问:
Decorating And Cutting A Cake Fairly
Imagine an N x N square cake you are tasked with decorating with N types of toppings (colors). You then with to share the cake equally amongst N friends. All pieces must be rectangular and have same the area. It's also only fair that no matter what kind of piece your friends have, they each have the exact same type of toppings. How should you decorate the cake so that no matter how you cut it, it's always a fair division?

For example below, the 4 x 4 case of 4 different colors for 4 friends:

Take any equal area rectangular cut and each will contain all four different colors:

It's not hard to see that any cut that's equal is a fair division by the definition above. Call any N x N coloring of this kind a fair division coloring.

Below are fair division colorings for 2 x 2, 3 x 3, and 5 x 5, 7 x7, and 9 x 9:

For N=p, p a prime, the coloring is trivial. However, I found it is impossible for N=6 and N=8. I used brute force methods and found that any coloring of 6 different colors for the 6 x 6 grid, there will exist a cut that doesn't create a "fair" division. The same is true for N=8.

My question: is there a general rule to determine for what N a fair division coloring of the square exists?

EDIT: I am absolutely confident this kind of problem has been studied before, but I'm not sure where to find a paper on this. I appreciate any and all thoughts on where I can look.

答:
Any square of size $n=pq$ with $p<q$ and $q$ not divisible by $p$ has no fair division colouring (so the only $n$ that have colourings are the primes and possibly some prime powers).

Let me first prove the case with $p=2$.

Proof:

Suppose $n=2m$, with $m$ odd. There are dissections which have the following rectangles at the top left corner: (shown for $m=5$, but holds for general $m$)

```<code>a                     b                     c
 X X X X X . . . . .   X X X X X X X X X X   O X X X X X . . . .
 X X X X X . . . . .   . . . . . . . . . .   O X X X X X . . . .
 . . . . . . . . . .   . . . . . . . . . .   O . . . . . . . . .
 . . . . . . . . . .   . . . . . . . . . .   O . . . . . . . . .
 . . . . . . . . . .   . . . . . . . . . .   O . . . . . . . . .
 . . . . . . . . . .   . . . . . . . . . .   O . . . . . . . . .
</code>```

Let the colour of the top left corner square be red. There can be no other red square at any of the X's in dissections a and b above. Therefore the X rectangle in dissection c must have its red square at its bottom right corner (row $2$, column $m+1$).

Consider the dissections d and e below where there are vertical rectangles along the top row.

```<code>d                     e                     f
 A A B B C C D D E E   A B B C C D D E E F   R X X X X X X X X X
 A A B B C C D D E E   A B B C C D D E E F   X X X X X R X X X X
 A A B B C C D D E E   A B B C C D D E E F   1 2 3 4 5 6 . . . .
 A A B B C C D D E E   A B B C C D D E E F   1 2 3 4 5 6 . . . .
 A A B B C C D D E E   A B B C C D D E E F   1 2 3 4 5 6 . . . .
 . . . . . . . . . .   A . . . . . . . . F   . . . . . . . . . .
 . . . . . . . . . .   A . . . . . . . . F   . . . . . . . . . .
 . . . . . . . . . .   A . . . . . . . . F   . . . . . . . . . .
</code>```

In figure f, there can be no red at locations marked 1 or 2 because of dissection d.

Dissection e forces a red at a location marked 3.

Dissection d eliminates red from locations marked 4.

Dissection e forces a red at a location marked 5.

Dissection d causes a contradiction cause it has a rectangle containing reds in column 5 and column 6.

This generalises to all odd $m$ -- the reds are forced to occur in the odd-numbered columns, and the last one clashes with the red in column $m+1$.

This proof generalises further. Suppose $n=pq$, with $p<q$.

The first part still works, and shows that there must be a red in column q+1, on one of the rows 2 to p.

The second part generalises too, and instead of two dissections you use $p$ dissections, establishing $q\times p$ rectangles at any x-coordinate along the top row, to prove that every column with a number $1$ mod $p$ must contain a red square in one of the top $q$ rows, which will conflict with the red in column $q+1$, unless $p|q$.

Therefore, any square of size $n=pq$ with $p<q$ and $p\not|q$ has no fair division colouring.

So this proves that any composite that is not a prime power has no fair division colouring (such $n$ can always be written as a product of coprime numbers, where one is smaller than the other).

There are clearly fair division colourings when $p$ is prime, and the fair division colouring given for $n=9$ should generalise to $n=p^2$.
The only undecided cases left are the prime powers $p^k$ with $k>2$. You've already shown that $n=8$ fails, so not all prime powers have colourings.

Edit: I'm fairly sure that prime powers $p^k$ with $k>2$ will not have fair division colourings, and that it can be proved in a similar way as the above proof using only the four rectangle shapes $1\times p^k$ and $p\times p^{k-1}$ in both orientations. I may try to construct the proof later.

Title: Decorating And Cutting A Cake Fairly

Body:
Any square of size $n=pq$ with $p<q$ and $q$ not divisible by $p$ has no fair division colouring (so the only $n$ that have colourings are the primes and possibly some prime powers).

Let me first prove the case with $p=2$.

Proof:

Suppose $n=2m$, with $m$ odd. There are dissections which have the following rectangles at the top left corner: (shown for $m=5$, but holds for general $m$)

a                     b                     c
 X X X X X . . . . .   X X X X X X X X X X   O X X X X X . . . .
 X X X X X . . . . .   . . . . . . . . . .   O X X X X X . . . .
 . . . . . . . . . .   . . . . . . . . . .   O . . . . . . . . .
 . . . . . . . . . .   . . . . . . . . . .   O . . . . . . . . .
 . . . . . . . . . .   . . . . . . . . . .   O . . . . . . . . .
 . . . . . . . . . .   . . . . . . . . . .   O . . . . . . . . .

Let the colour of the top left corner square be red. There can be no other red square at any of the X's in dissections a and b above. Therefore the X rectangle in dissection c must have its red square at its bottom right corner (row $2$, column $m+1$).

Consider the dissections d and e below where there are vertical rectangles along the top row.

d                     e                     f
 A A B B C C D D E E   A B B C C D D E E F   R X X X X X X X X X
 A A B B C C D D E E   A B B C C D D E E F   X X X X X R X X X X
 A A B B C C D D E E   A B B C C D D E E F   1 2 3 4 5 6 . . . .
 A A B B C C D D E E   A B B C C D D E E F   1 2 3 4 5 6 . . . .
 A A B B C C D D E E   A B B C C D D E E F   1 2 3 4 5 6 . . . .
 . . . . . . . . . .   A . . . . . . . . F   . . . . . . . . . .
 . . . . . . . . . .   A . . . . . . . . F   . . . . . . . . . .
 . . . . . . . . . .   A . . . . . . . . F   . . . . . . . . . .

In figure f, there can be no red at locations marked 1 or 2 because of dissection d.

Dissection e forces a red at a location marked 3.

Dissection d eliminates red from locations marked 4.

Dissection e forces a red at a location marked 5.

Dissection d causes a contradiction cause it has a rectangle containing reds in column 5 and column 6.

This generalises to all odd $m$ -- the reds are forced to occur in the odd-numbered columns, and the last one clashes with the red in column $m+1$.

This proof generalises further. Suppose $n=pq$, with $p<q$.

The first part still works, and shows that there must be a red in column q+1, on one of the rows 2 to p.

The second part generalises too, and instead of two dissections you use $p$ dissections, establishing $q\times p$ rectangles at any x-coordinate along the top row, to prove that every column with a number $1$ mod $p$ must contain a red square in one of the top $q$ rows, which will conflict with the red in column $q+1$, unless $p|q$.

Therefore, any square of size $n=pq$ with $p<q$ and $p\not|q$ has no fair division colouring.

So this proves that any composite that is not a prime power has no fair division colouring (such $n$ can always be written as a product of coprime numbers, where one is smaller than the other).

There are clearly fair division colourings when $p$ is prime, and the fair division colouring given for $n=9$ should generalise to $n=p^2$.
The only undecided cases left are the prime powers $p^k$ with $k>2$. You've already shown that $n=8$ fails, so not all prime powers have colourings.

Edit: I'm fairly sure that prime powers $p^k$ with $k>2$ will not have fair division colourings, and that it can be proved in a similar way as the above proof using only the four rectangle shapes $1\times p^k$ and $p\times p^{k-1}$ in both orientations. I may try to construct the proof later.


问:
How to prove that $1/n!$ is less than $1/n^2$?
I want to prove
$$\sum_{n=0}^{\infty} \frac{1}{n!}$$ is a converging series. So I want to compare it with $\sum_{n=0}^{\infty} \frac{1}{n^2}$. I want to do direct comparison test.

How to prove $n^2 < n! $ ?

答:
Most likely this is proved first time with induction. You  could also take the root test, or see that
\[ \sum_{n=1}^\infty \frac{1}{n!} = e-1 \]</p>

ContentLicense: CC BY-SA 3.0


问:
Linear combination of trigonometric functions
Consider the following equality:

$$a_1\sin(x) + a_2\sin^2(x) + ... +a_n\sin^n(x) = a_1\sin(x) + a_2\sin(2x) + ... + a_n\sin(nx)$$

where $a_1, a_2\dots, a_n \in \mathbb{C}, \;n \in \mathbb{N}$.

This equality clearly holds for $n = 1$ and for any $a_1 \in \mathbb{C}$. The trivial equality ($a_1 = a_2 = \cdots = a_n = 0$) also holds.

Do there exists any other solutions? If so, are there infinitely many of such solutions?

Do there exist other solutions if we constrain the domain of coefficients to $\mathbb{Q}$, $\mathbb{R}$ or $\mathbb{Z}$?

答:
With $z:=e^{ix}$, you want to establish the identity

$$\sum_k a_k\left(\frac{z-z^{-1}}{2i}\right)^k=\sum_k a_k\frac{kz-(kz)^{-1}}{2i}.$$

By identification of the polynomial coefficients, only $a_1$ can be nonzero.

ContentLicense: CC BY-SA 4.0


问:
Complex analysis book with a view toward Riemann surfaces?
I am considering complex analysis as my next area of study.  There are already a few threads asking about complex analysis texts (see Complex Analysis Book and What is a good complex analysis textbook?).  However, I'm looking for something a little more specific, if such a thing exists.

Is there a nice, slow-paced introductory complex analysis text that features at least some (introductory) material on Riemann surfaces?

A look through texts mentioned in the pages linked above did not yield any.  I am not big on analysis and tend to favor more algebraic, topological, and geometric-flavored areas of mathematics.  I am however trying to learn at least at a basic level the core disciplines of mathematics, and I feel I would be amiss if I did not study complex analysis.  For background: I have a basic knowledge of real analysis, algebra (group, ring, and field theory), linear algebra, and will have knowledge of topology.

In addition to my above desire in a complex analysis text: is there one you would recommend for its view toward algebraic, topological, or geometric applications of complex analysis?

Any online lecture notes (or inexpensive book) on Riemann surfaces that would be accessible after or along with an introductory look at complex analysis would be welcome as well.

EDIT: After what has developed, I feel this question is now appropriate: Is there a complex analysis text that would be particularly recommended if one wishes to study Riemann surfaces?  What topics in particular is it important to develop a good grasp of?

答: Jones and Singerman, Complex Functions: An algebraic and geometric viewpoint.


问:
Solve integrals $\int_{\alpha}^{\infty}e^{-v}e^{-e^{-\beta v}}v~\mathrm{d}v$ and $\int_{\alpha}^{\infty}e^{-v}e^{-e^{-\beta v}}~\mathrm{d}v$?
Can the solution to the integrals

$$\int_{\alpha}^{\infty}e^{-v}e^{-e^{-\beta v}}v~\mathrm{d}v$$

and

$$\int_{\alpha}^{\infty}e^{-v}e^{-e^{-\beta v}}~\mathrm{d}v$$

be computed analytically - or approximated? I have only figured out the trivial solution for $\beta=1$ where I can compute the second integral by

$$\int_{\alpha}^{\infty}e^{-v}e^{-e^{-v}}~\mathrm{d}v = \left[e^{-e^{-v}}\right]_{\alpha}^{\infty} = 1 - e^{-e^{-a}}.$$

答:
For $\beta=1$,
$$I=\int v \,e^{-v}e^{-e^{- v}}dv$$
Let $v=-\log(t)$
$$I=\int e^{-t} \log (t)\,dt$$ One integration by parts
$$I=\text{Ei}(-t)-e^{-t} \log (t)$$ where appears the exponential integral function. So,
$$J=\int_\alpha^\infty v \,e^{-v}e^{-e^{- v}}dv=\gamma-\alpha e^{-e^{-\alpha }}  -\text{Ei}\left(-e^{-\alpha }\right)$$ which is an interesting function.


问:
Are there names for function which doesn't have all items in it's domain as defined?

Consider a function similar to the one above where except for a few points in the domain, the function is bijective. Is there a name given for this kind of function similar to bijectivity?

My initial guess was discontinuous functions but I'm not sure how well that concept works for discrete functions.

答: A discontinuous function is still defined on every point in its domain, discontinuity is defined in terms of limits. What you’re describing is called a ‘partial function’, I believe the term sees some use in computer science, but you’re unlikely to find it in a standard mathematics course.</p>


答:
here is a hint. $n! = 1\times2\times...\times(n-2)\times(n-1)\times n$, where as $n^2 = n\times n$. so you have effectively as your step one:
 $1\times2\times...\times (n-1)\times n =n\times n $.
</p>

<p>now you can cancel things out and....well I think that is hint enough as far as the question of proving $n^2 < n!$ goes. </p>

Title: How to prove that $1/n!$ is less than $1/n^2$?

Body:
I want to prove
$$\sum_{n=0}^{\infty} \frac{1}{n!}$$ is a converging series. So I want to compare it with $\sum_{n=0}^{\infty} \frac{1}{n^2}$. I want to do direct comparison test.

How to prove $n^2 < n! $ ?

Body:
here is a hint. $n! = 1\times2\times...\times(n-2)\times(n-1)\times n$, where as $n^2 = n\times n$. so you have effectively as your step one:
 $1\times2\times...\times (n-1)\times n =n\times n $.

now you can cancel things out and....well I think that is hint enough as far as the question of proving $n^2 < n!$ goes.


问:
Simplistic simulation of double pendulum doesn't work. Is this approach valid?

Introduction

I'd like to preface this by saying that while I'm a reasonably competent developer, my calculus skills leave a lot to be desired.

I've been reading up on basic physics, and I thought I had a grip on the behaviour of the Lagrangian formulation of physics, but it turns out there is something I just don't understand, which is why I'm turning to this place for assistance.

