y'(x)=\lambda y(x)-\lambda\int_{-1}^xy(t)dt-\lambda y(x)+\lambda\int_x^1y(t)dt=-\lambda\int_{-1}^xy(t)dt+\lambda\int_x^1y(t)dt. $$
So
$$ y''(x)=-2\lambda y(x). $$
If $x\ge1$, then
$$y(x)=\lambda \int_{-1}^1 (1-(x-t))y(t)dt=\lambda \int_{-1}^1 (1-x+t)y(t)dt$$
and hence
$$ y'(x)=-\lambda \int_{-1}^1 y(t)dt. $$
Thus the equation is
$$ y''=-2\lambda y $$
with boundary condition
$$ y'(-1)=\lambda\int_{-1}^1y(t)dt, y'(1)=-\lambda\int_{-1}^1y(t)dt. $$


问:
$\int_0^{\pi/4}\!\frac{\mathrm dx}{2+\sin x}$ , $\int_0^{2\pi}\!\frac{\mathrm dx}{2+\sin x}$
Please help me integrate

$$\int_0^{\pi/4}\!\frac{\mathrm dx}{2+\sin x}$$

and

$$\int_0^{2\pi}\!\frac{\mathrm dx}{2+\sin x}$$

I've tried the standard $u = \tan \frac{x}{2}$ substitution but it looks horrible.

Thanks in advance!

答:
Let's give another try to your failed technique...

$$\displaystyle\int \frac{dx}{2+\sin x}$$

Let $u = \tan \frac{x}{2}$

$$\int \frac{du}{u^2+u+1} = \int \frac{du}{\left(u+\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}$$

Let $s=u+\frac{1}{2}$

$$\begin{align}\int \frac{ds}{s^2 + \left(\frac{\sqrt{3}}{2}\right)^2} &= \frac{2}{\sqrt{3}}\arctan\left(\frac{2s}{\sqrt{3}}\right)\\ &=\frac{2}{\sqrt{3}}\arctan\left(\frac{2u+1}{\sqrt{3}}\right)\\&=\frac{2}{\sqrt{3}}\arctan\left(\frac{2\tan\frac{x}{2}+1}{\sqrt{3}}\right)\end{align}$$

Evaluating the above from $0$ to $\dfrac{\pi}{4}$ yields approximately $0.33355$ while $0$ to $2\pi$ gives $\dfrac{2\pi}{\sqrt{3}}$.


问:
Prove $x^2-\sqrt2$ is irreducible over $\mathbb{Z}[\sqrt 2]$
Dummit and Foote Q9.4.9

Prove that the polynomial $x^2-\sqrt2$ is irreducible over $\mathbb{Z}[\sqrt 2]$. (you may use the fact that $\mathbb{Z}[\sqrt 2]$ is a U.F.D.)

In a UFD

every element can be written as a product of irreducibles

this decomposition is unique.

Do I use 1)? But I do not know the irreducibles in $\mathbb{Z}[\sqrt 2]$.

Or do I use 2)? or something else?

I tried assuming factors of degrees 0 and 2, and of degrees 1 and 1. And solve for their coefficients. But since the factors may not be monic, the equations are complicated.

Please give a hint.

Thanks!

答:
Something to help you get started:

An irreducible polynomial is one that cannot be factored into the product of two non-constant polynomials. To show this is irreducible, you would write $x^2-\sqrt{2}=(ax+b)(cx+d)$ for some $a,b,c,d\in\mathbb{Z}[\sqrt{2}]$ and hope to arrive at a contradiction.

Two polynomials are equal if and only if the coefficients of each $x^i$ are equal. So we would need $1=ac$, $0=ad+bc$, and $-\sqrt{2}=bd$.

The interesting equation there is $1=ac$. What values in $\mathbb{Z}[\sqrt{2}]$ have a multiplicative inverse, i.e. what are the options for $a$ and $c$?


问:
Does boundness in Topological vector space depend on the base point we choose?
Let X be a topological vector space,we know bounded set $B$ defined as for any neiborhood of origin,there exist some $\lambda$ such that $B\subset \lambda U$.

Does this depend on the base point we choose (for example here we choose definition for the neiborhood at origin.) Can we show this definition is equivalent to given point $x\ne 0 $ any neiborhood $U_x$ exist $\lambda$ such that $V\subset \lambda U$?

No since bounded set under translation(which is continuous linear map) is also bounded

答:
If you choose $x=(1,0)$ in $\mathbb R^2$, then the unit ball $B=B_1(0,0)$ will not be bounded according to your definition.  This is because if $U=B_{1/2}(x)$, then one would never have
$$
B\subseteq\lambda U,
$$
because $(0,0)$ is not in $\lambda U$, unless $\lambda=0$, which clearly doesn't work either.

As an alternative, instead of $\lambda U$, one may consider  homothety relative to $x$, meaning the map
$$
  h_\lambda(u) = \lambda(u-x) + x = \lambda u + (1-\lambda)x.
$$

This is like multiplying by $\lambda$ when the origin is changed to $x$.  It is now easy to see that a set $B$ is bounded if and only if, for any neighborhood $U$ of $x$, there exists some $\lambda$ such that $B\subseteq h_\lambda(U)$.


问:
Finding all polynomials $f(X), g(X)\in \mathbb{F}_5[X]$ such that $f(X)\neq g(X)$ and $f(x)=g(x)\neq 0$ for $x\in\mathbb{F_5}$.
I have to find all polynomials $f(X),g(X)\in\mathbb{F}_5[X]$ such that $f(X)\neq g(X)$ and $f(x)=g(x)\neq 0$ for $x\in\mathbb{F}_5$ where $\mathbb{F}_5=\{0,1,2,3,4\}$.

For example: Let $f(X)=X^5$ and $g(X)=X$ where $f(X)\neq g(X)$ in $\mathbb{F}_5[X]$, but if I put $x\in\mathbb{F}_5$ in $f(x), g(x)$ then we get as follows:

$x=0: f(0)=0^5 = 0 = g(0)$

$x=1: f(1)=1^5=1 = g(1)$

$x=2: f(2)= 2^5 = 32\equiv_5 2 = g(2)$

$x=3: f(3)=3^5 \equiv_5 (-2)^5 = -32 \equiv_5 3 = g(3)$

$x=4: f(4)=4^5 \equiv_5 (-1)^5 = -1 \equiv_5 4 = g(4)$.

So, for $x\in\mathbb{F}_5$, we get $x^5 =x$.
How can I find other polynomials?

答:
Just converting the comment to an answer :

Suppose $f(a)=g(a)$ , then $(f−g)(a)=0$ so $(x−a)$ divides $(f−g)$ as a polynomial. If $f(x)=g(x)$ for all $x \in \mathbb F_5$ then the above should tell you something about $f−g$ that turns out to be necessary and sufficient.

Basically, $f-g$ is a multiple of $\prod_{a \in \mathbb F_5} (x-a)$. Prove that this is sufficient and necessary.


问:
Given a Geodesic square what is the area of the square?
I need help with this question

Given a Geodesic square on unit ball, The interiors angle are 2 radians each,
What is the area of the square?

答:
Divide the square into two triangles by its diagonal.

Now use the formula for area of geodesic triangle where $A, B, C$ are angles of triangle $$\text{area} = r^2( A+B+C -\pi)$$

Summing up ($r=1$), $$\text{area of square} = 2+2+2+2-\pi-\pi=8-2\pi$$


问:
Is this function from $[-1,1] \rightarrow \mathbb{R} \cup \infty$ continuous?
just a short question. So I was wondering about functions from compact sets into
$\mathbb{R} \cup \{+\infty\}$.

Let's say we have a function $f : [-1,1] \rightarrow \mathbb{R} \cup \{+\infty\}$,
defined by

$f(x) = -1/x$ if $x \in [-1,0)$,

$f(0) = +\infty$,

$f(x) = 1/x$ if $x \in (0,1]$.

I think this should be continuous, although somehow my brain doesn't want
to believe that it is.

答: This is clearly continuous everywhere on $[-1,0)\cup(0,1]$, so we need only check continuity at $x = 0$. To talk about continuity, we need a topology, but taking a lead from the $\epsilon - \delta$ definition of continuity in real analysis, we can define $f(x)$ to be continuous with value of $\infty$ at $x = a$ if for all $N > 0$, there exists $\delta > 0$ such that $\left|\,f(x)\right| > N$ whenever $\left|x - a\right| < \delta$. In this case, we have $\left|\frac{1}{x}\right| > N$ whenever $\left|x\right| < \frac{1}{N}$, so that $f$ is indeed continuous with value $\infty$ at $x = 0$ (you would have to check that this definition of continuity agrees with your topology on $\Bbb R\cup \{\infty\}$).</p>


答:
If $f(x)=g(x)$ for all $x \in \mathbb F_5$, then $x, x-1,x-2,x-3, x-4$ divide $f-g$ and conversely.

So for example take any non vanishing polynomial for $f$ and $g(x) = f(x) + x^a (x-1)^b (x-2)^c (x-3)^d (x-4)^e$. This gives you an infinite number of examples.

Like

$$\begin{cases}
f(x) &= x^2\\
g(x) &= x^2 + x(x-1)(x-2)^2(x-3)(x-4)
\end{cases}$$

Title: Finding all polynomials $f(X), g(X)\in \mathbb{F}_5[X]$ such that $f(X)\neq g(X)$ and $f(x)=g(x)\neq 0$ for $x\in\mathbb{F_5}$.

Body:
I have to find all polynomials $f(X),g(X)\in\mathbb{F}_5[X]$ such that $f(X)\neq g(X)$ and $f(x)=g(x)\neq 0$ for $x\in\mathbb{F}_5$ where $\mathbb{F}_5=\{0,1,2,3,4\}$.

For example: Let $f(X)=X^5$ and $g(X)=X$ where $f(X)\neq g(X)$ in $\mathbb{F}_5[X]$, but if I put $x\in\mathbb{F}_5$ in $f(x), g(x)$ then we get as follows:

$x=0: f(0)=0^5 = 0 = g(0)$

$x=1: f(1)=1^5=1 = g(1)$

$x=2: f(2)= 2^5 = 32\equiv_5 2 = g(2)$

$x=3: f(3)=3^5 \equiv_5 (-2)^5 = -32 \equiv_5 3 = g(3)$

$x=4: f(4)=4^5 \equiv_5 (-1)^5 = -1 \equiv_5 4 = g(4)$.

So, for $x\in\mathbb{F}_5$, we get $x^5 =x$.
How can I find other polynomials?


问:
$\lim _{x\rightarrow 0^{-}}e^\left(1/x\right)=0$ ($\delta-\varepsilon$ definition)
$\lim _{x\rightarrow 0^{-}}e^\left(1/x\right)=0$ So I wrote the definition:
$\forall \varepsilon>0, \exists \delta_{\varepsilon}$ such that if $x_{0}-\delta<x<x_{0}$ then $\left|f(x)-L\right|<\varepsilon$
So let $\varepsilon>0$ be given. Choose $\delta=-\dfrac{1}{\ln(\varepsilon)}$ if $-\delta<x<0$ implies that $\left|e^\left(1/x\right)\right|<\varepsilon$. So did I make a mistake? Because when I take $\varepsilon=0.001$ I didn't get closer to $0$
$\varepsilon=0.001$ then $\delta=0.144$ then the interval is $0.144<x<0$. So for example if I take $0.140$ I can't even get closer to $0$. For instance: $e^\left(1/0.140\right)=1265.03...$

答: If $\varepsilon\geqslant1$, any $\delta$ will do. Otherwise, you have\begin{align}\left|e^{1/x}\right|<\varepsilon&\iff e^{1/x}<\varepsilon\\&\iff\frac1x<\ln\varepsilon\\&\iff x>\frac1{\ln\varepsilon}.\end{align}So, take $\delta=-\frac1{\ln\varepsilon.}$


问:
Quick question about the definition of non-degenerate $*$-morphism $A \to \mathcal{L}(E)$.
Consider the following definition in Lance's book on Hilbert $C^*$-modules. Let $A$ be a $C^*$-algebra.

How is $\alpha(a_i) x_i$ defined? Is this a function evalutation?  I.e. does one evaluate $\alpha(a_i) : E \to E$ in $x_i \in E$? Or is this some sort of multiplication? How does one multiply elements in $E$ and in $\mathcal{L}(E)$ which is the set of adjointable maps $E \to E$.

答: Each $\alpha(a_i)$ is a linear map from $E$ to itself, and $\alpha(a_i)x_i$ should be thought of as evaluation of the function $\alpha(a_i)$ at $x_i$.  The notation stems from linear algebra, where $\alpha(a_i)$ is treated like a matrix, $x_i$ is treated like a vector, and $\alpha(a_i)x_i$ is thought of as the product of the "matrix" $\alpha(a_i)$ and the "vector" $x_i$.


问:
question about the closure of subset
A subset $A$  of space $  (X,\tau)$ has $\beta$-property if  every $\tau$-open cover of $A$ has a finite subfamily whose $\tau$-closures cover $int A$.

