问:
Direct cut-elimination for natural deduction.
Just reading:

"Gentzen was motivated by a desire to establish the consistency of number theory. He was unable to prove the main result required for the consistency result, the cut elimination theorem - the Hauptsatz - directly for Natural Deduction. For this reason he introduced his alternative system, the sequent calculus for which he proves the Hauptsatz both for classical and intuitionistic logic."
http://en.wikipedia.org/wiki/Natural_deduction

Any direct proofs of cut-elimination for natural deduction

around? Whats their underlying substantial idea exactly?

Best Regards

答:
I guess cut elimination from natural deduction is rather the rule
than the exception. If I remember well Gentzens paper itself goes
from Hilbert style Hx, to Natural Deduction Nx and then to
sequent calculus Lx.

The first step requires to convert each Hilbert style axiom to
be converted into a Natural Deduction derivation, using already sequents. Here are some examples, from minimal logic implication fragment only:

Hilbert Style axiom: $A \rightarrow A$

Natural Deduction derivation: $\begin{array}{c} A \vdash A \\
            \hline
            \vdash A \rightarrow A
            \end{array}$

Hilbert Style axiom: $A \rightarrow (B \rightarrow A)$

Natural Deduction derivation: $\begin{array}{c} A \vdash A \\
            \hline
            A \vdash B \rightarrow A \\
            \hline
            \vdash A \rightarrow (B \rightarrow A)
            \end{array}$

Hilbert Style axiom: $(A \rightarrow (B \rightarrow C)) \rightarrow (A \rightarrow B) \rightarrow A \rightarrow C$

Natural Deduction derivation: $\begin{array}{c} \begin{array}{c} \begin{array}{c}
\hline
A \rightarrow (B \rightarrow C) \vdash A \rightarrow (B \rightarrow C) \\
\end{array}
\begin{array}{c}
\hline
A \vdash A \\
\end{array} \\
\hline
A \rightarrow (B \rightarrow C), A \vdash B \rightarrow C
\end{array}
\begin{array}{c} \begin{array}{c}
\hline
A \rightarrow B \vdash A \rightarrow B \\
\end{array}
\begin{array}{c}
\hline
A \vdash A \\
\end{array} \\
\hline
A \rightarrow B, A \vdash B
\end{array}
 \\
            \hline
            A \rightarrow (B \rightarrow C), A \rightarrow B, A \vdash C \\
            \hline
            A \rightarrow (B \rightarrow C), A \rightarrow B \vdash A \rightarrow C \\
             \hline
            A \rightarrow (B \rightarrow C) \vdash (A \rightarrow B) \rightarrow A \rightarrow C \\
            \hline
            \vdash (A \rightarrow (B \rightarrow C)) \rightarrow (A \rightarrow B) \rightarrow A \rightarrow C
           \end{array}$

Since Natural Deduction has also modus ponens, like Hilbert style, we are already done with the transformation from Hx to Nx. The cut elimination can now begin, taking the Nx derivation, converting it to a Lx derivation.

On can put this transformation on the computer, see here.


问:
Baire's theorem from a point of view of measure theory
According to Baire's theorem, for each countable collection of open dense subsets of $[0,1]$, their intersection $A$ is dense. Are we able to say something about the Lebegue's measure of $A$? Must it be positive? Of full measure?

Thank you for help.

答: Let $q_1,q_2,\ldots$ be an enumeration of the rationals. Let $I_n^m$ be an open interval centered at $q_n$ with length at most $1/m 1/2^n$. Then $\bigcup_n I_n^m$ is an open dense set with Lebesgue measure at most $1/m$. The intersection of these open dense sets has Lebesgue measure zero.</p>

ContentLicense: CC BY-SA 3.0


问:
Analyzing the Coefficient of a generating function and its asymptotic
The generating function
$\sum_{n\geq 0} D(n) x^n = \frac{1}{\sqrt{1-6x+x^2}}$ is the gf of the Delannoy number.

See last paragraph in https://en.wikipedia.org/wiki/Delannoy_number

In this link, they mentioned that the coefficient of this gf  is $D(n) = \sum_{k=0}^{n} {n \choose k} { n+k \choose k}$.

My question is how can this coefficient be calculated/extracted from its gf.And how they conclude that it behaves asymptotically as

${\displaystyle D(n)={\frac {c\,\alpha ^{n}}{\sqrt {n}}}\,(1+O(n^{-1}))}$

where
$\alpha =3+2{\sqrt {2}}\approx 5.828$

and

$c=(4\pi (3{\sqrt {2}}-4))^{-1/2}\approx 0.5727$.

How to calculate/extract $D(n)$ from its gf and how to analyze it asymptotically then?
Is that possible to do that with a direct calculations?

Edit
I tried to find the asymptotic of the codfficidnt in the above gf this way:

The gf has two poles: $r_1=3+\sqrt{8}$, $r_2=3-\sqrt{8}$. So the smallest/dominant pole is $r_2=3-\sqrt{8}$.
We can write the gf as:
$A(x)=1/ \sqrt{r_2-x} * 1/ \sqrt{r_1-x}$
Then A(x) behaves as:

$1/ \sqrt {r_1-r_2} * 1/ \sqrt{r_2-x}=
1/ \sqrt {2 * 8^{1/4}} * \sum_{n\geq 0} {-1/2 \choose n} (-1)^n (r_1)^{n+1}x^n$
Here I used the expansion's formula  of $(1+x)^a$ for $a=-1/2$ and $x=-x/r_2$.
So the coefficient behaves asymptotically as:
$1/ \sqrt 2 * 1/ 8^{1/4} * (3+\sqrt 8)^{n+1/2} {-1/2 \choose n} (-1)^n$.

We can simplify it more, by using the asymptotic of ${-1/2 \choose n}$.

Is that approach fine?

答:
For the asymptotics we use the Wilf text, Theorem 5.3.1 (page 179)
(Darboux) as  suggested in the comments. We seek the asymptotics of

$$[z^n] \frac{1}{\sqrt{1-6z+z^2}}
= [z^n] \frac{1}{\sqrt{(z-(3+2\sqrt{2}))(z-(3-2\sqrt{2}))}}
\\ = \frac{1}{(3-2\sqrt{2})^n} (3-2\sqrt{2})^n
[z^n] \frac{1}{\sqrt{(z-(3+2\sqrt{2}))(z-(3-2\sqrt{2}))}}
\\ = (3+2\sqrt{2})^n
[z^n] \frac{1}{\sqrt{((3-2\sqrt{2})z-(3+2\sqrt{2}))
((3-2\sqrt{2})z-(3-2\sqrt{2}))}}
\\ = (3+2\sqrt{2})^n
[z^n] \frac{1}{\sqrt{((17-12\sqrt{2})z-1)
(z-1)}}
\\ = (3+2\sqrt{2})^n
[z^n] \frac{1}{\sqrt{(1-(17-12\sqrt{2})z)
(1-z)}}.$$

Now here we have one as the dominant singularity on the circle of
convergence and the theorem  applies, taking the parameter $\beta =
-1/2.$ Expanding the term  containing the subdominant  singularity around
one we get for the first (constant) term

$$\frac{1}{\sqrt{1-(17-12\sqrt{2})\times 1}}
= \frac{1}{\sqrt{12\sqrt{2}-16}}
= \frac{1}{2\sqrt{3\sqrt{2}-4}}.$$

This gives the asymptotic

$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2\sqrt{3\sqrt{2}-4}}
(3+2\sqrt{2})^n
{n-1/2\choose n}.}$$

The Wilf text gives $O(n^{-3/2})$ for the error in this approximation,
in sync with Wikipedia.

Wilf quotes on the same page an asymptotic for the remaining binomial
coefficient namely

$${n-\alpha-1\choose n} \sim \frac{n^{-\alpha-1}}{\Gamma(-\alpha)}$$

with $\alpha$ not a nonnegative integer. In the present case we have
$\alpha = -1/2$ so we obtain

$$\frac{1}{\sqrt{n}} \frac{1}{\Gamma(1/2)} =
\frac{1}{\sqrt{n}} \frac{1}{\sqrt{\pi}}.$$

This gives the form from the Wikipedia entry which is

$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2\sqrt{\pi(3\sqrt{2}-4)}}
\frac{1}{\sqrt{n}}
(3+2\sqrt{2})^n.}$$

ContentLicense: CC BY-SA 4.0


问:
Showing that the maximum value of $\sin x+\sin y\sin z$, where $x+y+z=\pi$, is the golden ratio

Find the maximum of
$$\sin x+\sin y\sin z$$
if  $x+y+z=\pi$.

By using Lagrange multipliers, concluded that $y=z$, further plug $x=\pi-2y$, I've reduced the problem to single-variable expression.
$$\sin^2 y+\sin(2y)$$
Then by taking first derivative and using formulas for $\sin(\arctan x)$ and $\cos(\arctan x)$, finally we can obtain the maximum
$$\frac{\sqrt5+1}{2}$$
As one can see this is precisely the golden ratio $\phi$!

But this solution takes some time, so I'm interested in different no calculus solution of this problem, especially considering that it is related to the golden ratio.