I've been simulating various systems like pendulums, gravitational systems, etc and my very simplistic approach has worked fine. However, when I tried it on the double pendulum, I'm not getting the expected results.

Calculating the Lagrangian

I'm assuming the coordinates of the masses:

$$ \begin{align}
x_1 &= sin(θ_1)\,l_1 \\
x_2 &= x_1 + sin(θ_2)\,l_2 \\
y_1 &= -cos(θ_1)\,l_1 \\
y_2 &= y_1 - cos(θ_2)\,l_2
\end{align} $$

I then use Maxima to compute the Lagrangian, which is:

$$ {{2\,l_{1}\,l_{2}\,m_{2}\,\dot{θ}_{1}\,\dot{θ}_{2}\,\sin θ_{1}\,\sin θ_{2}+
 \left(2\,l_{1}\,l_{2}\,m_{2}\,\dot{θ}_{1}\,\dot{θ}_{2}\,\cos θ_{1}+2\,g\,l_{2}\,
 m_{2}\right)\,\cos θ_{2}+\left(2\,g\,l_{1}\,m_{2}+2\,g\,l_{1}\,m_{1}
 \right)\,\cos θ_{1}+l_{2}^2\,m_{2}\,\dot{θ}_{2}^2+\left(l_{1}^2\,m_{2}+
 l_{1}^2\,m_{1}\right)\,\dot{θ}_{1}^2}\over{2}} $$

I then use the Euler-Lagrange equation to give me the solution for the first coordinate:

$$ \ddot{θ}_1 = -{{\left(\ddot{θ}_{2}\,l_{2}\,m_{2}\,\sin θ_{1}-l_{2}\,m_{2}\,\dot{θ}_{2}^2\,
 \cos θ_{1}\right)\,\sin θ_{2}+\left(l_{2}\,m_{2}\,\dot{θ}_{2}^2\,\sin
 θ_{1}+\ddot{θ}_{2}\,l_{2}\,m_{2}\,\cos θ_{1}\right)\,\cos θ_{2}+\left(g\,
 m_{2}+g\,m_{1}\right)\,\sin θ_{1}}\over{l_{1}\,m_{2}+l_{1}\,m_{1}}} $$

And the solution for the second coordinate:

$$ \ddot{θ}_2 = -{{\left(\ddot{θ}_{1}\,l_{1}\,\sin θ_{1}+l_{1}\,\dot{θ}_{1}^2\,\cos θ_{1}+g
 \right)\,\sin θ_{2}+\left(\ddot{θ}_{1}\,l_{1}\,\cos θ_{1}-l_{1}\,\dot{θ}_{1}^2\,
 \sin θ_{1}\right)\,\cos θ_{2}}\over{l_{2}}} $$

While my Lagrangian looks a bit different from the typical one I can find when searching online for the Lagrangian for a double pendulum, they do seem to be equivalent, and I did try to enter the one I found online and I get the same numbers from that one.

Trying to simulate the motion

Once I got this, I wanted to simulate it in the same simplistic manner as I have have simulated other physical systems, by accumulating the positions and velocities.

For each iteration, I compute 6 variables like so:

$$ \texttt{thetadotdot_1 = } \ddot{\theta}_1 \Delta t \texttt{    /* The result above */} \\
   \texttt{thetadotdot_2 = } \ddot{\theta}_2 \Delta t \texttt{    /* The result above */} \\
   \texttt{thetadot_1 = thetadot_1 + thetadotdot_1} \\
   \texttt{thetadot_2 = thetadot_2 + thetadotdot_2} \\
   \texttt{theta_1 = theta_1 + thetadot_1} \\
   \texttt{theta_2 = theta_2 + thetadot_1} $$

I then draw the two objects based on theta_1 and theta_2 alogn with the values of $l_1$ and $l_2$.

The problem I am facing is that while the two objects swings back and forth, they do so completely independently. Each one swings like a regular pendulum, without being influenced by the other. The fact that they swing at all suggests that the way I approach the problem isn't entirely wrong, but I clearly made a mistake somewhere.

I'd be very happy is someone could point out where my mistake is?

答:
The Lagrangian has the traditional form $L=T-V$ of difference of kinetic and potential energy.
$\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$
\begin{align}
T&=\frac{m_1}{2}(\dot x_1^2+\dot y_1^2)+\frac{m_2}{2}(\dot x_2^2+\dot y_2^2)
\\
&=\frac{m_1}{2}l_1^2\dot\theta_1^2+\frac{m_2}{2}\Bigl(l_1^2\dot\theta_1^2+2l_1l_2\cos(\theta_1-\theta_2)\dot\theta_1\dot\theta_2+l_2^2\dot\theta_2^2\Bigr)
\\
V&=m_1gy_1+m_2gy_2
\\
&=-(m_1+m_2)l_1g\cos\theta_1-m_2l_2\cos\theta_2
\end{align}
with partial derivatives
\begin{align}
\pd{L}{\dot\theta_1}=\pd{T}{\dot\theta_1}
&=(m_1+m_2)l_1^2\dot\theta_1+m_2l_1l_2\cos(\theta_1-\theta_2)\dot\theta_2
\\
\pd{L}{\dot\theta_2}=\pd{T}{\dot\theta_2}
&=m_2l_1l_2\cos(\theta_1-\theta_2)\dot\theta_1+m_2l_2^2\dot\theta_2
\\
\pd{L}{\theta_1}&=-m_2l_1l_2\sin(\theta_1-\theta_2)\dot\theta_1\dot\theta_2-(m_1+m_2)l_1g\sin\theta_1
\\
\pd{L}{\theta_2}&=m_2l_1l_2\sin(\theta_1-\theta_2)\dot\theta_1\dot\theta_2-m_2l_2g\sin\theta_2
\end{align}
and Euler-Lagrange equations
\begin{align}
\frac{d}{dt}\pd{L}{\dot\theta_1}=\pd{L}{\theta_1}&\implies
\\
(m_1+m_2)l_1^2\ddot\theta_1+m_2l_1l_2\cos(\theta_1-\theta_2)\ddot\theta_2
&=-m_2l_1l_2\sin(\theta_1-\theta_2)\dot\theta_2^2-(m_1+m_2)l_1g\sin\theta_1
\\
\frac{d}{dt}\pd{L}{\dot\theta_2}=\pd{L}{\theta_2}&\implies
\\
m_2l_1l_2\cos(\theta_1-\theta_2)\ddot\theta_1+m_2l_2^2\ddot\theta_2
&=m_2l_1l_2\sin(\theta_1-\theta_2)\dot\theta_1^2-m_2l_2g\sin\theta_2
\end{align}

This now is a nice linear system for the second derivatives
$$
\begin{bmatrix}
(m_1+m_2)&m_2\cos(\theta_1-\theta_2)\\
m_2\cos(\theta_1-\theta_2)&m_2
\end{bmatrix}
\begin{bmatrix}
l_1\ddot\theta_1\\l_2\ddot\theta_2
\end{bmatrix}
=
\begin{bmatrix}
-m_2l_2\sin(\theta_1-\theta_2)\dot\theta_2^2-(m_1+m_2)g\sin\theta_1
\\
m_2l_1\sin(\theta_1-\theta_2)\dot\theta_1^2-m_2g\sin\theta_2
\end{bmatrix}
$$
or after division by $m_2$, setting $\mu=1+\frac{m_1}{m_2}$,
$$
\begin{bmatrix}
\mu&\cos(\theta_1-\theta_2)\\
\cos(\theta_1-\theta_2)&1
\end{bmatrix}
\begin{bmatrix}
l_1\ddot\theta_1\\l_2\ddot\theta_2
\end{bmatrix}
=-
\begin{bmatrix}
\mu g\sin\theta_1+l_2\sin(\theta_1-\theta_2)\dot\theta_2^2
\\
g\sin\theta_2-l_1\sin(\theta_1-\theta_2)\dot\theta_1^2
\end{bmatrix}
$$
This can be solved via manual elimination, the Cramer formula/multiplication with the adjoint matrix, or a linear system solver.


问:
What is the sum of series $1+\frac1{3\cdot4^1}+\frac1{5\cdot4^2}+\frac1{7\cdot4^3}+\cdots$
I am trying to solve this series question which is

$$1+\frac1{3\cdot4^1}+\frac1{5\cdot4^2}+\frac1{7\cdot4^3}+\cdots$$

Till now I've only been able to write the general form of this series which is

$$\sum_{n=1}^\infty\frac1{(2n-1)4^{n-1}}$$

but I don't know how to proceed.

答:
Hint Reindexing your expression for the series for convenience gives
$$\sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\frac{1}{4}\right)^k .$$

Now, the factors $\frac{1}{2 k + 1}$ suggest considering the power series
$$\operatorname{artanh} x \sim \sum_{k = 0}^\infty \frac{1}{2 k + 1} x^{2 k + 1} .$$

Rewrite the given series to get $$\sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\frac{1}{4}\right)^k = 2 \sum_{k = 0}^\infty \frac{1}{2 k + 1} \left(\color{red}{\frac{1}{2}}\right)^{2 k + 1} = 2 \operatorname{artanh} \color{red}{\frac{1}{2}} = 2 \cdot \frac{1}{2}\log \frac{1 + \color{red}{\frac{1}{2}}}{1 - \color{red}{\frac{1}{2}}} = \boxed{\log 3} .$$


Body: Use induction. Note $k^2<k!$$\implies(k+1)!$$=k!(k+1)$$>k^2(k+1)$$>(k+1)^2.$ That $k^2>k+1$ follows from $k\geq2.$


问:
Computation Itô integral
I am studying the Itô-integral and I am trying to solve the following with the definition of the integral (so $\int_{0}^{t}u_{s}\,dB_{s}$ is the $L^2$-limit of $\sum_{t_j} u^n_{t_j}(B_{t_j}-B_{t_{j-1}})$ for a partition $0=t_1<...< t_n= t$ and some step (or simple) process $u^n$, which is approximating $u_s$).

Let $X_{t}=\int_{0}^{t}e^{s}\,dB_{s}$, and $Y_{t}=\int_{0}^{t}X_{s}\,dB_{s}$. I want to compute $E(Y_{t})$ and $E(Y_{t}^{2})$. I know $E(Y_{t})=0$ since stochastic integrals are centered random variables, but how do I get $E(Y_{t}^{2})$? Can somebody give me a hint?

答:
Using Ito Isometry, we have:

$$\mathbb{E}[X_t^2]=\mathbb{E}\left[\left(\int_0^t e^s dW_s\right)^2\right]=\mathbb{E}\left[\int_0^t (e^s)^2 ds\right]=\frac{1}{2}\int_0^t2e^{2s}ds=0.5(e^{2t}-1)$$

So the variance of $X_t$ is $\mathbb{E}[X_t^2]-\mathbb{E}[X_t]^2=0.5(e^{2t}-1)$ and the expected value of $X_t$ is $\mathbb{E}[X_t]=0$ by the martingale property of Ito Integral. So $X_t$ is normally distributed with mean zero and variance $0.5(e^{2t}-1)$.

Now, again using Ito Isometry:

$$\mathbb{E}[Y_t^2]=\mathbb{E}\left[\left(\int_0^t X_s dW_s\right)^2\right]=\mathbb{E}\left[\int_0^t (X_s)^2 ds\right]=\int_0^t\mathbb{E}[X_s^2]ds=0.5\int_0^t(e^{2s}-1)ds=0.5[0.5e^{2s}-s]_0^t=0.5[0.5e^{2t}-1-t]$$

Title: Computation Itô integral


问:
How to solve $\left(\frac{1}{2}\right)^{100}\sum_{k=0}^{59}{100 \choose k}$?
Part of an exercise, I got the following sum:
$$
\left(\frac{1}{2}\right)^{100}\sum_{k=0}^{59}{100 \choose k}
$$
I'm trying to evaluate it and get the answer. What is the easiest way to do it?

答:
Take it as a joke (but the answer is correct)
$$\frac 1{2^n}\sum_{k=0}^p \binom{n}{k}=1-\frac { \binom{n}{p+1}}{2^n} \, _2F_1(1,p-n+1;p+2;-1)$$ where appears the gaussian hypergeometric function.

Back to serious, as @Buraian answered, you cannot do much (except using Wolfram Alpha !!).


问:
Expected maximum absolute value of $n$ iid standard Gaussians?
I have a problem where my errors are normally distributed and I want to know what the expected maximum error is if I repeat the process $n$ times.

What is the smallest constant $C$ such that the following statement is true for all $n\geq 2$?

Let $X_1, X_2, \cdots, X_n$ be independent standard Gaussian random variables. Then $$\mathbb{E}\left[\max_{i=1}^n \left|X_i\right| \right] \leq C \sqrt{\log_e n}.$$

I can show that the answer is between $1.35$ and $2$.

答:
I believe the bound is $\sqrt(2\log(2n))$, when the quantity is the maximum absolute value of $n$ independent standard normal Gaussians.

A trick is to consider the maximum of $\{X_1,-X_{1},...,X_n,-X_n\}$ and use the maximum of $2n$ standard normal Gaussians (see  https://math.stackexchange.com/q/89147, for example).

Note that the bound holds even though the $2n$ random variables are not mutually independent.


问:
One algorithm example and the probability calculation of 1/n?
We have one algorithm that gets the input from {$k_1,...,k_n$} ($n$ - distinct numbers) and we want to read ($k_n$) by filling the variable $z$ with $k_n$ with probability $\frac{1}{i}$. At the final step of the algorithm, print $z$ as the output; in which the probability  $k_n$ is ($\frac{1}{n}$).

This is one example.

My question is, how is ($\frac{1}{n}$) calculated?

答: As I understood the problem, we have $n$ distinct numbers $k_i$, $i=1,\dots,n$ and a variable $z$. At each $i$-th step we read $k_i$ and with probability $\tfrac 1i$ put $z:=k_i$. Then $k_n$ is read at $n$-th step, so the probability that we put $z:=k_n$ is $\tfrac 1n$.


问:
Isometric mapping of two subsets in a metric space
Let (M,d) be a metric space, and M1, M2 be two subspace of M.

Edit: I forgot to mension that both M1 and M2 are dense.

Suppose we have

$f: M \rightarrow M$
and we know f is an isometric mapping from M1 to M1, i.e.,
$$
d(f(a),f(b))=d(a,b)
$$
for any $a,b \in M1$.

Suppose we also have
$$
d(f(a),f(b))=d(a,b)
$$
for any $a \in M1$ and $b \in M2$.