Auppose the subset  $ A \subseteq X $has the $\beta$-property, does the closure of$A$ have$\beta$-property?

答: Let $X=\Bbb R$ with the usual topology, and let $A=\Bbb Q$. $A$ trivially has the $\beta$-property, since its interior is empty, but $\operatorname{cl}A=\Bbb R$ does not.

Title: question about the closure of subset


问:
$z_1,z_2,z_3\in \mathbb{S} (1)$ form an equilateral triangle iff the sum is zero
Let $z_1,z_2,z_3$ be complex numbers of module 1. Prove that they form an equilateral triangle if, and only if, $z_1+z_2+z_3=0$.

I know that if they form am equilateral triangle, then
$$ \frac{z_3-z_2}{z_2-z_1}=
   \frac{z_1-z_3}{z_3-z_2}.
 $$
I would appreciate any hint or idea to show that the sum is 0. Thank you.

答:
HINT

The numbers $z_1,\,z_2,\,z_3\,$ represent vertices of a triangle, and also vectors which start at $0$.

If the triangle is equilateral, then $0$ is its center of gravity and $z_1+z_2+z_3=0.$

EDIT


答:
Your mistake was that you took $0\lt x\lt \delta$, when you took $x=0.140$.

You should take $-\delta\lt x\lt 0$, and then your requirement will be satisfied.

E.g., $e^{1/-0.140}\approx0.00079<0.001$.

Title: $\lim _{x\rightarrow 0^{-}}e^\left(1/x\right)=0$ ($\delta-\varepsilon$ definition)

Body:
$\lim _{x\rightarrow 0^{-}}e^\left(1/x\right)=0$ So I wrote the definition:
$\forall \varepsilon>0, \exists \delta_{\varepsilon}$ such that if $x_{0}-\delta<x<x_{0}$ then $\left|f(x)-L\right|<\varepsilon$
So let $\varepsilon>0$ be given. Choose $\delta=-\dfrac{1}{\ln(\varepsilon)}$ if $-\delta<x<0$ implies that $\left|e^\left(1/x\right)\right|<\varepsilon$. So did I make a mistake? Because when I take $\varepsilon=0.001$ I didn't get closer to $0$
$\varepsilon=0.001$ then $\delta=0.144$ then the interval is $0.144<x<0$. So for example if I take $0.140$ I can't even get closer to $0$. For instance: $e^\left(1/0.140\right)=1265.03...$


问:
proof verification on separable metric space
let $(X,d)$ and $(Y,p)$ be metric spaces and $f:X\rightarrow Y$ continuous, if $A\subseteq X$  and $(A,d|_{A})$ is separable show that $$(f(A),p|_{f(A)})$$ is also separable.

Since $f$ is continuous and $A$ is countable then $f(A)$ is countable.
because $\overline{A}= X\Rightarrow B(x,\delta  )\cap X\neq \varnothing (x\in A)$

$\Rightarrow  f(B(x,\delta  ))\cap f(X)\neq \varnothing $ (can i do that?)

$f$ is continuous so $B(f(x),\varepsilon  )\cap Y\neq \varnothing $

答:
You cannot assume that $A$ is countable: it is separable, so it has a countable dense subset $D$, but it need not itself be countable.

Let $Z=f[A]$; we want to show that $Z$ is separable. Let $E=f[D]$; $E$ is countable, so we’ll be done if we can show that $E$ is dense in $Z$. Let $z\in Z$ and $\epsilon>0$; then $z=f(x)$ for some $x\in A$, and since $f$ is continuous there is a $\delta>0$ such that $f[B(x,\delta)]\subseteq B(z,\epsilon)$. $D$ is dense in $A$, so there is a $p\in D\cap B(x,\delta)$, and clearly $f(p)\in E\cap B(z,\epsilon)$. Thus, $E$ is dense in $Z$, and $Z$ is separable.


问:
How to solve this improper integral using Simpson's rule
I'm trying to calculate value of $\int_{0}^{\infty}{e^{-2x}(1+e^{-x})}dx$ using Simpson 1/3 rule.
I know the solution is substitution $x$ by $g(t)$ but every $g(t)$ I've tested the result was incorrect.

Body:
Substitute $$x\to \frac{1-t}{t+1};\;dx=-\frac{2dt}{(t+1)^2};\;t\to \frac{1-x}{x+1}$$
Integral becomes
$$\int_1^{-1} \frac{(-2) e^{-\frac{2 (1-t)}{t+1}} \left(e^{-\frac{1-t}{t+1}}+1\right)}{(t+1)^2} \, dt=\int_{-1}^1 \frac{2 e^{\frac{2 (t-1)}{t+1}} \left(e^{\frac{t-1}{t+1}}+1\right)}{(t+1)^2} \, dt$$
Notice that $\underset{t\to -1}{\text{lim}}f(t)=0$ so you can set $f(-1)=0$


问:
Gauss lemma for Irreducibility over $\mathbb{Z}\bigl[\sqrt{13}\mkern2mu\bigr]$
Gauss lemma states that a monic polynomial over $\mathbb{Z}[x]$ is irreducibile iff it is Irreducibile over $\mathbb{Q}[x].$

My concern is whether this statement is valid for monic polynomials over $\mathbb{Z}\bigl[\sqrt{13}\mkern2mu\bigr]$ and $\mathbb{Q}\bigl[\sqrt{13}\mkern2mu\bigr].$

答:
A version of Gauss's lemma is true in far more generality. Here's the most general version I know: If $A \subset B$ is an inclusion of integral domains and $K$ is the field of fractions of $A$, then $B \otimes_A K$ is an integral domain. (One can prove this by checking that $B \otimes_A K$ is isomorphic to the localization of $B$ at the multiplicative subset $A \setminus \{0\}$, so $B \otimes_A K$ is naturally a subring of the field of fractions of $B$.)

Here's how we relate this back to polynomials: Let $R$ be an integral domain with field of fractions $F$, and let $g \in R[x]$ be a monic prime element. We have an injective homomorphism $R \to R[x]/(g)$, so applying the above theorem, we conclude that $R[x]/(g) \otimes_R F \cong F[x]/(g)$ is an integral domain, hence $g$ is prime in $F[x]$.

If $R$ happens to be a UFD, then "prime" is equivalent to "irreducible", so we have the familiar statement about irreducible polynomials. In general, the analogous statement holds for prime polynomials. So you don't need to check whether the ring is a UFD, you just need to check whether the irreducible polynomial in question is prime.


问:
Where does (n+1) come from when finding quantiles
So many textbooks suggest a similar formula like the following for finding quantiles $q$ is the quantile, $p$ a number between 0 and 1 and $n$ the total number of our samples:

$q(p) = x_{(k)} + \alpha(x_{(k+1)} - x_{(k)})$,

where $k = [p(n+1)]$ and $\alpha = p(n+1) - [p(n+1)]$.

In which I don't understand the reasoning behind $(n+1)$ in $k = [p(n+1)]$ and subsequently other statements.

If we're going for the median, $n+1$ would make sense considering $m-1=n-m$, but shouldn't this differ now that we're technically multiplying one side?

答:
If I understand the notation correctly, the idea is that the $n$ samples partition all possible values of $X$ into $(n+1)$ equally probable ranges:

range $0$: $X < x_1$;

range $1$: $x_1 < X < x_2$;

range $2$: $x_2 < X < x_3$; $\ldots$

range $(n-1)$: $x_{n-1} < X < x_n$;

range $n$: $x_n < X$,

and some boundary handling when $X \in \{x_i\}$. That's where the $(n+1)$ originates from.

If $p(n+1)$ is an integer and $1 \le p(n+1) \le n$, the $p$-quantile is the start of range $k = p(n+1)$. So for example, when $p = \frac 2{n+1}$, the $\frac 2{n+1}$-quantile is the start of range $2$, i.e. $x_2$, and two of the $(n+1)$ ranges are before $x_2$.

If $p(n+1)$ is not an integer but still $1 < p(n+1) < n$, the formula of $q(p)$ considers inside range $k = \lfloor p(n+1)\rfloor$, and just performs linear interpolation between the two ends $x_{\lfloor p(n+1)\rfloor}$ and $x_{\lceil p(n+1)\rceil}$.


问:
Is my justification right to evaluate the limit?
I want to evaluate :

$$\lim_{x\to 0^+}\left(\frac1e(1+x)^{\tfrac1x}\right)^{\tfrac{1}{x^2}}$$

We know $\lim_{x\to 0^+}(1+x)^{\tfrac1x}$  is other form of $\lim_{u\to\infty}(1+\frac1u)^u=e$ which is increasing on $(0,\infty)$ it means at the infinity we get $e^-$ (the value is a litter less than $e$) therefor in the original limit we have $(1^-)^{\infty}$ therefore the limit equals to $0$.
Is my justification right?

答:
No, it is not. The limit $\lim_{u\to\infty}\left(1+\frac1u\right)^u$ is not “a litter less than $e$”; it is equal to $e$. And therefore $\lim_{x\to0^+}\frac1e(1+x)^{1/x}$ is exactly $1$. So, your limit is an indeterminate form and you will have to work harder to find its value.

Note: Your limit is indeed $0$, but, if your argument was correct, it would also prove that$$\lim_{x\to0^+}\left(\frac1e(1+x)^{1/x}\right)^{1/x}=0.$$However, this is not true; actually,$$\lim_{x\to0^+}\left(\frac1e(1+x)^{1/x}\right)^{1/x}=\frac1{\sqrt e}.$$


问:
What's the meaning behind annuity coefficient?
There is a formula for mortgage month payment calculation:
$$
A=P\cdot {\frac  {r(1+r)^{n}}{(1+r)^{n}-1}}
$$
where:

${\displaystyle A}$ is the periodic amortization payment;

${\displaystyle P}$ is the principal amount borrowed;

${\displaystyle r}$ is the rate of interest expressed as a fraction (per month);

${\displaystyle n}$ is the number of payments;

What exactly does the coefficient after ${\displaystyle P}$ represent?
Why don't banks use a much simpler formula: $A=P\cdot {\frac {(1+r)^{n}}{n}}$ ?

答:
About the "why":

Simple interests are not viable in front of compound ones. If compound interests did not exist, everyone would withdraw their deposits after one year and re-
deposit the capital + interest on the next day, so achieving compound interests.


问:
Solution of an ordinary differential equation of the form $f' = f (f-1)$
How can I find the function $f: [0, \infty] \rightarrow [0, \infty)$, satisfying,

$f^\prime = f (f-1)   $

such that f(0) = c>0, where $f^\prime$ is the first derivative?

答:
Hint:
$$f^\prime = f (f-1)$$
Is a separable differential equation. It's also A Bernoulli's differential equation:
$$f'+f= f^2$$
$$-\left (\dfrac 1 f \right)'+\dfrac 1 f=1$$
$$u'-u=-1$$
Then use integrating factor:
$$(ue^{-x})'=-e^{-x}$$
Inetgrate and unsibstitute $u=\dfrac 1f$.


问:
Proving that $F[X]/(f(X)) \cong F(\alpha).$
So I've been trying to prove that, for a field $F$ and an element $\alpha$ in a field extension of $F$, $F[X]/(f(X)) \cong F(\alpha)$ where $f(X) \in F[X]$ is the minimal polynomial over $\alpha$. Since I'm still in high school I don't have a certain textbook about field theory that I can follow along with so I've been searching for various PDFs here and there and hope that I learn something from them. I've seen this theorem in a lot of PDFs but I can't seem to fully understand the proof. Here's what I know so far:

$\varphi: F[X] \rightarrow F(\alpha)$ defined by $g(X) = g(\alpha)$ is a homomorphism. Furthermore, $F[X]$ is a PID and since $ker(\varphi)$ is an ideal to $F[X]$ then $(f(X)) = ker(\varphi)$. What's left to prove is that $Im(\varphi) = F(\alpha)$ by the first isomorphism theorem. What I can't prove is that $Im(\varphi)$ is a field, i.e it has inverses. If I manage to prove that then it's quite trivial that $Im(\varphi) = F(\alpha)$ by the definition of $F(\alpha)$. I hope that someone is kind enough to help me!

答:
The key in proving that the image is a field is Bézout's Identity.

If $f, g$ are relatively prime polynomials then there exists polynomials $a, b$ such that $$af + bg = 1.$$

Another way to write this is $b \equiv g^{-1} \pmod{f}$.

So if $g(\alpha) \in \operatorname{Im}(\varphi)$, and is non-zero, then by division, we can find a quotient and remainder such that $g = qf + r$ with $\deg r < \deg f$. So now $g(\alpha) = r(\alpha)$ and since $\deg r < \deg f$, we know that $r$ and $f$ are relatively prime.

Now, by Bézout's Identity, we find that
$$a(\alpha)f(\alpha) + b(\alpha)r(\alpha) = b(\alpha)r(\alpha) = 1.$$
So $b(\alpha) = g(\alpha)^{-1}$.