答:
Using algebra and trigonometry only:

First observe that since the angles sum to $\pi$ radians, $\sin x=\sin(y+z)$.  Then, apply the product-sum relation

$\sin y\sin z=(1/2)[\cos(y-z)-\cos(y+z)]$

Then the objective function may then be rendered as

$\sin x+\sin y\sin z =\color{blue}{\sin(y+z)-(1/2)\cos(y+z)}+\color{brown}{(1/2)\cos(y-z)}$

The blue expression has the form $a\sin\theta-b\cos\theta$ for positive $a$ and $b$, therefore $\sqrt{a^2+b^2}\sin[\theta-\arccos(a/\sqrt{a^2+b^2})]$.  Thus the blue expression is maximized at $\sqrt5/2$ whenever $y+z=\pi/2+\arccos(2/\sqrt5)=\pi-\arcsin(2/\sqrt5)$.  If we then select each of $y$ and $z$ as half of this sum, then the brown term is simultaneously maximized at $1/2$.  Thereby a maximum value of $(\sqrt5+1)/2$ is achieved.


问:
Series convergence or divergence $\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(3n)}{\sqrt{n^2 + 2}}$
Can you give me some hint for the series
\begin{align*}
\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(3n)}{\sqrt{n^2 + 2}}
\end{align*} I have tried Leibniz test. It didn't help me.

答:
$\sum_\limits{n=1}^\infty \frac {1}{\sqrt{n^2 + 2}}$ can be compared to the harmonic series.  The harmonic series diverges... slowly.

The series is not absolutely convergent.  But, it still could be conditionally covergent.

The alternating series test doesn't help as $(-1)^n\cos 3n$ is not necessarily alternating.

That leaves Dirichlet's test.

If $\{a_n\}$ is a sequence of real numbers and $\{b_n\}$ is a sequence of complex numbers, such that:

$\{a_n\}$ is monotonic and $\lim_\limits{n\to\infty} a_n = 0$

And

$\forall N\in \mathbb N : \left|\sum_\limits{n=1}^N b_n \right|< M$ for some constant $M.$

$\sum_\limits{n=1}^\infty a_nb_n$ converges.

The first parts of the test, are easy enough and I will leave to you.

$(-1)^n\cos 3n = Re(e^{(3+\pi)in})$

And

$\left|\sum_\limits{n=1}^N Re(e^{(3+\pi)ni})\right| < \left|\sum_\limits{n=1}^N e^{(3+\pi)ni}\right|$

We still need to show that $\left|\sum_\limits{n=1}^N e^{(3+\pi)ni}\right|$ is bounded for all $N.$

$\sum_\limits{n=1}^N e^{(3+\pi)ni}$ is a geometric series.

$\sum_\limits{n=1}^N e^{(3+\pi)ni} =  e^{(3+\pi)i}\frac {1-e^{(3+\pi)Ni}}{1-e^{(3+\pi)i}}\\
\left| e^{(3+\pi)i}\frac {1-e^{(3+\pi)Ni}}{1-e^{(3+\pi)i}}\right| <
\frac {2}{1+\cos 3}$


问:
Can travelling salesman be translated into a SAT problem?
I read that all NP-complete problems can be translated into one another.

I can't see how the travelling salesman problem which involves distances which are real numbers can be translated into a boolean satisfaction problem? Do you have to limit it to integer distances?

Is there a way to do it?

答:
I don't know what all the comments are talking about, this is definitely a question in Mathematics. To answer your question, we note that real numbers aren't actually encodable by Turing Machines (if they're infinitely long), so the problem of TSP over arbitrary real numbers is not really applicable (how would you even express the input?). Over rational numbers (or any other countable subset of real numbers that is closed under addition) it can still be done since they're countable, so they can be encoded in similar ways as integers.

Also if you look at Cook's proof that SAT is NP-Complete, you'll see that he uses a reduction from the non-deterministic polynomial time turing machine that solves an NP problem, so the structure of the problem isn't really relevant.


问:
Are terminating decimals dense in the reals?
I have read that rationals are dense in the reals, but I've been curious if the terminating decimals are as well. Since, in applied math, all effectively "real" numbers are approximated with terminating decimals, I would expect the terminating decimals to have this property, and if not I'm curious why. Here I'm defining the terminating decimals like this:

$$D = \{\frac{m}{10^n} \in \mathbb{Q} : m \in \mathbb{Z} \wedge n \in \mathbb{N}_0\}$$.

答:
Yes, they are.

First, think informally (and you do this, with your sentence beginning "since"). A set $X$ is dense iff every real number $r$ can be "well-approximated" by an element of $X$. For terminating decimals, this approximation amounts to simply ... truncating at a certain point! E.g. $\pi$ is the limit of the sequence $$3.0000..., 3.1000..., 3.1400..., ...$$ of reals with terminating decimal expansions.

OK, now we need to make this rigorous. Fix reals $a<b$.

First, show that there is some $n$ such that $10^{-n}<b-a$. (This is actually the most substantial step, but at the same time may be the intuitively clearest one.)

Let $X=\{z\cdot 10^{-n}: z\in\mathbb{Z}\}$. Show that $X\cap (-\infty,a]$ has a greatest element - that is, there is a largest "$n$-place decimal" which is $\le a$.

It may be easiest to first assume $a>0$ here.

Calling that number "$\alpha$," what can you say about $\alpha+10^{-n}$?

Note that this works for any base, not just $10$.


问:
How to reduce lower one number, while another number increases goes up increments
This is such a basic question I'm sure, but I've been trying to find a robust solution to it for a while.

Lets say I have the numbers 0-100 in series.

As the number increases, I want another number say 1000, to decrease. so that when you got to 100, it was at 0

0 = 1000

50 = 500

100 = 0

What is this kind of series called?

答:
The function you want is $y=10(100-x)$. This was obtained by attempting to compute the quadratic function that passes through the three points ($n+1$ points uniquely determine an nth-degree curve), and then noting that the interpolating polynomial is actually a linear function.

Alternatively, computing the determinant

$\begin{vmatrix}1&0&1000\\1&50&500\\1&100&0\end{vmatrix}$

(the area of the "triangle" formed by the three points)

gives a value of 0, and thus the three points are collinear. Take any two of the three points, and you can derive the equation of the line passing through them.

ContentLicense: CC BY-SA 2.5


问:
Exercise about prime ideals
Let $A$ be a ring. Prove that the following conditions are equivalent:

$i)$  All ideals $I \subsetneq A$ are prime.

$ii)$ The set of all ideals of $A$ is totally ordered by
inclusion and all ideals of $A$ are idempotent.

Please give me a hint. I dont see the relation between these two statemens.

答:
Assume (i).
Let $\mathfrak a,\mathfrak b\subsetneq A$ be two ideals and assume $\mathfrak a$ is not contained in $\mathfrak b$, i.e. there exists $a\in\mathfrak a\setminus \mathfrak b$. Then for $b\in\mathfrak b$, we have $aAb\subseteq\mathfrak a\cap \mathfrak b$. As the latter is a prime ideal and $a\notin\mathfrak a\cap \mathfrak b$, we conclude $b\in\mathfrak a\cap \mathfrak b\subseteq \mathfrak a$, in other words $\mathfrak b\subseteq \mathfrak a$.
Similarly, if $a\in \mathfrak a$, then $aAa\in\mathfrak a^2$ implies $a\in\mathfrak a^2$, i.e. $\mathfrak a=\mathfrak a^2$.</p>

Title: Exercise about prime ideals


问:
Multivariable Calculus - is this function continuous?
$f(x,y) = \left\{\begin{matrix}
\frac{|x|sin(y^{2})}{\sqrt{x^{2}+y^{2}}},(x,y)\ne (0,0) & \\
 0 , (x,y) = (0,0)&
\end{matrix}\right.$

is it correct to change the expression to this way $\frac{|r*cos(a)|sin(r^2sin(a)^2)}{\sqrt{r^2}}$? if so and we want to check the limit when $f$ approaches zero I used the squeeze theorem so I got that, $\lim_{r\rightarrow 0^{+}}\frac{|r*cos(a)|sin(r^2sin^2)}{r}\leq \ \lim_{r\rightarrow 0^{+}}\frac{r\sin(r^2sin^2)}{r} =0$ since $ -r \leq rcos(a) \leq r$

so the function is continuous ?

答:
for $x \neq 0$ we have
$$  \frac{|x|}{\sqrt{x^2 + y^2}} \leq 1  $$
Next,
$$ \sin y^2 \leq y^2  $$
by Taylor series, obvious when $y^2 < 1$


问:
Can we always numerically integrate?
I posted this question in physics SE, as few suggested me to post it here: when I watch online lectures (usually of physics) most lecturers say that if we cant solve a integral exactly we can always numerically integrate it. (E.g. in central force problem the orbit where we find $\theta(r)$ or $r(\theta)$ is pretty difficult to compute and professor said that if needed we can always integrate it numerically if we cannot do it exactly)

So is it always possible to numerically integrate? Is there any example in physics where it not possible to integrate numerically or any integral which cannot be even integrated numerically? Is there any limitation of numerical integration?

答:
The answer by @GEdgar considers numerical integration in exact arithmetic. This answer considers numerical integration using finite precision arithmetic.

So is it always possible to numerically integrate?

No. It is essentially impossible to evaluate an integral if the exact value is sufficiently close to zero. It is a question of conditioning. Finite sums that contain both negative and positive terms and sum to nearly zero are ill-conditioned. Our inability to compute the necessary function values with the required accuracy will destroy the accuracy of the computed approximation of the integral.