The question is : can we say that f is an isometric mapping of $M1\cup M2$?

In other words, is it possible to show that
$$
d(f(a),f(b))=d(a,b)
$$
for any $a,b \in M1\cup M2$.

I think it is true in Euclidean space, but I have no idea how to do it in a general metric space.

Any suggestion would be greatly appreciated.

Body: $M_1=\{a\}$, $M_2=\{b,c\}$, $f(a)=a, f(b)=f(c)=b$, $d(x,y)=1$ if $x\neq y.$


问:
Are there infinitely $M$ such that $\{\sqrt{M+2}\}=\{\sqrt{M}\}+1$? (where $\{x\}$ is the closest integer function)
I was trying to solve the following problem, but realized that I probably reached a dead end:

Prove that the $n$th positive nonsquare integer is $n+\{ \sqrt{n} \}$ where $\{x\}$ is denotes the integer closest to $x$.

Nevertheless, I have a question related to the above about the closest integer function: $\{x\}$. On close inspection it is not obvious to me that there is a nice limititing behavior for this function.

One argument I was working on relied on showing that

$\{\sqrt{M+2}\}>\{\sqrt{M}\}$, or $\{\sqrt{M+2}\}=\{\sqrt{M}\}+1$ for finitely many $M$, which I believe is false but I am not sure if my idea is correct.

I think one way to disprove this is to show that given any $N \in \mathbb{N}$ we can find an integer $M$ such that $N \leq M$ and $\{\sqrt{M+2}\}=\{\sqrt{M}\}+1$ , (which I think is equivalent to the following) meaning we can find an integer $k$

such that $\sqrt{M}\ < k+\frac{1}{2}$, and $\sqrt{M+2}>k+\frac{1}{2} \implies $

$M < k^{2}+k+\frac{1}{4}$, and $M>k^{2}+k-\frac{3}{4} \implies$

$k^{2}+k-\frac{3}{4}<M<k^{2}+k+\frac{1}{4}$.

It looks like this is trivial choosing some very large $k > N$ where the steps can be reversed. Is this thinking correct?

I am also interested if the generalization for

$\{\sqrt{M+n}\}=\{\sqrt{M}\}+1$ where $n\in \mathbb{N}$ follows as easily.

答: You just want to have $$k^2\leq M<(k+1)^2\leq M+n<(k+2)^2$$ so essentially $(k+1)^2-n\leq M<(k+1)^2$ for $k>n$ so you have plenty solutions for large enough $k$


问:
Smooth invariance of domain
This is Theorem 22.3 in Tu's textbook. There are a multiple of similar questions, but all answers seemed unsatisfactory for me. (One of them used algebraic topology result, but I am unfamiliar with the subject.)

Theorem: Let $U$ be an open subset, $S \subset \mathbb R^n$ an arbitrary, and $f: U \rightarrow S$ a diffeomorphism. Then $S$ is open in $\mathbb R^n$.

Proof: Let $f(p)$ be an arbitrary point in $S$ with $p \in U$. Since $f$ is a diffeomorphism, there is an open set $V \subset \mathbb R^n$ containing $S$ and a smooth map $g: V \rightarrow \mathbb R^n$ such that $g|_S = f^{-1}$.

I am not sure why the bolded statement is true. I agree that $f(U) \subset S$ is open in $S$, but why does it guarantee the existence of such $g$? If $S$ were a closed set, I may extend $f^{-1}: S \rightarrow U$ to $\tilde g: V \supset S \rightarrow U$ with $\tilde{g}|_S = f^{-1}$ using partition of unity, but there is no assumption about $S$.

答:
This is a comment but its to long for a comment.
You have to ask yourself what does it mean for a function to be smooth for arbitrary sets of $\mathbb{R}^n$? one possible definition given in the appendix of Lee introduction to smooth manifolds goes as follows

We sometimes need to consider smoothness of functions whose domains are subsets of
$\mathbb{R}^{n}$ that are not open. If $A \subseteq \mathbb{R}^{n}$ is an arbitrary subset, a function $F: A \rightarrow \mathbb{R}^{m}$ is said to be smooth on $A$ if it admits a smooth extension to an open neighborhood of each point, or more precisely, if for every $x \in A,$ there exist an open subset $U_{x} \subseteq \mathbb{R}^{n}$ containing $x$ and a smooth function $\widetilde{F}: U_{x} \rightarrow \mathbb{R}^{m}$ that agrees with $F$ on $U_{x} \cap A .$ The notion of diffeomorphism extends to arbitrary subsets in the obvious way: given arbitrary subsets $A, B \subseteq \mathbb{R}^{n},$ a diffeomorphism from $A$ to $B$ is a smooth bijective map $f: A \rightarrow B$ with smooth inverse.

You can use partition of unity to construct $g$

Title: Smooth invariance of domain


问:
Finding the number of solutions for $e^{\tan x}=\sin x+ \cos x$ graphically
The number of solutions of the equation
$$e^{\tan x}=\sin x+ \cos x$$ In the interval $[-\pi,\pi]$

I tried to approach this question graphically but hit a dead end when trying to graph $e^{tanx}$

For graphing sinx+cosx I graphed $\sqrt 2(sin(x+\frac{\pi}{4}))$

I still feel this question can be solved graphically,Is there a method to graph $e^{tanx}$ to some accuracy ?

答:
As you can see from the graph below  there are two roots in the given interval $[-\pi,\pi]$.

Given $$f(x)=e^{\tan x}-\sin x-\cos x$$
One root is quite obvious since $f(0)=0$.

To find the other we use the Newton method. Compute the derivative
$$f'(x)=\sin x-\cos x+e^{\tan x} \sec ^2 x$$

starting from $x_0=2$ we define
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

$$
\begin{array}{c|r|r}
 n & x_n & \text{error}\\
\hline
 0 & 2 & - \\
 1 & 2.1487490 & 0.851251 \\
 2 & 2.1845423 & 0.0357933 \\
 3 & 2.1844057 & 0.000136542 \\
 4 & 2.1844057 & 1.625\times 10^{-9} \\
\end{array}
$$

$$
...
$$


答:
You could reduce the number of binomial coefficients you need to calculate by using:

$$\frac12\binom{100}{50}+\sum_{k=0}^{49}\binom{100}{k} = \frac12\sum_{k=0}^{100}\binom{100}k = 2^{99}$$

hence:

$$\left(\frac12\right)^{100}\sum_{k=0}^{59}\binom{100}k=\frac12+\left(\frac12\right)^{100}\left(\frac12\binom{100}{50}+\sum_{k=51}^{59}\binom{100}k\right)$$

so you "only" need to calculate $10$ of them.

Title: How to solve $\left(\frac{1}{2}\right)^{100}\sum_{k=0}^{59}{100 \choose k}$?

Body:
Part of an exercise, I got the following sum:
$$
\left(\frac{1}{2}\right)^{100}\sum_{k=0}^{59}{100 \choose k}
$$
I'm trying to evaluate it and get the answer. What is the easiest way to do it?


问:
An application of Weierstrass M-Test to determine sets of uniform convergence.
let $\sum_{n=1}^{\infty}fn(x) $ = $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}(\frac{x}{1+x})^n$

I am attempting to apply Weierstrass M-Test on [a,$\infty$)

I get that

$|\frac{1}{\sqrt{n}}(\frac{x}{1+x})^n|$<$|\frac{1}{\sqrt{n}}|$

but the RHS doesnt converge, so we cant use M-Test.

What other methods can I use?

答:
You cannot apply on $[a,+\infty)$, because $$M_n:=\sup_{x\ge a} |f_n(x)|=\lim_{x\to+\infty} f_n(x)=\frac{1}{\sqrt n}.$$ So you this is teh best bound and Weierstrasss M-Test cannot be applied.

But you can use it in $[a,b]$ for all $b>0$, since the series of genreral term $\frac{1}{\sqrt n} \left(\frac{b}{b+1}\right)^n$ is convergent.

With this method, you can prove that this series is uniform convergent on any compact set of $[a,+\infty)$.


问:
Modules and prime numbers
Let's say that $p$ is prime number, $z$ is natural number and our form $\frac{p-1}{2^z}=n$ and is even. $p>n$. From binomial theroem $n^x$ ($x$ is natural number) = form in photo. So is $$n^x \equiv \bigg(\frac{1}{2^z}\bigg)^x \pmod{p}$$ the truth ? Better to look at photo so you can properly see what I want to show you.

I don't know why, beacuse it makes kinda sense to me, it is not true. For example $p=13$ and $z=1$ and $x=p-2$ or any other I think. Also I don't know if mod can be written as fraction and if it is written like that what I need to do to make it  natural number?

..

答:
That is not necessarily true. $$\bigg(\frac{p-1}{2^z}\bigg)^x\equiv\bigg(\frac{-1}{2^z}\bigg)^x\pmod{p}$$

So only if $x$ is even you have $$\bigg(\frac{p-1}{2^z}\bigg)^x\equiv\bigg(\frac{1}{2^z}\bigg)^x\pmod{p}$$

EDIT: you cannot always find the residue. $\frac{1}{2^z}$ is the modular inverse of $2^z$, ususally denoted with $2^{-z}$ Those do not have a formula. Read more here.

Title: Modules and prime numbers


问:
Is this proof right?
Let $f_n(x) = n^2 x e^{- n^2 x}$, $n=1,2,3,...$.

Show that $\sum_{n=1}^{\infty} f_n(x) $ convergeces uniformly on $[a,\infty]$, where $a>0$.

Show that $\sum_{n=1}^{\infty} f_n(x) $ does not convergece uniformly on $[0,\infty]$.

Proof:

First, I found the maximum of $f_n(x)$ which occured at $x=1/2n$, from which I constructed the upper bound for $f_n(x)$ for all $n$. Namely

$$|f_n (x) | \leq M_n = f_n(1/2n),\,\,\,\, x \neq 1/2n$$

I claimed that $\sum_1^\infty M_n$ is convergent because $\int_1^\infty M_n dn = 3 e^{-1/2}$.

Therefore, since we have the requirements to apply the M-test theorem, we can say that $\sum f_n(x)$ is uniformly convergent on $[a, \infty],\,\,\, a>0$. While $\sum f_n(x)$ is not uniformly confergennt because $f_n(x)$ is not bounded at $x=0$. (there are no integer $n$ that makes $1/2n$ vanishies).

答:
The Right Solution to My Problem.

First, we check the pointwise convergence using the ratio test.

\begin{align}
L &= \lim_{n \to \infty} \bigg|  \frac{(n+1)^2 x e^{- (n+1)^2 x}}{n^2 x e^{- n^2 x}} \bigg|\\
L&= \lim_{n \to \infty} \frac{(n+1)^2}{n^2} \times \lim_{n \to \infty} e^{- x (1+2n)}\\
L&= 0 < 1
\end{align}
Thus, the series $\sum_{n=1}^{\infty} f_n (x)$ is pointwise convergent on $(0,\infty)$.

\emph{Note}: We can omit the pointwise convergent if we are aiming to prove the uniform convergence.

Since it is clear that $\lim_{n \to \infty} n^2 x e^{- n^2 x} =0$ as $n \to \infty$, then we will find $\|f_n(x) -0 \|_{\infty}$, where $\|.\|_{\infty}$ is the uniform norm, (it is also called supremum norm, the Chebyshev norm, the infinity norm and the maximum norm where the maximum value of the function is included)

$$\|f\|_{\infty} = \max_{a\leq x<\infty}{|f(x)|}$$

\begin{align}
\|f_n(x)\|_{\infty}   &= \max_{a \leq x <\infty)} |f_n(x)|\\
\|f_n(x)\|_{\infty}  &=\max_{a \leq x <\infty} \bigg| \frac{n^2 x}{e^{n^2 x}}\bigg|\\
\|f_n(x)\|_{\infty}  &=  \max_{a \leq x <\infty} \bigg\{ \frac{n^2 x}{1 +n^2 x + n^4 x^2 /2 + n^6 x^3 /6+...+ (n^2 x)^k/k!+...} \bigg\} ,\,\,\,\, m=1,2,3,...\\
\|f_n(x)|_{\infty} & < \max_{a \leq x <\infty} \bigg\{ \frac{n^2 x}{1 + \frac{1}{2} n^4 x^2} \bigg\},\\
\|f_n(x)\|_{\infty}&< \frac{n^ 2  }{\frac{1}{2} a n^4}\\
\|f_n(x)\|_{\infty} & < \frac{2}{a n^2}
\end{align}

Let $M_n = \frac{2}{a \, n^2}$, since we found that $\|f_n(x)\|_\infty < M_n$ for all $n$, and $\sum_{n=1}^{\infty} M_n$ is convergent, for

$$\sum_{n=1}^{\infty} M_n = \frac{2}{a}\sum_{n=1}^{\infty} \frac{1}{n^2}.$$

Thus, the series $\sum_{n=1}^{\infty} n^2 x e^{- n^2 x}$ converges uniformly on $[a,\infty)$ for $a>0$. To show that the series is not uniformly convergent on $[0,\infty)$.