问:
Doubts about directional derivative in one variable
Sometimes I read that directional derivatives for function in one variables are right and left derivatives. But this doesn't make sense to me. The only unit vector in $\mathbb{R}$ are $\pm1 $ , so we are saying that:
$$D_1f(x)=f'_+(x)=\lim_{t\to 0^+} \frac{f(x+t)-f(x)}{t}$$
$$D_{-1}f(x)=f'_-(x)=\lim_{t\to 0^-} \frac{f(x+t)-f(x)}{t}$$
But by definition:
$$D_{\mathbf{v}} f(\mathbf{x})=\lim_{t\to 0} \frac{f(\mathbf{x}+t\mathbf{v})-f(\mathbf{x})}{t}$$
By this logic:
$$D_{\mathbf{-v}} f(\mathbf{x})=\lim_{t\to 0} \frac{f(\mathbf{x}-t\mathbf{v})-f(\mathbf{x})}{t}$$
I can substitute $-t=u$ ($-t$ doesn't assume infinite times the value $0$ so I can apply composite function limit theorem), so:
$$D_{\mathbf{-v}} f(\mathbf{x})=\lim_{u\to 0} \frac{f(\mathbf{x}+u\mathbf{v})-f(\mathbf{x})}{-u}=-\lim_{u\to 0} \frac{f(\mathbf{x}+u\mathbf{v})-f(\mathbf{x})}{u}=-D_{\mathbf{v}} f(\mathbf{x})$$
This should mean that:
$$f'_+(x)=-f'_-(x)$$
That in general is false!
Wouldn't be more correct to say that by convention in one variable we use derivative along the unit vector $1$ and that right and left derivatives are simply the right and left directional derivatives along the unit vector $1$.
Thanks in advance.

答:
The directional derivative is not the same as the limit approaching from the left or the right.  If you use the negative unit vector, you're asking what the slope is if you're traveling to the right vs. to the left.  You get the negative of the slope if you use the negative unit vector.  But that is correct. The divide by the
magnitude of v ($|v|$) is superfluous if you restrict v to be a unit vector.

$$\nabla_v f(x) = \nabla f(x) . \frac{v}{|v|}$$


问:
Continuous piecewise function integration from domain of area to domain of boundary
σ: stress, ε: strain

a, b, c, d, $ε_2$, $ε_3$ are constants.

σ is a piecewise CONTINUOUS and SMOOTH function:

$σ(ε)= d\cdot \begin{cases}
     1-(1-\frac{ε}{ε_2})^2 & 0<=ε<ε_2 \\
     1 & ε_2<=ε<=ε_3
   \end{cases}$

Cross section always are planar, so $ε = a x + b y + c$

To get the axial force I must double integrate the $σ$ on area domain.

Now, I want to avoid double integration of $σ$ on area domain, and I want to use half of Green theorem $\int_S L(x,y)dy=\iint_A σ(x,y) dx dy$ where $L(x,y)=\int σ(x,y)dx$

But piecewise function L(x,y) IS NOT CONTINUOUS in boundary domain (nor the area domain).

So this is a forbidance to use Green theorem for piecewise functions?

I really need to use Green Theorem. Is there a workaround to make $L(x,y)$ continuous and valid for Green theorem?

答:
Ok. My mistake.
$L(x,y)=\int σ(x,y)dx+f(y)$

$f(y)$ must be chosen wisely, to make piecewise function continuous.

If this cannot give an analytical function $L(x,y)$ then the area domain must be splitted into 2 parts and Green theorem must be used on boundary of each of them.


答:
Hint: You have
$$\frac{1}{f(f-1)}\,f'=1,$$
which becomes
$$\int\frac1{f(f-1)}\,df=\int1\,dx.$$

Title: Solution of an ordinary differential equation of the form $f' = f (f-1)$

Body:
How can I find the function $f: [0, \infty] \rightarrow [0, \infty)$, satisfying,

$f^\prime = f (f-1)   $

such that f(0) = c>0, where $f^\prime$ is the first derivative?


问:
Induction over a range of values
In the induction proof below where $[a_0; a_1,\ldots, a_h]$ is a continued fraction, could someone please clarify which induction method is being used here? Is it:

an induction over $i$ and $h$ i.e. show for each $h\in \mathbb{N}_0$ that it the result holds for all $-1\leqslant i \leqslant h$.

an induction only over $i$. Where $h$ is just a given value in $\mathbb{N}_0$

I initially assumed it was the first option but I struggled a lot and wasn’t able to do the ‘straightforward’ induction. If it is the second option then why for the base step is the result proven for $i=-1$ and $i=0$ and not just for one value?

答:
Option 2:  It's definitely option 2.

This is induction over $i$ for a fixed $h$.

Of course $h$ could be any value as there is nothing it the proof that requires any specific qualities of $h$ (other than $h\in \mathbb N_0$).

Don't be confused in that this is a proof by induction on one variable, $i$, but a proof by generality on another, $h$.  Don't forget a "proof by generality"--- to prove something is true for all values of a set $G$, which can be $\mathbb N_0$ or any other set, by arbitrarily picking a value, $w\in G$, and showing it is true for $w$, therefore it is true for all elements in $G$ because ... there was nothing special about $w$--- is still valid.

so the statement is: For any $h\in \mathbb N_0$ and for an $i: 1\le i \le h$ then .... something.

And the proof goes like:

Fix an arbitrary $h_0 \in \mathbb N_0$ (the subscript $_0$ is my idea; the author didn't think it was necessary-- it wasn't but it illustrates my point).

Prove via induction on $i$ that the statement is true for all $-1 \le i \le h_0$.

Then conclude that because $h_0$ was arbitrary, it must be true for all $h \in \mathbb N_0$.

Title: Induction over a range of values


答:
In order to achieve the induction step here, you need to assume the induction hypothesis is true for the previous two values of $i$, since $d_i$ depends on $d_{i-1}$ and $d_{i-2}$.

I think you could view this either as strong induction -- like your option (1), where the induction hypothesis is that the statement holds for all values from $-1$ to $i$ -- or you could view it as regular induction (like your option (2)), where the proposition $P(i)$ is that it holds for $i$ and $i-1$, and that's why you would need to prove the base step for two values.

Title: Induction over a range of values

Body:
In the induction proof below where $[a_0; a_1,\ldots, a_h]$ is a continued fraction, could someone please clarify which induction method is being used here? Is it:

an induction over $i$ and $h$ i.e. show for each $h\in \mathbb{N}_0$ that it the result holds for all $-1\leqslant i \leqslant h$.

an induction only over $i$. Where $h$ is just a given value in $\mathbb{N}_0$

I initially assumed it was the first option but I struggled a lot and wasn’t able to do the ‘straightforward’ induction. If it is the second option then why for the base step is the result proven for $i=-1$ and $i=0$ and not just for one value?


问:
Finding the flow lines of a given vector field on $S^2$
Good morning!

I have a quick question which I couldn't find the answer to.

Say we have a vector field $\eta : S^2 \to TS^2$ which takes a point $(x,y,z) \in S^2$ to the tangent vector $(-y, x, 0)$.

I am trying to figure out the flow lines of $\eta$, ie, fixing $(x,y,z)$, I'm trying to find a curve $\gamma$ through $(x,y,z$) such that $\gamma'(0) = (-y, x, 0)$.

I was just thinking to set: $\gamma : t \mapsto (x - yt, y + xt, z)$, which does satisfy the property above... but I want it to be a curve in $S^2$, and I do not know how to enforce that, because unless $x \ne y$ or $t = 0$, $||\gamma(t)||^2 \ne 1$.

答:
In addition to Lee Mosher's answer here is another way of finding the flow line of this particular vector field:
You wish to find $\gamma\colon (-\varepsilon,\varepsilon) \to \mathbb{S}^2$ such that $\gamma(0)=p_0=(x_0,y_0,z_0)^T$ and
$$
\begin{align*}
  \dot{\gamma}(0)
    = \begin{pmatrix}
        -y_0 \\ x_0 \\ 0
      \end{pmatrix}
    = \begin{pmatrix}
        0 & -1 & 0 \\
        1 &  0 & 0 \\
        0 &  0 & 0 \\
      \end{pmatrix}
      \begin{pmatrix}
        x_0 \\ y_0 \\ z_0
      \end{pmatrix}
    = L_z \cdot p_0.
\end{align*}
$$
Now why did I write this as a matrix-vector multiplication? The reason is the following:

The special orthogonal group $\mathrm{SO}(3)$ operates on $\mathbb{S}^2$ by left-multiplication. Its Lie algebra $\mathfrak{so}(3)$ is the set of skew-symmetric $(3\times 3)$-matrices. One basis is given by
$$
\begin{align*}
  \mathfrak{so}(3)
    =\mathrm{span}\{L_x,L_y,L_z\}
    = \mathrm{span} \left\{
      \begin{pmatrix}
        0 &  0 &  0 \\
        0 &  0 & -1 \\
        0 &  1 &  0 \\
      \end{pmatrix},
      \begin{pmatrix}
        0  &  0 & 1 \\
        0  &  0 & 0 \\
        -1 &  0 & 0 \\
      \end{pmatrix},
      \begin{pmatrix}
        0 & -1 & 0 \\
        1 &  0 & 0 \\
        0 &  0 & 0 \\
      \end{pmatrix}
      \right\}.
\end{align*}
$$
Here $L_x, L_y, L_z$ are the infinitesimal generators of rotations about the $x$-, $y$- and $z$-axis.

(I wrote earlier that $\mathfrak{so}(3)$ is isomorphic to $\mathbb{R}^3$. This is not only the case as same-dimensional vector spaces but as Lie algebras. This means that the Lie bracket needs to transform naturally. Choosing the above basis you can directly compute that $\mathfrak{so}(3)$ with the Lie bracket $[A,B]_{\mathfrak{so}(3)}=AB-BA$ is isomorphic to the Lie algebra $\mathbb{R}^3$ with the Lie bracket given by the cross product, i.e. $[X,Y]_{\mathbb{R}^3}=X\times Y$. Note that this is a special feature of $\mathbb{R}^3$ and doesn't hold in different dimensions!)

You can now use the exponential map $\exp\colon \mathfrak{so}(3)\to \mathrm{SO}(3), A \mapsto \sum_{k=0}^{\infty} \tfrac{1}{k!}A^k$ to explicitly calculate the rotations around the corresponding axes about an angle $\theta$ as $R_i(\theta)=\exp(\theta L_i)$ for $i\in\{x,y,z\}$. For example,
$$
\begin{align*}
  R_z(\theta)= \exp(\theta L_z) =
      \begin{pmatrix}
        \cos(\theta) & -\sin(\theta) & 0 \\
        \sin(\theta) & \cos(\theta) & 0 \\
        0 &  0 & 1 \\
      \end{pmatrix}.
\end{align*}
$$
You can check here directly that $\dot{R_z}(0) = L_z$. More generally, this is because the differential of the exponential map at the origin of the Lie algebra is the identity map.

The flow line of your vector field (which corresponds to $L_z$) is then given by the action of $\mathrm{SO}(3)$ on $\mathbb{S}^2$, i.e. left-multiplication:
$$
\begin{align*}
  \gamma\colon (-\pi, \pi) \to \mathbb{S}^2,
  \theta \mapsto
    R_z(\theta)\cdot p_0=
    \begin{pmatrix}
        \cos(\theta)x_0-\sin(\theta)y_0 \\ \sin(\theta)x_0+\cos(\theta)y_0 \\ z_0
    \end{pmatrix}
\end{align*}
$$
which is of course a parametrisation of the circle of latitude in Lee Mosher's answer :)

Edit:
You might notice that if you calculate the first-order Taylor approximation of the flow line which I wrote down then you will get the one that you wrote down in your original question. Maybe some geometric intuition here is that the exponential map of the group action guarantees that you "stay on the manifold" which is being acted on (here $\mathbb{S}^2$). Linear approximation of the exponential map will make you "leave your manifold" straight away. Of course, this only makes sense if your manifold is embedded in some higher-dimensional ambient manifold. I hope that makes sense.


问:
Show $\mathbb{Z}[X]/(2,X) \simeq \mathbb{Z}/2\mathbb{Z}$
I am a beginner in abstract algebra and I am trying to prove $\mathbb{Z}[X]/(2,X) \simeq \mathbb{Z}/2\mathbb{Z}$.

Let $\varphi : \mathbb{Z}[X] \rightarrow \mathbb{Z}/2\mathbb{Z}$. $\varphi$ is a homomorphism ring. We get the isomorphism if :

$\ker \varphi = (2,X)$ and

$\mathrm{Im} \varphi = \mathbb{Z}/2\mathbb{Z}$.

But I had a few problems. I don't know if what I did is correct.

Proof of 1.

Let $P(X) : = \sum_{i = 0}^{+ \infty} a_iX^i \in \ker \varphi$ with $a_i \in \mathbb{Z}$. So $\varphi(P(X)) = \overline{0}$ because $\overline{0}$ is the zero-element of $\mathbb{Z} / 2 \mathbb{Z}$.

Since $\varphi$ is a homomorphism ring, then we get $\varphi(a_i X^i) = \overline{a_i X^i} = \overline{0}$. Here we get $\overline{a_i} = \overline{0}$ or $\overline{X^i} = \overline{0}$ i.e $ \ker \phi$ is generated by $0$ and $X$.

