问:
Need help understanding computation of area and volume using Matrices and linear algebra?
I am currently studying linear algebra from Gilbert R Strang's book. I have a few questions regarding computing area using matrices.

In the book, for a triangle with coordinates:
$(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$

The area is given as 1/2*det($$
    \begin{matrix}
    x_1 & y_1 & 1 \\
    x_2 & y_2 & 1 \\
    x_3 & y_3 & 1 \\
    \end{matrix}
$$)

1. Is the intuition behind adding the column of 1's just to make it into a square matrix?

Moving onto calculating volume of a rectangle with sides of length (r,s,t).
The coordinate matrix will be:
$$
    \begin{matrix}
    0 & 0 & 0 \\
    r & 0 & 0 \\
    0 & s & 0 \\
    0 & 0 & t \\
    r & s & 0 \\
    0 & s & t \\
    r & 0 & t \\
    r & s & t \\
    \end{matrix}
$$

2. Going by the logic applied to compute area of triangle, will I need to add 5 columns of 1's in order to get the volume.

Because in the book, the author mentions that computing the determinant of the diagonal matrix A with diagonal entries r,s,t would give the volume as rst (which is true by the Lbh formula)

3. But, what I fail to understand is how does a computing the determinant of a subset of three vectors give me the volume?

Also if I use the three vectors (r,s,0) , (r, 0 ,t) and (0, s, t) to do the same, the value of the determinant is -2rst which is the incorrect volume. Do the three vectors chosen need to be perpendicular?

4. Basically I am having trouble wrapping my head around how you need all three coordinates to compute the area but only 3 coordinates (vectors?) to compute the volume despite there being 8 coordinates that bound the rectangle in 3D space.

答:
1. Why a Column of Ones is Added

The volume of the parallelepiped that is generated by the vectors $u,v,w$ is given by
$$
\det\begin{bmatrix}
u_x&u_y&u_z\\
v_x&v_y&v_z\\
w_x&w_y&w_z
\end{bmatrix}\tag1
$$
The volume of the tetrahedron whose vertices are $0,u,v,w$ is $\frac16$ of the volume of the parallelepiped.

Furthermore, the volume of any pyramid is $\frac13\times|\text{base}|\times\text{altitude}$. By setting the $z$-coordinate of each vertex of the triangle to $1$, we have essentially set the altitude to $1$ (the dotted line in the rotating image) and the base to the triangle. Thus, the area of the triangle is
$$
3\times\text{volume}=\frac12\det\begin{bmatrix}
u_x&u_y&1\\
v_x&v_y&1\\
w_x&w_y&1
\end{bmatrix}\tag2
$$
2. Rectangle With Given Measurements

A rectangle with edges $r,s,t$ is a parallelepiped generated by $(r,0,0),(0,s,0),(0,0,t)$, and so its volume is
$$
\det\begin{bmatrix}r&0&0\\0&s&0\\0&0&t\end{bmatrix}=rst\tag3
$$
3. Why the Determinant of Three Vectors Gives the Volume of a Rectangle

A rectangle is a parallelepiped generated by the three vectors used.


问:
Limit of $\frac{1}{\sqrt[n] n}$
I would like to show that the limit of the sequence $(\frac{1}{\sqrt[n] n})_{n=1}^{\infty}$ is equal to 1 using the definition of the limit of a sequence. (Another way to say this is that it converges to 1).

Definition: A sequence $(a_n)_{n=1}^{\infty}$ converges to a real number A iff for each $\epsilon > 0$ there is a positive integer N such that for all $n \geq N$ we have $|a_n - A| < \epsilon$.

Edit: Here is my work.

Let $\epsilon > 0$. There exists $N$ such that for $n \geq N$,

$$|\frac{1}{\sqrt[n] n}-1| = |\frac{n^{\frac{n-1}{n}}}{n} - 1| = |\frac{n^{\frac{n-1}{n}} - n}{n}| = \frac{n-n^{\frac{n-1}{n}}}{n}$$

Here is where I am stuck.

ContentLicense: CC BY-SA 4.0

Title: Limit of $\frac{1}{\sqrt[n] n}$

Body:
An old standby:

By Bernoulli's inequality,
$(1+n^{-1/2})^n
\ge 1+n^{1/2}
\gt n^{1/2}
$.

Raising to the $2/n$ power,
$n^{1/n}
\lt (1+n^{-1/2})^2
=1+2n^{-1/2}+n^{-1}
\lt 1+3n^{-1/2}
\to 1
$.


问:
Exponential Generating function for inhomogenous recurrence relation
I want to determine all solutions to the recurrence relation

$$
a_n−2na_{n−1}+n(n−1)a_{n−2} = 2n·n!, \ \ \ n≥2, a_0 = a_1 = 1.
$$

by using exponential generating functions.

My idea was something like that:
$$
\sum_{n\geq2} a_n \frac{z^n}{n!}−\sum_{n\geq2}2na_{n−1} \frac{z^n}{n!}+\sum_{n\geq2}n(n−1)a_{n−2} \frac{z^n}{n!} - \sum_{n\geq2}2n\frac{z^n}{n!} = 0
\\
$$

Denoting

$$Â(z) =\sum_{n\geq0} a_n \frac{z^n}{n!}$$

which is the exponential generating function, I can derive that $\sum_{n\geq2}2na_{n−1} \frac{z^n}{n!}$ is $z^2 Â'(z)$ and for $\sum_{n\geq2}2n\frac{z^n}{n!}$, that $zÂ(2z)$ (which I am also not 100% sure of). However, I get stuck for the third summation term. How would I continue, am I even on the right path?

答:
For the first term $$\sum_{n\geq2} a_n \frac{z^n}{n!}$$ I get $Â(z)-1-z$ by taking care of the first two initial conditions. For the second term $$\sum_{n\geq2}2na_{n−1} \frac{z^n}{n!}$$
I get $2(zÂ(z)-z)$, notice $\frac{n}{n!}=\frac{1}{(n-1)!}$ and that taking derivatives shift the index of $a_n$ up, multiplying by z shifts the index of $a_n$ down. The -z is obtained by accounting for initial conditions.

For the third term $$\sum_{n\geq2}n(n−1)a_{n−2} \frac{z^n}{n!}$$
I get $z^2Â(z)$. Notice that $\frac{n(n-1)}{n!}=\frac{1}{(n-2)!}$.

For the fourth term is actually $$ \sum_{n\geq 2}2nz^n$$
I get $\frac{2(z-2)z^2}{(z-1)^2}$. Because $\sum_{n\geq 1}nz^n=\frac{z}{(1-z)^2}$ and taking out the first term we get $\sum_{n\geq 2}nz^n=\frac{z}{(1-z)^2}-z=\frac{z(1-(1-z)^2)}{(1-z)^2}=\frac{z^2(2-z)}{(1-z)^2}$.


问:
Finding constants for second order diff. equation with repeated real root
http://personal.psu.edu/sxt104/class/Math251/Notes-2nd%20order%20ODE%20pt1.pdf

On Page 28 of the above link it shows example of how to find general form of second order diff. equation with repeated root.

The equation is $y″ − 4y′ + 4y = 0$ with condition $y(0) = 4$, $y′(0) = 5$.

But how they find the constants using initial conditions?

答:
Once you have
$$
y(t)=C_1e^{2t}+C_2te^{2t}
$$
you can differentiate and
$$
y'(t)=2C_1e^{2t}+C_2(2te^{2t}+e^{2t}).
$$
Now, let's plug in the conditions $y(0)=4, y'(0)=5$.
$$
\begin{cases}
4=y(0)=C_1e^0+C_2e^0 \cdot 0=C_1\\
5=y'(0)=2C_1e^0+C_2(2e^0 \cdot 0+e^0)=2C_1+C_2.
\end{cases}
$$
From the first equation $C_1=4$ and from the second equation $C_2=5-2C_1=5-8=-3$.


问:
Uniform Convergence of $\sum_{n=1}^{\infty}x^ne^{-nx}$ on $[0,\infty]$
Prove that

$\sum_{n=1}^{\infty}x^ne^{-nx}$  is uniform convergent on  $[0,\infty]$,

I have started by considering the two cases $x\geq0,$ and $x \in [-\alpha,0]$ and applying the M-Test,
for $x \geq 0$ I have $\frac{x^n}{e^{nx}}$, but I dont know how to find a series larger than this.

答:
For the case where $x \in [0,\infty)$ show using calculus that $xe^{-x} \leqslant e^{-1}$ and the M-test is applicable to prove uniform convergence since $x^n e^{nx} \leqslant e^{-n}$.

When $x < 0$ write the general term as $x e^{-nx} = (-1)^n (|x|e^{|x|})^n $ and apply the Dirichlet test for uniform convergence. Since $\sum_{n=1}^m (-1)^n$ is uniformly bounded for all $m$, the series converges uniformly if $(|x|e^{|x|})^n \downarrow 0$ monotonically and uniformly as $n \to \infty$.

If $-\alpha \leqslant x \leqslant 0$, then $|x| e^{|x|} \leqslant \alpha e^{\alpha} < 1$ when $\alpha \in (0,\lambda)$ where $\lambda e^{\lambda} = 1$. This ensures the monotonic and uniform convergence condition of the Dirichlet test and we have uniform convergence of $\sum x^n e^{-nx} = \sum (-1)^n (|x|e^{|x|})^n$ for $-\alpha \leqslant x \leqslant 0$.


问:
Can you solve for x for the equation $25+2^{\log_{10} x}=x$?

Solve for $x$ : $$25+2^{\log_{10} x}=x$$

My work

Well, I could not figure out an algebraic solution to this problem.
$$25+2^{\log_{10} x}=x \implies 5^2+x^{\log_{10} 2}=x$$
$$\implies x^{\log_{10}2}-x-25=0$$
which does not seem to be solved further.

I have solved this by using the graphical method by plotting both sides of this equation. And the answer comes near to  $27.7$. I have also verified it by using the desmos graph calculator according to which the answer is $27.718$.

How can I solve this question by the algebraic method?

答:
Solve for x :  \begin{align} 25+2^{\log_{10} x}=x \tag{1}\label{1} \end{align}

Note that \eqref{1} is equivalent to

\begin{align}
x^{\log_{10}2}-x+25&=0
\tag{2}\label{2}
\\
\text{or }\quad
x^a-x+b&=0
\tag{3}\label{3}
\end{align}
with non-rational $a$.
It is known that such equations
don't have an algebraic solution and
can be solved only by means of numerical methods.

For example, we can use Halley's method
to iteratively find the approximation of the root as

\begin{align}
x_{n+1}&=F(x_n)
,\\
F(x)&=x-\frac{2\,f(x)\,f'(x)}{2f'(x)^2-f(x)\,f''(x)}
,\\
f(x)&=x^{\log_{10}(2)}-x+25
,\\
f'(x)&=\log_{10}(2)\cdot x^{\log_{10}(2)-1}-1
,\\
f''(x)&=\log_{10}(2)\log_{10}(\tfrac15)\cdot x^{\log_{10}(2)-2}
.
\end{align}

For example, starting with $x_0=1$, we get

\begin{align}
x_1&=6.60306336935\\
x_2&=26.5079884286\\
x_3&=27.7184046785\\
x_4&=27.7184201926\\
x_5&=27.7184201926\\
\end{align}

Edit

The rate of convergence to the root is cubic,
compare for example
to the Newton's method:
starting with the same $x_0$, the Halley's approximations would be

\begin{align}
x_0&=\color{blue}{ 27}.386363636363636363636 \\
x_1&=\color{blue}{ 27.7184}19892956254689994 \\
x_2&=\color{blue}{ 27.718420192574854316455}
\end{align}