Let's take $x=\frac{1}{m^2} \to 0$ as $m \to \infty$, and let $g(x) = \sum_{n=1}^{\infty} n^2 x e^{n^2 x}$.

```<code><span class="math-container">$$g(\frac{1}{m^2}) \geq \sum_{n=1}^{m} \frac{n^2}{m^2} e^{\frac{-n^2}{m^2}} \ge \sum_{n=1}^{m} \frac{1}{e} $$</span>

<span class="math-container">$$g(\frac{1}{m^2}) \ge \frac{m}{e}.$$</span>
</code>```

We can see that $g$ is unbounded on $0<x<a$ for any $a>0$. Which brakes the condition of the uniform convergence i.e.((If $\sum_{n=1}^{\infty} S_n(x)$ converges uniformly to $S$, then $S$ is bounded.)). Thus, the series $\sum_{n=1}^{\infty} n^2 x e^{-n^2 x}$ does not converge uniformly on $[0,\infty)$.

Title: Is this proof right?

Body:
The Right Solution to My Problem.

First, we check the pointwise convergence using the ratio test.

\begin{align}
L &= \lim_{n \to \infty} \bigg|  \frac{(n+1)^2 x e^{- (n+1)^2 x}}{n^2 x e^{- n^2 x}} \bigg|\\
L&= \lim_{n \to \infty} \frac{(n+1)^2}{n^2} \times \lim_{n \to \infty} e^{- x (1+2n)}\\
L&= 0 < 1
\end{align}
Thus, the series $\sum_{n=1}^{\infty} f_n (x)$ is pointwise convergent on $(0,\infty)$.

\emph{Note}: We can omit the pointwise convergent if we are aiming to prove the uniform convergence.

Since it is clear that $\lim_{n \to \infty} n^2 x e^{- n^2 x} =0$ as $n \to \infty$, then we will find $\|f_n(x) -0 \|_{\infty}$, where $\|.\|_{\infty}$ is the uniform norm, (it is also called supremum norm, the Chebyshev norm, the infinity norm and the maximum norm where the maximum value of the function is included)

$$\|f\|_{\infty} = \max_{a\leq x<\infty}{|f(x)|}$$

\begin{align}
\|f_n(x)\|_{\infty}   &= \max_{a \leq x <\infty)} |f_n(x)|\\
\|f_n(x)\|_{\infty}  &=\max_{a \leq x <\infty} \bigg| \frac{n^2 x}{e^{n^2 x}}\bigg|\\
\|f_n(x)\|_{\infty}  &=  \max_{a \leq x <\infty} \bigg\{ \frac{n^2 x}{1 +n^2 x + n^4 x^2 /2 + n^6 x^3 /6+...+ (n^2 x)^k/k!+...} \bigg\} ,\,\,\,\, m=1,2,3,...\\
\|f_n(x)|_{\infty} & < \max_{a \leq x <\infty} \bigg\{ \frac{n^2 x}{1 + \frac{1}{2} n^4 x^2} \bigg\},\\
\|f_n(x)\|_{\infty}&< \frac{n^ 2  }{\frac{1}{2} a n^4}\\
\|f_n(x)\|_{\infty} & < \frac{2}{a n^2}
\end{align}

Let $M_n = \frac{2}{a \, n^2}$, since we found that $\|f_n(x)\|_\infty < M_n$ for all $n$, and $\sum_{n=1}^{\infty} M_n$ is convergent, for

$$\sum_{n=1}^{\infty} M_n = \frac{2}{a}\sum_{n=1}^{\infty} \frac{1}{n^2}.$$

Thus, the series $\sum_{n=1}^{\infty} n^2 x e^{- n^2 x}$ converges uniformly on $[a,\infty)$ for $a>0$. To show that the series is not uniformly convergent on $[0,\infty)$.

Let's take $x=\frac{1}{m^2} \to 0$ as $m \to \infty$, and let $g(x) = \sum_{n=1}^{\infty} n^2 x e^{n^2 x}$.

$$g(\frac{1}{m^2}) \geq \sum_{n=1}^{m} \frac{n^2}{m^2} e^{\frac{-n^2}{m^2}} \ge \sum_{n=1}^{m} \frac{1}{e} $$

$$g(\frac{1}{m^2}) \ge \frac{m}{e}.$$

We can see that $g$ is unbounded on $0<x<a$ for any $a>0$. Which brakes the condition of the uniform convergence i.e.((If $\sum_{n=1}^{\infty} S_n(x)$ converges uniformly to $S$, then $S$ is bounded.)). Thus, the series $\sum_{n=1}^{\infty} n^2 x e^{-n^2 x}$ does not converge uniformly on $[0,\infty)$.


问:
Hausdorff space and disjoint open sets
Let $X$ be a Hausdorff space and $x_1,...,x_n$ points of $X$. Then there exist open pairwise disjoint sets $V_1,...,V_n$ such that $x_i$ is in $V_i$ for every $i$.

How can I prove this?

答:
You know that each pair $i \neq j$ there are open sets $V_{ij}$ and $V_{ji}$ so that $x_i \in V_{ij}$ and $x_j \in V_{ji}$ and $x_i \notin V_{ji}, x_j \notin V_{ij}$.  For convenience of notation, define $V_{ii} = X$.

Now consider the intersection $U_i = \bigcap_{j=1}^n V_{ij}$.  This set is open (why?) and non-empty (why?) and contains $x_i$ but not $x_j$ for any $j \neq i$.

Title: Hausdorff space and disjoint open sets


问:
Closed form of $\sum_{k=1}^{n}k!$
I was trying to get a closed form of the sum $$\sum_{k=1}^{n}k!$$
Mathematica gives the answer
$$(-1)^n\Gamma(n+2)(!(-n-2))-!(-1)-1$$
Here the ! sign before a number is the subfactorial. Here is my try:
\begin{align}
\sum_{k=1}^{n}k!&=\sum_{k=1}^{n}\Gamma(k+1)\\
&=\sum_{k=1}^{n}\int_{0}^{\infty}x^{k}e^{-x}dx\\
&=\int_{0}^{\infty}e^{-x}\sum_{k=1}^{n}x^kdx\\
&=\int_{0}^{\infty}\frac{e^{-x}(x-x^{n+1})}{1-x}dx
\end{align}
But I don't know what to do further. So, how can that closed form given by mathematica be derived?

Also, that closed form expression looks complex for some numbers to me. Is it even right?

答:
I believe that this is possibly how Mathematica got its answer.

Extend the definition of the sum to negative upper indices:
$$
\sum_{k=0}^nf(k)=-\sum_{k=n+1}^{-1}f(k)\tag1
$$
Extend the definition of factorial to negative numbers:
$$
n!=\frac{(-1)^{-n-1}}{(-n-1)!}(-1)!\tag2
$$
Applying these extensions to the definition for $!n$, we get
$$
\begin{align}
!n
&=\sum_{k=0}^n(-1)^k\frac{n!}{k!}\tag{3a}\\
&=-\sum_{k=n+1}^{-1}(-1)^k\frac{n!}{k!}\tag{3b}\\
&=-\sum_{k=n+1}^{-1}(-1)^n\frac{(-k-1)!}{(-n-1)!}\tag{3c}
\end{align}
$$
Explanation:

$\text{(3a)}$: normal definition of $!n$

$\text{(3b)}$: apply $(1)$

$\text{(3c)}$: apply $(2)$

Substituting $n\mapsto-n$ and $k\mapsto-k$, we get
$$
!(-n)=\frac{(-1)^{n-1}}{(n-1)!}\sum_{k=1}^{n-1}(k-1)!\tag4
$$
Finally, substituing $n\mapsto n+2$ and $k\mapsto k+1$, we get
$$
!(-n-2)=\frac{(-1)^{n+1}}{(n+1)!}\sum_{k=0}^nk!\tag5
$$
which then leads immediately to
$$
\sum_{k=0}^nk!=(-1)^{n+1}!(-n-2)\overbrace{\ (n+1)!\ }^{\Gamma(n+2)}\tag6
$$
Formula $(6)$ agrees with what Mathematica gives, using $!(-1)=0$.

I note that my answer and Mathematica's gives a different exponent of $-1$ than in the question. Also, my answer includes $k=0$, so it is $1$ greater.

How Mathematica Actually Computes Subfactorial

Let $m\in\mathbb{Z}^+$ and $n+m\gt-1$.

Integrating by parts, we get
$$
\frac1{n!}\int_{-1}^\infty x^ne^{-x}\,\mathrm{d}x=\frac{(-1)^n}{n!}e+\frac1{(n-1)!}\int_{-1}^\infty x^{n-1}e^{-x}\,\mathrm{d}x\tag7
$$
Thus, by induction we get
$$
\frac1{n!}\int_{-1}^\infty x^ne^{-x}\,\mathrm{d}x=\sum_{k=0}^n\frac{(-1)^k}{k!}e\tag8
$$
and therefore,
$$
\begin{align}
!n
&=\sum_{k=0}^n(-1)^k\frac{n!}{k!}\tag{9a}\\
&=\frac1e\int_{-1}^\infty x^ne^{-x}\,\mathrm{d}x\tag{9b}\\
&=\frac1e\,\Gamma(n+1,-1)\tag{9c}\\
&=\frac1e\,\Gamma(n+1)+\frac{e^{\pi in}}e\int_0^1x^ne^x\,\mathrm{d}x\tag{9d}
\end{align}
$$
Explanation:

$\text{(9a)}$: equation $(5)$ from this answer

$\text{(9b)}$: apply $(8)$

$\text{(9c)}$: definition of the Incomplete Gamma Function

$\text{(9d)}$: write $(-1)^n=e^{\pi in}$

I think $(-1)^n=e^{-\pi in}$ is just as reasonable a replacement, and better is to average the two, getting $(-1)^n=\cos(\pi n)$. This also has the benefit of keeping $!n\in\mathbb{R}$ for $n\in\mathbb{R}$. However, we will continue with this, as this seems to be what Mathematica does.

First, a recurrence for the integral in $\text{(9d)}$:
$$
\begin{align}
\frac1{n!}\int_0^1x^ne^x\,\mathrm{d}x
&=e\frac1{(n+1)!}-\frac1{(n+1)!}\int_0^1x^{n+1}e^x\,\mathrm{d}x\tag{10a}\\
&=-e\sum_{k=1}^m\frac{(-1)^k}{(n+k)!}+\frac{(-1)^m}{(n+m)!}\int_0^1x^{n+m}e^x\,\mathrm{d}x\tag{10b}
\end{align}
$$
Explanation:

$\text{(10a)}$: integration by parts

$\text{(10b)}$: induction on $m$ using $\text{(10a)}$
$$
\begin{align}
\frac1e\int_0^1x^ne^x\,\mathrm{d}x
&=-\sum_{k=1}^m\frac1{k!\binom{-n-1}{k}}+\frac1{m!\binom{-n-1}{m}}\frac1e\int_0^1x^{n+m}e^x\,\mathrm{d}x\tag{11a}\\
&=-\sum_{k=1}^{m-1}\frac1{k!\binom{-n-1}{k}}-\frac1{m!\binom{-n-1}{m}}\left(1-\frac1e\int_0^1x^{n+m}e^x\,\mathrm{d}x\right)\tag{11b}
\end{align}
$$
Explanation:

$\text{(11a)}$: multiply $(10)$ by $n!/e$ and use $\frac{(-1)^k(n+k)!}{n!}=k!\binom{-n-1}{k}$

$\text{(11b)}$: move the $k=m$ term from the sum to the right

$$
\begin{align}
!n
&=\overbrace{\frac{(-1)^m\Gamma(n+m+1)}{e\,m!\binom{-n-1}{m}}}^{\frac1e\Gamma(n+1)}\\
&-\sum_{k=1}^{m-1}\frac{e^{\pi in}}{k!\binom{-n-1}{k}}-\frac{e^{\pi in}}{m!\binom{-n-1}{m}}\left(1-\frac1e\int_0^1x^{n+m}e^x\,\mathrm{d}x\right)\tag{12a}\\[6pt]
&=\frac{e^{\pi in}\left(e-\int_0^1x^{n+m}e^x\,\mathrm{d}x\right)-(-1)^m\Gamma(n+m+1)}{e\,(m-1)!\binom{-n-1}{m-1}(n+m)}\\
&-\sum_{k=1}^{m-1}\frac{e^{\pi in}}{k!\binom{-n-1}{k}}\tag{12b}
\end{align}
$$
Explanation:

$\text{(12a)}$: $\text{(9d)}$, $\text{(11b)}$, and $\Gamma(z+1)=z\,\Gamma(z)$

$\text{(12b)}$: commute terms and use $m!\binom{-n-1}{m}=-(m-1)!\binom{-n-1}{m-1}(n+m)$

Let $n\to-m^+$, then $\text{(12b)}$ and L'Hôpital yield
$$
\begin{align}
!(-m)
&=(-1)^m\left[\frac{\gamma-\int_0^1\log(x)\,e^x\,\mathrm{d}x}{e\,(m-1)!}-\sum_{k=1}^{m-1}\frac1{k!\binom{m-1}{k}}\right]+\frac{(-1)^m\pi i}{e\,(m-1)!}\tag{13a}\\
&=\frac{(-1)^m}{(m-1)!}\left[\frac{\gamma+\pi i-\int_0^1\log(x)\,e^x\,\mathrm{d}x}e-\sum_{k=1}^{m-1}(k-1)!\right]\tag{13b}
\end{align}
$$
where $\gamma$ is the Euler-Mascheroni Constant, which is computed in this answer, and
$$
\begin{align}
\int_0^1\log(x)\,e^x\,\mathrm{d}x
&=\gamma-\operatorname{Ei}(1)\tag{14a}\\
&=-\sum_{k=1}^\infty\frac1{k\,k!}\tag{14b}\\
\end{align}
$$
where $\operatorname{Ei}(x)$ is the Exponential Integral.

The values given in $\text{(13b)}$ match those returned by Mathematica for Subfactorial.

Thus, substituting $m\mapsto m+2$ and $k\mapsto k+1$ in $\text{(13b)}$, we get
$$
\begin{align}
\sum_{k=0}^mk!
&=(-1)^{m+1}!(-m-2)(m+1)!+\frac{\gamma+\pi i-\int_0^1\log(x)\,e^x\,\mathrm{d}x}e\tag{15a}\\
&=(-1)^{m+1}!(-m-2)(m+1)!\,-\,!(-1)\tag{15b}
\end{align}
$$

Title: Closed form of $\sum_{k=1}^{n}k!$


问:
Expected value of a biased game
I heard this puzzle a while back and wasn't sure how to answer it:

You play a game whose cost to play goes up at every round, and there are 2 possible outcomes.
Winning and losing.
You win with 60% likelyhood and lose with 40% chance at all rounds.

Round 1 - Costs \$1 to play
On the first round, if you win, you win another \$1.
If you lose, you lose your \$1.

Round 2 - Cost \$2 to play
On the second round, if you win, you win another \$1.
If you lose, you lose your \$2.

Round 3 - Cost \$3 to play
On the Thirdround, if you win, you win another \$1.
If you lose, you lose your \$3.

What are the expected values at each round, and how do you calculate them?

答:
Cost \$2 and you win another \$1 means that if you win you cash \$3?, Right? This because usually the cost of a game is never reimbursed. Have I understood well? And you want to know the expected gain at EACH round? This independently about what happened before?

Under these assumptions, the three expectations are the following:

$$\mathbb{E}[R_1]=1\times0.6-1\times0.4$$

$$\mathbb{E}[R_2]=1\times0.6-2\times0.4$$

$$\mathbb{E}[R_3]=1\times0.6-3\times0.4$$

Title: Expected value of a biased game


问:
Variants of Fatou's Lemma
Fatou's Lemma is

Let $(\Omega, \Sigma, \mu)$ be a measure space and $X \in \Sigma$. Let $f_n; f_n : X \to [0, +\infty]$ be a sequence of $(\Sigma, \mathcal{B}_{\mathbb{R}_{\geq 0}})$-measurable functions. Define $f(x) = \lim\inf_{n\to\infty} f_n$. Then $f$ is $(\Sigma, \mathcal{B}_{\mathbb{R}_{\geq 0}})$-measurable and

$\begin{align} \int_X f d\mu \leq \liminf_{n\to\infty} \int_X f_n d\mu \end{align}$

Is this true if you generalize the codomain of $f$ and $f_n$ to be all of $\mathbb{R}$ and $(\Sigma, \mathcal{B}_{\mathbb{R}})$-measurable?

Is this true if you replace $\liminf$ with $\limsup$?

Provide counterexamples please if possible. No this is not a homework question.

答:
The answers to both questions are no. Here's a counterexample for both cases.

Let $X = \mathbb{R}$. Let $f_{n}(x)=-1$ if $x \in [n, n+1)$, and $0$ otherwise. Then
$$
0=\int f d\mu > \liminf_{n} \int_{n} f_{n} d\mu = \limsup_{n} \int f_{n} d\mu =-1.
$$

Title: Variants of Fatou's Lemma


问:
Is the space of signed Radon measures a locally convex TVS?
Let $X$ be a compact Hausdorff space. In https://www.math.uni-bonn.de/people/scholze/Analytic.pdf (Ch. 3) Scholze considers the space of signed Radon measures on $X$ equipped with the filtered colimit (aka inductive limit) topology of the (in the weak$^*$-topology) compact absolutely convex subsets $\mathcal{M}(X)_{\leq c}$. Here $\mathcal{M}(X)_{\leq c}$ denotes the subset of measures with total variation norm less or equal than $c$ with $c>0$.

Then he states that the resulting topology is a locally convex vector topology. I was wondering if the subsets $\mathcal{M}(X)_{\leq c}$ form a neighborhood basis of the origin. If the answer is yes, then I do not see why the resulting topology is not the same as the one induced by the total variation norm. If the answer is no, then I do not see how to show that this topology is a locally convex vector topology.

Any clarification on this would be really appreciated.

答:
I followed @Kevin Arlin's advice and posted the question on MO
https://mathoverflow.net/questions/377896/is-the-filtered-colimit-topology-on-the-space-of-signed-radon-measures-linear-an.