I get $0$ instead of $2$, I don't know why.

Proof of 2.

Let $\overline{k} \in \mathrm{Im} \varphi$, with $k = 0,1$, then $\exists P(X) \in \mathbb{Z}[X]$ s.t $\varphi(P(X)) = \overline{k}$

And I don't know how to pursue. Thank you.

答:
Hint

The ideal

$$I=(2,x)=\{2n +xq(x) \mid n \in \mathbb Z, \, q \in \mathbb Z[x]\}$$

is the kernel of the ring homeomorphism that maps $p \in \mathbb Z[x]$ to $\overline{p_0} \in \mathbb Z/2 \mathbb Z$ where $p_0$ is the coefficient of degree $0$ of $p$.


Title: Differentiate $y = \sqrt {{{1 + 2x} \over {1 - 2x}}} $ logarithmically

Body:
$\eqalign{
  & y = \sqrt {{{1 + 2x} \over {1 - 2x}}}   \cr
  & \ln y = {1 \over 2}\ln (1 + 2x) - {1 \over 2}\ln (1 - 2x)  \cr
  & {1 \over y}{{dy} \over {dx}} = {1 \over 2} \times {2 \over {(1 + 2x)}} - {1 \over 2} \times {{ - 2} \over {(1 - 2x)}}  \cr
  & {1 \over y}{{dy} \over {dx}} = {1 \over {(1 + 2x)}} + {1 \over {(1 - 2x)}}  \cr
  & {{dy} \over {dx}} = {{(1 - 2x) + (1 + 2x)} \over {(1 + 2x)(1 - 2x)}} \times y  \cr
  & {{dy} \over {dx}} = {{2{{(1 + 2x)}^{{1 \over 2}}}} \over {(1 + 2x)(1 - 2x){{(1 - 2x)}^{{1 \over 2}}}}}  \cr
  & {{dy} \over {dx}} = {2 \over {{{(1 + 2x)}^{{1 \over 2}}}{{(1 - 2x)}^{{3 \over 2}}}}} \cr} $

However the answer is:

${{dy} \over {dx}} = {2 \over {(1 - 2x)\sqrt {(1 - 4{x^2})} }}$

Where have I went wrong?

Body: $$\frac{2}{(1+2x)^\frac{1}{2}(1-2x)^\frac{3}{2}} = \frac{2}{(1+2x)^\frac{1}{2}(1-2x)^\frac{1}{2}(1-2x)} = \frac{2}{(1-4x^2)^\frac{1}{2}(1-2x)}$$


问:
Rejection region: the probability udner true parameter
I would like to understand in the third snippet below in the context of statistical hypotheses testing how was obtained the formula (namely the numbers $1-,22.5$ and $-3\theta$) in the example 8.2.3 originated from the example 8.2.1:

$$Q(\theta)=1-\Psi(22.5-3\theta)$$

Book extract 1

Book extract 2

Book extract 3

答:
It actually says $Q(\theta)=1-\Phi(22.5-3\theta)$ where $\Phi$ is the cumulative distribution function of a standard normal distribution.

Here the underlying population is $\mathcal N(\theta, 1)$ so the average of a sample size $9$ has mean $\theta$ and standard deviation $1/3$.

So the probability the sample average is less than $7.5$ is $\Phi\left(\dfrac{7.5 - \theta}{1/3}\right) = \Phi\left({22.5 - 3\theta}\right)$

and the probability the sample average is greater than $7.5$ is $1-\Phi(22.5 - 3\theta)$.

This is the power of a test that rejects $H_0$ iff $\bar X > 7.5$


问:
Where is the mistake in this proof related to linear maps
I have the following statement to prove:

Let $T$ a linear map such that $T:U\to V$ and $\{u_i\}_{i=1}^n$ a
linearly independent subset of $U$. Prove that if $T$ is injective
then $\{T(u_i)\}_{i=1}^n$ is also linearly independent.

Well, my proof was:

Since $\{u_i\}_{i=1}^n$ is linearly independent, then we have $\sum_{i=1}^n\alpha_iu_i=0$ with unique scalars $a_i=0 \forall i \in [1,n]$.

So, applying $T$ to both sides, we have:

$$T\left(\sum_{i=1}^n\alpha_iu_i\right) = \sum_{i=1}^n\alpha_iT(u_i)=T(0)=0$$ so, since the zero vector written as linear combination of $T(u_i)$ have unique scalars $a_i=0 \forall i \in [1,n]$ implies that $\{T(u_i)\}_{i=1}^n$ is linearly independent.

I know that this is false, since we can create an unique linear map $T:\mathbb{R}^3 \to \mathbb{R}^3$ such that

$T(1,0,0)=(2,2,2), T(0,1,0)=(2,2,2), T(0,0,1)=(2,2,2)$ that contradicts my result. So, where is my mistake?

答: It does not follow from the fact that $\alpha_1,\ldots,\alpha_n=0$ are the unique scalars such that $\sum_{i=1}^n\alpha_iu_i=0$ that they are the unique scalars such that $\sum_{i=1}^n\alpha_iT(u_i)=0$. If, say, $T\colon\Bbb R^2\longrightarrow\Bbb R^2$ is defined by $T(x,y)=(x,0)$, if $u_1=(1,0)$ and if $u_2=(0,1)$, then indedd $\alpha_1=\alpha_2=0$ are the unique scalars such that $\alpha_1u_1+\alpha_2u_2=0$, but if you take $\alpha_1=0$ and $\alpha_1=1$, you will also have $\alpha_1T(u_1)+\alpha_2T(u_2)=0$. Of course, this map $T$ is not injective.


问:
On the trace of Sobolev function
Let $\Omega \subset \mathbb{R}^n$ be a $C^2$ domain. Consider any $f \in H^1 (\Omega)$. Does there exist some $ g \in H^2(\Omega)$ such that trace of $f$ and $g$ are same?

答:
This depends on $f$. A basic example is any function in $H^1_0$, which means that the trace vanishes. For these functions the answer is trivially affirmative; $g=0$.

However, the range of the trace operator on $H^1(\Omega)$ is exactly the space $H^{1/2}(\partial \Omega)$. On the other hand, the range of the trace operator on $H^2(\Omega)$ is the space $H^{3/2}(\partial \Omega)$. These are not straigtforward theorems, and frankly I don't know the proof in detail, but if needed I could try and find some reference.

Anyway, the bottom line is that there are functions $f\in H^1(\Omega)$ such that their trace is in $H^{1/2}(\partial \Omega)$ and not in $H^{3/2}(\partial \Omega)$. For these functions, the answer is negative.

Title: On the trace of Sobolev function


问:
Possible subspace of $\mathbb{R}^2$
Which is the possible subspace of  $\mathbb{R}^2$

Any line passing through $\begin{bmatrix}1 & 1 \end{bmatrix}^T$

Any line passing through $\begin{bmatrix}0 & 0 \end{bmatrix}^T$

It is given that the answer is option 2, but I am unable to get that. Can someone help?

答: A line passing through $(1,1)$ may not include the origin; that line is not a vector space by way of not including the origin. A line passing through $(0,0)$ is a vector space; it may be written as $\{rx:r\in\mathbb R\}$ for some nonzero vector $x$.

Title: Possible subspace of $\mathbb{R}^2$


问:
Which of the following statements are true about finite cyclic groups?
$1.$ If $G$ is a finite cyclic group, then for every subgroup $H,K$ of $G,$ we have either $H\subset K$ or $K\subset H.$

$2.$ If for every subgroup $H,K$ of a finite group $G,$ we have either $H\subset K$ or $K\subset H,$ then $G$ is cyclic.

I have figured a counterexample for the first statement:
Let $G=\mathbb{Z}_{6}.$ Then $H=2\mathbb{Z}_{6}$ and $K=3\mathbb{Z}_{6}$ are subgroups of $\mathbb{Z}_{6}$ but neither $H\subset K$ nor $K\subset H.$ So the first statement is wrong.

However, I am not sure about the 2nd statement.

答:
The statement is true. Suppose $G$ is not cyclic, and take an element $g\in G$ of maximal order. Such an element exists because the group is finite. By assumption $\langle g\rangle\ne G$, and so there is some element $x\in G\setminus\langle g\rangle$. Now let $H=\langle g\rangle$ and $K=\langle x\rangle$. We will show that none of these subgroups is contained in the other, and this will be a contradiction.

Since $x\in G\setminus\langle g\rangle=G\setminus H$ it follows that $x\notin H$, and so $K$ is not contained in $H$. So we just have to show $H$ is not contained in $K$. Let's assume $H$ is contained in $K$. Then there is some $n\in\mathbb{N}$ such that $g=x^n$. It is well known that if $ord(x)=m$ then $ord(g)=ord(x^n)=\frac{m}{\gcd(m,n)}$. If $\gcd(m,n)>1$ then it would mean that $ord(g)<ord(x)$, contradicting the assumption that $g$ has maximal order. So we must have $\gcd(m,n)=1$. Thus there are some $k,l\in\mathbb{Z}$ such that:

$$km+ln=1$$

But then note that $x=x^{km+ln}=(x^m)^k(x^n)^l=g^l$, and so $x\in\langle g\rangle$, a contradiction.


问:
Characteristic of a ring: intuitive explanation
I know the following definition of characteristic of a ring: it is the smallest positive $n$ such that $$\underbrace{a+\cdots+a}_{n \text{ summands}} = 0$$
for every element a of the ring, if $n$ exists, otherwise it is zero. However, I don't understand what is the "intuitive" meaning of this. Could you give a physical analogy of anything that may help to see what it is.</p>

答:
I agree with the above answer completely and would like to add little bit more to it.

Most familiar rings to us are $\mathbb{Z}, \mathbb{Q}$ and $\mathbb{Z}/n\mathbb{Z}$ and once derived from them (function rings, polynomial rings, matrix rings etc). It's not difficult to see that all of these are somehow connected to (or obtain from) $\mathbb{Z}.$ We can make this comparison mathematically precise by observing that for any ring $R$ (with unity) there is a canonical ring homomorphism $\varphi : \mathbb{Z}\to R.$

The kernel of this map (the data that we lost during the transformation) is an ideal of (the principal ideal domain) $\mathbb{Z}.$ Therefore we have a smallest (non-negative) integer that generates it, say $m$. This is exactly the characteristic of $R.$

Moreover, by the first isomorphism theorem we have that $$\mathbb{Z}/m\mathbb{Z}\cong \operatorname{im}\varphi,$$ which is a sub-ring of $R.$  Also since $$m.1_R=0$$ the ring $R$ become a $\mathbb{Z}/m\mathbb{Z}$-algebra. An instructive example for this is Boolean rings. Furthermore, if $R$ is an integral domain then $m$ must be a prime number or zero. So, this number $m$ (characteristic) carries some important information about the ring $R.$


问:
$f_n\xrightarrow{wk\star}f$ then $\int_{\mathbb{R}^d}f_n g \to \int_{\mathbb{R}^d}f g$?
I know that if $(f_n)_{n\in\mathbb{N}}$ is a sequence of $L^p(\mathbb{R}^d)$
functions than if $f_n\xrightarrow{wk}f$ (weak convergence), we also have $\int_{\mathbb{R}^d}f_n g \to \int_{\mathbb{R}^d}f g$. But is the same true for weak-$\star$ convergence? If $f_n\xrightarrow{wk\star}f$ then $\int_{\mathbb{R}^d}f_n g \to \int_{\mathbb{R}^d}f g$? (here $1/p+1/q=1$).

答:
Note When $p >1$ then $L^p$ is a reflexive Banach space, and therefore the weak and weak-* topologies coincide.

For $p=1$, $L^1$ is not the dual of anything, see for example the link @rubikscube09 suggested here or this MO post. This means we cannot talk about the weak* topology on $L^1$.


问:
Are direct sums necessarily invertible?
In my linear algebra text, the example used to introduce direct sums (shown below) results in an invertible vector space. I am struggling to understand whether this happened to be the case in this instance, or if this is necessarily true for all direct sums.