Corresponding python code:

```<code>import decimal
decimal.getcontext().prec = 23

lg2 = decimal.Decimal(2).log10()

def f(x):
  return 25+x**lg2-x

def df(x):
  return lg2*x**(lg2-1)-1

def ddf(x):
  return lg2*(lg2-1)*x**(lg2-2)

def F(x):
  fx=f(x)
  dfx=df(x)
  ddfx=ddf(x)
  return x-2*fx*dfx/(2*dfx**2-fx*ddfx)

x=decimal.getcontext().divide(1205,44); print(x)
x=F(x); print(x)
x=F(x); print(x)
x=F(x); print(x)

# 27.386363636363636363636
# 27.718419892956254689994
# 27.718420192574854316455
# 27.718420192574854316455
</code>```

Body:
Solve for x :  \begin{align} 25+2^{\log_{10} x}=x \tag{1}\label{1} \end{align}

Note that \eqref{1} is equivalent to

\begin{align}
x^{\log_{10}2}-x+25&=0
\tag{2}\label{2}
\\
\text{or }\quad
x^a-x+b&=0
\tag{3}\label{3}
\end{align}
with non-rational $a$.
It is known that such equations
don't have an algebraic solution and
can be solved only by means of numerical methods.

For example, we can use Halley's method
to iteratively find the approximation of the root as

\begin{align}
x_{n+1}&=F(x_n)
,\\
F(x)&=x-\frac{2\,f(x)\,f'(x)}{2f'(x)^2-f(x)\,f''(x)}
,\\
f(x)&=x^{\log_{10}(2)}-x+25
,\\
f'(x)&=\log_{10}(2)\cdot x^{\log_{10}(2)-1}-1
,\\
f''(x)&=\log_{10}(2)\log_{10}(\tfrac15)\cdot x^{\log_{10}(2)-2}
.
\end{align}

For example, starting with $x_0=1$, we get

\begin{align}
x_1&=6.60306336935\\
x_2&=26.5079884286\\
x_3&=27.7184046785\\
x_4&=27.7184201926\\
x_5&=27.7184201926\\
\end{align}

Edit

The rate of convergence to the root is cubic,
compare for example
to the Newton's method:
starting with the same $x_0$, the Halley's approximations would be

\begin{align}
x_0&=\color{blue}{ 27}.386363636363636363636 \\
x_1&=\color{blue}{ 27.7184}19892956254689994 \\
x_2&=\color{blue}{ 27.718420192574854316455}
\end{align}

Corresponding python code:

import decimal
decimal.getcontext().prec = 23

lg2 = decimal.Decimal(2).log10()

def f(x):
  return 25+x**lg2-x

def df(x):
  return lg2*x**(lg2-1)-1

def ddf(x):
  return lg2*(lg2-1)*x**(lg2-2)

def F(x):
  fx=f(x)
  dfx=df(x)
  ddfx=ddf(x)
  return x-2*fx*dfx/(2*dfx**2-fx*ddfx)

x=decimal.getcontext().divide(1205,44); print(x)
x=F(x); print(x)
x=F(x); print(x)
x=F(x); print(x)

# 27.386363636363636363636
# 27.718419892956254689994
# 27.718420192574854316455
# 27.718420192574854316455


问:
Determine one-sided limits of a function given its graph (no analytical formula)
I need some help figuring this out.

So this time I have a problem which requires us to determine some of the one-sided limits based on the graph.

I deduced the values for some of the limits (whereas I'm not entirely sure for the last one), and for the others I don't even know where to start. Here are the ones I could find a value for:

$$\lim_{x \to -1-} f(x) = 1$$
$$\lim_{x \to 0+} f(x) = -\infty$$
$$\lim_{x \to -2-} f(x) = 2$$

I am not sure about the last one, because if we are approaching $x$ from both sides we end up at $(-2,2)$ but the function is defined for $x=-2$ by $y=1$ so I am not sure.

Now here are the ones I couldn't find (to be quite frank I don't even understand them):

$$\lim_{x \to 1-} ((x-1)^2f(x) $$
$$\lim_{x \to 2+} (f(x)f(-x)) $$

Could anyone clarify what happens when x approaches $-2^-$ and why?

Could anyone explain the two limits I have not been able to comprehend?

Keep in mind that we have not yet covered the subject of derivatives and continuity (although I am familiar with it).

Thanks!

答:
Explanation

So sided limits are a bit tricky, you have to always remember that there is two ways of seeing a number or approaching it, from the left (or coming from $- \infty$ or from the right coming from $+\infty$.

Therefore,

Could anyone clarify what happens when x approaches $−2^−$ and why?

Now we are approaching the value $-2$ from the $-$ side which is $- \infty$ , therefore you can imagine that $-2^-$ is something close to $-2.0000000000001$ .

On the other hand if we were approaching $-2$ from the $+$ side, we would be coming from $+ \infty$ which yields something close to $-1.9999999999$.

Now the trick in this particular exercise, is that on $-2$ there's one hell of a mess, but i will make it easier for you.

Coming from the left or right side of $-2$, we have a discontinuity as you can see. And $EXACTLY$ on $x=-2$ we have the black dot which is a value and not a limit for $f$.

In other words $lim -2^{-+} f(x) = 2$ and $f(-2)=1$

Help for the last 2 parts

As you were able to find $(x-1)f(x)$ yields a $0*f(x)$ and since $f(1^-)=1$. Therefore, we end up having a 0.


问:
Hilbert space $\mathcal{H}$ is separable if and only if $\mathcal{B}(\mathcal{H})$ is separable

Let $\mathcal{H}$ be a Hilbert space. Then $\mathcal{H}$ is separable if and only if $\mathcal{B}(\mathcal{H})$, the space of all bounded linear operators on $\mathcal{H}$, is separable.

I am looking to determine whether or not the above statement is true, and if it is, how to prove it. Honestly, I do not know how to even approach the question.

If $\mathcal{H}$ is separable, then it has an orthonormal basis $(e_n)_{n \in \mathbb{N}}$ such that
$$
x = \sum_{n=1}^\infty \langle x, e_n \rangle e_n.
$$

Then, for every $T \in \mathcal{B}(\mathcal{H})$, we have that
$$
Tx = \sum_{n=1}^\infty \langle x, e_n \rangle Te_n.
$$

How can the above help us in defining a countably dense subset in $\mathcal{B}(\mathcal{H})$? For the converse implication, I do not know how to start.

答:
The statement can't be true: let $\mathcal{H}$ be a separable Hilbert space. Since it is separable, consider a countable orthonormal basis $\{\alpha_1,\alpha_2,\dots\}$ for $\mathcal{H}$. For each bounded sequence $x=(x_n)$ of real numbers, that is an element of $l_{\infty}$, consider the diagonal operator on $\mathcal{H}$, defined by
$$X_{x}(\alpha_n)= x_n\alpha_n$$
for $\alpha_n$ an orthonormal basis vector.

Then $\|X_x\|_\text{op} = \|x\|_{l_\infty}$ (check!), and this induces an isometric inclusion of $l_{\infty}$ into the space of all linear bounded operators on $\mathcal{H}$, and $l_{\infty}$ is not separable!


问:
How to integrate $\int_0^\infty e^{-ty^2} \sin t dt$
My book suggests that I do some sort of limiting

$\lim_{A \to \infty} \int_0^A e^{-ty^2} \sin t d t$

But I'm not getting anywhere.

答:
Well, I guess there some other interesting ways to solve the problem.
One more try.
You can look at that integral as the Laplace transform:
$$F(s) =\int_0^{\infty} e^{-st} f(t) \,dt$$
obtained for $s=y^2$.

 So you have $f(t)=\sin(t), \ s=y^2$. It is well known that the Laplace transform of $ \sin(\omega t)$ is $ { \omega \over s^2 + \omega^2  } $ . So just set $\omega=1$ and one will obtain:
$$\int_0^\infty e^{-ty^2} \sin t dt=\frac{1}{y^4+1}$$

ContentLicense: CC BY-SA 3.0


问:
friend combination
A class has 100 students.Let $a_i$,1$\leq$i$\leq100$,denotes the number of friends the i-th student has in the class.For each 0$\leq$j$\leq$99,let $c_j$ denotes the number of students having at least j friends.Show that $\sum_{i=1}^{100}${$a_i$}
=$\sum_{j=0}^{99}${$c_j$}

答:
There's an issue with your summation indices. You likely want

 $$ \sum_{i=0}^{100} a_i = \sum_{j=1}^{99} c_j $$

(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)

Hint: If a person has exactly $k$ friends, then they have at least $1, 2, \ldots k$ friends.

Title: friend combination


问:
parameter estimation by the method of moments
Let $X_1,X_2,\dots X_n$ be a random sample from the density $$f(x;\theta)=\theta x^{-2}, \quad 0<\theta \le  x<\infty$$
Find the method of moments estimator of $\theta$.

My attempt

$ E(X) = \int_{\theta}^{\infty}\theta x^{-2} = 1 $

I think that the integral bounds aren't correct.

答:
The density is nice because

$$\int_{\theta}^{\infty}f(x)dx=1$$

But

$$E(X^n)=\int_{\theta}^{\infty}x^nf(x)dx=\infty$$

That means your density does not have finite moments and thus it is not possibile to estimate the parameter with moments

... but you can estimate it with Max likelihood


问:
An isolated point in spectrum of a normal operator is an eigenvalue.
Suppose $T$ is a normal operator on a Hilbert space and $\lambda$ is an isolated point in $\sigma(T)$.

Assume $\lambda$ isn't an eigenvalue.

Since $T-\lambda$ and $T^*-\lambda^*$ have the same kernel,  $T^*-\lambda^*$ is also injective. Then by the non-invertibility, $T-\lambda$ must not be surjective. Therefore there are unit vectors $x_j$ such that  $\|(T-\lambda)x_j\|<1/n$.

Then I got  stuck. What should I do next?

答:
I might be wrong, but I don't see how you could use that $\lambda$ is an approximate eigenvalue to conclude that it is an eigenvalue.

What you can use, due to $T$ being normal, is the continuous funcional calculus. Take $f$ to be continuous on $\sigma(T)$, with $f(\lambda)=1$ and $f=0$ on $\sigma(T)\setminus\{\lambda\}$. Then $P=f(T)$ is a nonzero projection, and since $$t\,f(t)=\lambda\,f(t),$$
you get
$$
TP=\lambda P.
$$


问:
Find a polynomial function given local extrema
I'm struggling to find a polynomial function with given local extrema.

For instance, consider the problem:

Find a function with a local maximum at $(0,0)$ and a local minimum at $(1,−5)$.

I start with $f(0)=0$, $f(1)=−5$, $f′(0)=0$ (or is it undefined?), $f′(1)=0$ and then try to solve $f(x)=ax^3+bx^2+cx+d$ with the above values.

So far I have $f(1)=-5$ gives $a+b+c+d=-5$ and $f'(1)=0$ gives $3a+2b+c=0$, but I'm unsure how to go about working with $f(0)$ and $f'(0)$.

Is it $f(0)=0$ so $a+b+c=d=0$ so $d=0$ and $f'(0)=a+b=c=0$ so $c=0$? This seems incorrect, but I'm not sure else I can do here.

Body: We have $f'(x)=kx(x-1)$, hence $f(x)=k(x^3/3-x^2/2)+c$.  From $f(0)=0$ it follows that $c=0$. As $-5=f(1)=k(1/3-1/2)=-k/6$ we know $k=30$, hence $f(x)=10x^3-15x^2$.


问:
$\int \frac{dx}{(x^4 + 1)^2}$
What would be a relatively simple method for computing the indefinite integral below?

$\displaystyle \int \frac{dx}{(x^4+1)^2}$

Furthermore, how would one evaluate the following, possibly by detouring the computation of the indefinite integral?

$\displaystyle\int\nolimits_{-\infty}^\infty \frac{dx}{(x^4+1)^2}$

Note: The resolved value of the definite integral (according to wolfram alpha) is $\displaystyle \frac{3\pi}{4\sqrt{2}}$

答:
For the second, definite integral, assuming you have computed the first, indefinite integral; you can do a change of variables to obtain a form of the integral where you can use
differentiation under the integral sign to compute its value.

More explicitly, let
$$f(\alpha) = \int_{-\infty}^\infty \frac{1}{(x^4+\alpha)}\mathrm dx ,$$
then the value you want to compute is
$$\int_{-\infty}^\infty \frac{1}{(x^4+1)^2}\mathrm dx = - \left.\frac{\mathrm d f}{\mathrm d\alpha}\right|_{\alpha = 1}.$$

Title: $\int \frac{dx}{(x^4 + 1)^2}$


问:
Does the intersection of two subsets (with a maximum) necessarily have a maximum too?
let A and B be sets with a maximum.

I want to know if:

$$
A \cap B
$$

also has a maximum then.

I know that $$A \cap B $$

has a supremum and I already proofed that $$ sup(A \cap B) \le min\{supA,supB\} $$

Can someone help me? ;)

答: You can take $A= (0,1) \cup (3,4]$ and $B= (0,2]$ as an counterexample. A has the maximum 4 and B has the maximum 2. The intersection is $A\cap B= (0,1)$ which doesn't attain its maximum.


问:
Show that the direct product of a family indexed by a partition of set must have one-one mapping
So this is an exercise I stumbled upon in an introductory book to the field of topology, in a preface chapter that discusses arbitrary products and it is phrased exactly as follows:

Let {$X_\alpha$}$_{\alpha\in I}$ be an indexed family of sets
and let $I = I_1{\cup}I_2$, where $I_1{\cap}I_2 = \emptyset$.
Show that there is a one-one mapping between $(\prod_{{\alpha}{\in}I_1}X_\alpha) \times (\prod_{{\alpha}{\in}I_2}X_\alpha)$ onto $\prod_{{\alpha}{\in}I}X_\alpha$.

Now, I am not really sure where to begin with this one and I maybe misunderstanding some basic concept but I don't see how does the indexing set affects the product of the actual family it indexes.

Any directions, guidance or hints regarding this exercise will be greatly appreciated.

答:
By definition, $\prod_{\alpha \in I} X_\alpha$ is set of functions $I \to \bigcup_{\alpha \in I} X_\alpha$ s.t. image of $\alpha \in I$ is in $X_\alpha$.

Define function $F: \prod_{\alpha \in I} X_\alpha \to (\prod_{\alpha \in I_1} X_\alpha) \times (\prod_{\alpha \in I_2} X_\alpha)$ by rule $F(f) = \langle f|_{I_1}, f|_{I_2}\rangle$ - restriction of $f$ to $I_1$ and $I_2$.