Is there any example in physics where it not possible to integrate
numerically

Yes. The solution of the initial value problem
\begin{align} y'(t) &= y(t) \\ y(0) &= 1 \end{align}
will rapidly exceed the representational range of any finite number system.

or any integral which cannot be even integrated numerically?

Yes. Consider a polynomial $p$ that is zero for all machine numbers. Then $|p|$ is continuous and
$$ \int_a^b |p(x)|dx > 0,$$
for any interval $[a,b]$, but the computed value of any Riemann sum or Gauss quadrature will be zero.

Is there any limitation of numerical integration?

Yes. You must provide the user with

an approximation of the integral

an error estimate, and

an argument that explains why the error estimate is reliable.

If the truncation error obeys an asymptotic error expansion, then this is theoretically possible. However, if the order of accuracy is too high, then the asymptotic range is too short and it is essentially impossible to recognize that the error estimates are reliable.

Title: Can we always numerically integrate?


答:
Hint: use Dirichlet's test. See more at Wikipedia

It's clear that $\left\{\dfrac 1 {\sqrt{n^2 + 2}}\right\}$ is monotonic and converges to 0

Now the task is to prove $-\cos 3 + \cos 6 - \cos 9 + \cos 12 + ...$ is bounded

Recognize that $\cos 6 - \cos 3 = -2\sin \frac 9 2 \sin \frac 3 2$

$\cos 12 - \cos 9 = -2\sin \frac {21} 2 \sin \frac 3 2$

$\cos 18 - \cos 15 = -2\sin \frac {33} 2 \sin \frac 3 2$

and $-2 \sin 3\left(\sin \frac 9 2 + \sin \frac {21} 2 + \sin \frac {33} 2 ... \right)= \cos \frac {15} 2 - \cos \frac 3 2 + \cos \frac {27} 2 - \cos \frac {15} 2 +...$

You can see that $\sum\limits_{i=1}^{2n} (-1)^k \cos {3k} = \dfrac{\sin \frac 3 2}{\sin 3} \left|\cos(2n + 1.5) - \cos 1.5 \right|$, which is always bounded

Therefore the series converges

Title: Series convergence or divergence $\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(3n)}{\sqrt{n^2 + 2}}$

Body:
Can you give me some hint for the series
\begin{align*}
\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(3n)}{\sqrt{n^2 + 2}}
\end{align*} I have tried Leibniz test. It didn't help me.


问:
What is derivative of Matrix with respect to a Matrix?
I want to calculate the derivative of dot product of two Matrices of not the same order.

$X = \begin{bmatrix}x_{11} & x_{12} & x_{13}\\x_{21} & x_{22} & x_{23}\\x_{31} & x_{32} & x_{32}\end{bmatrix}$

$y= \begin{bmatrix}y_{11} & y_{12}\\y_{21} & y_{22}\\y_{31} & y_{32}\end{bmatrix}$

Problem is I can't figure out What does it mean to derivative of matrix with respect of matrix individual elements.
I tried to use the sum notation to calculate derivative of a single element of the resultant matrix.

$c_{i,j} = \sum_{k=1}^na_{i,k}\cdot b_{k,j}$

$\frac{\partial (X y)_{11}}{\partial X} =
\begin{bmatrix}y_{11} & y_{12} & y_{21}\\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$

and the other partial Derivatives are similar to this.
I want to know that what is

$\frac{\partial Xy}{\partial X} = ?$

I can't figure out how to get this when the element derivative itself is a matrix.
and the matrices as inputs are not even of the same order.

答:
Based on the comments, it sounds like you have a scalar function $(\phi)$ defined as follows
$$\eqalign{
C_{ij} &= \sum_{k=1}^p X_{ik}Y_{kj} \\
\phi &= \sum_{i=1}^m \sum_{j=1}^n C_{ij}
  \;=\; \sum_{i=1}^m \sum_{j=1}^n \sum_{k=1}^p X_{ik}Y_{kj} \\
}$$
This can be written in matrix notation using an all-ones matrix $J$ the same size as $C$.

In this form, the gradient is very easy to calculate
$$\eqalign{
\phi &= J:XY \\
d\phi &= J:dX\,Y \;=\; JY^T:dX \\
\frac{\partial\phi}{\partial X} &= JY^T \\
}$$

In the above, a colon is used as a convenient product notation
for the trace function
$$A:B = \sum_{i=1}^m\sum_{j=1}^n A_{ij}B_{ij} = {\rm Tr}(A^TB)$$
The product can be applied to vectors by treating them as rectangular matrices (set $n=1$) in which case it's just the dot product.
The terms in such a product can be rearranged in a number of ways, e.g.
$$\eqalign{
 A:B &= B:A = B^T:A^T \\
CA:B &= C:BA^T = A:C^TB  \\
}$$
due to the properties of the underlying trace function.


问:
Where can I find theorems on the maximum of real valued concave functions over integers?
I was analyzing a bunch of games and often I wanted to find the integer(s) such that $f(i)=\max_{j\in Z\cap[a,b]} f(j)$ where $f:[a,b]\rightarrow R$ is concave.  I proved a few simple results myself, and I imagine that my theorems are trivial consequences of published theorems. Has anybody here seen theorems proven that are similar to the simple theorems shown below?  The Discrete Convex Analysis papers seem to be much more focused on finding the minimum value of $f(x)$ where $f(x)$ is convex and $x$ is confined to a subset of $Z^n$ with $n>1$.

I would appreciate it if anyone can give me a reference for convex analysis that has theorems similar to the ones below, or better yet a reference that contains theorems which are more general than the ones I give below and that I could be use to prove the theorems below.

Here is a list of the simple theorems that I found useful for maximizing concave $f(x)$ over integers.  There are also bounds that you can put on the integer solutions if the function is thrice differentiable.  I haven't worked those out yet.

Theorem 1:  Suppose that $f:[a,b]\rightarrow R$ is strictly concave, $a$ and $b$ are real numbers, $a<b$, and there exists a real $x$ satisfying $a\leq x<x+1 \leq b$ and $f(x)=f(x+1)$.  Then, for any integer $z\in [a,b]$,

$x<z<x+1$ if and only if $ f(z)>f(x)$ and

$x\leq z\leq x+1$ if and only if $ f(z)\geq f(x)$.

Furthermore,  $\max_{i\in Z\cap [a,b]} f(i)$ is well defined and for any integer $j$, and
$\max_{i\in Z\cap [a,b]} f(i) = f(j)$  if and only if $j\in\{\mathrm{ceil}(x), \mathrm{floor}(x+1)\}$.

Corollary 1:  Suppose that $f(x) = a x^2 +b x + c$ with $a<0$ and $\{b,c\}\subset R$. Then for any integer $j$, $\max_{i\in Z\cap [a,b]} f(i) = f(j)$ if and only if $j\in\{\mathrm{ceil}(x)), \mathrm{floor}(x+1)\}$ with $x=-\frac{b}{2 a}-1/2.$

Theorem 2:  Assume that $f:[a,b]\rightarrow R$ is strictly concave with $a$ and $b$ real satisfying $a+1<b$. Let $x\in[a,b]$ such that $f(x) = \max_{y\in[a,b]} f(y)$ and let $m =\max_{j\in[a,b]\cap Z} f(j)$. Then exactly one of the following is true:

there exists a unique integer $i$ such that  $f(i) = m$ and for that unique integer $i$, $|i-x|<1$, or

there exists a unique integer $i$ such that $\{j\in Z |\ f(j)=m\}=\{i,i+1\}$ and that $i$ satisfies $x-1 \leq i \leq x$.

答:
Theorem 1 until 'Furthermore' is well known and does not need $z$ to be integer: the 'only if' follows directly from the definition of a concave function, and the 'if' is, e.g., (2) in this paper.

These theorems remind me of the news vendor problem with discrete demand, and marginal analysis.


问:
The Basel problem
As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.

I believe many of you know some nice proofs of this, can you please share it with us?

Title: The Basel problem

Body:
Since $\int_0^1 \frac{dx}{1+x^2}=\frac{\pi}{4}$, we have

$$\frac{\pi^2}{16}=\int_0^1\int_0^1\frac{dydx}{(1+x^2)(1+y^2)}\overset{t=xy}{=}\int_0^1\int_0^x\frac{dtdx}{x(1+x^2)(1+t^2/x^2)}$$

$$=\frac12\int_0^1\int_t^1\frac{dxdt}{x(1+x^2)(1+t^2/x^2)}\overset{x^2\to x}{=}\frac12\int_0^1\left(\int_{t^2}^1\frac{dx}{(1+x)(x+t^2)}\right)dt$$

$$=-\frac12\int_0^1\frac{\ln\left(\frac{4t^2}{(1+t^2)^2}\right)}{1-t^2}dt\overset{t=\frac{1-x}{1+x}}{=}-\frac12\int_0^1\frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{x}dx$$

$$\overset{x^2\to x}{=}-\frac14\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{x}dx=-\frac14\int_0^1\frac{\ln\left(\frac{(1-x)^2}{1-x^2}\right)}{x}dx$$

$$=-\frac12\int_0^1\frac{\ln(1-x)}{x}dx+\frac14\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}dx}_{x^2\to x}$$

$$=-\frac38\int_0^1\frac{\ln(1-x)}{x}dx\Longrightarrow \int_0^1\frac{-\ln(1-x)}{x}dx=\frac{\pi^2}{6}$$

Remark:

This solution can be considered a proof that $\zeta(2)=\frac{\pi^2}{6}$ as we have $\int_0^1\frac{-\ln(1-x)}{x}dx=\text{Li}_2(x)|_0^1=\text{Li}_2(1)=\zeta(2)$


问:
Pattern for cubed values
We now have a pattern for squared values, but what about cubed ones? Basically it is $ x^3 $.