It already got answered.</p>


问:
Is Secp256k1's prime prime?
Bitcoin protocol relies on the elliptic curve secp256k1 for its cryptographic security.
For that purpose the integer number $p = 2^{256}-2^{32}-977$ must be prime.
How do they know $p$ is actually prime? I mean, which primality test can be used to prove it?

答:
As pointed out in the comments, this is not a large number to factorise with modern methods.

The Magma online calculator http://magma.maths.usyd.edu.au/calc/ verifies that it is prime, when asked to factor it, almost instantly.

time Factorisation(2^256-2^32-977);

returns

[ <115792089237316195423570985008687907853269984665640564039457584007908834671663, 1> ]
Time: 0.070

as the factorisation into powers of primes; the number $2^{256}-2^{32}-977$ is a single prime to the power $1$. It takes 0.07 seconds to get this result.

Primality testing is consistent as well.

time IsPrime(2^256-2^32-977);

yields

true
Time: 0.060

Title: Is Secp256k1's prime prime?


问:
Whats the revenue value
Let's say the following scenario:

$4A + 311B  = \text{ SUM}$

Where

$4A$ contributes to $40\%$ of the total sum.

$311B$ contributes to $60\%$ of the total sum.

Question:

If we want to get the $A$ portion to be $20\%$ instead of $40\%$ of the new sum,
and $4A$ stays the same.
How many of $B$ should it be in order to make this happen?

My Approach:

Let $x$ = the old sum

$n$ = is the number of $B$ required.

$\frac{2}{10}(\frac{40}{100}x+n*\frac{60}{100}*x) = \frac{40}{100}x$

By solving the equation,

I found that $n = 829.3333333$
which is wrong, can I know what I'm I doing wrong here?

答:
I'll simplify the problem and make the solution as concise as I can. Let's first get rid of the numerical constants and just say that we have a sum
$$
A + B = X
$$
It is given that $B=0.6X$ and $A=0.4X$. Then we want a new sum so that $A$ is equal to $0.2$ times the right-hand side; it's quite clear that in the new equation, the right-hand side should be equal to $2X$. Therefore, we have the following new sum:
$$
A+kB = 2X
$$
where $k$ is some scaling constant. Now, we want the part $kB$ to be equal to $0.8$ times the right-hand side, or
$$
kB = 0.8\cdot 2X
$$
Earlier, we learned that $B=0.6X$. Let's plug that in and we get
$$
k \cdot 0.6X = 0.8 \cdot 2X
$$
As long as $X\neq 0$, we get $k = \frac{0.8\cdot 2}{0.6}= \frac{8}{3}$.

Title: Whats the revenue value


问:
Can I write $\prod_{i=1}^k$ if $k$ starts at 0?
I'm hoping that $\prod_{i=1}^0i=1$.

答: Yes, it's called the empty product. The aim is to ensure $\prod_{i=1}^{n+1}a_i=a_{n+1}\prod_{i=1}^na_i$ still works when $n=0$. Similarly, an empty sum $\sum_{i=1}^0a_i=0$.

Body: I'm hoping that $\prod_{i=1}^0i=1$.


问:
Definitions of relative compactness
Per definition, a subset $A$ of a metric space $(X,d)$ is relatively compact if and only if its closure $\overline{A}$ is compact.
That means that any sequence in $\overline{A}$ contains a subsequence that converges in $\overline{A}$. Does that imply that $\partial A$ is compact?

According to an alternative definition, $A$ is relatively compact if any sequence in $A$ contains a subsequence convergent in $X$. How do you derive from this that $\partial A$ is compact?

In other words, how can you show that both definitions are equivalent?

答:
Yes.

The boundary of $A$ is defined as
$$
\partial A=\overline{A}\cap\overline{A^c}.
$$
So $\partial A$ is a closed subset of a compact set, and hence compact.

Let $\{x_n\}\subset\partial A$. Since $\partial A\subset\overline{A}$ and $\overline{A}$ is compact, then there exists a subsequence $x_{n_k}\to x\in\overline{A}$. However,
$\{x_{n_k}\}\subset\partial A$, and since $\partial A$ is closed and $\{x_{n_k}\}$ convergent to $x$, then $x\in \partial A$. Thus every sequence in $\partial A$ possesses a convergent subsequence with limit in $\partial A$. Hence, $\partial A$ is compact.

Title: Definitions of relative compactness


问:
A proof on the existence of an injective function
In La matematica della verità (p.65-66, theorem 3.5.3), Ettore Casari proves the following theorem:

$f : A \to B$ is injective iff  for all $g : C \to A, h : C \to A$ and all $C$:

$$f \circ g \equiv f \circ h \to g \equiv h$$

His proof for the right-to-left hand is unclear to me. He writes:

Consider the set C and the functions $g$ and $h$ defined as follows: C: = { z } (where $z$ is some individual), $g,h : C \to A$, $g(z) := x$ and $h(z) := y$. It follows that $f(g(z)) = f(h(z))$, i.e that $(f \circ g)\,z = (f \circ h)\,z$, and since $z$ is the only argument of $f \circ g$ and $f \circ h$, $f \circ g \equiv f \circ h$ and therefore, by hypothesis, $g \equiv h$, so that $g(z) = h(z)$ and $x = y$.

I don't understand how he argues on the basis of a very specific set $C$ and a very specific stipulation (that $g(z) := x$ and $h(z) := y$) to the claim that for any $C$, and any functions $g,h : C \to A$, it follows that $f$ is injective.

How does this proof generalise to other cases where $C$ is not the set as defined above, and $g$ and $h$ are not as defined)?

答:
The hypothesis is that we are given that for any $C$ we have some condition, then prove $f$ is injective. We don't need to prove that for every $C$ with some condition, then $f$ is injective. This is why considering a singleton set is sufficient.

Since $g,h$ are functions from $C$ to $A$, we have that $g(z):=x$ and $h(z):=y$ for some elements $x,y \in A$. $f(g(z)) = f(h(z))$  comes from the fact that when proving the injectivity of some function $f$, one can assume $f(x) = f(y)$. Then as $z$ is the only argument , we must have $f \circ g \equiv f \circ h$ so that $ g \equiv h$ by hypothesis.

Thus, $g(z) = h(z)$ which means $x=y$.


问:
A problema about gamma distribution and convergence in distribution
How can I solve this problem?

Let $Y_{n}\sim \textbf{Gamma}(n,1)$, $n\geq 1$. Prove that $$\frac{Y_{n}-n}{\sqrt{Y_{n}}}\overset{D}{\to}N(0,1)$$
My approach: First I think that since $Y_{n}$ is a sample so I can suppose that $Y_{i}$, $i=1,\ldots, n$ are independent random variables. Now, I think I need to prove that $Y_{n}-n \overset{D}{\to} N(0,1)$ also $\sqrt{Y_n} \overset{D}{\to} 1$. So, by Slutsky' theorem we can conclude that $\frac{Y_{n}-n}{\sqrt{Y_{n}}}\overset{D}{\to} N(0,1)$.

But, can I continue from here? I think the central limit theorem is helful , but I don't know how to use the central limit theorem in this problem.

答:
First observe that $Y_n=\Sigma_n X_n$ where $X_n\sim Exp(1)$, thus $E(X_i)=V(X_i)=1$

then re-write your sequence in the following way

$$\frac{\Sigma_n X_n-n}{\sqrt{n}}\cdot \sqrt{\frac{n}{\Sigma_n X_n}}=A\times \sqrt{B}$$

Now, $A\xrightarrow{d}N(0;1)$ because it is exactly the defintion of CLT.

Note also that $B=\frac{1}{\overline{X}_n}$ and thus, by Law of Large Numbers and Continuous mapping theorem, $\sqrt{B}\xrightarrow{\mathcal{P}}1$

Finally apply Slutsky  to conclude your proof.

$\sqrt{n}$ is needed in the denominator...that was the aim of my question before: if you have not it...get it re-writing your sequence!


问:
What is the value of $\log5-\log9+\log10$
Sorry it might be a stupid question but I am confused why

$\log5-\log9+\log10=\log(50/9)$?
By bodmas rule first if we add then it should be $\log(5/90)$?

Body:
$$\log 5=\log\left(\frac{10}{2}\right)$$
$$=\log 10-\log 2$$
Therefore $$\log 5-\log 9+\log 10=2\log 10-\log 2-\log 9$$
$$=\log \left(\frac{100}{2\times 9}\right)$$
$$=\log\left(\frac{50}{9}\right)$$


答:
We have three statments here, $P_1$, $P_2$ and $Q$ which are given by

$P_1$:= for all sets $C$ and for all $g,h:C \rightarrow A$ we have $f \circ g \equiv f \circ h \implies g \equiv h$

$P_2$:= for the set $C = \{z\}$ and for all $g,h:C \rightarrow A$ we have $f \circ g \equiv f \circ h \implies g \equiv h$

$Q$:= $f : A \rightarrow B$ is injective.

In his proof he basically proves $P_2 \implies Q$.

It is also immediate that $P_1 \implies P_2$ (check this!, this is what you are asking).

So by transitivity we have $P_1 \implies Q$, which is what we wanted.

Finally a remark about how to interpret this theorem:

As you probably know we have that a funciton is invertible iff it is bijective. Note here that invertible means left and right invertible (which is almost never mentioned) and it follows without assumption that the left and the right inverse are equal if they exist.

This theorem just tells us that a function is left invertible iff it is injective. Indeed suppose $f$ is left invertible with function $\tilde{f}$ then we can say that $f \circ g \equiv f \circ h \implies \tilde{f} \circ f \circ g \equiv \tilde{f} \circ f \circ h \implies g \equiv h$, so we can see that the statment $Q$ above implies that $f$ is left-invertible (and the converse is also true).

Similarly he will likely prove that right invertible iff it is surjective.

So, what I wnat to say is that the right hand side $Q$ seems kind of hard to interpret at first. But the idea is just that $Q$ is equivalent to $f$ being left-invertible, whis is much more intuitive in my opinion. In general, these kinds of theorems give characterisations between kinds of invertibility and properties of functions like injectiveness.

Title: A proof on the existence of an injective function

Body:
In La matematica della verità (p.65-66, theorem 3.5.3), Ettore Casari proves the following theorem:

$f : A \to B$ is injective iff  for all $g : C \to A, h : C \to A$ and all $C$:

$$f \circ g \equiv f \circ h \to g \equiv h$$

His proof for the right-to-left hand is unclear to me. He writes:

Consider the set C and the functions $g$ and $h$ defined as follows: C: = { z } (where $z$ is some individual), $g,h : C \to A$, $g(z) := x$ and $h(z) := y$. It follows that $f(g(z)) = f(h(z))$, i.e that $(f \circ g)\,z = (f \circ h)\,z$, and since $z$ is the only argument of $f \circ g$ and $f \circ h$, $f \circ g \equiv f \circ h$ and therefore, by hypothesis, $g \equiv h$, so that $g(z) = h(z)$ and $x = y$.

I don't understand how he argues on the basis of a very specific set $C$ and a very specific stipulation (that $g(z) := x$ and $h(z) := y$) to the claim that for any $C$, and any functions $g,h : C \to A$, it follows that $f$ is injective.

How does this proof generalise to other cases where $C$ is not the set as defined above, and $g$ and $h$ are not as defined)?


问:
Which sets admit a countably additive measure defined on their powerset?
I want to know when a cardinality $\kappa$ admits a (countably additive) probability measure on its powerset that gives probability zero to subsets of cardinality strictly less than $\kappa$.

I am aware of some results in the vicinity. We can strengthen the last condition by requiring that the measure also be $\kappa$-additive. In that case $\kappa$ is real-valued-measurable and various textbooks lay out interesting restrictions on what those can be like. Or we can weaken the last condition to the measure's giving probability zero to all singletons. In that case my question would be easily answered: $\kappa$ has that property if and only if it exceeds the least such cardinal with that property, $\lambda$, because if $\mu$ is a measure on $\lambda$, then we can define a measure $\nu$ on $\kappa>\lambda$ by setting $\nu(X)=\mu(X\cap\lambda)$. I believe Ulam showed that the least cardinal with this property is the least real-valued measurable cardinal.

So what happens when we require that the probability measure give probability zero to subsets smaller than $\kappa$, but allow that it is not $\kappa$-additive? Is it easy to reduce the problem to either of the above two cases?

答:
If you want your measure to only be finitely additive, then this is a theorem of $\sf ZFC$. For any infinite cardinal $\kappa$, consider the filter of co-small subsets (i.e. $\{A\subseteq\kappa\mid |\kappa\setminus A|<\kappa\}$), extend it to an ultrafilter, and that is a finitely additive $0$-$1$ measure.

If you want more substance, and that the measure is $\sigma$-additive, then for the least such $\kappa$, it will necessarily be $\kappa$-additive again, and we're back with real-valued measurable. You can find this in Jech's "Set Theory" (3rd Millennium edition) as Lemma 10.5(i).


问:
(2n + 1) + (2n) is odd?
I'm looking at basic proofs in Lang and in Downing. Here's the proof in question in Downing:

$m$ and $n$ are natural numbers $\ne 0$.

$$s = (2n) + (2m + 1)$$
$$s = 2(m + n) + 1$$
Thus the answer is odd by an earlier definition of odd as $2n + 1$.
This seems to me like just selectively applying the rules to get an answer you want. Why can't I just rewrite this as