$F$ is injection: if $f_1(\alpha) \neq f_2(\alpha)$ then either $\alpha \in I_1$ and $f_1|_{I_1}(\alpha) \neq f_2|_{I_1}(\alpha)$, or $\alpha \in I_2$ and $f_1|_{I_2}(\alpha) \neq f_2|_{I_2}(\alpha)$.
In both cases, $\langle f_2|_{I_1}, f_2|_{I_2}\rangle \neq \langle f_2|_{I_1}, f_2|_{I_2}\rangle$.

$F$ is also surjection: preimage of pair $\langle g, h\rangle$ is function $f$ defined by $$f(\alpha) = \begin{cases} g(\alpha), \alpha \in I_1\\ h(\alpha), \alpha \in I_2\end{cases}$$

Thus $F$ is one-one mapping.


问:
states in the cross product $C^*$-algebras
Let $\tau$ be  a state of cross product $C^*$-algebra $A\rtimes G$, where $A$ is a $C^*$-algebra and $G$ is a discrete group. Is $\tau $ also a state of $A$?

答:
Yes.  Let $\phi:A\hookrightarrow A\rtimes G$ denote the canonical embedding, and let $\tau_A=\tau\circ\phi$, so that $\tau_A$ is the restriction in question.  As a composition of two positive linear maps, we have that $\tau_A$ is positive and linear.  To show that $\tau_A$ is a state,  let $(u_\lambda)$ be an approximate unit for $A$. Then $(\phi(u_\lambda))$ is an approximate unit for $A\rtimes G$, so we have
$$\|\tau_A\|=\lim_\lambda\tau_A(u_\lambda)=\lim_\lambda\tau(\phi(u_\lambda))=\|\tau\|=1$$
and therefore $\tau_A$ is a state.


问:
Conditions for f(x,y) to coincide with some g(x) + h(y)

Given a function of two variables, say f(x,y), how to know/check whether it can be simplified into some g(x) + h(y)?

Some property of some condition that condition that f(x,y) satisfies to know this.

答:
For purposes of analysis proofs, and algebraicly, I think the most useful way of formulating that property is how you already wrote it, i.e. that it can be written as the sum of two functions, one of which doesn't depend on $x$ and the other of which doesn't depend on $y$.

Assuming you're looking for a different formulation, I don't know one in general, but for differentiable $f$
$$
\frac{\partial f}{\partial x}(x,y) \text{ is independent of } y
$$
$$
\frac{\partial f}{\partial y}(x,y) \text{ is independent of } x
$$

Also, if $f$ is twice differentiable, and the second derivative is continuous, a nice formulation is the condition that
$$
\frac{\partial f}{\partial x \partial y} (x,y) = 0.
$$

(Can you prove that the above is both necessary and sufficient in the continuous, twice differentiable case?)


问:
Can this be proved using Combinatorics or generating functions?
What is the proof for this identity?

$\displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{m+r}{n}\ = \displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{m}{r}2^{r}$

I thought of changing $r$ to $n-r$ on left and right both. Then, I thought of building a set with atleast $n$ elements and atmost $2n$ elements with atmost $n$ elements from $m$ element set, and we can do this on LHS by choosing $n-r$ elements from n and out of remaining $r+m$ elements we can choose n more elements. On RHS, we can again choose $n-r$ elements from $n$ and $r$ elements from $m$, then remaining $r$ elements from n element set have 2 choices each. But I think it has a flaw. Can someone please help? If my proof is not correct, please suggest a combinatorial or a generating function proof.

答:
The story would be like(any resemblance to reality is pure coincidence) :

Suppose you want to build a committee of $n$ mathematicians. You have a pool of $n$ graduate students and $m$ full professors. The rule is that you have to double vote the graduate students. That means, first you choose the graduate students(first filter) and then you choose from the pool of professors and selected graduate students(second filter).

The LHS is you choose $r$ graduate students and then you choose from $m+r$ candidates, the $n$ mathematicians.

The right hand side is you choose the amount of professors that are going to be in the final committee, say $r,$ in $\binom{m}{r}$ and so there should be $n-r$ graduate students. You choose them in $\binom{n}{n-r}$ but notice that the remaining $r$ graduate students, perhaps, pass the first filter so you are going to mark the ones that pass the first filter, you can do that in $2^r$ ways.

This should be reminescent of the proof of $\binom{a}{b}\binom{b}{c}=\binom{a}{c}\binom{a-c}{b-c}.$


问:
Fundamental Forms and Connect Forms Question
Given $I = du^2 + cos^2(u) dv^2 $, $K = 1$ and $II = a^{-1} du^2 + acos^2(u)$ for some function a

Use the structure equations to show that this is only a surface if $a = +/-1$

Earlier I was asked to find $\theta_{1}, \theta_{2}$, $w_{13}, w_{23}$, and $w_{12}$. Assuming a was constant

$I = \theta_{1}^2 + \theta_{2}^2 $

So $\theta_{1} = du$ and $\theta_{2} = cos(u)dv$

And from here we can also see that

$d\theta_{1} = 0$ and $d\theta_{2} = -sin(u) du \wedge dv$

Then we have an equation that says

$II = -\theta_{1}*w_{13} - \theta_{2}*w_{23}$

So

$a^{-1} du ^2 + acos^2(u) dv^2 = -du*w_{13} - cos(u)*w_{23}$

From here you can see that

$w_{13} = -a^{-1} du = \frac{-1}{a} du$ and $w_{23} = -acos(u) dv$

To find $w_{12}$ we can use the following equations

$d\theta_{1} = \theta{2} \wedge w_{12}$

$d\theta_{2} = -\theta{1} \wedge w_{12}$

$d\theta_{1} = 0$ so you know $w_{12}$ will have a $dv$ piece.

$d\theta_{2} = -sin(u) du \wedge dv$

Substituting

$-sin(u) du \wedge dv = -du \wedge w_{12}$

From here you can see that $w_{12} = sin(u) dv$

But know I need to show that this is only a surface if $a = +/- 1$

Here is what I've tried so far

A) $dw_{12} - w_{13} \wedge w_{23} = 0$

B ) $dw_{13} + w_{12} \wedge w_{23} = 0$

C) $dw_{23} - w_{12} \wedge w_{13} = 0$

A) $dw_{12} = w_{13} \wedge w_{23}$
$cos(u) du \wedge dv = -du/a \wedge -acos(u) dv$
$cos(u) du \wedge dv = a/a cos(u) du \wedge dv$

$+/1$ certainly work here, but so does $+/-2, +/-3, ...$

B ) $dw_{13} = 0$ since $w_{13} = \frac{-1}{a} du$ and a is constant, so

$w_{12} \wedge w_{23} = 0$

Nothing surprising here they both have a $dv$ term

C) $w_{23} = -acos(u) dv$
$dw_{23} = asin(u) du \wedge dv$
$dw_{23} = w_{12} \wedge w_{13}$
$-asin(u) du \wedge dv = cos(u) dv \wedge -1/a du$

this also does not add anything to the discussion as you will get $-asin(u) du ^\wedge dv = acos(u) du \wedge dv$ and this is not even true.

So none of these seem to work, have I done a miscalculation somewhere? or am I just using the wrong equations?

Thanks!

答:
Your sign conventions are different from ones I'm used to, and I would write $\otimes$ instead of $*$.

You forgot the most basic intrinsic structure equation. From
$$d\theta_1 = \theta_2\wedge\omega_{12} \quad\text{and}\quad d\theta_2 = -\theta_1\wedge\omega_{12}$$
you should find immediately that $\omega_{12}=\sin u\,dv$. As a check, we have $d\omega_{12} = \cos u\,du\wedge dv = \theta_1\wedge\theta_2$.

Your computation of $d\omega_{13}$ is flawed: You should have $d\omega_{13}=0$. You should go back and finish correctly. Looking at $d\omega_{23}$, you should find immediately that $a=a^{-1}$, and so $a=\pm 1$.


问:
Approximating function by its step approximation
$f$ be any bounded measurable function on $[0,1]$. Define the step function $f_n(x)=f(\frac{i}{n})$, if $x \in [\frac{i-1}{n},\frac{i}{n}]$. Is it always true that
$$\int |f(x)- f_n(x)|dx \rightarrow 0?$$
Or do we need some extra condition.

答: This is not true in general. Let $f = \chi_{\mathbb{Q}^c \cap [0,1]}$. Then $f_n(x) = 0$ everywhere on $[0,1]$, for all $n$. So $\int |f(x) - f_n(x)|\,dx = \int |f(x)|\,dx = 1$.


答:
Using coefficient extractors we get for the LHS

$$\sum_{r=0}^n {n\choose r} {m+r\choose n}
= [z^n] (1+z)^m \sum_{r=0}^n {n\choose r} (1+z)^r
= [z^n] (1+z)^m (2+z)^n.$$

We get for the RHS

$$\sum_{r=0}^n {n\choose r} {m\choose r} 2^r
= \sum_{r=0}^n {n\choose r} {m\choose m-r} 2^r
\\ = [z^m] (1+z)^m \sum_{r=0}^n {n\choose r} 2^r z^r
= [z^m] (1+z)^m (1+2z)^n.$$

This last term is

$$\mathrm{Res}_{z=0} \frac{1}{z^{m+1}} (1+z)^m (1+2z)^n.$$

There is no pole other than the one at zero and residues sum to zero so
this is

$$-\mathrm{Res}_{z=\infty} \frac{1}{z^{m+1}} (1+z)^m (1+2z)^n
\\ = \mathrm{Res}_{z=0} \frac{1}{z^2}
z^{m+1} \frac{(1+z)^m}{z^m} \frac{(2+z)^n}{z^n}
\\ = \mathrm{Res}_{z=0} \frac{1}{z^{n+1}} (1+z)^m (2+z)^n
= [z^n] (1+z)^m (2+z)^n.$$

This is the claim.

Title: Can this be proved using Combinatorics or generating functions?

Body:
What is the proof for this identity?

$\displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{m+r}{n}\ = \displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{m}{r}2^{r}$

I thought of changing $r$ to $n-r$ on left and right both. Then, I thought of building a set with atleast $n$ elements and atmost $2n$ elements with atmost $n$ elements from $m$ element set, and we can do this on LHS by choosing $n-r$ elements from n and out of remaining $r+m$ elements we can choose n more elements. On RHS, we can again choose $n-r$ elements from $n$ and $r$ elements from $m$, then remaining $r$ elements from n element set have 2 choices each. But I think it has a flaw. Can someone please help? If my proof is not correct, please suggest a combinatorial or a generating function proof.


问:
Derivative of trace function
$\DeclareMathOperator{\tr}{tr}$
Let $A,B$ be self-adjoint matrices and $f$ be a real differentiable function on $\mathbb{R}$ with derivative $f'$. Then why is it true that
$$
\left.\ \frac{d}{dt}\right|_0 \tr f(A+tB)=\tr (f'(A)B)
$$

This is used in the Klein's inequality. However, I'm not sure why exactly this is true in general. It's pretty clear why it's true for polnomials since we can use the commutation relation of the trace function, but it's harder to justify in general. I also checked the linked reference (E. Carlen, Trace Inequalities and Quantum Entropy: An Introductory Course, Contemp. Math. 529 (2010) 73–140) with no luck, as the author didn't give much explanation.

EDIT: After some further thought, let me provide an incomplete proof of what I got so far. Hopefully someone with better knowledge can finish the proof.

For simplicity, let $\lambda_i(A)$ denote the eigenvalues of $A$ in descending order, i.e., $\lambda_1(A) \ge \cdots \ge \lambda_d (A)$. Then
$$
\tr \left( \frac{f(A+tB)-f(A)}{t}\right) = \sum_i \frac{1}{t}[f(\lambda_i(A+tB)-f(\lambda_i(A))]
$$
Notice that by Weyl's inequality (stability of eigenvalues), we see that $|\lambda_i(A+tB)-\lambda_i(A)|\le t||B||$. Hence, using an $\epsilon,\delta$ arguement, we can replace the above with
$$
\sum_i \frac{1}{t}(\lambda_i(A+tB)-\lambda_i(A)) f'(\lambda_i(A))
$$
Now first assume that $A$ has a simple spectrum, then $A+tB$ is also simple for sufficiently small $t$. Then by Hadarmard's variation formula, we see that
$$
\frac{1}{t}(\lambda_i(A+tB)-\lambda_i(A)) \to \langle i|B| i\rangle
$$
where $|i\rangle$ is the corresponding eigenvector (unique up to phase since we are assuming that $A$ is simple) to $\lambda_i(A)$. Plugging all this back in, we see that the formula at least holds when $A$ is simple.

EDIT 2. I think I now have a way of dealing with degenerate eigenvalues. I will provide a sketch and fill in the details later (if someone else doesn't point out an error).

Let $\lambda_1 (A)=\cdots =\lambda_r(A)$ be the degenerate eigenvalues. Then for sufficiently small $t$, the eigenvalues $\lambda_i (A+tB),i=1,...,r$ will not touch the other eigenvalues (Weyl's inequality again). Let us use the Riesz projector
$$
P_A =\frac{1}{2\pi i} \oint_\Gamma \frac{dz}{A-z}
$$
where $\Gamma$ is some "smooth" contour around the $\lambda_1 (A)=\cdots =\lambda_r(A)$ and its interior does not contain any other eigenvalues. By Weyl's inequality, we can assume that $\lambda_i(A+tB),i=1,...,r$ are still in the interior of $\Gamma$ for sufficiently small $t$. Notice that
$$
\frac{d}{dt} \Big|_0 \tr {((A+tB)P_{A+tB})} = \tr(BP_A)
$$
where I got some inspiration from @Ruy's comment and used the fact that
\begin{align}
\frac{d}{dt}\Big|_0 \tr{(A(P_{A+tB}-P_A))}&=\tr A\oint_\Gamma \frac{dz}{(z-A)^2}B \\
&= \sum_{i=1}^r \oint_\Gamma \lambda_i(A)\frac{1}{(z-\lambda_i(A))^2} dz \langle i|B|i\rangle \\
&=0
\end{align}
Hence, if we combine this with the previous part, we see that the equality holds.

My proof is a little convoluted, so I would still hope to see a more straightforward approach

答:
Lemma 1.  Let $f$ and $g$ be real valued functions of class $C^1$ defined on a neighborhood $(a-\delta , a+\delta )$
of the real number $a$, such that
$$
  f(a)=g(a),\quad \text{and}\quad  f'(a)=g'(a).
  $$
Also let $\lambda :U\to (a-\delta , a+\delta )$ be any function defined on a neighborhood $U$  of zero
such that
$$
  |\lambda (t)-a|\leq c|t|, \quad \forall t\in  U,
  $$