I wasn't able to find a precise pattern that's similar to my answer on the linked post, as the result contains:

```<code>1^3 = 1
7
2^3 = 8
19
3^3 = 27
37
4^3 = 64
61
5^3 = 125
91
6^3 = 216
127
7^3 = 343
169
8^3 = 512
217
9^3 = 729
271
10^3 = 1,000
</code>```

And I could barely make an equation basing it.

Without the help of a calculator, can I get the cubed values via patterns?

答:
Well, the  difference between consecutive cubes is

$$x^3-(x-1)^3= x^3-x^3+3x^2-3x+1=3x(x-1)+1$$

Three times the product of the bases plus one.

Title: Pattern for cubed values


问:
Proving $T$ and $U$ are equal if I know $(X,T)$ and $(X,U)$ are hausdorff and compact
This question is from Wayne Patty Exercise 4.2

Let $T$ and $U$ are Topologies on set X. Suppose $ (X,T) $ and $(X,U)$ are compact Haurdsorff Spaces. Prove that $T=U$ or $T$ is not a subset of $U$ or $U$ is not a subset of T.

I thought of proving it by this way: Assuming that if $T$ is a subset of $U$ or $U$ is a subset of $ T$ then $T=U$ must be true.
But I was unable to think on how to use compactness or hausdorff property and was unable to proceed.

Well, also this question was asked before on this website. But I am unable to understand how the solution will work despite trying to understand it for a fair amount of time as it was just outlined.

So, I am looking for a detailed solution.

答:
The continuous image of a compact space is a compact space.

A  subset of a compact Hausdorff space is closed iff it is compact.

Suppose $T\subseteq U.$

Then the function $f=id_X$ is  continuous from $(X,U)$ to $(X,T)$ because if $S$ is $T$-open then $S\in T$ so $S\in U$ so $f^{-1}S=S$ is $U$-open.

Consider any $Y\in U.$

Then $X$ \ $Y$ is $U$-closed,

so by 2., $X$ \ $Y$ is compact in $(X,U),$

so by 1., its image $f[X$ \ $Y]=X$ \ $Y$ is compact in $(X,T),$

so by 2., $X$  \ $Y $ is $T$-closed,

so $Y\in T.$

In summary, $T\subseteq U\implies \forall Y\,(Y\in U\implies Y\in T)\implies U\subseteq T.$


答:
No, not really. For $\varepsilon\in[0,1)$ let $C_\varepsilon$ be a fat Cantor set whose measure is $\varepsilon$. We know that such sets exist for every $\varepsilon$ in the given range.

The fat Cantor sets are nowhere dense, and closed, so their complement is open and dense. Pick $\varepsilon_n$ to be a sequence which approaches $1$, and let $O_n$ be the complement of $C_{\varepsilon_n}$, note that the measures of $O_n$ approach zero. Then $\bigcap O_n$ is the intersection of dense open sets, but its measure is zero.

Title: Baire's theorem from a point of view of measure theory

Body:
According to Baire's theorem, for each countable collection of open dense subsets of $[0,1]$, their intersection $A$ is dense. Are we able to say something about the Lebegue's measure of $A$? Must it be positive? Of full measure?

Thank you for help.


问:
Combinatorial sums, a possible formula for inserting a dummy variable in the front?
I've recently been working with combinatorial sums (I love the mix of algebra and combinatorics!). When working with some basic problems, I have found a pattern that may or may not be true.

Example 1: Consider $\sum_{k=0}^n\binom{n}{k}$. It is well known that this is equivalent to $2^n$. Now what happens when we insert a $k$ in the front? It becomes $\sum_{k=0}^nk\binom{n}{k}$. This is actually not too difficult to evaluate if we do some manipulation. It turns out to be $n\cdot 2^{n-1}$.

Example 2: Consider $\sum_{k=0}^{n}\binom{n}{k}a^{n-k}$. Let's try to evaluate it. Observing the $a^{n-k}$, we think of the binomial theorem. In fact, this is the expansion for $(1+a)^n$. Now what happens if we insert the $k$ in there again? I checked this with Mathematica and it seems  as if it evaluates to $n(1+a)^{n-1}$.

There are multiple other examples. My  hypothesis is as follows:

Given a function $f:\mathbb{Z}\to\mathbb{Z}$ such that $\sum_{k=0}^{n}f(k)=x^n$, where $x$ is some constant, we have that $\sum_{k=0}^{n}kf(k)=nx^{n-1}$.

Is this even true? If it is, how do we prove this?

Note: Since $\int nx^{n-1}=x^n+C$, and integrals have a relationship with summations, I think calculus may be needed to prove/disprove the hypothesis.

答:
The only function that satisfies $ \sum_{k=0}^n f(k) = x^n$ is $ f(n) = x^n - (x-1)^n$.

So, is $ \sum kf(k) = \sum k ( x^k - (x-1)^k)  = nx^{n-1}$?

Unfortunately, no.


问:
Intuition behind sums of sums of whole numbers
So I was playing around, and all this is just a curiosity and nothing serious.

Anyway, most readers probably know:
$$1+2+3+4+5+...+(n-1)+n=\frac{1}{2}n^{2}+\frac{1}{2}n=\binom{n+1}{n-1}$$

I started playing around, adding the individual sums of whole numbers instead of whole numbers alone. Words aren't very helpful to describe this process, instead consider the sum of sums for $n=4$, which we shall call $N_2(4)$ for simplicity:
$$\left ( 1+2+3+4 \right ) + \left ( 1+2+3 \right ) + \left ( 1+2 \right ) + \left ( 1 \right ) = 20$$

Remarkably, there is a simple formula (I did the math):
$$N_{2}(n)=\binom{n+2}{n-1}$$

Where $N_2(n)$ is the sum of sums as above. Formally, $N_2(n)=\sum_{1\leq i}^{n}\sum_{1\leq j\leq i}j$.

Now imagine going further, with sums of sums of sums, for instance:
$$N_3(4) = \left ( \left ( 1+2+3+4 \right ) + \left ( 1+2+3 \right ) + \left ( 1+2 \right ) + \left ( 1 \right ) \right ) + \left ( \left ( 1+2+3 \right ) + \left ( 1+2 \right ) + \left ( 1 \right ) \right ) + \left ( \left ( 1+2 \right ) + \left ( 1 \right ) \right ) + \left ( \left ( 1 \right ) \right ) = 35$$

Again, this seems to follow the pattern (I haven't explicitly checked):
$$N_3(n)=\binom{n+3}{n-1}$$

And we might conjecture:
$$N_k(n)=\binom{n+k}{n-1}$$

One angle of attack is this: realizing that the previous series always adds up to that of the differences between successive elements of the next series, and so verifying that:

$$\binom{n+k}{n-1} - \binom{(n-1)+k}{(n-1)-1}=\binom{n+(k-1)}{n-1}$$

I.e. that $N_{k}(n)-N_{k}(n-1)=N_{k-1}(n)$ for any suitable $n$ and $k$.

My question is if there's some intuition behind all this. Maybe an alternative way of looking at this, or proving it. Why are the sums so neatly expressible?

答:
I like to  look at things like this in term of matrices.

Let the elements of a sequence to be summed $a_0,a_1,a_2...,a_{n-1}$ form a columnvector $A$ .

Then consider the operator (=matrix) for the partial sums
$$ D = \small \begin{bmatrix}
 1 & . & . & . & . & . \\
 1 & 1 & . & . & . & . \\
 1 & 1 & 1 & . & . & . \\
 1 & 1 & 1 & 1 & . & . \\
 1 & 1 & 1 & 1 & 1 & . \\
 1 & 1 & 1 & 1 & 1 & 1 \\
 \end{bmatrix}$$
(Of course the size must agree with the number of elements in your sequence/vector $A$).

Then $$D \cdot A = S_1$$ gives the first-order (partial) sums; $$D^2 \cdot A = S_2$$ the second order (partial) sums and so on.

Interestingly, using the matrix-logarithm on $D$ , we can even define fractional orders, because we can define fractional powers of  $D$:
$$ D^h =  \exp  (h \cdot \log(D))$$
where we need a software which is able to compute the matrix-logarithm and the matrix-exponential (I do this in Pari/GP using the according powerseries definitions) .

Finally, we can see the $h$'th power of $D$ with the parameter $h$ symbolic:

$\qquad  \Large D^h = $

$ \qquad \qquad $ Here the middle coefficients are the binomial-coefficients, as for instance in the Pascal-matrix. The factorials come from the row- and column-indexes (each beginning at $0$).

With this one can compute positive,negative, fractional and even complex orders of the generalized harmonic sums (or "hyper-harmonic sums" as Conway/Guy have christened them)


问:
If the operator $T$ is defined by $Tf(x)=\int_0^xf(t)\,dt$, show that $Tf \in C[0,1]$

Consider the operator $T$ on $L^2[0,1]$ defined by $Tf(x)=\displaystyle \int_0^xf(t)\,dt.$ Show that $Tf \in C[0,1].$

I have one question before this:

What are the implications between $L^p$ spaces, i.e if $f \in L^p$ does this imply that $f \in L^{p+1}$?

My attempt:

Assume $Tf$ is not in $C[0,1]$, so there exists $y \in [0,1]$ and $\epsilon>0$ such that for all $\delta >0$, we can find $x_0$, with $|x-x_0|<\delta$ but

$$\bigg\rVert\int_0^x f-\int_0^{x_0} f\,\bigg\rVert>\epsilon$$

WLOG assume $x>x_0$, so $$\bigg\lVert \int_{x_0}^x f\,\bigg\rVert \geq \epsilon.$$
So for $\delta_n=\frac{1}{n},$ we can find $x_n \in [0,1]$ such that $|x-x_0|<\frac{1}{n}$ and $$\bigg\lVert\int_{x_n}^xf\,\bigg\rVert \geq \epsilon$$
I don't know if that will lead me to a contradiction.

I would appreciate any help or hints with that.

答:
Since a bounded linear functional is continuous, it suffices to show that $Tf$ is bounded. Set $k(x, t) = \chi_{[0, x]}(t) $. We have:

$||Tf||_{L^{2}}^{2} = \langle Tf, Tf \rangle = \int_{0}^{1} \left[ \int_{0}^{1} k(x,t) f(t)dt\right]^{2}dx = \int_{0}^{1} \left[ \int_{0}^{1} |k(x,t)| |f(t)|dt\right]^{2}dx $

$\leq \int_{0}^{1} \left[ \left( \int_{0}^{1} |k(x,t)|^{2} dt\right)^{\frac{1}{2}} \cdot \left( \int_{0}^{1} |f(t)|^{2} dt \right)^{\frac{1}{2}}\right]^{2}dx  = \int_{0}^{1} \int_{0}^{1} |k(x, t)|^{2} dt dx \cdot ||f||_{L^{2}}^{2} = \frac{1}{2} ||f||_{L^{2}}^{2}$.

We get $||Tf||_{L^{2}} \leq \sqrt{\frac{1}{2}} ||f||_{L^{2}}$ and we have $Tf$ is bounded and so it must be continuous.


答:
Another approach is to show that the dyadic rationals (IE numbers of the form $\frac{ a}{ 2^b}$, where $a$ is an integer and $b$ is a natural number) are dense.

Another number can be approximated by a dyadic sequence by bisecting the interval that it is in.

The 2-adic rationals terminates with at most $b$ decimal places. (In fact, it needs $b$ decimal places if $a$ is odd.)