$s = 2(m + n + 1)$ by associativity, thus producing an even number? I suspect I'm misunderstanding something, but I can't figure it out.

答:
Associativity does not mean "move the parentheses wherever you want".

Note that associativity is the axiom that $(a+b)+c = a+(b+c)$.

If you tried rewriting $2(m+n)+1$ in the way you suggest, you would first have to reduce $2(m+n)+1$ to $(2m + 2n)+1$ by the distributive property,

which would stop you from writing the expression as $2(m + n + 1)$.

Title: (2n + 1) + (2n) is odd?


问:
How to prove that for any partial order R exists S partial order so that R is a subset of S. S not equals R.
How to prove that for any partial order $R$ on $A$ exists $S$ partial order on $A$ so that $R$ is a subset of $S$.
$S \ne R.$

I have tried to create a new relation S which is A Union of R with a pair $(a,b)$ that $(a,b),(b,a)$ not belong to R. $(a,b)\in $ A. but its become a little bit to complicated to make $S$ be a partial order.

thaks

答: If $A=\{0\}$ and $R=\{(0,0)\}$ you obviously can't construct a strictly bigger partial order on $A$.


问:
Boundedness of the first Fréchet derivative
Let $(E,\Vert\cdot\Vert)$ be a normed vector space and $f:E \rightarrow \mathbb{R}$ be a twice Fréchet differentiable function with $\sup_{x \in E} \frac{\vert f(x)\vert}{1+\Vert x\Vert^3} < \infty$ as well as
\begin{align}
\Vert D^2f(x+h) - D^2f(x) \Vert_L < C_1 \Vert h\Vert \qquad \forall x,h \in E
\end{align}
for some constant $C_1$. For a multilinear form $B$ on $E$ we used here the notation $\Vert B \Vert_L:=\sup_{\Vert x \Vert=1} \vert B[x,\ldots,x]\vert $.

Why does this imply
\begin{align}
\Vert Df(x)\Vert_L < C_2 (1+\Vert x\Vert^2) \qquad \forall x \in E
\end{align}
for some constant $C_2$?

Body:
Let $y\in E$ arbitrary. Then
$$
F(t)=Df(tx)(y), \quad t\in\mathbb R,
$$
is a real valued differentiable function and $F'(t)=D^2f(tx)(x,y)$. Mean Value theorem implies that, there exists a $t^*\in (0,1)$, such that
$$
Df(x)(y)-Df(0)(y)=F(1)-F(0)=F'(t^*)=D^2f(t^*x)(x,y)
$$
Hence
$$
\| Df(x)(y) \| =\|D^2f(t^*x)(x,y)+Df(0)(y)\|\le \|D^2f(t^*x)\|\|x\|\|y\|+\|Df(0)\|\|y\|
$$
and thus
$$
\| Df(x)\| \le \|D^2f(t^*x)\|\|x\|+\|Df(0)\| \\
\le \|D^2f(t^*x)-D^2f(0)\|\|x\|+\|D^2f(0)\|\|x\|+\|Df(0)\|
\\ \le C_1 \|t^*x\|\|x\|+\|D^2f(0)\|\|x\|+\|Df(0)\| \\ \le C_1\|x\|^2+\|D^2f(0)\|\|x\|+\|Df(0)\|
$$


问:
Discrete Math, Injective, Surjective
I am trying to understand a question I got:

Let $U$ be a set.

For every $A\subseteq U$ we define the function: $f_{A}:P(U) \to P(U)$ such as $f_{A}(B)=(A\cap B)$.

Prove: $f$ is injective and surjective if and only if $A=U$

First, I do not understand what $f_{A}$ and $f_{A}(B)$ mean.

I think that if I will understand this, I will be able to start solving the question.

Any tips will be wonderful!

Thanks!

答:
Let's take an example, maybe that helps.

Start with $U=\Bbb N$ and the subset $\{1,2,3,4,5\}$. Now we have a function $f_{\{1,2,3,4,5\}}$ that takes sets of natural numbers as input, and yields sets of natural numbers as output. It works by taking the input set, and intersecting it with $\{1,2,3,4,5\}$. So, for instance:
$$
f_{\{1,2,3,4,5\}}(\Bbb N)=\{1,2,3,4,5\}\\
f_{\{1,2,3,4,5\}}(\text{the odd natural numbers})=\{1,3,5\}\\
f_{\{1,2,3,4,5\}}(\text{the primes})=\{2,3,5\}\\
f_{\{1,2,3,4,5\}}(\text{the perfect squares})=\{1,4\}\\
f_{\{1,2,3,4,5\}}(\text{the two-digit numbers})=\varnothing
$$

Title: Discrete Math, Injective, Surjective


问:
Points in plane with every pair having at least two equidistant points?
I was given this question in person by a fellow trainee at the downtime of an IMO training session, which made me think this problem is Olympiad related. I am interested in the solution as much as the origin!

$\textbf{Problem:}$Does there exist a non-empty finite set of points $P$ in $\mathbb{R}^2$ such that for each pair of points $x$, $y$ in $P$ there exists at least two distinct points $a$ and $b$ in $P$ such that both $a$ and $b$ are equidistant from $x$ and $y$.

To be honest, in over a decade and getting a degree in Mathematics, I have not made much headway into this problem. I tried to do some counting based on the number of pairs and pigeonhole to get the conclusion that there must be a point that is the circumcenter of a triangle made up of points in $P$. Assuming $P$ is a minimal solution (in terms of size of $P$) does not seem to help that much.

EDIT: originally the question left open whether it was in the plane or space, but the space case is trivial.

EDIT EDIT: added a note to say the set should be non-empty. Note that we do not need at least two points, because when we say "pairs of points" we don't say "distinct pair of points"!

答:
Before I begin, a bit of philosophy on this problem:

I sincerely think there is no such set. I tried some onfigurations and I cannot seem to find one that works. So from now on, I will just talk about disproving the problem

This problem is clearly a geometric-combitoric style question. As far as I see, I don't think the geometric part of the problem is very important; I think only the simple notions of perpendicular bisectors and maybe polygons is enough. The important part is the combinatorical part and here are the main ideas that come in mind:

extremal principle

pigeonhole principle

double countings of any useful thing (angles, triangles, edges etc.)

considering a graph or the set of lines which pass through (at least) $2$ points or maybe the set of perpendicular bisectors of the set of segments

However, here is the big big issue:

We were never told that the points cannot be collinear or form parralel lines

As a bit of theory, we say that some points are in general position when no $3$ are collinear and no $2$ lines formed by points from our set are parralel. So we were never told that the set of points we are looking for is a general position! This means that we cannot use most of the ideas listed above as counterexamples are easy to find.

And yet is seems so easy.

I have tried myself various applications of the extremal principle and double countings but didnt actually manage to get a contradiction. However, I did manage to get a contradiction assuming that the points are in general position! So here is the solution:

Let $P$ be the set of points. Let $S$ be the set of lines that pass through exactly $2$ points of $P$. Observe that $|S|=\binom{n}{2}$ because the points are in general position.

Consider $S'$ the set of the perpendicular bisectors of the segments whose endpoints are in $P$. Because there are $\binom{n}{2}$ segments whose endpoints are in $P$ and the points are in general position, there are $\binom{n}{2}$ perpendicular bisectors, so $|S'|=\binom{n}{2}$.

BUT, for any segment $AB$ there are at least $2$ points $X,Y$ which both lie on the perpendicular bisector of the segment, thus every perpendicular bisector in $S'$ is actually one of the lines in $S$, so there exists an injection from $S'$ to $S$.

But we have shown that $|S|=|S'|$, so there must be a bijection from $S'$ to $S$, so every line that passes through $2$ points from $P$ is the perpendicular bisector of a segment whose endpoints are in $P$.

Take the convex hull of the points in $P$ and take any line formed by $2$ points on the convex hull. That line clearly cannot be the perpendicular bisector of a segment whose endpoints are in $P$

$\mathcal{Q.E.D}$

The fact that the points are in general position let us prove that there is an injection from $S'$ to $S$. If the points can actually be collinear or form parralel lines, then the above argument is false.

To conclude, there is no such set assuming the points are in general position. As for the general case, I am not sure. To be honest, the original statement might involve this detail, and Cryvate might have forgotten to add it, because in my experience, $99\%$ of geometric-combinatorical problems involve points in general position.

EDIT: $2$ people have pointed out in the comments that the question did not involve general position.

Yes, I know, I simply wanted to show a solution for the case in which the points are indeed in general position and highlight why the problem is much harder when they are not.

I discussed some approcahes and this was purely theoretical on te main issue, while providing the solution for the special case.

Moreover, as I said above, it is highly possible that if this was at an olympiad preparation camp, the original question actually involved general position. I am myself an olympiad student and for all I know, $99\%$ of these problems involve a general position, and when they don't they usually involve osmething else, like colorings or supplementary conditions.


答: I’m not sure what your definition of a partial order is, but this is false for any total order. Take $(\mathbb{R}, \leq)$ as an (counter-)example. Suppose there exists a partial order $R$ that strictly contains $\leq$. Then there is a pair of real numbers $(a,b)\in R$ that is not in $\leq$. This is only possible if $a>b$ and $(b,a)\in \phantom{}\leq$. Since $R$ contains $\leq$, it follows that $(b,a)\in R$ too. Recall that $(a,b)\in R$, therefore by anti-symmetry we have that $a=b$. But this contradicts our assumption that $(a,b)\notin \phantom{}\leq$, so no such pair $(a,b)$ exists. (Excuse the clumsy notation, but I feel this makes the proof easier to follow)

Title: How to prove that for any partial order R exists S partial order so that R is a subset of S. S not equals R.

Body:
How to prove that for any partial order $R$ on $A$ exists $S$ partial order on $A$ so that $R$ is a subset of $S$.
$S \ne R.$

I have tried to create a new relation S which is A Union of R with a pair $(a,b)$ that $(a,b),(b,a)$ not belong to R. $(a,b)\in $ A. but its become a little bit to complicated to make $S$ be a partial order.

thaks


问:
Show $\int_{-\infty}^\infty f^2 dt$ converges when $f$ is integrable.

A function $f:\Bbb R \to \Bbb R$ is integrable on $\Bbb R$ if $f$ is Riemann integrable on all intervals and the improper integrable $$\int_{-\infty}^\infty|f|dt$$ exists and is finite.
Show that if $f$ is bounded and integrable on $\Bbb R$, then $f^2$ is integrable on $\Bbb R$.

I have shown that $f^2$ is Riemann integrable on all intervals, but I have no idea how to show that $$\int_{-\infty}^\infty f^2 dt$$ converges.

Can I write $$\int_{-\infty}^\infty f^2 dt=\Bigl(\int_{-\infty}^\infty |f| dt\Bigr)\Bigl(\int_{-\infty}^\infty |f| dt\Bigr)$$?

Thanks for suggestion.

答:
There are very simple examples to show that $$\int_{-\infty}^\infty f^2 dt=\Bigl(\int_{-\infty}^\infty |f| dt\Bigr)\Bigl(\int_{-\infty}^\infty |f| dt\Bigr)$$ is not correct.

Answer for the old version of the question: This statement is false (unless you assume that $f$ is bounded on the whole real line). For example if $f(x)=n$ on $(n,n+\frac 1 {n^{3}})$ for $n=2,3,...$ and $0$ elsewhere then $f$ is integrable but $f^{2}$ is not.

Answer for the new version: Suppose $f$ is bounded, say $|f|\leq M$. Then $\int_x^{y} f^{2}(t) dt \leq M\int_x^{y}|f(t)| dt \to 0$ as $y>x \to \infty$. A similar result holds on the negative side. This implies that $\int_{0}^{y_n}f^{2}(x)dx $ is Cauchy sequence and hence it has a finite limit whenever  $y_n \to \infty$. Similarly for $\int_{x_n}^{0} f^{2}(t)dt$ when $x_n \to -\infty$. Hence $f^{2}$ is integrable.


问:
Prove: $\alpha \beta = 1 = \beta \alpha$
The following question is from an introduction to group theory  by Rotman:

For each $\alpha \in S_X$, prove that there is $\beta\in S_X$ with $\alpha \beta = 1 = \beta \alpha$ (Hint: let $\beta$ be the inverse function go the bijection $\alpha$)

Answer attempt: Let there be $\alpha \in S_X$ this is a permutation, this is a bijection, and for all bijection there is an inverse function in this case we will call it $\beta\in S_X$

Let $k\in X$ therefore:
$\alpha \beta(i_k)=\alpha(k)=i_k$

On the other hand:

$\beta \alpha(k)=\beta(i_k)=k$

But this does not show that $\alpha \beta = \beta \alpha$, looking in an example: if $\alpha(1)=5$ so $\beta(5)=1$ but

$$\alpha\beta(5) \neq \beta\alpha(1)$$

What am I missing?

答:
Of course
$$
5=\alpha\beta(5)\ne \beta\alpha(1)=1,
$$
since $1\ne 5$

However,
$$
1=\alpha\beta(1)= \beta\alpha(1)=1, \quad \text{and}\quad
5=\alpha\beta(5)= \beta\alpha(5)=5.
$$


问:
Controllability and eigenveectors
Let $\frac{dx}{dt}=Ax(t)+Bu(t)$ be a control system and suppose that A has two or more linearly independent eigenvectors associated to the same eigenvalue. Show that the system is not controlable.

My attempt: I thought about using the Kalman rank condition but couldn't get to any solid result

答:
No, The system is controllable iff the minimal polynomial of $A$ has degree $n$ and $B$ has a component in the direction of every eigenvector.