答:
The question boils down to this:

```<code>What possible dimension can the sum <span class="math-container">$U_1 + \dots + U_m$</span> have?
</code>```

It is bounded by the sum of the dimensions of the $U_i$, i.e.,
$\dim (U_1 + \dots + U_m) \leq \sum_{i = 1}^m \dim U_i$.
This is essentially true by definition of the (internal) sum: $U_1 + \dots + U_m$ is the subspace generated by the $U_i$, hence the dimension cannot be greater than the sum of the individual dimensions. If you prefer, the internal sum is by definition the image of the map given in your source (another way to see that the inequality holds, see next part).

On the other hand, $\dim (U_1 \times \dots \times U_m) = \sum_i \dim U_i$.

Thus, if the map is injective, the dimension of the sum $U_1 + \dots + U_m$
is at least $\sum_i \dim U_i$. This enforces equality in the inequality.

Then by the usual theorem that for a linear map between vector spaces of equal dimension, being injective and surjective and bijective are equivalent, you get your desired result (the converse of course is trivial).

Small remark, the direct sum itself is not invertible (which makes not a lot of sense in this context).

Title: Are direct sums necessarily invertible?

Body:
The question boils down to this:

What possible dimension can the sum $U_1 + \dots + U_m$ have?

It is bounded by the sum of the dimensions of the $U_i$, i.e.,
$\dim (U_1 + \dots + U_m) \leq \sum_{i = 1}^m \dim U_i$.
This is essentially true by definition of the (internal) sum: $U_1 + \dots + U_m$ is the subspace generated by the $U_i$, hence the dimension cannot be greater than the sum of the individual dimensions. If you prefer, the internal sum is by definition the image of the map given in your source (another way to see that the inequality holds, see next part).

On the other hand, $\dim (U_1 \times \dots \times U_m) = \sum_i \dim U_i$.

Thus, if the map is injective, the dimension of the sum $U_1 + \dots + U_m$
is at least $\sum_i \dim U_i$. This enforces equality in the inequality.

Then by the usual theorem that for a linear map between vector spaces of equal dimension, being injective and surjective and bijective are equivalent, you get your desired result (the converse of course is trivial).

Small remark, the direct sum itself is not invertible (which makes not a lot of sense in this context).


问:
PNT can't be applied Legendre's Conjecture as it can be to Bertrand's Postulate?
1. Background

Bertrand's Postulate says that there is at least one prime between an integer $n$ and its double $2n$.

$$n < p < 2n$$

Although not a proof, the Prime Number Theorem (PNT), that the number of primes up to $x$ denoted $\pi(x)$ is asymptotically equivalent to $x/\ln(x)$, gives us a clue that this is true.

\begin{align}
\frac{\pi(2n)}{\pi(n)} &\sim \frac{2n}{\ln(2n)} \cdot \frac{\ln(n)}{n} \\
& = \frac{2 \ln(n)}{\ln(2)+\ln(n)} \\
& \to 2 \text{ as } n \to \infty
\end{align}

Which says that for large enough $n$, on average we expect there to be $n/\ln(n)$ primes between $n$ and $2n$.

This logic is also presented by the Wikipedia article on Bertrand's postulate, read on 28th Nov 2020.

2. Question

That same wikipedia article also suggests we can't apply similar logic to Legendre's Conjecture, which states there is at least one prime between two consecutive squares:

$$n^2 < p < (n+1)^2$$

Here is the analogous argument, an elaboration of the more concise description in that wikipedia article.

\begin{align}
\frac{\pi((n+1)^2)}{\pi(n^2)} &\sim \frac{(n+1)^2}{\ln(n+1)^2} \cdot \frac{\ln(n^2)}{n^2} \\
& = \frac{(n^2 + 2n + 1) \cdot 2\ln(n)}{2\ln(n+1) \cdot n^2} \\
& = \frac{(1+\frac{2}{n} + \frac{1}{n^2}) \cdot \ln(n)}{\ln(n+1)} \\
& \to 1 \text{ as } n \to \infty
\end{align}

How should I interpret this result?

Is the correct interpretation that this application of the PNT doesn't reveal any information about the number of primes between $n^2$ and $(n+1)^2$ because $\pi(n^2)$ grows too similarly to  $\pi((n+1)^2)$, where "grows too similarly" means asymptotic equivalence?

3. Alternative Conjecture?

We can state a related conjecture:

$$n < p < n^2$$

A similar application of the PNT:

\begin{align}
\frac{\pi(n^2)}{\pi(n)} &\sim \frac{n^2}{\ln(n^2)} \cdot \frac{\ln(n)}{n} \\
& = \frac{n \cdot ln(n)}{2\ln(n)} \\
& = \frac{n}{2} \\
& \to \infty \text{ as } n \to \infty
\end{align}

I'm not an expert, but this suggests that for large $n$, we expect on average more than one prime between $n$ and $n^2$.

What makes the $n^2$ and $(n+1)^2$ case less amenable to this kind of analysis?

答: $2n-n=n$ and $(n+1)^2-n^2=2n+1$ thus there is approximately the same number of primes between $n$ and $2n$, as in $n^2$ and $(n+1)^2$. When talking about equivalents, $\pi(2n)\sim 2\pi(n)$ means that $\pi(2n)-2\pi(n)=o(\pi(n))$ and $\pi((n+1)^2)\sim\pi(n^2)$ means that $\pi((n+1)^2)-\pi(n^2)=o(\pi(n^2))$. But $\pi(n^2)$ grows larger than $\pi(n)$ thus the remainder between $\pi((n+1)^2)$ and $\pi(n^2)$, and $\pi(2n)$ and $\pi(n)$ can be negligeable compared to $\pi(n^2)$, but not compared to $\pi(n)$, this is why it is not detected in your approach of Legendre's conjecture.


问:
Ordinal arithmetic $(2\cdot (\omega+1))\cdot \omega$ and $\gamma = \bigcup \{\omega, \omega^\omega, \omega^{\omega^\omega}\}$ Cantor Normalform
Ordinal arithmetic $(2\cdot (\omega+1))\cdot \omega$ and $\gamma = \bigcup \{\omega, \omega^\omega, \omega^{\omega^\omega}, ...\}$ Cantor Normalform

I understood that $(2\cdot (\omega+1))\cdot \omega=(\omega+1)\cdot \omega = \omega\cdot \omega$ for the first one. Is this correct?

I know that the epsilon nought is $\varepsilon _{0}=\omega ^{\omega ^{\omega ^{\cdot ^{\cdot ^{\cdot }}}}}=\sup\{\omega ,\omega ^{\omega },\omega ^{\omega ^{\omega }},\omega ^{\omega ^{\omega ^{\omega }}},\dots \}$, so $\gamma$ should equal $\varepsilon _{0}$.

Is the attempt correct like this? If so, is $\varepsilon _{0}$ in Cantor Normalform? How to proceed?

答: $2\cdot(\omega+1)=2\cdot\omega+2=\omega+2$, not $\omega+1$, but $(\omega+2)\cdot\omega=\omega\cdot\omega$, so you ended up in the right place. The Cantor normal form of $\epsilon_0$ is $\omega^{\epsilon_0}$.


问:
What's the meaning behind annuity coefficient?
There is a formula for mortgage month payment calculation:
$$
A=P\cdot {\frac  {r(1+r)^{n}}{(1+r)^{n}-1}}
$$
where:

${\displaystyle A}$ is the periodic amortization payment;

${\displaystyle P}$ is the principal amount borrowed;

${\displaystyle r}$ is the rate of interest expressed as a fraction (per month);

${\displaystyle n}$ is the number of payments;

What exactly does the coefficient after ${\displaystyle P}$ represent?
Why don't banks use a much simpler formula: $A=P\cdot {\frac {(1+r)^{n}}{n}}$ ?

答:
Consider the $k$th payment of $A$ in the future. That payment would correspond to a loan of $A(1+r)^{-k}$ in the present.

Summing the present values of the $n$ future payments,

$$\begin{align*}
P &= \sum _{k = 1}^n A(1+r)^{-k}\\
&= \frac A{1+r} \sum_{k=0}^{n-1} (1+r)^{-k}\\
&= \frac A{1+r}\cdot  \frac{1-(1+r)^{-n}}{1-(1+r)^{-1}}\\
&= A\cdot\frac{1-(1+r)^{-n}}{r}\\
&= A\cdot \frac{(1+r)^n - 1}{r(1+r)^n}\\
A &= P\cdot \frac{r(1+r)^n}{(1+r)^n - 1}\\
\end{align*}$$

For your proposed formula $A = \dfrac{P(1+r)^n}{n}$, consider an alternative payment schedule:

to borrow $P$ now,

and to pay $P(1+r)^n$ in full after $n$ months.

The total payment amount is the same as in the annuity payment schedule, yet the alternative allows a later payment.

This shows that either the annuity amount is not valued correctly, or the lender may be using the formula to promote a lower interest rate than the effective rate.


问:
Expected number of ones before the first 2
Let $(X_i)_{i\geq1}$ be a sequence of iid. random variables with

$$P(X_i=1)=p_1, P(X_i=2)=p_2, P(X_i=3)=p_3$$

and $p_1+p_2+p_3=1$.

We want to compute the expectation of the random number of 1s in a sample before the first 2 appears.

Let's call this random number $T$.

I found the solution using first step analysis:

$$E(T)=p_1\cdot(E(T)+1)+p_2\cdot 0+p_3\cdot E(T)\Leftrightarrow E(T)=\frac{p_1}{p_2}.$$

In class we discussed the following approach, but couldn't come to a conclusion. Let $N_2$ be the first time a 2 appears. Then

$$T=\sum_{k=1}^{N_2}1_{X_i=1}$$

where $1_{X_i=1}$ is the indicator function of the event, that $X_i$ takes the value 1.

If I slap the expectation onto both sides, i should be able to use the Wald-Identity (a priori without checking the assumptions for it). Then

$$E(T)=E(N_2)P(X_i=1)$$

$N_2$ is geometric with parameter $p_2$, so $E(N_2)=\frac{1}{p_2}$ and $P(X_i=1)=p_1$.

Although this gives the right solution, i don't see the independence of $N_2$ and $1_{X_i=1}$.

Is there a way to show this without the optimal stopping theorem?

答:
I finally got it. One can prove it directly, no Wald Identity needed explicitly.

$$E\left(\left.\sum_{i=1}^{N_2-1}1_{X_i=1}\right|N_2=k\right)=\sum_{i=1}^{k-1}E(1_{X_i=1}|N_2=k)=(k-1)\cdot P(X_1=1|X_1\neq 2).$$

Indeed $P(X_1=1,X_1\neq2)=\frac{p_1}{p_1+p_3}$ and $N_2$ ist geometric with parameter $p_2$, so $E(N_2)=\frac1{p_2}.$

So,
$$E\left(\left.\sum_{i=1}^{N_2-1}1_{X_i=1}\right|N_2\right)=(N_2-1)\cdot \frac{p_1}{p_1+p_3}=\frac{p_1}{p_2}$$


问:
$\lim _{x\rightarrow 0^{-}}e^\left(1/x\right)=0$ ($\delta-\varepsilon$ definition)
$\lim _{x\rightarrow 0^{-}}e^\left(1/x\right)=0$ So I wrote the definition:
$\forall \varepsilon>0, \exists \delta_{\varepsilon}$ such that if $x_{0}-\delta<x<x_{0}$ then $\left|f(x)-L\right|<\varepsilon$
So let $\varepsilon>0$ be given. Choose $\delta=-\dfrac{1}{\ln(\varepsilon)}$ if $-\delta<x<0$ implies that $\left|e^\left(1/x\right)\right|<\varepsilon$. So did I make a mistake? Because when I take $\varepsilon=0.001$ I didn't get closer to $0$
$\varepsilon=0.001$ then $\delta=0.144$ then the interval is $0.144<x<0$. So for example if I take $0.140$ I can't even get closer to $0$. For instance: $e^\left(1/0.140\right)=1265.03...$

答:
Option:

Consider $y:=-x$ and limit $y \rightarrow 0^{+}$.

$|e^{-1/y}|=|\dfrac{1}{e^{1/y}}|=|\dfrac{1}{1+1/y....}|\le |y|.$

Choose $\delta=\epsilon.$


问:
Solving dependent systems
When I'm solving a system of equations and realize that I have a dependent system, I need to express the answer in terms of y = {some value} where x is any real number, OR x = {some value} where y is any real number. How do I choose whether to

For instance, I have this problem in my homework:

```<code>x+4y = 5
5x + 20y = 25

x = 5-4y

5(5-4y) + 20 y = 25
25-20y+20y = 25
25 = 25 : DEPENDENT SYSTEM
</code>```

I then solve for x using the first equation:

```<code>x + 4y = 5
</code>```

And arrive at the answer of:

```<code>x = -4y + 5, where y is any real number
</code>```

However, the answer in the back of the packet is:

```<code>y = -(x/4) + (5/4) where x is any real number
</code>```

I see how they arrived at this, but my question is: Why did they choose to solve for y instead of solving for x? Is this choice arbitrary? Would my answer be correct as well?

答:
It's just a standard mathematical convention to express a function in 2D space as $x$ being the independent variable, and $y$ being the dependent variable (dependent on $x$). Your answer is still correct, just a different way of expressing it.

This is because $x$ is usually the horizontal axis, and $y$ is usually the vertical axis, and looking at a function from left to right is easier to interpret than when looking at it from top to bottom.

Title: Solving dependent systems


问:
Exposure At Default: Calculating the present value

In this numerical example, I can't figure out with which numbers (when using the PV formula) to calculate exposure at default (EAD) as shown in the table.

The EAD is the value of the discounted future cashflows (CF) at the time of default.

With my calculations I do not get the EAD shown there starting from t=2. How do I replicate the the EAD in the table?

The following parameters are given in the calculation:

```<code>nominal amount: 1000
Duration: 6 years
Interest rate: 10%.
Effective interest rate: 10%.
Date of payment of interest: Annual
Credit structure: maturing loan
</code>```

答:
They are computing the present value, at the time of the loan, of the defaulted payments, using the same rate as the rate of the loan.  If the debtor defaults on the last payment, the present value is $$1100*1.1^{-6}=620.92$$

If he default at the end of year $5$ that will add another $$100*1.1^{-5}=62.09$$ bringing the total to $683.01$.