Title: Are terminating decimals dense in the reals?

Body:
I have read that rationals are dense in the reals, but I've been curious if the terminating decimals are as well. Since, in applied math, all effectively "real" numbers are approximated with terminating decimals, I would expect the terminating decimals to have this property, and if not I'm curious why. Here I'm defining the terminating decimals like this:

$$D = \{\frac{m}{10^n} \in \mathbb{Q} : m \in \mathbb{Z} \wedge n \in \mathbb{N}_0\}$$.


问:
Equilibrium points and homoclinic paths of dynamical system
Exercise

Investigate the equilibrium points of

$\dot{x} = y \left( 16 \left(2x^2+2y^2-1 \right) -1 \right)$

$\dot{y} = x - \left(2x^2+2y^2-1 \right)\left(16x-4\right)$

and classify them according to their linear approximations.

Show that the homoclinic paths through $(0,0)$ are given by

$\left(x^2 + y^2 -\frac{1}{2}x \right)^2 - \frac{1}{16}\left(x^2+y^2\right)=0$

and that one homoclinic path lies within the order. Draw the phase portrait.

(Hint: Find the Hamiltonian function of the system)

Attempt

I have calculated the equilibrium points to be $(0,0), (\frac{3+\sqrt{3}}{8},0),(\frac{3-\sqrt{3}}{8},0)$

For the classifications of the equilibrium points I calculated the linearized system around each point and found that

The point $(\frac{3+\sqrt{3}}{8},0)$ is unstable saddle point for the almost linear system (original system)

The point $(\frac{3-\sqrt{3}}{8},0)$ is stable center point for the almost linear system.

The point $(0,0)$ is stable center point for the linearized system. This means that the Hartman-Grobman theorem for system linearization cannot be applied. Q1: What can I do to determine the classification of this point for the almost linear system?

For the homoclinic path I calculated the Hamiltonian function of the original system to be

$H(x,y) = 8x^4 + 8y^4 -8x^3 + \frac{3}{2}x^2 - \frac{1}{2}y^2 +16x^2y^2 -8xy^2$

If I set $H(x,y)=0$ then I get the equation that is supposed to represent the homoclinic paths but I do not understand why that is true. Q2: Can someone explain?

Q3: I also don't understand what the phrase "one homoclinic path lies within the order" means.

答:
Q1: For the Jacobian computed on $(0,0)$, it holds

$$\left| \begin{matrix}
0 & -1 \\ -3 & 0
\end{matrix} \right|
<-\infty$$
so $(0,0)$ is a saddle point of the linearized.

Q2: $H$ is continuous. Assume that $H(\phi(t,t_0,x^0)) = c$, for every $t\in\mathbb{R}$ ($H$ is a constant of motion for the Hamiltonian system), where $x^0$ lies on a homoclinic path of $(0,0)$. By definition of homoclinic path, $\lim_{t\rightarrow\pm\infty}\phi(t,t_0,x^0)=(0,0)$. Under the continuous function $H$, preimage of a closed set in $\mathbb{R}$ (e.g. $\{0\}$) is closed in $\mathbb{R}^2$ (e.g. $H^{-1}(\{0\})$). Let $t_k$ be a sequence such that $t_k\rightarrow\infty$. Then
$$
\begin{array}{rcl}
\phi(t,t_0,x^0) \in H^{-1}(\{c\}) & \Rightarrow & \lim\limits_{k\rightarrow\infty}\phi(t_k,t_0,x^0) \in H^{-1}(\{c\}) \\
& \Rightarrow & (0,0) \in H^{-1}(\{c\}) \\
& \Rightarrow & c=0
\end{array}
$$
because $H(0,0)=0$.

Also, $H^{-1}(\{0\})$ is an 1-dimensional regular submanifold of $\mathbb{R}^2$ (I don't know if this answers Q3).


问:
Distribution of the sum of the number of rolls that result in an even number
I'd like to ask for some help in finding the distribution of the number of rolls that result in an even number. The problem goes like this:

Consider a loaded die where the probability that a roll results in $I$ is $p_i = I/21$, $i = 1,\ldots,6$. Suppose the die is rolled $35$ times and let $N_i$ be the number of times that the roll results in $i$. Then $N = (N_1, \ldots , N_6)$ is multinomial distributed with parameters $35$, $\left(\frac1{21}, \ldots, \frac6{21}\right)$, find the distribution of $X = N_2+N_4+N_6$.

In my attempt, knowing that $N_i$'s follow a $\text{Binomial}(35,p_i)$ distribution, I conditioned on $N_6$, and then on $ N_4$ but my expression was too complicated to simplify. I was wondering if there is an easier way to approach this problem? This is just a practice problem as I study for my exam.

答:
Hint:

What is the probability a particular single roll is even?

Are rolls independent?

What is the probability exactly $X$ rolls are even out of $35$ total rolls?


问:
Are there Olympiad problems that can be solved using split-complex numbers?
There are various uses for complex numbers in Math Olympiads. In addition, quadratic number fields are sometimes useful, for instance for Pell equations.

Are there any Olympiad/contest problems that can be solved using split-complex numbers ($a+jb$, $j^2=1$)?

答:
Just made this up.

Show that the product of 2 integers, each of which can be written as the difference of 2 squares, can again be written as the difference of 2 squares.

The algebraic approach would involve showing

$$ ( a + bj) ( c + dj) = (ac+ bd) + (ad + bc) j. $$

Of course, the number theoretic approach is slightly more direct. We use the classification that a number $n$ can be written as the difference of 2 squares iff it is either 1) odd, or 2) a multiple of 4.


问:
Inequality relating the Limit Superior and Limit Inferior of a bounded sequence
I'm doing exercise 2.4.7 in the book Understanding Analysis by Stephen Abbott. I'd like to ask, how to go about part (d) of the proof (please don't give away the complete proof). Additionally, is the proof to (a),(b),(c) rigorous and technically correct?

$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$

Let $(a_n)$ be a bounded sequence.

(a) Prove that the sequence defined by $y_n = \sup \{a_k: k\ge n\}$ converges.

(b) The limit superior of $(a_n)$ or $\limsup a_n$ is defined by

\begin{align*}
 \limsup{a_n} = \lim y_n
\end{align*}

where $(y_n)$ is the sequence from part (a) of this exercise. Provide a reasonable definition for $\liminf a_n$ and briefly explain why it always exists for any bounded sequence.

(c) Prove that $\liminf a_n \le \limsup{a_n}$ for every bounded sequence, and given an example of a sequence where this inequality is strict.

(d) Show that $\liminf{a_n} = \limsup{a_n}$ if an only if $\lim a_n$ exists. In this case, all three share the same value.

I cross-reference some excellent answers on this site on the limit superior and limit inferior of a bounded sequence.

What is limit superior and limit inferior?

Limit superior and Limit inferior of a sequence of subsets of a set

Proof.

(a) The sequence $(a_n)$ is a bounded sequence. Hence, $\absval{a_n} \le M$ for all $n \in \mathbf{N}$.

Claim. The sequence $(y_n)$ is a monotonically decreasing sequence. For $n=1$, $y_1 >= y_2$ because $y_1 = \sup \{a_k:k\ge 1\} = \max \{a_1,y_2\} \ge y_2$. Assume that $y_{k} \ge y_{k+1}$. Then, $y_{k+1} = \sup \{a_k: k+1 \ge n\} = \max \{ a_{k+1},y_{k+2} \} \ge y_{k+2}$. Hence, $y_{k} \ge y_{k+1}$ for all $k \in \mathbf{N}$.

We can further show that $(y_n)$ is a bounded sequence. Since, $\absval{a_n} \le M$ for all $n \in \mathbf{N}$, the supremum of any of the subsequences $\{a_k:k\ge n\}$ no matter what $k \in \mathbf{N}$, is always less than or equal to $M$. Thus, $\absval{y_n} \le M$ for all $n \in \mathbf{N}$. Therefore, $(y_n)$ is a bounded sequence.

By the Monotone Convergence Theorem, the sequence $(y_n)$ is convergent.

(b) From part (a), we can deduce that, the limit superior of a sequence is,
\begin{align*}
 \limsup{a_n} = \lim_{n\to\infty} \left(\sup \{a_k:k \ge n\}\right) = \inf \{ \sup \{a_k:k \ge n\}\}
\end{align*}

whereas, the limit inferior of a sequence is,
\begin{align*}
 \liminf{a_n} = \lim_{n\to\infty} \left(\inf \{a_k:k \ge n\}\right) = \sup \{ \inf \{a_k:k \ge n\}\}
\end{align*}

The above definition is easy to verify.

Define $z_n := \inf \{a_k:k\ge n\}$. Let us prove that $(z_n)$ is a convergent sequence.

Claim. The sequence $(z_n)$ is a monotonically increasing sequence. For $n=1$, $z_1 \le z_2$ because $z_1 = \inf \{a_k:k \ge 1\} = \min \{a_1,z_2\} \le z_2$. Assume that $z_{k} \le z_{k+1}$. We have, $z_{k+1} = \inf \{a_{k+1} : k+1 \ge n\} = \min \{a_{k+1},z_{k+2}\} \le z_{k+2}$. Hence, by principle of mathematical induction, $z_{n} \le z_{n+1}$ for all $n \in \mathbf{N}$.

Claim. The sequence $(z_n)$ is bounded below. Since, $(a_n)$ is a bounded sequence, $a_n \ge -M$ for all $n \in \mathbf{N}$. So, given any $k \in \mathbf{N}$, the infimum of the subsequence $\{a_{k}: k\ge n\}$ will be greater than or equal to $-M$. So, $-M$ is a lower bound for the sequence $(z_n)$.

By the Monotone Convergence Theorem, $(z_n)$ is convergent sequence.

(c) $(y_n)$ and $(z_n)$ are convergent sequences; their limiting value is the infimum/supremum. So, we are justified in setting,
\begin{align*}
 y &= \inf \{y_n:n \in \mathbf{N}\}\\
 z &= \sup \{z_m:m \in \mathbf{N}\}
\end{align*}

Observe that, the infimum of $k$-tail of the sequence is always less than or equal to the supremum of the tail. Hence, $z_k \le y_k$ for all $k \in \mathbf{N}$.

The visual representation makes it clear that, every $y_n$ is an upper bound for the sequence $\{z_m:m\in\mathbf{N}\}$. So, $\inf \{y_n:n \in \mathbf{N}\}$ is also an upper bound for the set $Z$. Thus, $\liminf a_n \le \limsup a_n$.

Consider the sequence
\begin{align*}
 a_n := \left(\frac{n+1}{n}\right)\sin\left( \frac{n\pi}{8}\right)
\end{align*}

This sequence is bounded, but not convergent, and so a strict inequality holds.

Code Snippet.

(d) Assume that $\limsup{a_n} = \liminf{a_n} = l$. Pick an abritrary $\epsilon > 0$.