A counter-example is
$$
A=\begin{bmatrix}1&1\\0&1\end{bmatrix} ,B=\begin{bmatrix}1\\1\end{bmatrix}
$$
you can see that since $A$ is upper triangular matrix its eigenvalues are the diagonal elements which are $1$. Check the controllability matrix $R$
$$
R=[B\ \ |\ \ AB \ \ ]=\begin{bmatrix}1&2\\1&1\end{bmatrix}
$$
Which has a rank of 2 and the system is controllable.

Edit:
The controllablity condition is that the $Rank(R)=n=Rank(R^T)$ which means that the coulumn vectors of $R$ are linearly independent i.e.
$$
(\sum_{i=0}^{n-1} \alpha_i A^i) B=0 \implies \alpha_i=0 \ \ \forall i
$$
so if $A$ has a minimal polynomial of degree $n-1$ or less then you could find $\sum_{i=0}^{n-1} \alpha_i A^i=0$ with $\alpha_i \neq 0$ and so your system is uncontrollable.
If $B$ doesn't have an component in one of the eigenvectors without loss of generality let this eigenvector be $v_{n}$ then choose
$(\sum_{i=0}^{n-1} \alpha_i A^i) = \left(\Pi_{i=1}^{n-1}(A-\lambda_i I)\right)$ so
$$
(\sum_{i=0}^{n-1} \alpha_i A^i) B=\left(\Pi_{i=1}^{n-1}(A-\lambda_i I)\right).\sum_{k=1}^{n-1}a_k v_k=\sum_{k=1}^{n-1} a_k \left(\Pi_{i=1,i\neq k}^{n-1}(A-\lambda_i I)\right).(A-\lambda_k I) v_k=0
$$
since $(A-\lambda_k I) v_k=0 \ \ \forall k$.

Title: Controllability and eigenveectors


问:
How to construct a function from a pair of possibly empty sets?
I am stuck on an elementary proof on the cardinality of sets on the following point:

Given two possibly empty sets, $A$ and $B$, I need to prove the existence of any function $f:A\rightarrow B$. Is it possible? Perhaps using AC? I'm thinking you must have at least a non-empty $B$ (all elements of $A$ mapped to the same point). Any help would be appreciated.

答:
With the issue of possibly empty sets $A, B$, you might want to read up a bit on the existence of the Empty function.

In mathematics, an empty function is a function whose domain is the empty set. For each set A, there is exactly one such empty function

$$f_A: \varnothing \rightarrow A.$$

The graph of an empty function is a subset of the Cartesian product ∅ × A. Since the product is empty the only such subset is the empty set ∅. The empty subset is a valid graph since for every x in the domain ∅ there is a unique y in the codomain A such that (x,y) ∈ ∅ × A. This statement is an example of a vacuous truth since there is no x in the domain.

For perhaps a better explanation than that provided by Wikipedia, read this earlier post:

Why is the empty function considered a
function?.

See also: Does there exist a bijection between empty
sets?


问:
Find a numerical solution for an ODE with an initial condition which is not at the beginning
This may sound really weird...

Consider the following Cauchy problem:

Find $y(t)$ for all $t \geq 0$ such that:
$$\begin{cases} \dot{y} & = f(t, y)\\ y(T) &= Y \end{cases},$$
for some $T > 0$.

I know that, under suitable conditions, this Cauchy problem has exactly one solution.

Anyway, all numerical tools that I know are meant (or I believe so!) to solve problems like:

Find $y(t)$ for all $t \geq 0$ such that:
$$\begin{cases} \dot{y} & = f(t, y)\\ y(0) &= Y \end{cases},$$

How can I solve the first problem?

答:
You can follow the suggestion by @LutzLehmann or you can just perform a change of variables, $\tau = T - t$ and get something like
$$
\begin{cases}y'(\tau) = - f(\tau, y(\tau)) \\[0.7em] y(0)= Y \end{cases}.
$$


问:
Why can I not make this function equal zero and solve for x when finding y-intercept of a rational fraction w/ numerator (2x^2+9)
When graphing a rational fraction, $\frac{2x^2+9}{x}$, and I have to find the $y$-intercept, why can't I do this? Prof. Leonard from youtube says that if discriminant is negative, then there is no real answer, but why do I get a value for $x$ when I do this?

$$2x^2+9 = 0 \to 2x^2= -9 \to x^2 = \frac{9}{2} \to x = \pm\frac{3}{\sqrt{2}}$$

$$x = \frac{-3}{\sqrt2}, x=\frac{3}{\sqrt2}$$

What am I doing wrong?

答:
What am I doing wrong?

You mistakenly dropped a minus sign.

It would be $x^2=\color{red}-\dfrac92$, which has no real solutions.


答: You wrote $x^2 = \frac{9}{2}$. But this is false.  It should read $x^2 = -\frac{9}{2}$.

Title: Why can I not make this function equal zero and solve for x when finding y-intercept of a rational fraction w/ numerator (2x^2+9)

Body:
When graphing a rational fraction, $\frac{2x^2+9}{x}$, and I have to find the $y$-intercept, why can't I do this? Prof. Leonard from youtube says that if discriminant is negative, then there is no real answer, but why do I get a value for $x$ when I do this?

$$2x^2+9 = 0 \to 2x^2= -9 \to x^2 = \frac{9}{2} \to x = \pm\frac{3}{\sqrt{2}}$$

$$x = \frac{-3}{\sqrt2}, x=\frac{3}{\sqrt2}$$

What am I doing wrong?


问:
How do I write this in spherical coordinates?
$$z=\sqrt{2x^2+2y^2}$$
I know that this is $\tan\phi=\frac{1}{\sqrt{2}}$ but the arctan of that number is 0.6154797, so I'm confused with what to put as the spherical coordinate.

答:
Spherical co-ordinates are described by $$x=r\sin\theta\cos\phi$$ $$y=r\sin\theta\sin\phi$$ $$z=r\cos\theta$$
You have:$$2(x^2+y^2)=z^2\implies2(r^2\sin^2\theta(\cos^2\phi+\sin^2\phi))=r^2\cos^2\theta$$

$$\implies2r^2\sin^2\theta=r^2\cos^2\theta\implies\sin^2\theta=\frac13,\cos^2\theta=\frac23\implies\tan^2\theta=\frac12$$ as you got.


问:
Convergence's analysis Newton method on special function $f(x) = x/(x^{2} + 1)$
So I am trying to study the convergence analysis of this function $f(x) = \frac{x}{x^{2} + 1}$ with Newton-Ralphson method. I expect the root to be $x =0$. However, when applying Newton numerically (I use matlab) the method actually does not converge. The error for each iteration is quite large. I checked $f'(x)$ and $f''(x)$. These functions are continuously differentiable in the domain, and also $f'(x) \neq 0$ for an interval that containing $x = 0$. Is there something that contradicts the convergence property of Newton's method? How can I fix this?

答:
As @Professor Vector commented, you must start with $|x_0]< \frac 1 {\sqrt 3}$.

Suppose that we choose
$$x_0^{(n)}=\frac 1 {\sqrt 3}-10^{-n}$$ and that we count the number of required iterations to have $|x|<10^{-6}$. Here are the results
$$\left(
\begin{array}{cc}
n & \text{number of iterations} \\
 1 & 4 \\
 2 & 6 \\
 3 & 7 \\
 4 & 9 \\
 5 & 11 \\
 6 & 12 \\
 7 & 14 \\
 8 & 16 \\
 9 & 17 \\
 10 & 19
\end{array}
\right)$$


问:
Erdős-Rényi model, network reliability
I recently came across the Erdős-Rényi model random graph model, which indeed fascinated me with its beauty and simplicity. I've started to dig into it and saw the following theorem (listed on Wiki, for instance):

Let $G(n, p)$ be a random graph containing $n$ vertices, then the following holds, for $p=\cfrac{c\cdot \log{n}}{n}$:

if $c > 1$, then graph will asymptotically almost surely be connected;

if $c < 1$, then graph will asymptotically almost surely have some isolated vertices and asymptotically almost surely be disconnected

The question is, what can we say about $c=1$? Is there anything interesting to say about this parameter value?

I would appreciate any response.

答:
When
$$
p=\frac{\ln n}{n}+\frac{s+o(1)}{n}
$$
for some fixed $s\in \mathbb{R}$,
$$
\mathsf{P}(G(n,p)\text{ is connected})\to e^{-\exp(-s)}.
$$
(see, e.g., Theorem 7.3 here).

Title: Erdős-Rényi model, network reliability


问:
How to prove that $1/n!$ is less than $1/n^2$?
I want to prove
$$\sum_{n=0}^{\infty} \frac{1}{n!}$$ is a converging series. So I want to compare it with $\sum_{n=0}^{\infty} \frac{1}{n^2}$. I want to do direct comparison test.

How to prove $n^2 < n! $ ?

答:
Try the ratio test:

$$
a_n = \frac{1}{n!} \quad \implies \quad \left| \frac{a_{n+1}}{a_n} \right| = \frac{n!}{(n+1)!} = \frac{1}{n+1}
$$
so that
$$
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1,
$$
and thus the series converges absolutely.  Later in your studying series, you may see that
$$
\sum_{n =0}^{\infty} \frac{x^n}{n!} = e^x
$$
which implies
$$
\sum_{n =1}^{\infty} \frac{1}{n!} = \sum_{n = 0}^{\infty} \frac{1^n}{n!} - \frac{1^0}{0!} = e - 1.
$$


问:
Finding an example the fundamental theorem of calculus fails when $f$ is not continuous.
Is the following statement true or not?

If $f$ is Riemann integrable on $[a,b]$, then the function $F(x)=\int_a^xf(t)dt$ is differentiable on $(a,b)$ and $F'(x)=f(x)$ for $x\in (a,b)$

I think it is not. Since $F$ is differentiable at the point where $f$ is continuous, but I can not think a counter example, is anyone has a counterexample? Thanks a lot.

答:
Start with an arbitrary continuous function $f$ on $[a, b] $ (eg $f(x) =x$). Take any point $c\in(a, b) $ and define a function $g$ on $[a, b] $ such that $g(x) =f(x) $ for all $x\neq c$ and define $g(c)$ to be any number different from $f(c)$ (eg $g(c) =f(c) +1$).

Then $g$ is Riemann integrable on $[a, b] $ and moreover $$F(x) =\int_a^x f(t) \, dt=\int_a^x g(t) \, dt\, \forall x\in[a, b] $$ and by continuity of $f$ we have $F'(c) =f(c) \neq g(c) $. The function $g$ now serves as a simple and obvious counterexample you seek.

The above example gives you a case when $F$ remains differentiable at $c$ but it's derivative at $c$ does not match the value of integrand $g$ at $c$.

Other answers have tried to construct an example where $F$ is not differentiable at $c$. For such examples it is sufficient to consider integrands which have jump discontinuity at $c$ and then it will be guaranteed that the integral function $F$ will not be differentiable at $c$.


问:
Integral of Heaviside($R^2-x^2-y^2$)
I have this integral (where $R$ is a positive constant) $$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} H(R^2-x^2-y^2)dxdy$$
and I'm pretty lost when trying to calculate it.

I don't know if it's smart to do so, but I tried to convert it to polar coordinates and got
$$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}H(R^2-r^2)rdrd\theta$$ By definition the Heavside function should be one everywhere, where $r<R$ so i thought about writing the following
$$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{R}rdrd\theta$$
But now I don't know what to do next or if this has been correct so far. Cause the result of this would be just infinity I guess...
I would be really grateful if someone could give me a little help or a hint on how to approach this. Thanks!

答: When you pass to the polar coordinates, you have to change the limits appropriately: when $x$ and $y$ run from $-\infty$ to $\infty$, the radius $r$ runs from $0$ to $\infty$ and the polar angle $\theta$ runs from $0$ to $2\pi$ (or from $-\pi$ to $\pi$, say). Adjusting your calculations accordingly will allow you to get the correct answer, which will probably look somewhat familiar.

Title: Integral of Heaviside($R^2-x^2-y^2$)


答:
Thanks Arthur for your kind reply!

So, in the above question, $f$ takes inputs from $B$, intersect with $A$, and the output must be $(A\cap B)$?

Assuming (negative) $f$ is injective and surjective and $A\neq B$.

$A\subseteq U$ so $A\in P(U)$, and also $U\in P(U)$.

So $f_{A}(A)=(A\cap A)=A$ and also $f_{A}(U)=(A\cap U)=U$,

Am I going the right way?

Title: Discrete Math, Injective, Surjective

Body:
I am trying to understand a question I got:

Let $U$ be a set.

For every $A\subseteq U$ we define the function: $f_{A}:P(U) \to P(U)$ such as $f_{A}(B)=(A\cap B)$.

Prove: $f$ is injective and surjective if and only if $A=U$

First, I do not understand what $f_{A}$ and $f_{A}(B)$ mean.

I think that if I will understand this, I will be able to start solving the question.

Any tips will be wonderful!

Thanks!


问:
Determine the range of $\alpha \in \mathbb{R}$ of a for a weighted convergent sequence.
Determine the range of $\alpha \in \mathbb{R}$ for which the following statement holds. If a sequence $\{ x_n \}_{n=1}^{\infty}$ of real numbers satisties $\lim_{n \to \infty}{ n^\alpha \,\, |x_n - x_{n+1}|}=0$, Then it is convergent sequence.

答:
When $\alpha < 1$ consider the sequence $x_n = \ln(n)$. Then
\begin{align*}
\lim_{n \rightarrow \infty }n^{\alpha}|x_{n+1}-x_n| &= \lim_{n \rightarrow \infty} n^\alpha \ln(1+\frac{1}{n})\\ &= \lim_{n \rightarrow \infty}\frac{1}{n^{1-\alpha}} n\ln(1+\frac{1}{n})=0
\end{align*} but $(x_n)$ diverges.

When $\alpha > 1$ we will show convergence by cauchy criterion which is for a given $\epsilon >0$,there exist a $n_0 \in \mathbb{N}$ such that $\lvert x_m - x_n\rvert < \epsilon$ for all $m,n \geq n_0$. Wlog assume $m>n$. Choose $n_0$ such that for the given $\epsilon$, $n^\alpha |x_{n+1}-x_n|<\frac{\epsilon}{c_0}$ for all $n>n_0 \geq 1$ and where $c_0 = \sum_{k=1}^\infty \frac{1}{k^\alpha}$ (It is well known that this is a finite number whenever $\alpha >1$). Applying triangle inequality to $\lvert x_m - x_n \rvert$ we get,