问:
Probability of selecting identical objects from 2 partially overlapping groups
Suppose there are two groups of objects. Group $1$ has $50$ unique objects, and Group $2$ has $100$ unique objects. These groups share $40$ identical objects (i.e. $40/50$ for group $1$ and $40/100$ for group $2$). Given that $20$ objects are selected from group $1$ and $15$ objects are selected from group $2$, what is the probability that $10$ objects will be identical between the two selections.

My main issue with this problem is that the two groups are different sizes with partial overlap. I'm not sure how to take that into account when I calculate some sort of cumulative probability for each selection. Further, even if a shared object is selected from each group, how do I account for the fact that the same shared object may not be selected from both groups? What I've been doing now is just simulating random selections, but I'd like to know if there is some probability calculation that's applicable.

答:
I think this can be solved by elementary means, if I understand the problem correctly. This feels to simple to be right, so do correct me if I'm wrong.

There are ${50 \choose 20}$ ways to choose the objects from group $1$, and ${100 \choose 15}$ ways to choose objects from group $2$. Hence,  ${50 \choose 20} \cdot {100 \choose 15}$ ways to make the two choices.

How many ways are there to choose the objects with $10$ element overlap? First, we select the $10$ common elements in ${40 \choose 10}$ ways. Next, we choose the remaining $10$ out of the group $1$, only we need to keep track of how many are in the common part of the two groups. Say we want $k$ elements from there (and $10-k$ from outside). Then, there are $10 \choose 10 -k$ ways to select the ones from outside the common part, and $30 \choose k$ to choose the ones from within the common part, giving the total of ${10 \choose 10 -k} \cdot {30 \choose k}$. Finally, we want to choose the $5$ remaining elements in group $2$. There were $100$ elements there initially, but we already chose $10$ out of them, and there are $k$ that we don't want to choose (else, the overlap would be larger than $10$). This gives ${90 - k \choose 5}$ ways. It remains to muliply things out and sum over $k$, to get the total number of ways:
$$ {40 \choose 10} \sum_{k=0}^{10} {10 \choose 10 -k} \cdot {30 \choose k} \cdot {90 - k \choose 5}$$
I am not sure if this can be simplified.

To get the probablity, divide this value by the number of ways to make the choices. The numerical value is $0.00690417...$.


问:
Can you just assume without proof for coprime bases?
Let $x,y ∈ \mathbb{Z}$ and $a,b ∈ \mathbb{Z+}$, where $a\ne b$ and a,b are coprime

Let $a^x=b^y$

The only solution is when $x = 0$ and $y = 0$, but is it necessary to prove it?

I've got as far as saying:

$a^x$ has factors $a^n$, where $1 \le n \le x$

Likewise, $b^y$ has factors $b^m$, where $1 \le m \le y$

As $a,b$ are coprime, so are $a^n$ and $b^n$. In order for $a^x=b^y$, the "answer" ($a^x$ and $b^y$) must have:

(factors $a^n$) $\cap$ (factors $b^m$). These are mutually exclusive conditions, so it's impossible

Hence, $a^x$ and $b^y$ both cannot have factors $a$ or $b$. This only happens when $x, y = 0$

Is this sufficient, or indeed necessary? Thanks for any help

答:
That is not correct. You have no reason to assume that $a$ or $b$ is a prime number.

Besides, the statement is clearly false if $a=1$ or $b=1$.

Otherwise, let $p$ be a prime factor of $a$. Then $p$ is also a prime factor of $a^x$ and, if $a^x=b^y$, then $p$ will also be a prime factor of $b^y$, if $y>0$. But then it's a prime factor of $b$. This is impossible, since $a$ and $b$ are coprime. So, $y=0$. By the same argument, if $p$ is a prime factor of $b$ and $a^x=b^y$, then $x=0$. Since both $x$ and $y$ must have prime factors, this proves that $x=y=0$.


问:
$| (a+b)^n - a^n| \leq n |b|(|a| + |b|)^{n-1}$
Let $a,b \in \mathbb{C}$, prove that for $n\geq 1$

$$| (a+b)^n - a^n| \leq  n |b|\;(|a| + |b|)^{n-1}$$

I tried using induction

For $n = 1$, $\;| (a+b) - a| = |b| =  1\times |b|\times(|a| + |b|)^{1-1}$

Assume $| (a+b)^n - a^n| \leq  n |b|\;(|a| + |b|)^{n-1}$

$$| (a+b)^{n+1} - a^{n+1}|  = |b| \;| (a+b)^{n} + (a+b)^{n-1}a + \ldots + a^n| $$

But I am unable to figure out how to use the inductive step in this.

答:
Rewrite $\:(a+b)^{n+1}-a^{n+1}$ as
$$\bigl((a+b)^n-a^n\bigr)(a+b)+a^nb$$
and use   the inductive hypothesis and the triangle inequality:
\begin{align}
\bigl|(a+b)^{n+1}-a^{n+1}\bigr|& \le
\bigl|(a+b)^n-a^n\bigr||a+b|+|a^n||b|\\
&\le n|b|\bigl(|a|+|b|\bigr)^{n-1}\bigl(|a|+|b|\bigr)+|a|^n|b|\\
&\le n|b|\bigl(|a|+|b|\bigr)^n+\bigl(|a|+|b|\bigr)^n||b|\\
&=(n+1)|b|\bigl(|a|+|b|\bigr)^n
\end{align}


问:
Iterative matrix system - Convergence
Let $k = 0, 1 \dotsc,$ and consider the following system
\begin{align*}
\mathbf{u}^{(0)} &:= \mathbf{f}, \\
\mathbf{u}^{(k+1)} &:= \mathbf{H}\mathbf{u}^{(k)}.
\end{align*}
where $\mathbf{H} \in \mathbb{R}^{d \times d}$ is a symmetric Matrix such that $h_{ij} \geq 0$ and the sum of each row is equal to $1$: $\sum_{j=1}^{d} h_{ij} = 1$ for all $i$.

Assume that $\rho(\mathbf{H}) = \max |\lambda_i| = 1$ the biggest eigenvalue of $\mathbf{H}$. Then $\lim_{k \to \infty} \mathbf{u}^{(k)} = \mu \mathbf{1}$ where $\mu = \sum_{j=1}^{d} \frac{1}{d} f_j$.

Attempt: Because of symmetry, the columns of $\mathbf{H}$ also sum to $1$ and clearly also $\mathbf{H} \cdot \mathbf{1} = \mathbf{1}$ eigenvector and also $\mathbf{H}^k \cdot \mathbf{1} = \mathbf{1}$. Taking the two-norm on $\mathbb{R}^{d}$, we have
\begin{align*}
||\mathbf{u}^{(k)} - \mu \mathbf{1}||_2 &=  ||\mathbf{H}^k \cdot \mathbf{f} - \mathbf{H}^k \cdot  (\mu  \mathbf{1})||_2 \\[5 pt]
&\leq  ||\mathbf{H}^k||_{2}  \ ||\mathbf{f} - \mu  \mathbf{1}||_2 \\[5 pt]
&= \rho(\mathbf{H}) \ ||\mathbf{f} - \mu  \mathbf{1}||_2 = ||\mathbf{f} - \mu  \mathbf{1}||_2.
\end{align*}
I am not really sure what to conclude further. Also, component-wise, we have $$ |\mathbf{u}_i^{(k)} - \sum_{j=1}^{d} \frac{1}{d} f_j| \leq |\mathbf{u}_i^{(k)}| + f_{max} \leq 2 f_{max},$$
where $f_{max} = ||\mathbf{f}||_{\infty}$ is maximum component of $\mathbf{f}$ and the inequality follows because $|\mathbf{u}_{max}^{(k)}| \leq |\mathbf{u}_{max}^{(k-1)}|.$

答:
As noted in the comments, this assertion is not true if we only require that the entries be non-negative, just by taking $\mathbf{H} = \mathbf{I}_d$.

The easiest way to prove this assertion whenever $\mathbf{H}$ is strictly positive (or by extension, when it is non-negative and irreducible) is by using the Perron-Frobenius Theorem, which asserts that the eigenvalue of greatest magnitude (call it $\rho$) has an eigenspace that is one-dimensional, i.e. there is only one non-zero linearly independent vector $\mathbf{v}$ such that $\mathbf{H} \mathbf{v} = \rho \mathbf{v}$. It further asserts that all the other eigenvalues of $\mathbf{H}$ are strictly smaller than $\rho$ in magnitude. From the given information about $\mathbf{H}$, we can see that $\rho = 1$ and the corresponding eigenvector is $\mathbf{1}$.

Since in addition $\mathbf{H}$ is symmetric, we know from the Spectral Theorem that it admits an orthonormal basis of eigenvectors, of which $\mathbf{v}_1 = \frac{1}{\sqrt d} \mathbf{1}$ is one. Let this basis of eigenvectors be $\{\mathbf{v}_1, \mathbf{v}_2, \cdots, \mathbf{v}_d\}$, with corresponding eigenvalues $\lambda_1, \cdots, \lambda_d$. Then $\mathbf{f}$ can be written as a linear combination of these eigenvectors:
$$\mathbf{f} = c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_d \mathbf{v}_d$$
and note that $c_1 = \langle \mathbf{f}, \mathbf{v}_1 \rangle = \frac{1}{\sqrt d} \sum_{j = 1}^d f_j = \mu \sqrt{d}$ (why?). So we have that
$$\mathbf{H}^k \mathbf{f} = c_1 \lambda_1^k \mathbf{v}_1 + c_2 \lambda_2^k \mathbf{v}_2 +  \cdots + c_1 \lambda_d^k \mathbf{v}_d = \mu \mathbf{1} + c_2 \lambda_2^k \mathbf{v}_2 +  \cdots + c_1 \lambda_d^k \mathbf{v}_d$$
and you can see that
$$\| \mathbf{H}^k \mathbf{f} - \mu \mathbf{1}\| = \|c_2 \lambda_2^k \mathbf{v}_2 +  \cdots + c_1 \lambda_d^k \mathbf{v}_d\| \leq |c_2| \cdot |\lambda_2|^k + \cdots + |c_d| \cdot |\lambda_d|^k$$
(by the triangle inequality) and thus tends to zero as $k \to \infty$. This concludes the proof. $\square$

Title: Iterative matrix system - Convergence


问:
Proving if $b^k = a$ and $\text{ord}(a) = n$ then $\text{ord}(b) = kn$.
Let $(G,e)$ be a group and $a \in G$ have finite order

$\quad \text{ord}(a) = n$

and let $\langle a \rangle$ denote the cyclic group generated by $a$.

Suppose for $b \in G$ and $k \ge 2$

$\quad b,\dots, b^{k-1} \notin \langle a \rangle$ and $b^k = a$

Then the order of $b$ is $kn$.

Proof

The order of $b$ must be a multiple of $n$ since $\langle a \rangle \subset \langle b \rangle$.

The order of $b$ must divide $kn$ since $b^{kn} = e$.

All that remains is to identify $kn$ distinct elements in $\langle b \rangle$.

Consider the mapping

$\quad (u,v) \mapsto a^u b^v \quad \text{where } 0 \le u \lt n \land 0 \le v \lt k$

Our work will be complete if we can show that this mapping is injective. This is accomplished by using the fact that the $b^v$ can never be a non-trivial inverses for any elements in $\langle a \rangle$.

Suppose $a^u b^v = a^s b^t$ and $u = s$. Then $v$ must be equal to $t$.

So assume, without loss of generality, that $u \gt s$. Then we can write

$\quad a^w b^v = b^t$

with $0 \lt w \lt n$.

If $v = t$ we have a contradiction since $a$ has order $n$.

If $v \gt t$ we have a contradiction since we can't construct a non-trivial inverse.

If $v \lt t$ we have a contradiction since $b^{t-u} \notin \langle a \rangle$.

This completes the proof.

Is this a valid proof?

It seems fine to me but the reason for posting this question is I could not find this on the internet of math facts. I could not find this (fact?) as a duplicate question on this site or anywhere else.

So any links to the literature that uses this would be of interest.

答:
Your proof is correct. A far shorter proof follows.

For $1\le x<kn$, $b^x\ne e$ since
$$b^x=(b^k)^{\lfloor x/k\rfloor}b^{x\bmod k}=a^{\lfloor x/k\rfloor}b^{x\bmod k}$$
Either $x\bmod k>0$ and $b^x\not\in\langle a\rangle$ because the first factor is in $\langle a\rangle$ and the second is not, or $x\bmod k=0$ but $\lfloor x/k\rfloor\in[1,n)$ and $b^x=a^{\lfloor x/k\rfloor}\ne e$.

Yet it is easy to show that $b^{kn}=e$. Hence the order of $b$ is $kn$.