where $c$ is a given  positive  constant.
Then
$$
  \lim_{t\to 0} \frac{f(\lambda _t) - g(\lambda _t)}t = 0.
  $$

Proof.  By the mean value Theorem,  for each $t$ in $U$, we may write
$$
  f(\lambda _t) = f(a) + f'(\xi _t)(\lambda _t-a),
  $$
and
$$
  g(\lambda _t) = g(a) + g'(\eta _t)(\lambda _t-a),
  $$
where $\xi _t$ and $\eta _t$ lie in between $\lambda _t$ and $a$.  Therefore
$$
  |f(\lambda _t) - g(\lambda _t)| = $$ $$ =
  |f'(\xi _t)-g'(\eta _t)||\lambda _t-a| \leq
  c|f'(\xi _t)-g'(\eta _t)| |t|.
  $$
Since both $\xi _t$ and $\eta _t$ converge to $a$,  as $t\to 0$,  we get
$$
  \lim_{t\to 0} \left|\frac{f(\lambda _t) - g(\lambda _t)}t\right| \leq $$ $$ \leq
  \lim_{t\to 0} c|f'(\xi _t)-g'(\eta _t)| =
  c|f'(a)-g'(a)| = 0.
  $$
QED

Lemma 2.  Let $A$ and $B$ be $n\times n$ self-adjoint complex matrices and
let $f$ and $g$ be real valued functions of class $C^1$ defined on a neighborhood of $\sigma (A)$, such that
$$
  f(a)=g(a),\quad \text{and}\quad  f'(a)=g'(a), \quad \forall a\in \sigma (A).
  $$
Then
$$
  \lim_{t\to 0} \frac{f(A+tB)-g(A+tB)}t = 0.
  $$

Proof.  Let $a_1\leq a_2\leq \cdots \leq a_n$ be the eigenvalues of $A$, and let $\lambda _1(t)\leq \lambda _2(t)\leq \cdots \leq \lambda _n(t)$ be the eigenvalues of $A+tB$.
By Weyl's inequality (stability of eigenvalues), we have that
$$
  |\lambda _i(t)-a_i|\leq |t| \|B\|.
  $$
So
$$
  \lim_{t\to 0}\left\|\frac{f(A+tB)-g(A+tB)}t\right\| = $$ $$ =
  \lim_{t\to 0}\max_{1\leq i\leq n}  \left|\frac{f(\lambda _i(t))-g(\lambda _i(t))}t\right| = 0,
  $$
by
Lemma (1).  QED

Theorem 3. Let $A$ and $B$ be $n\times n$ self-adjoint complex matrices and
let $f$ be a real valued function of class $C^1$ defined on a neighborhood of $\sigma (A)$.  Then
$$
  \frac d{dt}\Big|_{t=0} \text{tr}(f(A+tB)) = \hbox{tr}(f'(A)B).
  $$

Proof.  Assume first that $f$ admits a  holomorphic extension to  a complex neighborhood of the spectrum of $A$.   Then
$$
  \frac d{dt}\Big|_{t=0} f(A+tB) =
  \frac d{dt}\Big|_{t=0} \frac{1}{2\pi}\oint f(z)(z-A-tB)^{-1}\,d z =
  $$
$$
  =   \frac{1}{2\pi}\oint f(z)(z-A)^{-1} B(z-A)^{-1}\,d z
  $$
... which has the same trace as ...
$$
  \frac{1}{2\pi}\oint f(z)(z-A)^{-2} B\,d z =  $$ $$ =
 \left(\frac{1}{2\pi}\oint f(z)(z-A)^{-2} \,d z\right) B  = f'(A)B.
  $$

Back to  the general case,  let $p$ be a real polynomial such that
$$
  p(a)=f(a),\quad \text{and}\quad  p'(a)=f'(a), \quad \forall a\in \sigma (A).
  $$
We then have that
$$
  \left\|\frac{f(A+tB) - f(A)}t  -\frac{p(A+tB) - p(A)}t   \right\| = $$ $$ =
  \left\|\frac{f(A+tB) - p(A+tB)}t   \right\|,
  $$
which converges to zero by Lemma (2).  Taking the limit as $t\to 0$,  it  then follows that
$$
  \frac d{dt}\Big|_{t=0} \text{tr}(f(A+tB)) =
  \frac d{dt}\Big|_{t=0} \text{tr}(p(A+tB)) = $$ $$ =
  \hbox{tr}(p'(A)B) =
  \hbox{tr}(f'(A)B).
  $$
QED

Title: Derivative of trace function


问:
A question related to $S^{\perp}$ and closure of span of $S$
This question was asked in my linear algebra quiz previous year exam and I was unable to solve it.

Let V be an inner ( in question it's written integer , but i think he means inner) product space and S be a subset of V. Let $\bar S$denote the closure of S in V with respect to topology induced by the metric given by inner product. Which of the following statements are  true?

A $ S $=$ (S^{\bot})^{\bot}$

B $ \overline S$= $(S^{\perp})^{\perp}$

C $\overline {\text{span}(S)}$=$(S^{\bot})^{\bot}$

D $ S^{\bot} $=$ ((S^{\bot})^{\bot})^{\bot}$

I was completely blank on how can I approach this problem although I have studied linear algebra carefully. Can you please tell on how I should approach the problem .

Edit : I tried It again . I marked A ,D but answer is C,D. If A is false I don't see why D must be true. So, I think I am missing some concepts.

答:
$A,B,C$ are false, in general, and $D$ is true. If $V$ is Hilbert, then also $C$ is true.

Let $S$ be any set, then $S^\perp$ is the set of all vectors that are perpendicular to all elements of $S$. That's usually a definition, $S^\perp := \{v\,|\,\,\forall\,s\in S\,\,g(v,s)=0\}$.

You can easily verify that this is a vector subspace. Namely, if $v,w\in S^\perp$ and $\alpha\in \mathbb{R}$, then also $v + \alpha w\in S^\perp$.

So also $(S^\perp)^\perp$ is a vector subspace.

This immediately excludes $A$ and $B$, in cases when $S$ (or $\overline{S}$) is just a subset of $V$ but not a vector subspace. You can find plenty of counter-examples, for instance  $S=\{v\}$ for one vector $v\neq 0$.

$C$ is harder. The easy part is $LHS \subseteq RHS$.

Let $v$ be from the LHS. That is, $v$ is a limit of some $v_i$ such that $v_i\in \text{span}(S)$. Each $v_i$ is a combination $\beta_j s_j$ of elements of $S$. If $w\in S^\perp$, then $w$ is orthogonal to all elements of $S$ and hence $g(w, s_i)=0$ and also $g(w, v_i)=0$. Using continuity of $g$, we have $g(w, v) = g(w, \lim_i v_i) = \lim g(w, v_i) = 0$, so $w$ is also orthogonal to $v$. So $v$ is in $(S^\perp)^\perp$.

For the other implication, you need to show that a closed subspace (closure of the span of $S$ here) has an orthogonal complement. This is not always the case: there are proper closed subspaces $Y$ such that $Y^\perp = 0$!

So take such $Y$ as your $S$, then the span of $S$ is $S$ and the closure of $S$ is $S$, and $(S^\perp)^\perp = V \neq S$.

However, this being said, $C$ holds under some additional assumptions, such as $V$ being a Hilbert space.

If $C$ is true, then $D$ follows easily, because it reduces to
$$
S^\perp = (\overline{span(S)})^\perp
$$
which is obvious.

In general, if $V$ is not complete, then the proof of $D$ is given by the answer of @Ruy.


问:
if you invest $ 500 at 6 % compounded annually,
Please help me with this problem. It needs to be done in the same format as below
if you invest $500$ dollars at $6$ percent compounded annually, how many years to the nearest tenth would it take your investment to grow to $\$1300$. MY answer was $16$.

please show solution and explain.

答:
Here is an answer done "their" way-
We have $$1300=500(1.06)^n$$ $$\implies 13=5(1.06)^n$$ $$\implies \frac{13}{5}=(1.06)^n$$ $$\implies \log_{1.06}\frac{13}5 =n$$ $$\implies n \approx 16.39830702$$ $$\implies n \approx 16.4$$ Rounded to one decimal place.

Please ask if you need any further clarification.


问:
Evaluate the price of asset in log Black and Scholes model
Context:

I'm given Black and Scholes model.

Problem:

For $T > 1$ the asset returns $$Y_T = \ln(S_T)\cdot 1_{S_T>1} $$
where $S_t$ is solution for Black and Scholes model:
$$S_t = S_0 \cdot exp[(\mu - \frac{\sigma^2}{2})t +\sigma W_t]$$
$W_t$ is Wiener process and $1_{S_T>1} = 1$ if $S_T>1$ and $1_{S_T>1} = 0$ otherwise.

It is given that today $t$ is equal to $1$.

Evaluate the current price $Y_1$.

Can you give some ideas?

答:
Ito's Lemma is the answer here. First note that $S_t$ is an Ito's Process:

$$ S_t:= S_0 + \int_{h=0}^{h=t} \left( \mu S_h \right)dh + \int_{h=0}^{h=t} \left( \sigma S_h \right)dW_h  $$

The solution to the above equation is indeed: $S_t=S_0e^{(\mu-0.5\sigma^2)t+\sigma W_t}$

Since $S_t$ is an Ito Process, Ito's Lemma states that for any ("well-behaved", twice differentiable) function $F()$ of $S_t$ and $t$, the function $F(S_t,t)$ follows the process:

$$F(S_t,t)= F(S_0,t_0) + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial t}+\frac{\partial F}{\partial S}a(S_h,h)_{=\mu S_h}+\frac{1}{2}\frac{\partial^2 F}{\partial S^2}b(S_h,h)^2_{=\sigma^2 S_h^2}\right)dh + \int_{h=0}^{h=t}\left(\frac{\partial F}{\partial S}b(S_h,h)_{=\sigma S_h}\right)dW_h $$

Taking $F(S_t,t)=ln(S_t)$, we get:

$$F(S_t,t)= ln(S_0) + \int_{h=0}^{h=t}\left(0+\frac{1}{S_h}a(S_h,h)_{=\mu S_h}-\frac{1}{2}\frac{1}{S_h^2}b(S_h,h)^2_{=\sigma^2 S_h^2}\right)dh + \int_{h=0}^{h=t}\left(\frac{1}{S_h}b(S_h,h)_{=\sigma S_h}\right)dW_h=\\=ln(S_0) + \int_{h=0}^{h=t}\left(\mu-\frac{1}{2}\sigma^2\right)dh + \int_{h=0}^{h=t}\left(\sigma \right)dW_h=\\=ln(S_0)+(\mu - 0.5 \sigma^2)t + \sigma W_t$$

(taking exp on both sides of the above will actually give the solution $S_t=S_0e^{(\mu-0.5\sigma^2)t+\sigma W_t}$ to the original GBM SDE for $S_t$.)

The equations above give valid price processes under the real world measure $\mathbb{P}$. If we want to price the claim under the risk-neutral measure $\mathbb{Q}$ (associated with the money market Numeraire), we'll need to perform a change of measure, using the following Radon-Nikodym derivative:

$$\frac{d\mathbb{Q}}{d\mathbb{P}}=exp\left\{0.5\left(\frac{\mu-r}{\sigma}\right)^2 t-W_t\left(\frac{\mu-r}{\sigma}\right)\right\}$$

Applying the above R-N to the stock price process under $\mathbb{P}$ will result in the change of the drift in the stock price process from $\mu$ to $r$ and will give the price process under $\mathbb{Q}$, which is what we need for pricing.

Now the actual problem (everything below is under the measure $\mathbb{Q}$, I assume we had applied the R-N derivative as above to the price process of the Stock):

$$Y_T = \ln(S_T) \mathbb{I}_{S_T>1}$$

Given some initial state $S_0$, the above can be re-written using what we derived above for $ln(S_t)$:

$$\ln(S_T) \mathbb{I}_{S_T>1}=\left(ln(S_0)+(r - 0.5 \sigma^2)T + \sigma W_T\right)\mathbb{I}_{S_T>1}$$

Given some initial state at time $t_1=1$, instead of $t_0 = 0$, we can rewrite the above simply as:

$$\ln(S_T) \mathbb{I}_{S_T>1}=\left(ln(S_1)+(r - 0.5 \sigma^2)(T-1) + \sigma W(T-1)\right)\mathbb{I}_{S_T>1}$$

We are interested in the (risk-neutral) price of this claim (given the initial state at $t=1$); the price at time $t_1$ is (by definition) computed as the discounted, risk-neutral expectation under the measure $\mathbb{Q}$ of the pay-off at maturity, as follows:

$$ Y(t_1):=e^{-r(T-1)}\mathbb{E}^{\mathbb{Q}}\left[\ln(S_T) \mathbb{I}_{S_T>1} | \mathcal{F}_{t_1} \right]$$

Substituting in for $ln(S_T)$, we get:

$$e^{-r(T-1)}\mathbb{E}^{\mathbb{Q}}\left[\left(ln(S_1)+(r - 0.5 \sigma^2)(T-1) + \sigma W(T-1)\right)\mathbb{I}_{S_T>1} | \mathcal{F}_{t_1} \right] =\\= e^{-r(T-1)}\mathbb{E}_1^{\mathbb{Q}}[ln(S_1)\mathbb{I}_{S_T>1}]+\mathbb{E}_1^{\mathbb{Q}}[(r - 0.5 \sigma^2)(T-1)\mathbb{I}_{S_T>1}] + \mathbb{E}_1^{\mathbb{Q}}[\sigma W(T-1)\mathbb{I}_{S_T>1}]$$

Now term by term:

$$ \mathbb{E}_1^{\mathbb{Q}}[ln(S_1)\mathbb{I}_{S_T>1}]=ln(S_1)\mathbb{P}^{\mathbb{Q}}(S_T>1)=ln(S_1)\mathbb{P}^{\mathbb{Q}}(S_1e^{(r-0.5 \sigma^2)(T-1)+\sigma W(T-1)}>1)=\\= ln(S_1)\mathbb{P}^{\mathbb{Q}}\left(Z>\frac{ln\left(\frac{1}{S_1}\right)-r(T-1)+0.5 \sigma^2 (T-1)}{\sigma \sqrt{T-1}}\right)=\\=ln(S_1)\mathbb{P}^{\mathbb{Q}}\left(Z<\frac{ln(S_1)+r(T-1)-0.5 \sigma^2 (T-1)}{\sigma \sqrt{T-1}}\right):=ln(S_1)\Phi(d)$$

Where: $d:=\frac{ln(S_1)+r(T-1)-0.5 \sigma^2 (T-1)}{\sigma \sqrt{T-1}}$ and $\Phi$ is the standard Normal CDF.

The second term then evaluates to:

$$\mathbb{E}_1^{\mathbb{Q}}[(r - 0.5 \sigma^2)(T-1)\mathbb{I}_{S_T>1}]=(r - 0.5 \sigma^2)(T-1)\Phi(d)$$

For the last term, I will use the fact that $$\mathbb{P}^{\mathbb{Q}}(S_T>1)=\Phi(d)$$ and  the fact that:

$$\mathbb{E}_1^{\mathbb{Q}}[\sigma W(T-1)\mathbb{I}_{S_T>1}]=\sigma \left(\sqrt{T-1}\right) \mathbb{E}_1^{\mathbb{Q}}[Z \mathbb{I}_{Z<d}]=\\=\sigma \left(\sqrt{T-1}\right)  \int_{-\infty}^{d}h\frac{1}{\sqrt{2\pi}}exp\left\{\frac{-h^2}{2} \right\}dh:=\sigma \left(\sqrt{T-1}\right) f(d) $$

The final answer should then be:

$$Y(t_1)=e^{-r(T-1)}\left(ln(S_1)\Phi(d)+(r - 0.5 \sigma^2)(T-1)\Phi(d)+\sigma \left(\sqrt{T-1}\right) f(d)\right)$$


问:
Combinatoric "Rotation" Cycle
Suppose we have a set of numbers, $C=\{1,2,3,\dots,n\}$, and we have the function $\pi(x)$ that moves all elements to right by $x$ places iteratively, starting in the first element. If there are not $x$ more elements to the right, it goes back to the start.

For example, if $C=\{1,2,3,4,5\}$, then $\pi(1)$ applied to $C$ is $\{2,3,4,5,1\}$

Now for example, $A=\{1,2,3,4,5,6\}$ then $\pi(4)$ applied to $A=\{5,3,1,5,\dots\}$ so $\pi(4)$ applied to $A$ is $\{5,3,1\}$.

Now consider the set $Y=\{1,2,\dots,n\}$, $n\ge3$. How many $x$'s there are such that $\pi(x)$ applied to $Y$ doesn't not have the elements $1,2,3$ at the same time and are not repeated sequences?

What I mean by repeated sequences:

$C=\{1,2,3,4,5\}$ , $\pi(1) = \pi(6)$, so it's counted as 1.

答:
I'm pretty sure the answer is $\infty\ \forall\ n\ge3$.

Reasoning:

First note that there always exists an $x$ for any values of $n\ge3$ such that an application of $\pi(x)$ will include $1$, $2$, and $3$, namely $x=1$. Then note that an application of $\pi(n+1)$ has the same effect as applying $\pi(1)$, and same with $\pi(2n+1)$, $\pi(3n+1)$, and so on. So we get that $x\equiv1\pmod n$, of which there are infinitely many $x$'s that satisfy this congruence.