```<code>import numpy as np
import matplotlib.pyplot as plt
from matplotlib import style
import math

# The sequence x_n = (n+1)/n sin(n/8)
n = np.linspace(1,500,500)
x_n = np.multiply(np.divide(n+10,n),np.sin(n/8))

# The supremum of a subsequence
def sup(x_m,n):
    return max(x_m[int(n-1):])

# The infimum of a subsequence
def inf(x_m,n):
    return min(x_m[int(n-1):])

limsup = [sup(x_n,m) for m in n]
liminf = [inf(x_n,m) for m in n]

plt.style.use('fivethirtyeight')
plt.figure(figsize=(20,10))
plt.xlim(0,450)
plt.text(250,1.2,r'<span class="math-container">$\sup \{x_m: m > n\}$</span>')
plt.text(250,-1.2,r'<span class="math-container">$\inf \{x_m: m > n\}$</span>')
plt.plot(n,x_n,'o')
plt.plot(n,limsup)
plt.plot(n,liminf)
plt.show()
</code>```

答:
For (d) with $U=\lim\sup a_n$ and $D=\lim\inf a_n:$

Show that for any $r>0$ the sets $\{n: a_n\ge U+r\}$ and $\{n:a_n\le D-r\}$ are finite.

So if $D=U=L$ then for any $r>0$ the set $\{n: |a_n-L|\ge r\}$ is finite, so $a_n$ converges to $L.$


问:
Why does the cumulative distribution function of the standard normal distribution have point symmetry in (0,0.5)?
The probability distribution function for the standard normal distribution, $\phi(z)=\frac{e^{-\frac{1}{2}z^2}}{\sqrt{2\pi}}$ is even, with line symmetry in the y-axis.

The cumulative distribution function of the standard normal distribution is $\Phi(z)=\int_{-\infty}^z\phi(t)dt$.

My textbook said that

Because $\phi(z)$ is even, $\Phi(z)$ has point symmetry in $(0,0.5)$.

Which makes sense to me by looking at their graphs (Since the PDF is the derivative of the CDF), but is there a more rigorous mathematical proof for why the CDF has point symmetry in (0,0.5)?

And more importantly, is there some general pattern/mathematical theory that I am missing here which I am supposed to know?

What I mean by this is, obviously, the primitive of $y=x^2$ (which is even) is $y=x^3$, which has point symmetry in the origin. But I was wondering if there is some greater mathematical theory that says something about 'if a function has line symmetry, its primitive will always have point symmetry' or something like that..

Thanks..

答: $$\Phi(x) = \int_{-\infty}^x \phi(z) dz \stackrel{z \to -z}{=} \int_{-x}^{+\infty} \phi(-z)dz = \int_{-x}^{+\infty} \phi(z)dz = 1 - \int_{-\infty}^{-x} \phi(z)dz = 1- \Phi(-x)$$


问:
Integrate: $\int_0^{\infty}\frac{\sinh (ax)}{\sinh x} \cos (bx) dx$

Q: If $|a|< 1$ and $b>0$, show that
  $$\int_0^{\infty}\frac{\sinh (ax)}{\sinh x} \cos (bx) dx = \frac{\pi  \sin (\pi  a)}{2 (\cos (\pi  a)+\cosh (\pi  b))}$$

I need to evaluate the above integral by method of contour. I tried to use this contour on this question but at $2\pi i$, $\sinh(ax)$ changes to $\sinh(ax+2a\pi i)$ and I have difficulty taking out $\sinh(ax)$. Please give hints on which contour to use.Thanks in advance!!

ADDED::

Considering $-R \to R \to R + \pi i \to -R + \pi i \to -R$ with a bump on $0$ and $\pi i$ to avoid singularity.

$$(1 + e^{(a+ib)\pi i}) \int_{-\infty}^{\infty}\frac{e^{(a+bi)x}}{\sinh x}dx = -\pi i(1 - e^{(a+ib)\pi i}) \hspace{1 cm}(1)$$
$$(1 + e^{(-a+ib)\pi i}) \int_{-\infty}^{\infty}\frac{e^{(-a+bi)x}}{\sinh x}dx = -\pi i(1 - e^{(-a+ib)\pi i}) \hspace{1 cm}(2)$$
With a bit of algebra, we get
\begin{align*}
\int_{-\infty}^{\infty}\frac{e^{ax}-e^{-ax}}{\sinh x}e^{ibx}dx &= 2\pi i \left( \frac{1}{1 + e^{(a+bi)\pi i}} -  \frac{1}{1 + e^{(-a+bi)\pi i}} \right)\\
 &= 2 \pi \frac{\sin (a\pi)}{\cosh (b\pi) + \sin(a\pi)}
\end{align*}
From which we get the desired result.

答:
Consider

$$f(z) = \frac{\sinh(ax)e^{ibx}}{\sinh(x)}$$

so that the function we want integrated is $\Re f$.  Let us find

$$I = \int_{-\infty}^\infty f(z)\,dz$$

which is twice the integral in question.  Let $C$ be this integral, indented to avoid the pole at $z=0$:

The integral around the large arc disappears as $R\to\infty$, and as $\epsilon\to 0$, the integral around the small arc becomes:

$$\pi i \operatorname*{Res}_{z=0} f(z) = 0$$

Then we have, seeing that $f$ has poles in $C$ at $z_n = \pi i n$ for $n \ge 1$.

$$\operatorname*{Res}_{z=\pi in} f(z) = \lim_{z\to \pi i n}\frac{\sinh(ax)e^{ibx}}{\cosh(x)} = i(-1)^n e^{-b\pi n}\sin(\pi a n)$$

Then summing:

$$
I=\oint_C f(z)\, dz =\\
\Re\, 2\pi i \sum_{n=1}^\infty \operatorname*{Res}_{z=\pi i n} f(z) = \\
-2\pi\Im \sum_{n=1}^\infty (-1)^n e^{-b\pi n+\pi i a n} = \\
-2 \pi \Im \frac{1}{1+e^{-b\pi +\pi i a }} = \\
\frac{\pi\sin(a\pi)}{\cos(a\pi)+\cosh(b\pi)}
$$

Divide by two and the answer is obtained.