问:
Checking independence of random variables.
I'm revisiting the coupon collector's problem and I'm not sure how to prove that my variables are independent. Here's what I have:

Let $X$ denote the number of tries required to collect all the coupons, and $X_i$ denote the number of tries required to collect a new coupon, after having collected $i-1$ coupons. Then $X_i \sim Geom(p_i)$ with $p_i= \frac{n-i+1}{n}$.

Intuitively they should be independent but how would one go about proving this? Would calculating $E[X_i X_j]$ and showing it equal to $E[X_i]E[X_j]$ be enough? Is this the right way to proceed?

答:
It's certainly not enough to prove that $E[X_iX_j] = E[X_i]E[X_j]$  It's quite easy to construct dependent random variables that satisfy the above.
Suppose $D$ is the result when I through I throw a single die and set
$$\begin{array}{r|cccccc}D&1&2&3&4&5&6 \\ \hline X & -5&-4&-3&3&4&5 \\Y & 5&-4&-3&3&4&-5 \end{array}$$

Clearly not independent but $E[XY] = E[X]E[Y]$ because that's how I chose my random variables, $X$ and $Y$ are independent if and only if for every pair $x,y$ we have $P\{X=x,Y=y\} = P\{X=x\}P\{Y=y\}$.

To show independence of an infinite list of random variables you need to do a bit more work.

We need to show that finite $n$ and any sequence $x_1,\dots,x_n$ we have
$$P\{X_1=x_1,\dots X_n = x_n\} = P\{X_1=x_1\}\times\dots\times P\{ X_n = x_n\} $$

This is obviously true for $n=1$.

Now assume inductively that the above is true for $n=j$ and set $s_j =\sum_{i=1}^j x_i$ and pick an arbirtary test value of $x_{j+1}$

The event $\{X_1=x_1,\dots X_j = x_j\}$ occurs in the first $s_j$ draws
Given that this does occur the event $\{X_{j+1}= x_{j+1}\}$ only involves the results of draws after $S_j$ so these two events are independent.

Therefore
$$\begin{array}{rcl}
P\{X_1=x_1,\dots X_{j+1} = x_{j+1}\} &=& P\{X_1=x_1,\dots X_{j} = x_{j}\}P\{X_{j+1}=x_{j+1}\} \\
&=& P\{X_1=x_1\}\times\dots\times P\{ X_{j+1} = x_{j+1}\}\end{array}$$

Therefore by induction the statement holds for every $n$ and the sequence of random variables $X_1,X_2,\dots$ is independent.


问:
Given a Geodesic square what is the area of the square?
I need help with this question

Given a Geodesic square on unit ball, The interiors angle are 2 radians each,
What is the area of the square?

答:
For a geodesic polygon on a ball with radius $r$, the area is $(2\pi - \sum_k {\alpha}_k)r^2$ where $\alpha_k$ is the $k^{th}$ signed exterior angle.
If one walks around the curve bounding the polygon in counterclockwise manner. If at $k^{th}$ corner the curve turns to the left, $\alpha_k$ will be positive. if the curve turns to the right, $\alpha_k$ will be negative.

This is a special case of Gauss Bonnet theorem. see wiki entry for more info.

For the polygon at hand, all $\alpha_k = \pi - 2$ with $r = 1$. This means the area of the geodesic square is

$$2\pi - 4(\pi - 2) = 8-2\pi$$


问:
sequence of bounded random variables form martingale then they are uncorrelated
Let $X_i$ be a sequence of bounded random variables such that $S_n=\sum\limits_{i=1}^{n} X_i$ is an martinagle w.r.t the filtration $\mathcal F$. Could anyone tell me how to show $$Cov(X_i, X_j)=0\quad \forall i\ne j$$

Thanks.

答:
Using the fact that $(S_n)$ is a martingale with respect to $(\mathcal{F}_n)$, you can show that $E[X_{n+1} \mid \mathcal{F}_n] = 0$.

So, if $i < j$, one can show that $E[X_j] = E[E[X_j \mid \mathcal{F}_i]] = 0$ and $E[X_i] = E[E[X_i \mid \mathcal{F}_0]] = 0$, so

$$\text{Cov}(X_i ,X_j) = E[X_i X_j] = E[X_i E[X_j \mid \mathcal{F}_i]] = 0.$$


问:
Proof verification of $6^n \geq n3^n$
n is a natural number. Prove $6^n \geq n3^n$ holds for every natural number.

$n = 1:$

$$6 \geq 3 $$

$n \rightarrow n + 1:$

$$6 ^{n+1} = (3*2)^{n+1} = 3^{n+1}2^{n+1} \geq 3^{n+1}(n+1)2\geq 3^{n+1}(n+1) $$

Is my proof correct?

EDIT:
$$2^{n+1} \geq(n+1)2 \text{ because of Bernoulli inequality}$$

答:
Let us prove that $$(\forall n\ge 1)\;\; P_n : 2^n\ge n$$

first step
$$2^1=2\ge 1 \implies P_1$$

second step

Let $ n\ge 1 $ such that $ 2^n\ge n$.

$$2^n\ge n \implies 2^{n+1}\ge 2n$$

$$\implies 2^{n+1}\ge n+ n$$
$$\implies 2^{n+1}\ge n+ 1$$

Now, just replace $$2^n=\frac{6^n}{3^n}$$

Title: Proof verification of $6^n \geq n3^n$


问:
Is the area of a line really $0$?
Let's take a square with side $A$. The area of this is defined as $A\times A$.
The way I explained this to myself is by reapeatedly deviding the square till I reached a single line.
Then stacking up lines one on top of the other $A$ times.

But if the area of a line was really $0$ no matter how many lines were stacked up, they would always have an area of $0$.(would exist in one dimension only)

Another way to see this is defining a function $V(F)$ which gives the volume of two dimensional figure $F$.
If this volume became $0$ , no matter how many such Figures I stack up on each other they would never have a volume.

When this is combined with physics it makes a little more sense.
Even if I were to chop up a cube infinitely I could never reduce its height to less than the Planck length. This means that the volume would be $Area \times  ℓP$ (Planck length)

So the question is for a three dimensional object to exist must every two dimensional object have a non zero volume?

答: Your interpretation of the area of the square makes sense if each of your "lines" has a width of $1$ (that is why you have exactly $A$ such lines). Usual lines have zero width and you need to have uncountably infinitely many of them to obtain a height of $A$.

Title: Is the area of a line really $0$?


问:
Proof on showing if F(x,y,z)=0 then product of partial derivatives (evaluated at an assigned coordinate) is -1
The task is as follows:

Given: $$F(x,y,z) = 0$$    Goal:  Show    $\frac{\partial
 z}{\partial y}|_x \frac{\partial  y}{\partial x}|_z \frac{\partial x}{\partial z} |_y = -1$

Here is my work so far:

(1) Differentiate with respect to y, I get:

$0 + F_2 + F_3  \frac{\partial z}{\partial y} = 0$

So $ F_3  \frac{\partial z}{\partial y} = - F_2$

(2) Differentiate with respect to x, I get:

$F_1 + F_2 \frac{\partial y}{\partial x}  + 0 = 0$

So $ F_2  \frac{\partial y}{\partial x} = - F_1$

(3) Differentiate with respect to z, I get:

$F_1  \frac{\partial x}{\partial z}    + 0  + F_3 = 0$

So $ F_1  \frac{\partial x}{\partial z} = - F_3$

(4) After some manipulations with the $F_i$, I get to the conclusion that $\frac{\partial z}{\partial y}* \frac{\partial y}{\partial x} * \frac{\partial x}{\partial z}  = -1$, so when evaluated with x, z, y respectively, conclusion is still true

My questions are:

(1) Is my proof correct?

(2) For example, when I differentiate with respect to y, I "let" $F_1$ be 0 and find partials for other coordinates.

I had a hard time trying to explain to my friend on the reason(s) why I can do such "let be 0" thing.  Although I think if I can't do that, then there is no way that I can reach the conclusion, but I somehow feel confused about the fact too.  Since my book is doing it that way, my understanding is that I can do such "let be 0" thing based on the independece of x with respect to y, when I differentiate with respect to y.  But is my thought ok?

Would someone please help me on this question?
Thank you very much ^_^

答:
A more simple way is the total differential
$$dF=\frac {\partial F}{\partial x} dx+\frac {\partial F}{\partial y} dy+\frac {\partial F}{\partial z} dz=0$$
If it says "evaluated at $x$" it means that $x$ is fixed and $dx=0$
$$\frac {\partial F}{\partial y} dy+\frac {\partial F}{\partial z} dz=0\Rightarrow \frac{dz}{dy}=-\frac{\partial F/\partial y}{\partial F/\partial z}$$
and similarly
$$dy=0\Rightarrow\frac {\partial F}{\partial x} dx+\frac {\partial F}{\partial z} dz=0\Rightarrow \frac{dx}{dz}=-\frac{\partial F/\partial z}{\partial F/\partial x}$$
$$dz=0\Rightarrow\frac {\partial F}{\partial x} dx+\frac {\partial F}{\partial y} dy=0\Rightarrow \frac{dy}{dx}=-\frac{\partial F/\partial x}{\partial F/\partial y}$$
And it follows
$$\frac{dz}{dy}\frac{dy}{dx}\frac{dx}{dz}=\bigg(-\frac{\partial F/\partial y}{\partial F/\partial z}\bigg)\bigg(-\frac{\partial F/\partial x}{\partial F/\partial y}\bigg)\bigg(-\frac{\partial F/\partial z}{\partial F/\partial x}\bigg)=-1$$</p>


答:
Suppose $6^n\ge n\cdot 3^n$ is true for $n$ and let's prove it for $(n+1)$.

$6^{n+1}=6\cdot 6^n\ge 6\left(n\cdot 3^n\right)=2\cdot 3\left(n\cdot 3^n\right)=2 n\cdot 3^{n+1}\ge (n+1)3^{n+1}$

proved.

Title: Proof verification of $6^n \geq n3^n$

Body:
n is a natural number. Prove $6^n \geq n3^n$ holds for every natural number.

$n = 1:$

$$6 \geq 3 $$

$n \rightarrow n + 1:$

$$6 ^{n+1} = (3*2)^{n+1} = 3^{n+1}2^{n+1} \geq 3^{n+1}(n+1)2\geq 3^{n+1}(n+1) $$

Is my proof correct?

EDIT:
$$2^{n+1} \geq(n+1)2 \text{ because of Bernoulli inequality}$$


问:
How do I show that the ideal is not principal?
Let $S$ be the set of polynomials $p(t)$ in the ring $\mathbb{Z}[t]$ for which $p(1)$ is even. Is $S$ an ideal, and if so, is it principal?

I proved the ideal part. Now, I suppose the ideal $S$ is principal, i.e $S=(f(x))$, then $f(x)+2 \in S$, there exists $g(x)$ such that $f(x)+2=f(x)g(x)$. I tried to conclude this, but not able to conclude it correctly. Can anyone suggest some point for the last step?

答: If $S=(f)$ then in particular $f|2$, because the constant polynomial $2$ belongs to $S$. This implies $f$ has to be a constant polynomial. $S$ is clearly not the whole ring, and so $f\notin\{1,-1\}$. So we must have $f\in\{2,-2\}$, and anyway $S=(2)$. But this is a contradiction, for example because $t+1\in S$, and this polynomial is not divisible by $2$.


问:
Show if $\operatorname{diam}(A) < r$ $\implies$ $A \subset B_{r}(a)$ for $a \in A$.
I've seen this problem mentioned on Math.SE before, but not proved thoroughly. With that said I wanted to attempt it for myself and have my work verified.

Let $(M,d)$ be a metric space and $A \subset M$. If $\operatorname{diam}(A) < r$, show that $A \subset B_{r}(a)$ for some $a \in A$.

The proof for this seems pretty short:

Let $a_{o}, b \in A$ and $r > 0$. Since $\operatorname{diam}(A) := \sup\{d(a,b): a, b \in A\} < r$, then $d(a_o,b) < \operatorname{diam}(A) < r$. So since $d(a_{o},b) <r$ by definition of an open ball of radius $r$, $b \in B_{r}(a_{o})$ for some $a_{o} \in A$. Therefore $A \subset B_{r}(a_{o})$.

Any criticism is welcome.

答: It is almost correct. The only (minor) problem is when you state that $d(a_0,b)<\operatorname{diam}(A)$. This is not true; all you can say is that $d(a_0,b)\leqslant\operatorname{diam}(A)$. It it still follow from this that $d(a_0,b)<r$.


问:
Linear Independence and proving Span (B1) = Span (B2)
We let $$X = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & -5 \\ 0 & 1 & -2 \\ \end{bmatrix}.$$Let $B_1 = \{ I_3,2X,3X^2\}$, $B_2 = \{X^2,(X-I_3)^2,X^2+I_3\}$. Show that $B_1$ and $B_2$ are linearly independent sets and $span(B_1)=span(B_2)$.