Title: Combinatoric "Rotation" Cycle


问:
Volume via Jacobi fields: Proof of Lemma 5.4 of Sakai's book "Riemannian Geometry"
I am self-teaching Riemannian geometry and am stuck with the proof of Lemma 5.4 on page 65 of Takashi Sakai's book "Riemannian Geometry". The lines where I am stuck with are highlighted in the attached picture. In the picture, the $$\Vert \cdot \Vert$$ denotes the (absolute value of the signed) volume of the parallelepiped spanned by the vectors it embraces (and so please ignore the wedge $\wedge$ operator. For example, $\Vert v \wedge w \Vert$ for two m-dimensional vectors $v$ and $w$ is the (absolute value of the signed) area of the parallelogram spanned by $v$ and $w$).

I got why there is factor $t^{1-m}$ there but did not get why $$\Vert Y_{1}(t) \wedge \cdots \wedge Y_{m-1}(t)\wedge \dot{\gamma}(t)\Vert = \Vert Y_{1}(t) \wedge \cdots \wedge Y_{m-1}(t)\Vert,$$ where $\gamma_{u}(t)$ is a unit speed geodesic with initial speed vector $u=e_m$ (when $t=0$). Note that the Jacobi fields are not necessarily orthogonal along $\gamma_u(t)$.

I am not able to see how $\dot{\gamma}(t)$ and $\Vert \dot{\gamma}(t) \Vert \equiv 1$ plays their roles in the height-times-base formula for the above identity. Could someone please explain why the above identity is true? Any help or hint is appreciated.

Added: I think I got it. By Gauss Lemma, $\dot{\gamma}_u(t)$ is orthogonal to each $Y_{i}(t)$ for $i=1,\cdots,m-1$. Since $\Vert \dot{\gamma}(t) \Vert \equiv 1$, the height of the parallelepiped is $1$. So, the base area, $\Vert Y_{1}(t) \wedge \cdots \wedge Y_{m-1}(t)\Vert$, equals the volume.

答:
If $u$ is a unit tangent vector to $p \in M$, then $\gamma(t) = \exp_p(tu)$ is a unit speed geodesic. If $v \in T_pM$ is orthogonal to $u$, the Gauss lemma shows that
$$
g_{\gamma(t)}\left(\mathrm{d}\exp_p(tu)\cdot u, \mathrm{d}\exp_p(tu)\cdot v\right) = g_p(u,v)
$$
and then, $\mathrm{d}\exp_p(tu)\cdot v$ is orthogonal to $\mathrm{d}\exp_p(tu)\cdot u = \gamma'(t)$.

If $Y_i(t) = t\mathrm{d}\exp_p(tu)\cdot e_i$, then $Y_i(t)$ is colinear to $\mathrm{d}\exp_p(tu)e_i$, and if $e_i$ is orthogonal to $u$, then $Y_i(t)$ is orthogonal to $\gamma'(t)$. thus, if $Y_1(t),\ldots,Y_{m-1}(t)$ are all orthogonal to $\gamma'(t)$, one has
\begin{align}
\left\|Y_1(t)\wedge\cdots\wedge Y_{m-1}(t)\wedge \gamma'(t) \right\| &= \left\| Y_1(t) \wedge \cdots \wedge Y_{m-1}(t)\right\| \left\|\gamma'(t) \right\| \\
&= \left\| Y_1(t) \wedge \cdots \wedge Y_{m-1}(t)\right\|
\end{align}
because $\gamma$ has unit speed.

The first equality may seem a bit obscur. Here is a reason why. If $(e_1,\ldots,e_{m-1},u)$ is an orthonormal basis of $T_pM$, one can transport it as a parallel frame along $\gamma(t)$, say $(E_1(t),\ldots,E_{m-1}(t),\gamma'(t))$. It is an orthonormal basis of $T_{\gamma(t)}M$. The norm $\|Y_1(t)\wedge \cdots \wedge Y_{m-1}(t) \wedge \gamma'(t)\|$ is the absolute value of the determinant $\det\left(Y_1(t),\ldots,Y_{m-1}(t),\gamma'(t)\right)$ in any orthonormal basis. Thus, in the orthonormal basis $E_1(t),\ldots,E_{m-1}(t),\gamma'(t)$, this determinant have the form
$$
\begin{vmatrix}
g(Y_1(t),E_1(t)) & \cdots & g(Y_{m-1}(t),E_{1}(t)) & 0 \\
\vdots &  & \vdots & 0 \\
g(Y_1(t),E_{m-1}(t)) & \cdots & g(Y_{m-1}(t),E_{m-1}(t))\\
0 & 0 & 0 & \|\gamma'(t)\|
\end{vmatrix}
$$
and the formula is thus straightforward.

In the proof you are showing, the author takes $(e_1,\ldots,e_{m-1},u)$ to be and orthonormal basis. Thus, it is clear that $e_i$ and $u$ are orthogonal.

Remark: You can prove that $Y_i$ are all orthogonal to $\gamma'$ without the Gauss lemma, if you know that $V_i(t)=\mathrm{d}\exp_p(tu)\cdot e_i$ are Jacobi fields. Knowing they are Jacobi fields, they satisfy
$$
V_i'' = -R(\gamma',V_i)\gamma'
$$
Let $f(t) = g_{\gamma(t)}\left(V_i(t),\gamma'(t)\right)$. It is a smooth function, and one easily shows that
$$
f''(t) = g_{\gamma(t)}\left( V_i''(t),\gamma'(t)\right) = -R\left(\gamma'(t),V_i(t),\gamma'(t),\gamma'(t)\right) = 0
$$
the last equality being a direct consequence of the fact that the Riemann tensor is skew symmetric in the two last variable. Thus, $f$ is an affine function, of the form $f(t) = f'(0)t + f(0)$. The initial data saying that $V_i(0)$ and $V'_i(0)$ are orthogonal to $\gamma'(0)$ show that $f(0)=f'(0)=0$, thus $f=0$ and $V_i$ is orthogonal to $\gamma'(t)$.

Comment: there is a general use of these expressions for studying the volume outside a submanifold in a riemannian manifold. They are named Heintze-Kärcher equalities / inequalities, named after Heintze and Kärcher. You can find online (for example here) how they seemed to have first appeared.


问:
Generating function with singularity at $x=0$
I came across this when trying to solve the following recurrence relation via generating function.
$$(4n+3)a_n+2na_{n-1}+(2n+3)a_{n+1}=0$$

If we define $A(x)=\sum_{n=0}^{\infty}a_nx^n$, we see that it satisfies the ODE
$$A'+\frac{2x+1}{2x(x+1)}A=\frac{a_0}{2x(x+1)^2}$$
This equation can be solved to find two generating functions corresponding to the two independent solutions to the recurrence relation,
$$A(x)=\frac{1}{\sqrt{x(1+x)}},\quad A(x)=\frac{1}{1+x}.$$

The second generating function listed tells us that $a_n=(-1)^n$ is a solution. The first generating function has a branch cut at $x=0$, it doesn't seem like there's a way to find the coefficients $a_n$ expanded near $x=0$ for this function.

Luckily, it is still possible to find the second solution: $a_n=(-1)^n\frac{(2n)!!}{(2n+1)!!}$. My question is, how are these coefficients related to the function $\frac{1}{\sqrt{x(1+x)}}$? In fact, the generating function for this solution is
$$A(x)=\sum_{n=0}^{\infty}(-1)^n\frac{(2n)!!}{(2n+1)!!}x^n=\frac{\text{arcsinh}(\sqrt{x})}{\sqrt{x(1+x)}}$$

The factor $\text{arcsinh}(\sqrt{x})$ appears, and gets rid of the branch point. I am very new to the field of generating functions, this is all very confusing!

Note: I would like to understand this so I may get insight into the related recurrence relation (apologies for the alliteration)

$$(4n+3)a_n+2na_{n-1}+(2n+3)a_{n+1}=\lambda a_n$$

Where the problematic generating function is now
$$A(x)=\frac{1}{\sqrt{x(1+x)}}e^{-\frac{\lambda}{2(1+x)}}$$
Here I am not so lucky as to already have the solution.

答:
It appears that the form of the generating function you derived is incomplete. In fact, careful calculation shows that it satisfies a first order ODE similar to yours with an extra term

$$A'(x)+\frac{2x+1}{2x(x+1)}A(x)=\frac{a_0}{2x(x+1)^2}+3\frac{a_1+a_0}{2(x+1)^2}$$

Multiply by the integrating factor $\sqrt{x(1+x)}$  and rewrite to obtain

$$\sqrt{x(1+x)}A(x)=C+a_0\int dx\frac{1}{2\sqrt{x}(x+1)^{3/2}}+3(a_0+a_1)\int dx\frac{\sqrt{x}}{2(x+1)^{3/2}}$$

which evaluate to the final result

$$A(x)=\frac{C}{\sqrt{x(1+x)}}+a_0\frac{1}{1+x}+3(a_1+a_0)\left(\frac{\text{arcsinh}(\sqrt{x})}{\sqrt{x(1+x})}-\frac{1}{1+x}\right)$$

Evidently, for the choice $a_1=-\frac{2}{3}a_0$ one obtains the solution desired, so there's no contradiction.


问:
expectations of limits of norms of convergent r.v.
Assume we have two sequences of random vectors $(X_{n})_{n > 1}$ and $(Y_{n})_{n > 1}$ taking values in $\mathbb{R}^{k}$ and defined on the same probability space. Also, we know that $(X_{n})_{n > 1}$ converges weakly to some random $X$ and $E[||X_{n}||_{2}^{2}] < \infty$ for all $n$.

Next, for some fixed continuous function $g: \mathbb{R}^{k}\times \mathbb{R}^{k} \to \mathbb{R}_{+}$ let $Z_{n}$ be the following random variable

$$
Z_{n} = g(X_{n}, Y_{n})
$$
and assume there exists $n_{0}$ such that for all $n>n_{0}$ the following holds almost surely
$$
Z_{n} \leq C ||X_{n}||_{2}^{2},
$$
for some constant $C$.

Is the following correct?

$$
E[\limsup_{n}g(X_{n}, Y_{n})] \leq C \, E[||X||_{2}^{2}].
$$

答:
No, this is not correct. Draw $X_n=n$ with probability $1/n$ and $0$ with probability $1-1/n$, independently for all $n$. Pick $Z_n=X_n$. Then:

$(X_n)$ converges in distribution to $X\equiv 0$.

$(X_n\ge n)$ is a sequence of independent events whose sum of probabilities diverges, so by the second Borel-Cantelli lemma $(X_n\ge n)$ occurs infinitely often almost surely. In particular $\lim\sup X_n=+\infty$ almost surely.

So $X_n\le X_n^2$, $\mathsf{E} \lim\sup X_n=+\infty$ and $\mathsf{E} X^2=0$.


问:
Inferring a recurrence relation, stairway climbing
Problem.  Suppose a dog can jump one or two steps of a stairway of $k$ steps.  How many different ways can this dog climb up?  Answer it with a recurrence relation.

Solution.  I calculated the number of ways for the first step $n_1 = 1$, and also for second $n_2 = 2$ and third step $n_3 = 3$.  From the pattern, I'm inferring that $n_k = n_{k-1} + n_{k-2}$ and so it seems to be a Fibonacci sequence with the first element chopped off --- 1, 1, 2, 3, ... But I don't consider this a solution because I'm sort of guessing.  What is the reasoning that I could apply here to naturally build this recurrence relation?  How should I think of this to be able to infer it myself?  Must I always guess and prove things by induction?

答:
How did dog get to step $k$? From either step $k-1$ or step $k-2$.

Can it get to stair number $k - 1$ the same way as to $k-2$? No, as it can't get to different steps the same way:)

So, number of ways to get to the step $k$ is sum of number of ways to get to steps $k - 1$ and $k - 2$. Or, in formula, $n_k = n_{k - 1} + n_{k - 2}$.


问:
Integral $\int\limits^{\infty}_0\frac{\tan^{-1}t }{(1+t)^{n+1}} dt$
I'm having a good amount of trouble evaluating this: $$\int\limits^{\infty}_0\frac{\tan^{-1}(t)dt}{(1+t)^{n+1}},\ n>0$$
Here are some methods I've tried:
$$\int\limits^{\infty}_0\frac{\tan^{-1}(t)dt}{(1+t)^{n+1}}=\frac1n\int\limits^{\infty}_0\frac{dt}{(1+t^2)(1+t)^{n}}$$
using integration by parts. I then tried more integration by parts, residue theorem, and expanding into a power series but failed. I did however use partial fractions for $n=2$ to get $1/4$.
$$\int\limits^{\infty}_0\frac{\tan^{-1}(t)dt}{(1+t)^{n+1}}=\frac{\pi}{2n}-\int\limits^{\infty}_0\frac1{(1+t)^{n+1}}\int\limits^{\infty}_0\frac{\sin(x)}xe^{-xt}dxdt=\frac{\pi}{2n}-\int\limits^{\infty}_0\frac{\sin(x)}xE_{n+1}(x)e^{-x}dx$$
using the Laplace Transform of $\text{sinc}(x)$ and the $E_n$-function.
$$\int\limits^{\infty}_0\frac{\tan^{-1}(t)dt}{(1+t)^{n+1}}=\int\limits^1_0\frac{\tan^{-1}(t)dt}{(1+t)^{n+1}}+\int\limits^1_0\frac{\cot^{-1}(t)t^{n-1}dt}{(1+t)^{n+1}}=\int\limits^1_0\frac{\tan^{-1}(t)\left(1-t^{n-1}\right)dt}{(1+t)^{n+1}}+\frac{\pi}{2^{n+1}n}$$
This one I felt the best about and its also where I got that at $n=1$ the integral is $\pi/4$, but I was unable to go further.

Update: I had a couple more attempts, one of which I posted as an answer, after Claude Leibovici's idea reminded me you can do partial fractions on $\frac1{(1+x^2)(1+x)^n}$.

Notice that if we write
$$\frac1{(1+x^2)(1+x)^n}=\frac{1+x}{1+x^2}-\frac{a_0+a_1x+\dots+a_{m-1}x^{m-1}}{(1+x)^n}$$
then the coefficients $a_k$ follow the pattern
$$a_0=0,\ a_1=C_1^{n+1},\ a_2=C_2^{n+1}-a_0,\ a_3=C_3^{n+1}-a_1,\ a_4=C_4^{n+1}-a_2\dots$$
The only problem is that this sequence is always infinite and the power series does not converge on all of $[0,\infty)$, so I believe the coefficients of $1$ and $x$ in the numerator of $\frac{1+x}{1+x^2}$ could be changed to avoid this.

答:
I was able to get some kind of recurrence relation for
$$I_n=\int\limits_0^{\infty}\frac{dt}{(1+t^2)(1+t)^n},$$
but I am not satisfied with it since you can't really do anything with it. It's still kind of an answer but I'll accept a better one.

First substitute $t\mapsto\frac1t$ so that
$$I_n=\int\limits_0^{\infty}\frac{t^ndt}{(1+t^2)(1+t)^n}.$$
Notice that you can modify the binomal expansion so it is similar to partial fractions:
$$\begin{align*}
(1+t)^n&=\sum_{k=0}^n\left(\begin{matrix}n\\k\end{matrix}\right)t^k\\
\left(1+\frac{-1}t\right)^n&=\sum_{k=0}^n\left(\begin{matrix}n\\k\end{matrix}\right)\left(\frac{-1}t\right)^k\\
\frac{t^n}{(1+t)^n}&=\sum_{k=0}^n\left(\begin{matrix}n\\k\end{matrix}\right)\left(\frac{-1}{1+t}\right)^k.\\
\end{align*}$$
It then follows that
$$I_n=\int\limits_0^{\infty}\frac1{(1+t^2)}\sum_{k=0}^n\left(\begin{matrix}n\\k\end{matrix}\right)\left(\frac{-1}{1+t}\right)^kdt
=\sum_{k=0}^n\left(\begin{matrix}n\\k\end{matrix}\right)(-1)^kI_k.$$