问:
How to to find the sol. By pascal's roule and mathematical induction
\begin{equation}
\begin{array}{l}
\text { - For } n \geqslant 2 \text { , prove that } \\
\left(\begin{array}{c}
2 \\
2
\end{array}\right)+\left(\begin{array}{c}
3 \\
2
\end{array}\right)+\left(\begin{array}{c}
4 \\
2
\end{array}\right)+\ldots+\left(\begin{array}{c}
n \\
2
\end{array}\right)=\left(\begin{array}{c}
n+1 \\
3
\end{array}\right)
\end{array}
\end{equation}

答:
By induction, since the case $n=2$ is straightforward, we have using Pascal's rule
$$ \sum_{p=2}^{n+1}\binom{p}{2}=\sum_{p=2}^n\binom{p}{2}+\binom{n+1}{2}=\binom{n+1}{3}+\binom{n+1}{2}=\binom{n+2}{3} $$
Another way to obtain this result is by simply using Pascal's rule
$$ \binom{p}{k}=\binom{p+1}{k+1}-\binom{p}{k+1} $$ therefore, your sum is equal to
$$ \sum_{p=2}^n\left[\binom{p+1}{3}-\binom{p}{3}\right]=\binom{n+1}{3}-\binom{2}{3}=\binom{n+1}{3} $$
since the terms cancel each other.


问:
Monomial order refines the partial order of divisibility
Background:

Let $X$ be a set of possibly infinitely many variables, and let $K[X]$ be the polynomial ring with variables in $X$ over some field $K$. Define a monomial order to be a total order $>$ on the monomials of $K[X]$ such that for any monomials $m_1$,$m_2$, and any monomial $n\not=1$, we have $m_1>m_2$ implies $nm_1>nm_2>m_2$.

Question:

I want to show that any monomial order $>$ refines the partial order of divisibility on the monomials of $K[X]$. More specifically, I want to show that if $u=vw$ for some monomials $u,v,w$ with $v\not=1$ and $w\not=1$, then $u>v$ and $u>w$.

My thoughts:

I'd like to make use of the part of the definition that says $nm_2>m_2$, for it would follow that $vw>v$ and $vw>w$. But I need to know there is some monomial strictly greater than $v$ and some monomial strictly greater than $w$. How do I know that $u$ or $v$ are not somehow a maximal element with respect to $>$?

Is this definition of monomial order just bad? Or am I missing something? I'm hoping that I am just missing something.

答:
Think I figured it out. Just work by cases.

First suppose $v=w$. Since $w\not=1$, we cannot have $u=v$. Thus either $u>v$ or $v>u$. If $u>v$ then also $u>w$ so we're done. Suppose now that $v>u$. Then since $w\not=1$ we have $wv>wu$ so $u>wu$, a contradiction. This handles the case of $v=w$.

Suppose now $v\not=w$. WLOG assume $v>w$. Then since $w\not=1$ we have $wv>w$, so $u>w$. Again, since $v,w\not=1$, we cannot have $u=v$, so either $u>v$ or $v>u$. If $u>v$ we are done so suppose $v>u$. Since $w\not=1$ we obtain $wv>wu$, so $u>wu$, a contradiction.

Side note: both times we obtain a contradiction with $u>wu$ because $u$ is not a maximal element ($v>u$) and therefore since $w\not=1$, we have by definition $wu>u$.


问:
$(X,\tau)$ a topological space with $X\neq \emptyset$ and $D = \{D \subseteq X : D$ is dense in $(X, \tau) \}$.
Be $(X,\tau)$ a topological space with $X\neq \emptyset$ and $D = \{D \subseteq
X : D$ is dense in $(X, \tau) \}$.

(a)Prove that it is a discrete topology over X if, and only if, $D = \{X\}$.

Definition. The discrete topology is formed by the parts of X, i.e, ℘(X).

From the imprint that only set X is dense in itself, and no other subset of $X$ is dense in $X$.

(b) Prove that τ is the chaotic topology on X if, and only if, $D$ =
℘(X) - {∅}.

Definition. The caotica topology consists of: $\{\emptyset, X\}$

Can you help me, please?

答:
A subset $D$ of $X$ is dense in $X$ if and only if $\operatorname{cl}D=X$.

Show that if $X$ has the discrete topology, $\operatorname{cl}A=A$ for every $A\subseteq X$.

Show that if $X$ has the indiscrete topology, $\operatorname{cl}A=X$ for every non-empty $A\subseteq X$.

Once you’ve done this, you’ll have shown $X$ is the only dense subset of $X$ in the discrete topology, and every non-empty subset of $X$ is dense in $X$ in the indiscrete topology. (Why?)

For the other direction, you want to show (a) that if $X$ is the only dense subset of $X$, then $X$ has the discrete topology, and (b) that if every non-empty subset of $X$ is dense in $X$, then $X$ has the indiscrete topology.

For (b), show that if every non-empty subset of $X$ is dense in $X$, then the only closed sets in $X$ are $\varnothing$ and $X$. HINT: If $F\subseteq X$ is closed, then $F=\operatorname{cl}F$.

For (a), note that if $X$ does not have the discrete topology, then there is an $x\in X$ such that $\{x\}$ is not open; show that $X\setminus\{x\}$ is dense in $X$.


问:
Prove that $5^n \equiv 8n^2-4n+1 \mod 64$ using induction
I am trying to prove that $5^n \equiv 8n^2-4n+1 \mod 64 $ using induction. I have seen the post which solves this problem using other methods, but I really need to be able to do it using induction.

Here is what I have so far:

Base step:
$5 \equiv 5 \mod 64$

Then, I assumed that $5^k \equiv 8k^2 -4k+1 \mod 64.$

I want to prove that $5^{k+1} \equiv 8(k+1)^2-4(k+1)+1 \mod 64$.

Note that $8(k+1)^2-4(k+1)+1=8k^2+12k+5$.

Hence,
$5^{k+1} \equiv 5^k *5 \mod 64
\equiv (8k^2-4k+1)*5 \mod 64 \equiv 40k^2 -20k+5 \mod 64.$

I am stuck here.
Any insight would be very useful!!
Thanks

答:
Using binomial expansion,

$$5^n=(1+4)^n=1+\binom n1 4 + \binom n2 4^2 + \binom n3 \color{blue}{4^3} + \binom n4 \color{blue}{4^4} +\cdots + \binom nn \color{blue}{4^n}.$$

Since the binomial coefficients are integers and $4^3=64$, we have

$$5^n\equiv1+4n+8n(n-1)=8n^2-4n+1\bmod64.$$


问:
A limit of $f(x)\ln(x)$

Consider a real $\alpha>0$ and a function $\phi$ defined in $(-\alpha,\alpha)$ satisfying $\displaystyle{\lim_{x\to0}\phi(x)=0}$. Does $\displaystyle{\lim_{x\to0}\phi(x)\ln(x)}$ exists?

We know that if $\phi$ is differentiable at $0$ then the limit exists and its value is $0$ since $\displaystyle{\lim_{x\to0}\phi(x)\ln(x)}= \displaystyle{\lim_{x\to0}\dfrac{\phi(x)}x\times x\ln(x)=\phi’(0)\times 0=0}$.

But if we suppose that $\phi$ not necessarily differentiable is there a way to find this limit?

I guessed that the limit does not necessarily exist but I could not find a counter example.

Tried $\phi(x)=x^r$ where $r$ is positive rational number, in that case the limit is 0.

Also tried $\phi(x)=x\sin\left(\dfrac1x\right)$ but $|\phi(x)\ln(x)|\le |x\ln(x)|$ and also 0.

Any idea?

答: Let $\phi(x)=\frac{\sin\left(\frac{1}{x}\right)}{\ln(x)}$, then $\lim\limits_{x\rightarrow 0}\phi(x)=0$ but $\lim\limits_{x\rightarrow 0}\phi(x)\ln(x)=\lim\limits_{x\rightarrow 0}\sin\left(\frac{1}{x}\right)$ does not exist.

Title: A limit of $f(x)\ln(x)$


问:
How do I find the series expansion of the meromorphic function $\frac{1}{e^z+1}$?
in a theoretical physics book, the author makes the following claim:

$$\frac{1}{e^z + 1} = \frac{1}{2} + \sum_{n=-\infty}^\infty \frac{1}{(2n+1) i\pi - z}$$
and justifies this as

These series can be derived from a theorem which states that any meromorphic function may be expanded as a summation over its poles and residues at those poles

What's the name of that theorem? It's not really a Laurent series, since the Laurent series is for an expansion around one particular point only. I can see that the poles occur whenever $z = (2n+1)i\pi$ for $n \in \mathbb{N}$, but then where does that constant $1/2$ come from?