The provided answer is:

Reviewing this answer I can't seem to understand how they concluded from $\{I, x , x^2\}$ that $B_1$ is also a linearly independent set.
Moreover, I also do not understand how they concluded that $span(B_1) = span(B_2)$.
Any help would be great, thank you so much!

答:
Let $V$ be the vector space of your exercise, generated by the basis $C$.

For the first question the reason is that $C=\{I_3,X,X^2\}$ is a basis of $V$ (wa (?)) and for definition its vectors generate the space and they are linearly independents.

If we now take any of these three matrices ($M(3,\mathbb{R})\cong \mathbb R^9$) multiplied by a scalar they still generate all the vector space because the direction of these three vectors  is the same of the vectors of $C$.

In your case, since the elements of $B_1=\{I_3,2X,3X^2\}$ are multiple of the elements of $C$, these keep being independent vectors and their linear combination generate the space $V$.
It's now easy to prove that $$\forall A_i \in B_2 \nexists \lambda_1,\lambda_2\in \mathbb R:A=\lambda_1 A_j+\lambda_2 A_k,$$ where $A_i,A_j,A_k$ are elements of $B_2$.

In conclusion,  $3=\text{dim(span}(B_1))=\text{dim(span}(B_2))=\text{dim(span}(V))$, so $B_1$ and $B_2$  generate $V$ because their number of linearly independent elements is the same of the dimension of $V$.


问:
Find the limit of $a_1 = 1 $ , $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$
$a_1 = 1 $ ,  $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$

It is easy to prove that the limit exists: the boundary of that expression is $0$ and it is monotonically decreasing.

The problem is to actually find the limit (it is $0$) because if I take arithmetic of limits:

$ \lim_{x \to +\infty} a_{n+1} =  \frac{\sqrt{1+ \lim_{n \to +\infty} a_n^2+1}}{\lim_{n \to +\infty} a_n}$

$g = \frac{\sqrt{1+g^2+1}}{g}  \implies g^2 = {\sqrt{1+g^2}-1} \implies g^2 + 1 = {\sqrt{1+g^2}} \implies 0 = 0$

Body:
$$a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}=\\ \frac{\sqrt{1+a_n^2}-1}{a_n}.\frac{\sqrt{1+a_n^2}+1}{\sqrt{1+a_n^2}+1}=\\\frac{1+a_n^2-1}{a_n(\sqrt{1+a_n^2}+1)}=\\
\frac{a_n}{(\sqrt{1+a_n^2}+1)}\leq \frac{a_n}{(\sqrt{1}+1)}=\frac{a_n}{2} $$so $$0<a_{n+1}\leq \frac 12a_n\leq\frac14 a_{n-1}\leq\frac18a_{n-2}\cdots<1$$


问:
Oscillatory integral with absolute value
Suppose $f:[0,1]\to[0,+\infty)$ is a continuous function. Are there general conditions under which the following limit exists?

$$\lim_{n\rightarrow \infty} \int_0^1 |\cos(nf(x))|dx$$

Obviously if $f$ is a constant that is not an integer multiple of $\pi$ the limit does not exist. Does the limit exist if $f'(x)\neq 0$ for all $x \in [0,1]$? If so, what if we drop differentiability and only require $f$  not be locally constant?

When $f$ is not differentiable, can one still show
$$\liminf_{n \rightarrow \infty} \int_0^1 |\cos(nf(x))|dx >0$$
under suitable conditions on $f$?

答:
Proposition. Let $f:[0,1]\to\Bbb R$ be a continuously differentiable function such that $f’(x)\ne 0$ for each $x\in [0,1]$. Then $\lim_{n\rightarrow \infty} \int_0^1 |\cos(nf(x))|dx=\tfrac {2}\pi.$

Proof. Since $f’$ is a continuous function on a compact set $[0,1]$, it is uniformly continuous and $\ell=\inf \{|f’(x)|:x\in [0,1]\}>0$. Fix any $\varepsilon>0$. Since $f’(x)$ is uniformly continuous, there exists a natural number $N$ such that
$|f’(y)-f’(x)|\le \varepsilon$ for any $x,y\in [0,1]$ such that  $|y-x|\le\tfrac \pi{N\ell}$.

Let $n\ge N$ be any natural number. Put $X_n=\left\{x\in [0,1]:nf(x)\in \tfrac \pi2+\pi\Bbb Z \right\}$. Since $\ell>0$, $f$ is strictly monotonic, so the set $X_n$ is finite.  Let $x_1<x_2<\dots<x_m$ be an enumeration of $X_n$. Put $x_0=0$ and $x_{m+1}=1$. Then $x_{i+1}-x_i\le\tfrac \pi{n\ell}$ for each $i=0,\dots m$. Put $I_i=\int_{x_i}^{x_{i+1}} |\cos(nf(x))|dx$.

Then $I=\int_0^1 |\cos(nf(x))|dx=\sum_{i=0}^m I_i$. Clearly, $|I_0|, |I_m|\le \tfrac \pi{n\ell}$. Let $i$ be any integer from $1$ to $m-1$. By Lagrange’s theorem there exists $z\in (x_i, x_{i+1})$ such that $\tfrac {f(x_{i+1})-f(x_i)} {x_{i+1}-x_i}=f’(c)$. For each $y\in [x_i, x_{i+1}]$ put $g(y)=f(y)-f(x_i)-(y-x_i)f’(c).$ It is easy to check that $g(x_i)=0$ and for each $y$
$$|g’(y)|=|f’(y)-f’(c)|\le\varepsilon.$$
By Lagrange’s theorem, there exists $z\in (x_i, y)$ such that

$$|g(y)|=|g(y)-g(x_i)|=|g’(z)|(y-x_i) \le \varepsilon\frac \pi{n\ell}.$$ Again by  Lagrange’s theorem, there exists $d$ between $f(y)$ and $(f(x_i)+(y-x_i)f’(c))$ such that

$$|\cos (nf(y))-\cos n(f(x_i)+(y-x_i)f’(c))|=n|\sin nd|| f(y)- f(x_i)-(y-x_i)f’(c)|\le n |g(y)|\le \varepsilon\frac \pi{\ell}.$$

We have $|I_i- I_i’|\le I_i’’$, where $I’_i=\int_{x_i}^{x_{i+1}} |\cos n(f(x_i)+(y-x_i)f’(c))| dy $ and
$$I_i’’=\int_{x_i}^{x_{i+1}} |\cos (nf(y))-\cos n(f(x_i)+(y-x_i)f’(c))| dy\le \int_{x_i}^{x_{i+1}} \varepsilon\frac \pi{\ell} dy\le
(x_{i+1}-x_i) \varepsilon\frac \pi{\ell}.$$

Let us calculate $I’_i$. We see that $h(y)=f(x_i)+(y-x_i)f’(c)$ is a linear function such that $h(x_i)=f(x_i)$ and
$$h(x_{i+1})= f(x_i)+(x_{i+1}-x_i)\tfrac {f(x_{i+1})-f(x_i)} {x_{i+1}-x_i}=f(x_{i+1})=f(x_i)\pm \frac {\pi}n,$$

by the construction of the sequence $(x_j)$. Since the primitive of $\cos n h(y)$ is $\frac 1{n f’(c)}\sin n h(y)$, we have

$$I’_i=\int_{x_i}^{x_{i+1}} |\cos n h(y)| dy=\left|\int_{x_i}^{x_{i+1}} \cos n h(y) dy \right|=
\left|\frac 1{n f’(c)}\sin n h(y){\Huge|}_{x_i}^{x_{i+1}}\right|=\left|\frac 2{n f’(c)}\right|=\frac {2(x_{i+1}-x_i)} {\pi}.$$

We have
$$\left|\sum_{i=1}^{m-1} I_i-\sum_{i=1}^{m-1} I’_i\right|\le
\sum_{i=1}^{m-1} \left| I_i- I’_i\right|\le \sum_{i=1}^{m-1} I’’_i\le \sum_{i=1}^{m-1}
(x_{i+1}-x_i) \varepsilon\frac \pi{\ell}=(x_{m}-x_1) \varepsilon\frac \pi{\ell}\le \varepsilon\frac \pi{\ell} .$$

But $$\sum_{i=1}^{m-1} I’_i=\sum_{i=1}^{m-1} \frac {2(x_{i+1}-x_i)} {\pi}=\frac {2(x_{m}-x_1)}{\pi}$$

Since $\varepsilon$ can be chosen arbitrarily small, and  $\left|  \frac {2(x_{m}-x_1)}{\pi}-\frac 2{\pi}\right|\le  \frac 2{\pi}\cdot \frac\pi{n\ell}=\frac 2{n\ell},$ the proposition follows. $\square$


问:
$f$ is bounded on $\sigma$ and holomorphic on $\sigma\setminus \{a\}$, where $a$ lies in $\sigma$. Show that $f$ is holomorphic.

Suppose $f$ is a bounded function on $\sigma$ and $f$ is holomorphic on $\sigma\setminus \{a\}$, where $a$ lies in $\sigma$. Show that $f$ is holomorphic.

Can you provide a hint to get started with? I tried defining $g(z) = (z-a)f(z)$ and then show its limit goes to 0 when z goes to a. But I can not get forward.

答: Hint: Since $f$ is given to be bounded on $\sigma,$ by Riemann's theorem on removable singularity, the desired result follows.


问:
Given $p \in \mathbb R^N$ and $X$ a symmetric matrix, find $\varphi \in C^2$ such that $D\varphi = p, D^2\varphi = X$ at a given point
In their paper User's guide to viscosity solutions of second order differential partial equations, Crandall, Ishii and Lions define the superjet of an upper semicontinuous function $u$ at $\hat x \in \mathcal O$, $\mathcal O$ a locally compact subset of $\mathbb R^N$, to be the set of $(p, X) \in \mathbb R^N \times \mathcal S_N$ ($\mathcal S_N$ the set of symmetric matrices) such that
\begin{equation}
u(x) \leq u(\hat x) + \langle p, x - \hat x \rangle + \frac 12 \langle X(x - \hat x), x - \hat x \rangle + o(|x - \hat x|^2) \quad \quad (*)
\end{equation}
holds. The superjet is denoted $J_{\mathcal O}^{2, +}u(\hat x)$. It is left as an exercise that
$$
J_{\mathcal O}^{2, +}u(\hat x) = \{(D\varphi(\hat x), D^2 \varphi(\hat x)) \ : \ \varphi \in C^2 \text{ and } u - \varphi \text{ has a local maximum at } \hat x\}.
$$

My question is: given $(p, X) \in \mathbb R^N \times \mathcal S_N$ such that $(*)$ holds, how to find such a $\varphi$?

Thanks in advance.

答:
Let $u \in USC(\Omega)$ and $x_0 \in \Omega$. If $(p, X) \in J^{2, +}u(x_0)$ then there exists $\varphi$ of class $C^2$ supertangent to $u$ at $x_0$ such that $p = D \varphi(x_0)$ and $X = D^2 \varphi(x_0)$.

Proof:
Take $(p, X) \in J^{2, +}u(x_0)$. Then, in a neighborhood of $x_0$,
$$
u(x) \leq \underbrace{u(x_0) + p \cdot (x - x_0) + \frac 12 \left(X (x - x_0) \right) \cdot (x - x_0)}_{:= P(x)} + o(|x - x_0|^2),
$$
so
$$
u(x) - P(x) \leq o(|x - x_0|^2) = |x - x_0|^2 o(1).
$$

Our mission is then to find a positive function smooth enough, of smaller order than $|x - x_0|^2$, such that the gradient and the Hessian at $x_0$ are null.

For $r \geq 0$ small we define
$$
\omega(r) = \max_{|x - x_0| \leq r}(u(x) - P(x)).
$$
Since $u - P$ is upper semicontinuous and we are taking maximuns in compact sets $\omega$ is well defined. It is increasing, since the sets where we take the maximum are getting bigger. It is clear that $w(0) = 0$. Moreover, for $r > 0$ and by our previous remark it holds that
$$
\frac{\omega(r)}{r^2} = o(1), \quad r \to 0.
$$
This implies that $\omega$ is of class $C^2$ up to $r = 0$. Next we define
$$
\varphi(r) = \frac{1}{r^2} \int_r^{2r} \int_s^{2s} w(t) \ dt \ ds
$$
Since $\omega$ is increasing, we can estimate:
\begin{align*}
\frac{1}{r^2} \omega(r) \frac 32 r^2 & \geq \frac{1}{r^2} \int_r^{2r} \omega(s) \ ds \\
& \leq \varphi(r) \\
& \leq \frac{1}{r^2} \int_r^{2r}\omega(2s) s \ ds \\
& \leq \frac{1}{r^2} \omega(4r) \frac 32 r^2,
\end{align*}
from which we conclude that $\varphi(r) = o(r^2)$ as $r \to 0$.
Then
$$
u(x) - P(x) \leq \omega(|x - x_0|) \leq \frac 23 \varphi(|x - x_0|).
$$
We conclude that
$$
P(x) + \varphi(|x - x_0|)
$$
is a supertangent which has the desired gradient and Hessian. Indeed, the derivatives of $\varphi$ decrease as the square of the argument. The proof is complete.