$$\implies\boxed{(1-(-1)^n)I_n=\sum_{k=0}^{n-1}\left(\begin{matrix}n\\k\end{matrix}\right)(-1)^kI_k}$$
Unfortunately it sucks and the most I was able to do with it was find $I_3=3/4-\pi/8$ from already knowing $I_0=\pi/2,$ $I_1=\pi/4,$ and $I_2=1/2$.

There may be some merit in the study of
$$I_n=\frac12\int\limits_0^{\infty}\frac{(1+t^n)dt}{(1+t^2)(1+t)^n}$$
or maybe splitting the interval into $[0,1]$ and $[1,\infty)$.


问:
How to find a path tangent to a distribution
Let $D$ be the smooth distribution on $\mathbb{R}^3$ such that
$$
D_{(a,b,c)}=\{\,(x,y,z)\in\mathbb{R}^3\,:\,z-bx=0\,\}.
$$
How to show that for any $p,\,q\in\mathbb{R}^3$, there exists a path $\alpha$ from $p$ to $q$ tangent to $D$?

答:
Differential forms are a powerful tool. Soldiering on without them, here's a suggestion:  The vector fields $\partial/\partial y$ and $\partial/\partial x+y\,\partial/\partial z$ span $D$. Let $p=(p_1,p_2,p_3)$ and $q=(q_1,q_2,q_3)$.

(1) If $p_1=q_1$ and $p_3=q_3$, then you have an easy path contained in a single plane. Indeed, you can always move freely in the $y$ direction.

(2) If $p_1\ne q_1$, choose $b$ so that $(q_3-p_3)=b(q_1-p_1)$. Consider the points $(p_1,b,p_3)$ and $(p_3,b,q_3)$ and join them by a path tangent to $D$.

(3) If $p_1=q_1$ and $p_3\ne q_3$, first follow $D$ from $(p_1,p_2,p_3)$ to $(p_1,0,p_3)$ and then follow $D$ to $(p_1',0,p_3)$ with $p_1'\ne p_1$. Now you're back in case (2).


问:
Conjugation of a product of disjoint cycles
For a certain question I have to do, I have to use the fact that if $a$ is any permutation in $S_n$ and $\gamma = (i_1 \cdots i_j)$ is a cycle, then $$a\gamma a^{-1} = (a(i_1) \cdots a(i_j)),$$ for some $K \leqslant S_4$ (I have to determine $aKa^{-1} $ for all $a \in S_4)$ where $K$ has some elements which are the product of 2-cycles. How would I apply the above to a product of 2-cycles (namely the right-hand side)?

答:
Just use the fact that $\gamma \mapsto \gamma a \gamma^{-1}$ is a group inner automorphism. Which means that for a product of cycles $c_1, \dots, c_m$ you have

$$\gamma (c_1 \dots c_m) \gamma^{-1} = (\gamma c_1 \gamma^{-1}) \dots (\gamma c_m \gamma^{-1}).$$

Applied to $2$-cycles:

$$\gamma ((a_1 \ a_2) \dots  \dots (r_1 \ r_2)) \gamma^{-1} = (\gamma(a_1) \ \gamma(a_2)) \dots (\gamma(r_1) \ \gamma(r_2)).$$


问:
Probability problem from CMU contest prep course
Your sports team is playing a best-of-15 series against a single opponent. Against this opponent, your
team wins with probability 40%, unless it is behind in the series (with strictly fewer wins than losses
so far in these 15-game finals), at which point your team wins with probability 60%. What is the
probability that your team wins?

I'm not great at math so I wrote a dynamic programming algorithm to solve it. If my algorithm is correct, the answer would be 40%. I found that if the number of matches is odd, the probability that you win is always 40%, for any number of matches.

Can anyone verify this and provide a probabilistic explanation of the solution?

Thanks!

答:
Here's another way to match up your wins with your losses to show that your probability of winning is 40%. Notice first that the question's descriptions of the probabilities can be stated as: Whenever the series isn't tied, the team that's behind has a 60% chance of winning, but when the series is tied your opponent has a 60% chance of winning. So the only difference between the two teams is when the series is tied.

As in Calvin Lin's answer, represent the possible histories of the series by 15-letter strings of L and W. Define a function $F$ from such histories to histories as follows. Given a history $h$, let $F(h)$ be the same as $h$ up to the last moment $M(h)$ when the series was tied. (There's at least one such moment, since it's tied at 0-0 at the start. So $M(h)$ exists.) After $M(h)$ replace every L in that part of $h$ by W and replace every W by L. So you win in $h$ iff you lose in $F(h)$. Notice also that $F(F(h))=h$ for all histories $h$. So $F$ matches up your winning histories with your losing ones.

Now compare the probability of $F(h)$ with that of $h$. Each is the product of probabilities of the individual games' outcomes conditional on the previous games. Before $M(h)$, $F(h)$ and $h$ are identical, so those conditional probabilities agree. At $M(h)$, whichever of $h$ and $F(h)$ has your team winning will have lower probability by a factor 2/3, since the series is tied at $M(h)$ (which is also $M(F(h))$).  After that, there are no more differences in the probabilities, because there are no more ties. (Remember that $M(h)$ was the last tie.) So among $h$ and $F(h)$, the one where you win at $M(h)$ has lower probability than the other, by a factor of 2/3.

But, since there are no ties after $M(h)$, the team that wins the game right after $M(h)$ will win the series, for the other team can never catch up.  So I can rewrite the last sentence of the preceding paragraph as: Among $h$ and $F(h)$, the one where you win the series has lower probability than the other, by a factor of 2/3.  That is, you win the series with probability 40%.


问:
Is the set of the empty set and omega a collection?
I've recently started to study measure theory and in this respect am reading up on sigma algebras. In this context I saw that sometimes collections are denoted as calligraphic letters (also called algebras) containing other sets. However, for $\mathcal{A} = \{\emptyset, \Omega\}$ a text I read has "The set $\mathcal{A} = \{\emptyset, \Omega\}$ for any set $\Omega$". This confused me aboout whether the term 'collection' has any specific restrictions to be used, i. e. whether the sets contained in a collection have to configured in a certain way; my question therefore:

Is the set $\mathcal{A} = \{\emptyset, \Omega\}$ a collection?

答:
A collection is a set in this context.

The set $\mathcal A$ is a set of set, so $\mathcal A$ is definitely a set.
Because it is a set of sets, it is sometimes informally called
a collection or a collection of sets

(sometimes collection refers to something slightly different, see the comment of Gae. S., but not in this context.)


问:
Find a lattice with exactly three congruence relations
A lattice $L$ is a partially ordered set such that any two elements $a$ and $b$ have a least upper bound $a\lor b$ and a greatest lower bound $a\land b$.  A congruence relation on $L$ is an equivalence relation $\sim$ on $L$ that is compatible with the lattice operation in the sense that $a_1\sim b_1$ and $a_2\sim b_2$ implies $a_1\lor a_2\sim b_1\lor b_2$ and similarly for $\land$.

In any lattice there are always two trivial congruence relations, the congruence relation where each element is its own equivalence class (block), and at the other extreme the congruence relation with a single block.

What's an example of a lattice with exactly three congruence relations?

This is Ex 3.40 in Graetzer (Lattice Theory: Foundation).  So it is asking to find a lattice with exactly one non-trivial congruence relation.  Any example I can think of always has at least two non-trivial congruence relations.

For example, for the linear poset with three elements $a\prec b\prec c$, one congruence relation has $a,b$ in one block and $c$ alone, and another has $a$ alone and $b,c$ together in one block.  Another example, the standard diamond lattice with four elements (consisting of all subsets of a two element set ordered by inclusion) also has two non-trivial congruence relations.

答:
Found the following example:

The only non-trivial congruence relation has $a$ and $b$ in one block and all the other elements alone.


答:
(Fill in the gaps as needed. If you're stuck, write out your working and thought process to demonstrate where you're at.)

Hint: Show that $\pi(k) (Y_n) = \left\{ 1 + t\gcd(k, n ) {\large{\mid}}  0 \leq t \leq \frac{n}{\gcd(k,n) - 1} \right\}$.

E.g. In your example, $ k = 4, n = 6, \gcd(4,6) = 2$ and we have $\pi(4) (Y_6) =\{ 1, 3, 5 \}$.

(depending on the intending phasing with the double negative)
Hence, if you want it to have element $2$, then $ \gcd(k, n) = 1$, so there are $X$ of them.

 $X = \phi (n)$.

Title: Combinatoric "Rotation" Cycle

Body:
Suppose we have a set of numbers, $C=\{1,2,3,\dots,n\}$, and we have the function $\pi(x)$ that moves all elements to right by $x$ places iteratively, starting in the first element. If there are not $x$ more elements to the right, it goes back to the start.

For example, if $C=\{1,2,3,4,5\}$, then $\pi(1)$ applied to $C$ is $\{2,3,4,5,1\}$

Now for example, $A=\{1,2,3,4,5,6\}$ then $\pi(4)$ applied to $A=\{5,3,1,5,\dots\}$ so $\pi(4)$ applied to $A$ is $\{5,3,1\}$.

Now consider the set $Y=\{1,2,\dots,n\}$, $n\ge3$. How many $x$'s there are such that $\pi(x)$ applied to $Y$ doesn't not have the elements $1,2,3$ at the same time and are not repeated sequences?

What I mean by repeated sequences:

$C=\{1,2,3,4,5\}$ , $\pi(1) = \pi(6)$, so it's counted as 1.


问:
Quasi-coherent sheaves $\supset$ locally free sheaves?
This question has been completely reformulated following the guidelines in the comments below, to make it clearer where I was able to get and where I can't get out of. Such comments helped me a lot.

Let $(X,\mathscr{O}_{X})$ be a ringed space. Let $\mathscr{F}$ be a sheaf of $\mathscr{O}_{X}$-modules locally free.

Show that $\mathscr{F}$ is a quasi-coherent sheaf of $\mathscr{O}_{X}$-modules.

I've been thinking about the following: $\mathscr{F}$ locally free $\Rightarrow$ $X$ can be covered by open sets $U$ for which $\mathscr{F}|_{U}\cong \bigoplus_{i \in I}\mathscr{O}_{X}|_{U}$, for some set $I$.

As with any $ X $ scheme, the structure sheaf $\mathscr{O}_{X}$ is quasi-coherent, then $X$ can be covered by open affine subsets $U_j=$Spec $A_j$, such that for each $j$ there is an $A_j$-module $M_j$ with $\mathscr{O}_{X}|_{U_j} \cong \widetilde{M_j}$.

So, we have to: $\mathscr{F}|_{U \cap U_j}\cong \bigoplus_{i \in I}\mathscr{O}_{X}|_{U\cap U_j}\cong \bigoplus_{j \in I} \widetilde{M_j}\cong \widetilde{(\bigoplus_{j \in I} M_j)}$. Thus, $\mathscr{F}$is quasi-coherent.

But now I have a question. A quasi-coherent sheaf $\mathscr{F}$ is primarily a sheaf of $\mathscr{O}_X$-modules. So so that I can see that $\mathscr{O}_X$ is quasi-coherent, I have to look at $ \mathscr{O}_X(U)$ as $ (\mathscr{O}_X(U), +)$ a group and therefore a $\mathscr{O}_X(U)$ - module, for each open $U$ in $X$. To make sense of the definition of quasi-coherent in sheaf of rings $\mathscr{O}_{X} $.

Is that correct?

答:
You say at first that $(X,\mathcal{O}_X)$ is a locally ringed space, but then talk about schemes afterwards. Note that there are many locally ringed spaces which are far from being schemes.

The most likely definition of quasicoherent sheaf on a locally ringed space is a sheaf $\mathcal{F}$ so that there is an open cover $\{U_i\}_{i\in I}$ of $X$ so that over each $U_i$, we have a (possibly infinite) presentation
$$ \mathcal{O}_{U_i}^{\oplus I}\to\mathcal{O}_{U_i}^{\oplus J}\to \mathcal{F}\to 0.$$
Notice that a locally free sheaf $\mathcal{E}$ is of course of this form as there is an open cover $\{U_i\}_{i\in I}$ of the space so that on each $U_i$
$$ 0\to \mathcal{O}_{U_i}^{\oplus J}\xrightarrow{\sim} \mathcal{F}\to 0.$$
In particular, locally free implies quasicoherent.

In the context of schemes, we can define being quasicoherent as being locally of the form $\widetilde{M}_i$ for $M_i$ an $A_i$-module with respect to an open cover $\{\operatorname{spec} A_i\}_{i\in I}$ of the scheme $X$. Then a locally free sheaf can be locally written as $\widetilde{A^{\oplus I}}=\widetilde{A}^{\oplus I}\cong \mathcal{O}_{\operatorname{spec}A_i}^{\oplus I}$ over $\operatorname{spec}A_i$. In particular, a locally free sheaf is a fortiori a quasicoherent sheaf.


问:
Is there always a homotopy taking a point to another in a connected manifold?
Let $M$ be a connected topological $n$-manifold. Note this implies $M$ is path-connected.

Let $a,b \in M$. Must there always exist a continuous $H:M \times [0,1] \to M$ such that $H(m,0)=m$ for all $m \in M$, and $H(a,1)=b$?

We can note that this is true for $\mathbb{R}^n$, by considering $(x,t) \mapsto x+t(b-a)$, and it is also true for $S^1$ by considering $(e^{i\theta}, t)\mapsto e^{i(\theta+t(\theta_b-\theta_a))}$ where $a=e^{i\theta_{a}}, b=e^{i\theta_b}$.

答: If $M$ is a topological manifold and $a\in M$ is any point, then the inclusion $\{a\}\hookrightarrow M$ is a cofibration (i.e. has the homotopy extension property). Take a path $\gamma:I\rightarrow M$ from $a$ to $b$ and use the HEP to get a homotopy $H:M\times I\rightarrow M$ with $H_0=id_M$ and $H_1(a)=\gamma(1)=b$.