\begin{align*}
\lvert x_m - x_n\rvert &\leq \lvert x_m - x_{m-1}\rvert+\cdots +\lvert x_{n+1}-x_n\rvert\\ &=\frac{1}{(m-1)^\alpha} (m-1)^\alpha|x_m-x_{m-1}|+\cdots+\frac{1}{n^\alpha}n^\alpha|x_{n+1}-x_n|
\\ &< \biggl(\sum_{k=n}^{m-1}\frac{1}{k^\alpha}\biggr)\frac{\epsilon}{c_0} <\biggl(\sum_{k=1}^{\infty}\frac{1}{k^\alpha}\biggr)\frac{\epsilon}{c_0}=c_0\frac{\epsilon}{c_0}=\epsilon
\end{align*}

$\mathbf{Edit}$: I forgot the case $\alpha = 1$. When $\alpha = 1 $ consider the sequence $x_n = \sin(1/n)$. It is well known that $(x_n)$ oscillates rapidly as $n \rightarrow \infty$. Hence the limit does not exist. But,
\begin{align*}
\lim_{n \rightarrow \infty}n|x_{n+1}-x_n|&=\lim_{n \rightarrow \infty} n\biggl|\sin\biggl(\frac{1}{n+1}\biggr)-\sin\biggl(\frac{1}{n}\biggr)\biggr|\\&=\lim_{n \rightarrow \infty} n \biggl| 2\cos\biggl(\frac{2n+1}{2n(n+1)}\biggr)\sin\biggl(\frac{-1}{2n(n+1}\biggr)\biggr| \\ &=\lim_{n \rightarrow \infty} \frac{2(n+1)}{2(n+1)}n \biggl| 2\cos\biggl(\frac{2n+1}{2n(n+1)}\biggr)\sin\biggl(\frac{-1}{2n(n+1)}\biggr)\biggr|\\&=\lim_{n \rightarrow \infty} \frac{1}{2(n+1)}\biggl| 2\cos\biggl( \frac{2+1/n}{2(n+1)}\biggr) \frac{\sin\biggl(\frac{-1}{2n(n+1)} \biggr)}{\frac{1}{2n(n+1)}} \biggr| = \lim_{n\rightarrow \infty}\frac{2}{2(n+1)}=0
\end{align*}
because $\lim_{n \rightarrow \infty}\cos(\frac{2+1/n}{2(n+1)}) =1$, $\lim_{n \rightarrow \infty} \frac{\sin(1/n)}{1/n} = 1 $

Hence $(x_n)_{n=1}^{\infty}$ is convergent whenever $\alpha >1$


答:
The function you are integrating is zero outside the circle centered in $(0,0)$ with radius $R$ and is one inside that circle. Therefore, the integral corresponds to the area of the circle: $\pi R^2$. No need for polar coordinates. But if you really want to use them,

$$
\iint_{\mathbb{R}^2} H(R^2-x^2-y^2) dxdy = \int_{-R}^R \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} dxdy = \int_{0}^{2\pi} \int_0^R \rho d\rho d\theta = 2 \pi [\frac 12 \rho^2]_0^R = \pi R^2.
$$

Title: Integral of Heaviside($R^2-x^2-y^2$)

Body:
I have this integral (where $R$ is a positive constant) $$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} H(R^2-x^2-y^2)dxdy$$
and I'm pretty lost when trying to calculate it.

I don't know if it's smart to do so, but I tried to convert it to polar coordinates and got
$$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}H(R^2-r^2)rdrd\theta$$ By definition the Heavside function should be one everywhere, where $r<R$ so i thought about writing the following
$$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{R}rdrd\theta$$
But now I don't know what to do next or if this has been correct so far. Cause the result of this would be just infinity I guess...
I would be really grateful if someone could give me a little help or a hint on how to approach this. Thanks!


问:
Conjectural closed-form representations of sums, products or integrals
What are some examples of infinite sums, products or definite integrals that have conjectural closed form representations that were confirmed by numerical calculations up to whatever maximum precision anybody tried but still remain unproved?

I am also interested in values of special functions at certain points that have conjectural representations in terms of simpler functions (e.g. special values of hypergeometric functions, Meijer G-function or Fox H-function that representable in terms of elementary functions and well-known constants like $\pi$, $e$, Catalan, Euler–Mascheroni, Glaisher–Kinkelin or Khinchin).

To give some examples:

Gourevitch conjecture mentioned at MathOverflow:
$$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7\stackrel{?}{=}\frac{32}{\pi^3}.$$

Riemann hypothesis (in an unusual form, also mentioned at MathOverflow)
$$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt\stackrel{?}{=}\frac{\pi(3-\gamma)}{32}.$$

Another conjecture from MathOverflow attributed to J. M. Borwein, D. H. Bailey, Mathematics by Experiment: Plausible Reasoning in the 21st Century:
$$\frac{\displaystyle\int_{\pi/3}^{\pi/2}\log\left|\frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}\right|dt}{\displaystyle\sum_{n=1}^\infty\left(\frac n7\right)\frac{1}{n^2}}\stackrel{?}{=}\frac{7\sqrt{7}}{24},$$
where $(\frac{n}{7})$ denotes the Legendre symbol.

答:
Many such conjectures can be seen in Borwein-Crandall's recent text, Closed Forms: What They Are and Why We Care.  One cited case is the identity
$$F(3,5)= \frac{15}{\pi^2} \sum_{n=0}^\infty \binom{2n}{n}^2 \frac{(1/16)^{2n+1}}{2n+1},$$
in which $F(x,y)$ denotes the $4$-dimensional lattice sum:
$$F(x,y)=(1+x)(1+y)\sum_{a,b,c,d} \frac{(-1)^{a+b+c+d}}{\left((6a+1)^2+x(6b+1)^2+y(6c+1)^2+xy(6d+1)^2\right)^2}.$$
Both expressions have been numerically estimated; each is approximately $0.0628326$.

A second example of historical interest is the well-known Wallis product:
$$\frac{\pi}{2} =\frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5}\cdot \frac{6}{7}\cdot\frac{8}{7}\cdots \frac{2n}{2n-1}\cdot \frac{2n}{2n+1}\cdots$$
Wallis conjectured the given value $\pi/2$ after lengthy calculation in 1655.  Methods did not exist to derive such a result at least until the days of Euler (about 80 years later).

Wallis' product is notable in its relation to Stirling's approximation (i.e. completing the transition between the weak and strong Stirling approximations), and for its role in determining $$\zeta'(0)=-\log(2\pi)/2.$$


答: $$I=\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} H(R^2-x^2-y^2)dx dy =\int_{0}^{2\pi} \int_{0}^{R}H(R^2-r^2) r dr d\theta=\int_{0}^{2\pi} \int_{0}^{R} r dr d\theta= \pi R^2$$

Title: Integral of Heaviside($R^2-x^2-y^2$)


问:
Exists a non-empty subset $K$ of $2005$ senators in senat such that for every senator, the number of his enemies in $K$ is an even number.

There are $2005$ senators in a senate. Each senator has enemies within the senate. Prove that there exists a non-empty subset $K$ of senators such that for every senator in the senate, the number of his enemies in $K$ is an even number.

Let $n=2005$ and let $s_i$ be an indicator vector for $i$-th senator. So we are working in $\mathbb{F}_2^n$ and we are looking for such a vector $u$ that $s_i\cdot u =0$ for all $i$.

Now we can make a matrix $M = [s_1,...,s_n]$ which is symmetric and thus it has $\det M =0$ since $n$ is odd. So it has nontrivial kernel so there exists $u\ne 0$ such that $Mu=0$. Clearly $u$ is a vector we are looking for and thus we are done.

Now, I wonder if it can be done without matrix theory, only in terms of linear independence and ''similar things''from linear algebra. My idea was to put $S := \langle s_1,s_2,...,s_n \rangle$ and prove that $S^{\bot} $ is nontrivial, which would be true if $s_1,s_2,...,s_n$ are lineary dependent.

All I can think is that by handshake lemma and since $n$ is odd, at least one senator has even number of enemies. Suppose it is $s_1$, so if $s_1\cdot s_j =0$ for all $j>1$ we are done. But what if that is not the case.

答:
Here is a possibility to "not use matrices, but linear algebra". (Bilinear forms, and use "symplectic" ideas.)

Let $n$ be an odd positive integer. Let $F$ be the field with two elements and $V=F^n$ the canonical vector space of dimension $n$ over $F$. (Elements are identified with column vectors with $n$ components in $F$.)

Let $e_1, e_2,\dots,e_n$ be the canonical basis of $V$.
Let $e_I$ be $\displaystyle e_I=\sum_{i\in I}e_i$ for some subset $I$ of the index set $\{1,2,\dots,n\}$. Each element of $V$ is of the shape $e_I$ for a suitable $I$.

Let $B$ be the symmetric bilinear form so that
$B(e_i,e_j)$ is the $j$'th component in $s_i$ and/or the $i$'th component in $s_j$.

(Comment: The associated matrix $S$ for this bilinear form is the matrix with columns $s_1,\dots,s_n$, and $B(v,w)=v'Sw$. Note that it is essential that no senator feels to be its own enemy, translated by the fact that the diagonal entries in $S$ are all zero. The OP should have mentioned this in the row that claims $\det S=0$. This comment is not part of the solution, we will tacitly use $B(e_i, e_i)=0$, equivalent to $S_{ii}$=0, a relation which is valid for all $i$.)

Assume that this bilinear form is non-degenerated.
(We try to develop a contradiction.)

This form is not only skew-symmetric (since it is symmetric, and we are in characteristic two), but also alternating:
$$
B(e_I,e_I)
=\sum_{i,j\in I}B(e_i, e_j)
=\sum_{i\in I}B(e_i, e_i)
+2
\sum_{\substack{i,j\in I\\i<j}}B(e_i, e_j)
=0\ .
$$
An alternating $B$ can live only in even dimension. Contradiction.

So $B$ has non-trivial kernel.
(From the degeneration, we obtain a non-zero $v\in V$ in the kernel of the canonical map $V\to V^*$ induced by $B$, so $w\to B(w,v)=w'Sv$ is the zero map in $V^*$, so $Sv$ is the zero vector.)

$\square$

Here is the argument involved (induction with step two on $n=\dim V$) for the even dimension above. ("Not using matrices", it is the usual argument, mentioned explicitly to show that the idea can be accepted in the thematic of the OP, passing to the orthogonal space is used for instance.)

Let $B$ be alternating, non-degenerated on $V$. If $\dim V=0$ there is nothing to show. If $\dim V=1$ then it has only one non-zero vector $v\in V\cong\Bbb F_2$, and from $B(v,v)=0$ we obtain $B=0$, contradiction. So the needed property holds for all vector spaces with dimension $<2$. Let $V$ be a space of dimension $\ge 2$ now. There is some $u\in V$, $u\ne 0$. Since $B(u,u)=0$, but $B(u,\cdot)\not \equiv 0$, there is some $v\in V$ with $B(u,v)=1$.
Consider now the subspace $U=\langle u,v\rangle$, and its orthogonal complement w.r.t. $B$,
$$
\tag{$\dagger$}
W=U^\perp\ .
$$
Then standard arguments show that $V=U\oplus W$, and consider for the inductive step the restiction of $B$ to $W$, again alternating and non-degenerated.

(Explicitly, and without matrices: Since $B$ is non-degenerated, $U\cap W=U\cap U^\perp=0$, since none of the only non-zero elements of $U$, enumerated as $u$, $v$, $u+v$, lies in $U^\perp$. We further need the property $U+U^\perp=V$. Consider for this the map into the dual $U\to U^*$, $y\to B(y,\cdot)\Big|_U$, given explicitly by
$0\to 0$, $u\to B(u,\cdot)\Big|_U$, $v\to B(v,\cdot)\Big|_U$, $u+v\to B(u+v,\cdot)\Big|_U$. It is injective, thus surjective, so an isomorphism of linear spaces of dimension two.

Then for each $x\in V$ the form $B(x,\cdot)\Big|_U$ can be realized with an element $y\in U$. So $B(x-y,y')=0$ for all $y'\in U$. This means $x-y\in U^\perp = W$, realizing $x\in y+W\subset U+W$.)

Note: A matrix $M$ as the given one is exactly a matrix for an alternating bilinear form.


问:
Adding two points on a elliptic curve over $\operatorname{GF}(16)$ constructed using $p(x) = x^4+x+1$
E is the elliptic curve over P(GF($16$)) satisfying

$$y^2 + xy = x^3+β^4x^2+1$$.
$$\text{Let} \quad β= x\pmod {p(x)}.$$

I know that $(β^5,β^3)$ is a point on the curve. I want to find $13(β^5,β^3)$.

I am using the minimal addition chain for $13 (1,2,4,8,9,13)$ to find this.

But how do I add $(β^5,β^3)$ to itself to find $2(β^5,β^3)$?

I do not understand how to add two points on elliptic curve.

答:
You can avoid adding points by making use of some coincidences.

If you replace $y$ by $y+\gamma x$, then the equation remains the same, except that $\beta^4$
is replaced by $\beta^4+\gamma+\gamma^2$. Since the trace of $\beta$ and hence of $\beta^4$ is
zero, the equation $T^2+T=\beta^4$ has solutions in ${\bf F}_{16}$. Indeed, dividing the equation $\beta^4=\beta+1$ by $\beta^2$ and squaring gives $\beta^4 = 1/\beta^2 +1/\beta^4$.

Applying the change of variable $y\leftarrow y+x/\beta^2$, the curve becomes $y^2+xy=x^3+1$ and the point becomes $P=(\beta^5,0)$.  So the curve is defined over ${\bf  F}_2$!  It is easy to see that it has $4$ points over ${\bf F}_2$, so that  the Frobenius endomorphism $\phi$ of the curve 'is'  $(-1+\sqrt{-7})/2$.

Since the $x$=coordinate $\beta^5=\beta\cdot\beta^4$ is a norm to the subfield ${\bf F}_4$, our point $P$ is defined over ${\bf F}_4$. Computing the trace of $\phi^2$, we find that the curve has $8$ points over ${\bf F}_4$. Since we are in characteristic $2$, the group $G$ of points  over ${\bf F}_4$ is cyclic. Therefore $\phi$ acts on $G$ via multiplication by an element in $a\in ({\bf Z}/8{\bf Z})^*$. Since $\phi$ fixes the four ${\bf F}_2$-points, we have $a=5$.

Since $P$ is in $G$, we have  $13P=5P=\phi(P)=(\beta^{10},0)$. Transforming this back into the original coordinates, we get  $(\beta^{10},\beta^8)$. Since $\beta^8=\beta^2+1$, this is easily seen to be equal to be the point $(\beta^2+\beta+1,\beta^2+1)$, in agreement with the computation by dan_fulea