问:
Prove by induction: Let p1, · · · , pk be odd positive integers. Prove Sigma (p_i -1)/2 (from I=1 to k) ≡ ((p_1***p_k)-1)/2 (mod 2).
Prove by induction: Let $p_1, \ldots , p_k$ be odd positive integers. Prove $$\sum_{I=0}^k \frac{p_i -1}{2} ≡ \frac{p_1\cdot \ldots \cdot (p_k -1)} {2}(\text{mod } 2)$$

I was able to verify the base case because when $p$ is odd and you let $p_1=2n_1+1$ and $p_2=2n_2+1$, you end up with $n_1 + n_2$ is congruent to $n_1+n_2 (\text{mod } 2)$, which is obviously true.

Then I assumed that the formula holds when $k=l$ and the goal is to prove that the formula holds when $k=l+1$.
Then I have when $k=l+1$,
$$(\frac{p_1-1}{2} + \frac{p_2-1}{2} + \ldots + \frac{p_l-1}{2} + \frac{p_{l+1} -1}{2} \text{ is congruent to  } \frac{p_1\cdot p_2 \cdot \ldots \cdot (p_l - 1)}{2} + \frac{p_{l+1} -1}{2} (\text{mod } 2) $$
I am not sure what to do from here to show that this is true.

答: So you are trying to prove that $\frac{p_1+p_2+...+p_k-k}{2}-\frac{p_1 \cdot p_2 \cdot... \cdot p_k -1}{2} \equiv 0 \ (mod \ 2) \iff A=p_1+p_2+...+p_k-k-(p_1 \cdot p_2 \cdot... \cdot p_k -1) \equiv 0 \ (mod \ 4)$.  Now, suppose that exactly $n$ of numbers $p_1,p_2...,p_k$ have remainder $1$ when divided by $4$ and exactly $m$ of numbers $p_1,p_2...,p_k$ have remainder $3 \equiv (-1)$ when divided by $4$. Therefore $k=n+m$ and $$A = p_1+p_2+...+p_k-p_1 \cdot p_2 \cdot... \cdot p_k +1-k\equiv a+(-b)-1^a\cdot(-1)^b+1-(a+b)=1-2b-(-1)^b .$$ If $b$ is even, then $1-2b-(-1)^b\equiv 1-0-1=0\ (mod \ 4)$, and if $b$ is odd, then $1-2b-(-1)^b\equiv 1-2+1=0\ (mod \ 4) \implies A \equiv 0 \  (mod \ 4)$


问:
Differentiating an integral correctness and conditions
The goal is to turn the integral below in a differential form

$$
\int_{-\infty}^{\infty} k(x)p(x,t)dx = H(t)
$$

Where $p(x,t)$ is a probability density function whether it matters or not I don't know. Anyways, I procced as follows:

$$
\frac{\partial H(t)}{\color{red}\partial \color{red}x} = k(x)p(x,t) \\
$$

$$
\require{cancel}
\frac{\partial H(t)}{\cancel{\color{red}\partial \color{red}t}}\frac{\cancel{\color{red}\partial \color{red}t}}{\color{red}\partial \color{red}x} = k(x)p(x,t) \\
$$

$$
{\partial H(t)} = k(x)p(x,t)\color{red}{\partial x}
$$

Is there anything I have to watch out for, is this allowed, or is this even close to being correct at all?

Thank you so much!

答:
You cannot differentiate with respect to $x$ like this, notice how the right is not a function of $x$ so its differential would be $0$, if you were to it would tell you:
$$k(\infty)p(\infty,t)-k(\infty)p(\infty,t)=0$$
however you can differentiate with respect to $t$ if that helps in this situation. Also notice above that $t$ is not defined as a function of $x$ and so: $\frac{\partial t}{\partial x}=0$


问:
Orbits of two-dimensional real upper-triangular matrices on $\mathbb{R}^2$
Let G be the group $\{ \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} :a,b,c \in \mathbb{R} \}$ and define the action $\pi$ on $\mathbb{R}^2$ by $\pi_g(x) = gx, \forall x \in \mathbb{R}^2$. How many orbits are there?

So what I thought is that $Orb( \begin{pmatrix} 0 \\ 0 \end{pmatrix}) = \{ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \}$ and $Orb( \begin{pmatrix} 0 \\ 1 \end{pmatrix}) = \mathbb{R}^2 \backslash \{\begin{pmatrix} 0 \\ 0 \end{pmatrix} \}$ (since any nonzero vector $\begin{pmatrix} a \\ b \end{pmatrix}$ can be written as $g \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, where $g=\begin{pmatrix} 0 & a \\ 0 & b \end{pmatrix}$ is upper triangular, so there are only the two aforementioned orbits. Is this correct? Thanks in advance!

答:
$G$ is not a group as written, because if $a = 0$ or $c = 0$ then the matrix does not have an inverse.

If you assume $a,c \neq 0$ then you should end up with more than two orbits under the action of $G$ on $\mathbb R^2$.


问:
$A_1,...A_n\;$ are independent events iff $A_1,...A_{n-1},A_{n}^{c}\;$ are independent events
I know that if $A_1,...A_n\;$ are independent events, then we have forall $I\subseteq[n]\;$ $P(\bigcap_{i\in I}A_{i})=\prod_{i=1}^{n}P(A_{i})$

I am kind of stuck, I guess I should be using de morgans laws or something of that sort.

Would love to get some help!

答:
Suppose $A$ and $B$ are independent. Then:

$$\Pr(A\cap B^c)=\Pr(B^c \vert A)\Pr(A)=(1-\Pr(B\vert A))\Pr(A)=(1-\Pr(B))\Pr(A)=\Pr(B^c)\Pr(A) $$

So for $n=2$ this is not a problem. For $n>2$, for each $I$ that contains $n$, you can denote $B=A_n$ and $A=\bigcap\limits_{i\in I\setminus \{n\} A_i} $ and use the above argument.


问:
Using Spherical coordinates find the volume:
Inside the surfaces $z=x^2+y^2$ and $z=\sqrt{2-x^2-y^2}$

I integrated over the ranges:

$0 \leq \theta \leq 2\pi$

$ 0 \leq \phi \leq \frac{\pi}{2}$

$0 \leq r \leq \sqrt{2}$

I get $\frac{\pi}{2}(4\sqrt{2} -4).$

There answer is the same except a $-\frac{7}{2}$ instead of the 4 at the end. Obviously I'm missing a 1\2 but I seems, I can not find it.

答:
This problem is actually better suited for cylindrical coordinates:

$$\begin{align}\int_{0}^{2\pi}\int_0^1\int_{r^2}^{\sqrt{2-r^2}}rdzdrd\theta&=2\pi\int_0^1(r\sqrt{2-r^2}-r^3)dr\\&=2\pi(-\frac{1}{3}(2-r^2)^{3/2}-\frac{r^4}{4})\mid_{0}^1\\&=2\pi(-\frac{1}{3}-\frac{1}{4}+\frac{2\sqrt{2}}{3})\\&=\frac{\pi}{3}(4\sqrt2-\frac{7}{2})\end{align}$$

The problem with spherical coordinates here is that the radius is a (piecewise) function of the azimuthal angle, which makes the integration a bit more difficult.


问:
Question about $\delta$'s in the $\epsilon$-$\delta$ definition of continuity
If a function $f:M\to\mathbb{R}$ is continuous at point $x_0$ we know that for an arbitrary $\epsilon>0$ there exists a $\delta>0$ such that for all $x\in M$ and $|x-x_0|<\delta\implies |f(x)-f(x_0)|<\epsilon$. Let's call those $\epsilon$ and $\delta(\epsilon)$ a pair, $(\epsilon,\delta)$.

What happens if I shrink the $\delta$? Does this imply that $|f(x)-f(x_0)|$ also shrinks?

My intuition says that we can't make any claim on the behaviour of $|f(x)-f(x_0)|$. Sure if $\delta$ attains a value which is very small then it will be smaller than another $\delta'$ which belongs to a pair $(\epsilon',\delta')$ where the $\epsilon'<\epsilon$. But if I shrink $\delta$ only a bit what happens then? How do I argue in a formal way?

答:
Nothing happens.

Continuity says that if you are given $\epsilon > 0$ then you can find $\delta > 0$ such that the following statement is true:

$(*)$ if $|x-x_0| < \delta$ then $|f(x)-f(x_0)| < \epsilon$.

So, let's suppose that the $\epsilon$ has been chosen, and let's fix its value. And having done that, let's choose a value of $\delta$ which makes statement $(*)$ true. Then any smaller value will also make statement $(*)$ true. In other words, if $0 < \delta' < \delta$ and if $(*)$ is true then the following statement (obtained by substituting $\delta'$ in place of $\delta$) is also true:

$(*')$ if $|x-x_0| < \delta'$ then $|f(x)-f(x_0)| < \epsilon$.

The reason for this is that $|x-x_0| < \delta'$ and $\delta' < \delta$ together imply $|x-x_0| < \delta$.


答:
For manifolds with boundary: no, you can't take boundary points to non-boundary points.

For manifolds without boundary: yes. For 1-manifolds, this is easy. For $n > 1$, take a neighbourhood of a (shortest) path between your points. After removing problems (by taking a subset), that's homeomorphic to a disk, with both points in the interior. Take any homeomorphism of the disk that interchanges the points while fixing the boundary, and pass it through the homeomorphism to the neighbourhood of the path.

Title: Is there always a homotopy taking a point to another in a connected manifold?

Body:
Let $M$ be a connected topological $n$-manifold. Note this implies $M$ is path-connected.

Let $a,b \in M$. Must there always exist a continuous $H:M \times [0,1] \to M$ such that $H(m,0)=m$ for all $m \in M$, and $H(a,1)=b$?

We can note that this is true for $\mathbb{R}^n$, by considering $(x,t) \mapsto x+t(b-a)$, and it is also true for $S^1$ by considering $(e^{i\theta}, t)\mapsto e^{i(\theta+t(\theta_b-\theta_a))}$ where $a=e^{i\theta_{a}}, b=e^{i\theta_b}$.


问:
Norm of $ T:(C([0,1]),||\cdot||_\infty)\rightarrow\mathbb{R}$ where $Tf=\sum_{k=1}^n a_kf(t_k)$?
I need to calculate the operator Norm of the linear operator defined as:
$$ T:(C([0,1]),||\cdot||_\infty)\rightarrow\mathbb{R} \text{ where } Tf=\sum_{k=1}^n a_kf(t_k)$$
for $0\leq t_1<t_2<...<t_n\leq 1$ and $a_1,...,a_n\in \mathbb{R}$.

I have been able to show that $||T||\geq\left|\sum_{k=1}^n a_k\right|$ and $||T||\leq \sum_{k=1}^n|a_k|$ but I don't seem to able to bound it further than that. Would appreciate any help. Thank you.

答:
WLOG, one can assume that $a_k\ne0$ for all $k$ (if not, just drop all the zero-values, and in the case that all are zero, the problem is trivial).

Take $f(x)$ a continuous function with these properties:

$f(t_k)=\frac{|a_k|}{a_k}$.

$|f(x)|\le 1$ for all $x\in[0,1]$.

For example, you can take the polygonal through the points $\{(t_k,f(t_k)),\,1\le k\le n\}\cup\{(s_k,0),1\le k\le n-1\}$, where $s_k=\frac{t_k+t_{k+1}}{2}$ (the mid point of the interval $[t_k,t_{k+1}]$).

Then it is clear that $\|f\|_\infty=1$ (observe that $|f(t_k)|=1$) and $Tf=\sum_{k=1}^n |a_k|$.

This together with your calculations leads to $\|T\|=\sum_{k=1}^n|a_k|$.


答:
Let’s prove something stronger: for each such manifold $M$ (without boundary), the path-connected component of the identity in the group of homeomorphisms of $M$ acts transitively on $M$.

Since it’s a continuous group action and $M$ is connected, it’s enough to show that any orbit is open. By considering homotopies that are the identity outside a ball, we can assume $M=B^n$, $a,b$ being interior points, and require that the homotopy must be the identity on the boundary.

But in this case, you can consider the flow of a compactly supported vector field in the right direction.


问:
Erdős-Straus conjecture
The Erdős-Straus Conjecture (ESC), states that for every natural number $n \geq 2$, there exists a set of natural numbers $a, b, c$ , such that the following equation is satisfied:

$$\frac{4}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\tag{1}$$

The basic approach to solving this problem outlined by Mordell [Ref1] is described below

By defining $t$ and $m$ as positive integers greater than zero and $q$ a positive integer greater than one we can observe that

a) There is always a solution for even $n$, since if $n=2^qt$ we have the trivial solution
$$\frac{4}{4t}=\frac{1}{t}$$

In the remaining case $n=2(2t+1)$, a solution in the form of two Egyptian fractions can always be found  e.g.
$$\frac{4}{2(2t+1)}=\frac{2}{2t+1}=\frac{1}{t+1}+\frac{1}{(t+1)(2t+1)}$$

b) If $(1)$ is a solution for some particular prime $n$ then all composite numbers $mn$ divisible by $n$ are also solutions, thus

$$\frac{4}{mn}=\frac{1}{ma}+\frac{1}{mb}+\frac{1}{mc}$$

will also be a solution. This means that we can simplify the analysis to the cases where $n$ is a prime greater than 2.

Using Mordell's approach we have just shown that we only need to consider the cases where $n$ is prime and where $n \equiv 1 \pmod{2} \;\;[meaning \;\;n=2t+1]$

The argument continues...

Mordell goes on to show in turn that the search can be reduced further to the cases when
$$n \equiv 1 \pmod{4} \;\;[meaning \;numbers \;\;n=4t+1]$$
$$n \equiv 1 \pmod{8} \;\;[meaning \;numbers \;\;n=8t+1]$$
$$n \equiv 1 \pmod{3} \;\;[meaning \;numbers \;\;n=3t+1]$$
$$n \equiv 1,2,4 \pmod{7} \;\;[meaning \;numbers \;\;n=7t+1,n=7t+2 \;or\;n=7t+4 ]$$
$$n \equiv 1,4 \pmod{5}  \;\;[meaning \;numbers \;\; n=5t+1 \;or\;n=5t+4]$$

Assembling these results together, Mordell showed that the conjecture can be proved in this context except for the cases when
$$n \equiv 1,11^2,13^2,17^2,19^2,23^2 \pmod{840}$$

Mordell stated that since the first prime meeting this condition is 1009, this is proof that the conjecture holds for $n<1009$.

This basic approach can be pursued further. Other workers have shown that the conjecture holds for much higher values of $n$ using similar methods as can be seen on the above Wikipedia page.

Note that other intermediate results can be constructed from the above congruence's, e.g. $n \equiv 1 \pmod{24}$.

The question is:

Are there any other elementary approaches to solving this problem than the one outlined by Mordell (and described above)?

[Ref1] Louis J. Mordell (1969) Diophantine Equations, Academic Press, London, pp. 287-290.

答:
This is a description of an approach to the problem I have not been able to get to work, but nevertheless may be of interest to others.

If one wants to study the Erdős–Straus conjecture there there are many examples of formulas giving Egyptian Fraction Triple Solutions given a fraction $\frac{4}{n}$ in the mathematical literature and on this site for example.

These formulas come with various numbers of free parameters and often have divisibility conditions/constraints associated with them that does not always guarantee correct solutions; especially in the case of the residue classes targeted by Mordell and of most interest in regard to the Erdős–Straus conjecture, that I have listed in the revised question above, for example $n \equiv 1 \pmod{4}$.

However I devised a formula (in 2016/17  when I became interested in Egyptian fractions), with 3 integer parameters, with no nasty divisibility conditions that lead to false solutions (non-triples), and that works with a selected residue class $n \equiv 1 \pmod{8}$ or $n \equiv 1 \pmod{24}$ for example.

$$\frac{4}{8 t+1}=\frac{1}{2 t+r}+\frac{1}{m (8 t+1) (2 r+a)}+\frac{1}{m (8 t+1) (2 r-a-1)} \tag{1}$$

Positive numbers, $8t+1$, in the desired residue class can be generated using a linked formula such as

$$8t+1=16mr^2 -4r(2m+1) -4m(a^2+a) +1 \tag{2}$$

(combinations of the 3 positive integer parameter values are not allowed when they give a negative number result).

As pointed out by Dávid Laczkó, $r$ is always even so if $r=2s$ (2) becomes

$$8t+1=64ms^2 -8s(2m+1) -4m(a^2+a) +1 \tag{3}$$

An allowed combination of the three parameters will always result in a number in the desired residue class, as well as an associated Egyptian Fraction Triple Solution.

If it could be proved that formula (2) has full coverage of the applicable residue class(es) to the problem, then this might provide an alternative way of proving the Erdős–Straus conjecture, given the direct one-to-one mapping between formulas (1) and (2) (given identical input parameter values).

Since the numbers $n$ are not neatly ordered along a 1-D number line, but ordered in a much more difficult fashion across a portion of a 3-D number space, it is difficult (for me) to find a mathematical proof that, at the very least, all the prime numbers in the selected residue class are there in the set of numbers generated by (2).

(Sorry if I'm not using the correct mathematical terminology to describe this, in the last paragraph above).

Title: Erdős-Straus conjecture


问:
The recursive formula for generalized pentagonal numbers characteristic function
I am investigating the pentagonal number characteristic function. For the generalized pentagonal numbers the formula is known: $$p_n=n(3n\pm1)/2$$
The first few generalized pentagonal numbers are:
$${0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77}$$
The characteristic function is:
$${1, -1, -1, 0, 0, 1, 0, 1, 0, 0, 0, 0, -1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 1}$$
I am trying to find a way to represent the characteristic function in a recursive way. Frankly saying I was not able even to find a way to start this. I have looked at the wolfram, wikipedia and oeis to find anything related to this. But nothing found.

答:
Generate the generalized pentagonal number pairs: $(3r^2-r)/2$ and $(3r^2+r)/2$ starting with $r=1$.

r| pair
1| 1,2
2| 5,7
3| 12,15
and continue

If $cr$ is the characteristic function let $cr(0) = 1$.
Each of the pair of pentagonal numbers are the index for the characteristic function and the sign is = $(-1)^r$ .
If $cr$ is the characteristic function:

$cr(1) = cr (2) = -1$

$cr(5) = cr (7) =  1$

$cr (12) = cr (15) =-1$
etc.

cr (Non Pentagonal #s > 0) = 0


问:
How to evaluate $\int_{0} ^{2}\frac{(x-⌊x⌋)dx}{2x+1-⌊x⌋}$
I want to evaluate $\int_{0} ^{2}\frac{(x-⌊x⌋)dx}{2x+1-⌊x⌋}$. How can I do this? Can it be done just with the graph of the function? We have $[x] =x-\{x\}$ and after this the graph would follow. But without that, how can it be solved?

Is there  a general way to approach such functions?

答:
$x-\lfloor x\rfloor=n+\{x\}$ where $n=\lfloor x\rfloor$

$2x+1-\lfloor x\rfloor=2n+\{x\}+1$

so we can split up your integral into two parts:
$$f(x)=\frac{x-\lfloor x\rfloor}{2x+1-\lfloor x\rfloor}=\begin{cases}\frac{y}{y+1} & 0\le x<1\\\frac{1+y}{3+y}&1\le x<2\end{cases}$$
where $y\in[0,1)$


问:
Prove $rs=sr^{-1}$ in ${\rm Dih}(2n)$

Let $r$ and $s$ be the rotation and reflection symmetries respectively in ${\rm Dih}(2n)$, the dihedral group of order $2n$. Show that $rs=sr^{-1}$.

I also need to show by induction that $r^js=sr^{-j}$ but I think that will follow easily.

答:
I think you can find it in many Group theory book as a Theorem or as an exercise. Now, let $G_1,G_2$ are two groups and let $\phi:G_2\to Aut(G_1),~~\phi(b)=\phi_b$ (I assume you know what is $\phi_b$). We can prove that $G_1\times G_2$ with the following operation:

$$(a,b)(a',b')=(a\phi_b(a'),bb')$$

is a new group and you know that it is $G_1\times_{\phi} G_2$ called the semi-direct product of $G_1$ by $G_2$. In fact, assuming that $\phi_b$  where $b\in G_2$ is essential here. Now, let $G_1=\mathbb Z_n=\langle a\rangle$ and $G_2=\mathbb Z_2=\langle b\rangle$. Consider: $$\phi:G_2\to Aut(G_1)\\\\ \phi(b)=\phi_b,~~\phi(1)=\phi_1=id_{\mathbb Z_n}$$ $\phi$ is a group homomorphism and we denote $\mathbb Z_n\times_{\phi}\mathbb Z_2$ as $D_{2n}$. Here, you can find out why that relation is regarded. Just consider the way  $\phi$ acts.

Title: Prove $rs=sr^{-1}$ in ${\rm Dih}(2n)$


问:
Given mean, standard deviation and skewness find the distribution
Given three non negative real numbers x, y, z for which I know their mean, standard deviation and skewness, is it possible to determine uniquely the three numbers?

It is quite straightforward for only a pair of numbers therefore I think it should be possible to expand to three numbers. However I wasn't able to come to a solution yet.

I have tried to solve the equations analytically, but I had no luck on finding a solutions.
I have also tried to standardise the distribution so that the solutions lie in the intersection of a plane (through the origin with norm (1,1,1)) and a shpere. Again l found it hard to visualize the equation $x^3+y^3+z^3=k$.

Has anyone came across this and would like to show me the solution?

答:
I solved this with Mathematica but the equations are mildly complex.

Here's just $x$, where $m$ is the mean, $s$ the standard deviation, and $k$ the skew:

$$x = \frac{1}{2} \left(\sqrt[3]{2 \sqrt{2} \sqrt{\left(2 k^2-27\right) s^6}+4 k s^3}+\frac{3\
   2^{2/3} s^2}{\sqrt[3]{\sqrt{2} \sqrt{\left(2 k^2-27\right) s^6}+2 k s^3}}+2 m\right)$$

and valid regardless of the sign on the data points.


问:
Fermat numbers question
I am looking for direction in my question.
The question is to prove, by long polynomial division, that:
\begin{equation}
2^{2^{n+1}}-1\ |\ 2^{F_n-1}-1
\end{equation}

I have tried a lot of things, like calling:\begin{equation}  t=2^{2^n} \end{equation} and trying long polynomial division right away, but none of this worked successfully.

Title: Fermat numbers question

Body:
With the tip given by lulu you have $gcd(2^{2^{n+1}}-1,2^{F_n-1}-1)=2^{gcd(2^{n+1},F_n-1)}-1=2^{gcd(2^{n+1},2^{2^{n}})}-1$. Notice that for all $n \in \mathbb{N}$ $n+1\leq 2^n$(use induction to see this). Hence $2^{n+1} \mid 2^{2^n}$ and $gcd(2^{n+1},2^{2^{n}})=2^{n+1}$. So, $gcd(2^{2^{n+1}}-1,2^{F_n-1}-1)=2^{2^{n+1}}-1$ and we have that $2^{2^{n+1}}-1 \mid 2^{F_n-1}-1$.

Another try
Notice (again) that $n+1\leq 2^n$. Hence there is some $k \in \mathbb{N}$ such that $(n+1)+k=2^n$. Then $2^{2^n}=2^{(n+1)+k}=2^{n+1} \cdot 2^k$. Notice then that
$$
2^{F_n-1}-1=2^{2^{2^n}}-1=2^{2^{n+1} \cdot 2^k}-1=(2^{2^{n+1}})^{2^k}-1.
$$
So, to prove that $2^{2^{n+1}}-1 \mid 2^{F_n-1}-1$  is equivalent to prove that $2^{2^{n+1}}-1 \mid (2^{2^{n+1}})^{2^k}-1$. Now call $t=2^{2^{n+1}}$ and you need to prove that $t-1 \mid t^{2^k}-1$. Can you use long division to finish this?


问:
Infinitely many positive integers $n$ such that $n$ divides $2^{2^{n}+1}+1$ and $n$ doesn’t divide $2^{n}+1$
Show that there are infinitely many positive integers $n$ such that $n$ divides $2^{2^{n}+1}+1$ and $n$ does not divide $2^{n}+1$

I think we should increase $n$ by induction:
$x_{k+1}$=$2^{x_{k}}+1$, then we have $x_{k}$ doesn’t divide $x_{k+1}$ and divide $x_{k+2}$.
Next we should show that if it allows for $k=1,2,3,..m$, then allows for $k=m+1$.

I am not getting if this approach is right.

Thanks in advance.

答:
I don't see any way to solve the problem using induction. Instead, with the Lifting-the-exponent lemma, using $x = 2$, $y = 1$ and $p = 3$, then since $3 \mid 2 + 1$, for any positive integer $j$ and odd integer $k$ where $3 \not\mid k$,

$$\nu_3(2^{k(3^j)} + 1) = \nu_3(2 + 1) + \nu_3(k(3^j)) = 1 + j \tag{1}\label{eq1A}$$

which means $3^j \mid 2^{k(3^j)} + 1$. Next, for any integer $j \ge 2$ and odd integer $k$,

$$2^9 + 1 = 513 = 19(3^3) \implies 19 \mid 2^{k(3^j)} + 1 \tag{2}\label{eq2A}$$

We also have

$$2^{19} + 1 = 524289 = 3(174763) \tag{3}\label{eq3A}$$

Using the multiplicative order, let

$$\operatorname{ord}_{174763}(2) = m \tag{4}\label{eq4A}$$

Since $2^{2(19)} \equiv 1 \pmod{174763}$, then $m \mid 2(19)$. As $m \neq 1, 2, 19$, then $m = 2(19)$. Thus, for a positive integer $q$ and some non-negative integer $r$,

$$2^q \equiv -1 \pmod{174763} \iff q = 19 + rm \tag{5}\label{eq5A}$$

i.e., $q$ is an odd integral multiple of $19$.

For any integer $s \ge 2$, with $n = 174763(3^s)$, then since $19 \not\mid n$ (actually, $174763$ is a prime number), \eqref{eq5A} indicates $174763 \not\mid 2^n + 1$, so $n \not\mid 2^n + 1$. However, \eqref{eq2A} shows $19 \mid 2^n + 1$, so \eqref{eq5A} gives that $174763 \mid 2^{2^{n} + 1} + 1$. Also, as shown in \eqref{eq1A}, $3^s \mid 2^{n} + 1$ so $3^s \mid 2^{2^{n} + 1} + 1$. Thus, $n \mid 2^{2^{n} + 1} + 1$. This shows there's an infinite number of $n$ which meet the question conditions.


问:
Convex Problems?
Which of the following are convex problems?

Minimum volume ellipsoid that encloses a convex-hull

Minimum volume ellipsoid that encloses a polyhedron

Maximum volume ellipsoid inscribed in a convex-hull

Maximum volume ellipsoid inscribed in a polyhedron

I know that first and second options are standard convex problem and can be solved with the Lowner-Ellipsoid algorithm. I am wondering about the last two options. I found that they are called John Ellipsoids.

答:
You should be more specific than 'is it a convex problem'. Convexity is often about the specific formulation of a problem. I think you are asking whether there is a known convex formulation of the listed problems.

Slide 3 here answers that question for the last two sets. Parameterize the ellipsoid as $\{Bu+d : ||u||_2 \leq 1 \}$ where $B$ is symmetric positive definite (wlog). For a generic convex set $C$, the maximum volume ellipsoid inscribed in $C$ can be found by solving:
$$\max_{B,d} \{ \log \det B : Bu+d \in C \quad \forall u : ||u||_2 \leq 1\} $$
or, with an indicator constraint:
$$\max_{B,d} \{ \log \det B : \sup_{||u||_2 \leq 1} I_C(Bu+d) \leq 0  $$
This is a convex problem. The next question is whether this problem can be solved practically. That requires finding a "simple" constraint set where the difficulties are not hidden in an abstract function $I_C$.

The formulation for the maximum volume ellipsoid inscribed in a polyhedron defined by $Ax \leq b$ is given on the slides (log det objective and second order cone constraint).

If $C$ is the convex hull of a finite number of points $c_i$, the feasible set is the polytope $\{(x,\lambda) : x=\sum_i \lambda_i c_i, \; \lambda \geq 0, \; e^T\lambda=1\}$. The constraint $Bu+d \in C \; \forall u : ||u||_2 \leq 1$ is intuitively difficult because for each $u$ it needs to find a matching $\lambda$. There are three solutions that I can think of:

Find an H-representation of the convex hull.

Replace the unit ball with a finite number of extreme points of the unit ball as an approximation. For an accurate representation, you need more points as the dimension increases.

Parameterize $\lambda = Eu+f$, which results in the constraints $Bu+d = \sum_i (Eu+f)_i c_i \; \forall u \in B$, $Eu+f \geq 0 \; \forall u \in B$ and $e^T(Eu+f) = 1 \; \forall u \in B$. You may be getting better results by lifting $\lambda$ to a higher dimension.

Title: Convex Problems?