EDIT: Well, it appears that the general claim isn't valid, so now I'd be interested in a justification for the expansion in my particular example...

答:
Taken literally, the claim is certainly not true. Meromorphic functions are not uniquely determined by their poles and corresponding residues. Take for example
$$ \frac{1}{z} \qquad\text{and}\qquad \frac{e^z}{z}, $$
which have the same poles and same residues.

However, Mittag-Leffler's theorem as noted by Random Variable is related.

For your particular example, you can start from a well known series expansion of $\cot z$:
$$\cot z = \lim_{N\to\infty} \sum_{n=-N}^N \frac{1}{z+n\pi} = \frac1z + \sum_{n=0}^\infty  \frac{2z}{n^2\pi^2-z^2},$$
see for example this for a derivation.

Let's interpret your series as a symmetric sum and rewrite it by summing the terms pairwise: $$(\operatorname{term} 0 + \operatorname{term} -1) + (\operatorname{term} 1 + \operatorname{term} -2) + \cdots $$

This will give us
$$
\sum_{n=-N-1}^N \frac{1}{(2n+1)i\pi - z} =
-i\sum_{n=0}^N \frac{2z}{(2n+1)^2\pi^2 + z^2},
$$
which is very similar to the series for $\cot z$, but where we have thrown away half the terms. Inspired by this observation, let's compute
$$
\cot z - \frac12\cot \frac z2 =
\frac1z + \sum_{n=0}^\infty  \frac{2z}{n^2\pi^2-z^2} - \frac1{z} - \sum_{n=0}^\infty  \frac{2z}{4n^2\pi^2-z^2} = \sum_{n=0}^\infty \frac{2z}{(2n+1)^2\pi^2-z^2}.
$$
Hence
$$
\cot iz - \frac12\cot \frac {iz}{2} =
\sum_{n=0}^\infty \frac{2z}{(2n+1)^2-(iz)^2} =
\sum_{n=0}^\infty \frac{2z}{(2n+1)^2\pi^2+z^2}.
$$
Consequently, your sum is
\begin{align}
\frac12 -i\sum_{n=0}^\infty \frac{2z}{(2n+1)^2\pi^2 + z^2} &=
\frac12+\frac i2\cot \frac {iz}{2} - i\cot iz \\
&= \frac{1}{e^z+1}.
\end{align}


问:
Possible Converse to Iwasawa Lemma
After looking at how Iwasawa's Lemma proves the simplicity of $PSL_n(q)$ and $PSp_{2n}(q)$ groups, I wondered how powerful this lemma is. As such, I thought about this possible converse.

For reference, Iwasawa's Lemma:

For group $G$,

$G$ has a primitive group action $\phi:G\rightarrow\text{Sym}(X)$ on set $X$ (transitive, and $\text{Stab}(x)$ is maximal subgroup of $G$)

$G=G'$ (commutator subgroup)

There is a soluble normal subgroup $A$ of $\text{Stab}(x)$ such that $G=\langle gAg^{-1}|g\in G\rangle$.

Then $G/\text{ker}(\phi)$ is simple.

A possible converse can be postulated:

For every simple group $G$, $G$ can be expressed as $H/N$, where

$H$ has a primitive group action $\phi:H\rightarrow\text{Sym}(X)$ on set $X$ (transitive, and $\text{Stab}(x)$ is maximal subgroup of $H$)

$H=H'$ (commutator subgroup)

There is a soluble normal subgroup $A$ of $\text{Stab}(x)$ such that $H=\langle hAh^{-1}|h\in H\rangle$.

$N=\text{ker}(\phi)$

The overarching question is: Which simple groups can possibly satisfy the converse of Iwasawa's Lemma?

You may assume Classification of Finite Simple Groups.

Here's a possible attempt:

As we know the alternating groups $A_n (n\geq5)$ and cyclic groups of prime order $C_p$ are simple, we can try to look at them first. However, the natural action $A_n$ acting on $\{1...n\}$ does not work because the stabilisers are isomorphic to $A_{n-1}$ with no nontrivial normal subgroups for $n\geq 6$.

On examination, it follows that $G$ must act transitively on $X$ too. However, for $G=C_p$, this means that $|X|=p$ otherwise the action is trivial. I'm not sure how to further analyse this case, but it doesn't seem possible for abelian simple groups.

It was observed that $A_6$ can act on left cosets of $S_4$ (with inclusion map $\sigma\mapsto\sigma$ if $\sigma$ is even and $\sigma(5,6)$ otherwise) in $A_6$ via left multiplication. $A_6$ be expressed as $A_6/\{e\}$ with a maximal subgroup isomorphic to $S_4$ (standard twisted $S_4$). $S_4$ trivially has a soluble normal subgroup $A_4$, containing the 3-cycle $(1,2,3)$, whose conjugates generate $A_6$. This doesn't seem to generalise easily to the other altermating groups.

答:
Let $G$ be a non-abelian finite simple group. Then $G = G'$ and there exists a maximal subgroup $M < G$ with a normal solvable subgroup $N \neq 1$. In other words, the converse to Iwasawa's theorem holds.

See the answer and comments to this question on MO: Maximal subgroups of simple groups with normal $2$-subgroups.

Title: Possible Converse to Iwasawa Lemma


问:
Stochastic Integration: Riemann Sums
I am reading this book:

Introduction to Stochastic Differential Equations

On page 63, in the lemma in the middle of the page the author has the following

$\displaystyle \sum_{k=0}^{m_n - 1}W(\tau_k^n)(W(t_{k+1}^n) - W(t_k^n)) = \frac{W^2(T)}{2} - \frac{1}{2}\sum_{k=0}^{m_n - 1}(W(t_{k+1}^n) - W(t_k^n))^2 + \sum_{k=0}^{m_n - 1}(W(\tau_{k}^n) - W(t_k^n))^2 + \sum_{k=0}^{m_n - 1}(W(t_{k+1}^n) - W(\tau_k^n))(W(\tau_{k}^n) - W(t_k^n))$

$\tau_k$ is a midpoint in $[t_k, t_{k+1}]$

WLOG we can assume $W(t_0) = 0$ as it is a standard Brownian motion. Let $m_n = 2$ then we get the following:

NOTE: I will omit the exponent $n$ and $W(t_2) = W(T)$ as this is on $[0,T]$

The LHS becomes :

$W(\tau_0)(W(t_1) - 0) + W(\tau_1)(W(T) - W(t_1))$

The RHS side is then:

$\frac{1}{2}W^2(T) - \frac{1}{2}[W^2(t_1) + W^2(T) - 2W(T)W(t_1) + W^2(t_1)] + [W^2(\tau_0) + W^2(\tau_1) - 2W(\tau_1)W(t_1) + W^2(t_1)] + ...$

$= W(\tau_0)(W(t_1) - 0) + W(\tau_1)(W(T) - W(t_1)) + \bigg ((W^2(t_1) - \frac{1}{2}W^2(t_1) \bigg )$

So they are almost similar except the term $\bigg ((W^2(t_1) - \frac{1}{2}W^2(t_1) \bigg )$. And it seems there would be more of these terms left over for $m_n = j, j > 2$.

Possibly I made a mistake in my algebra, easy to misplace terms in expanding. If I did not make a mistake though could someone clarify because as it stands it seems incorrect.

答:
For ease of notation we replace $m_n,t_k^n,\tau_k^n$ with $m,t_k,\tau_k$ respectively. Starting on the right, we have the following:
\begin{align}
\text{RHS } & = \frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} -\frac{1}{2}W(t_{k+1})^2 + W(t_{k+1})W(t_k) - \frac{1}{2}W(t_k)^2 + W(\tau_k)^2 - 2 W(\tau_k)W(t_k)+W(t_k)^2 + W(t_{k+1})W(\tau_k) - W(t_{k+1})W(t_k) - W(\tau_k)^2 + W(\tau_k)W(t_k) \\
\end{align}

We cancel $W(t_{k+1})W(t_k) - W(t_{k+1})W(t_k)$ to get
\begin{equation}
\frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} -\frac{1}{2}W(t_{k+1})^2 - \frac{1}{2}W(t_k)^2 + W(\tau_k)^2 - 2 W(\tau_k)W(t_k)+W(t_k)^2 + W(t_{k+1})W(\tau_k) - W(\tau_k)^2 + W(\tau_k)W(t_k).
\end{equation}
Then we cancel $W(\tau_k)^2 - W(\tau_k)^2$ to get
\begin{equation}
\frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} -\frac{1}{2}W(t_{k+1})^2 - \frac{1}{2}W(t_k)^2 - 2 W(\tau_k)W(t_k)+W(t_k)^2 + W(t_{k+1})W(\tau_k)  + W(\tau_k)W(t_k).
\end{equation}
Then $W(\tau_k)W(t_k) - 2W(\tau_k)W(t_k) = -W(\tau_k)W(t_k)$, and we have
\begin{equation}
\frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} -\frac{1}{2}W(t_{k+1})^2 - \frac{1}{2}W(t_k)^2 - W(\tau_k)W(t_k)+W(t_k)^2 + W(t_{k+1})W(\tau_k) .
\end{equation}
We then combine $W(t_k)^2 - \frac{1}{2}W(t_k)^2 = \frac{1}{2}W(t_k)^2 $ to get
\begin{equation}
\frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} -\frac{1}{2}W(t_{k+1})^2 + \frac{1}{2}W(t_k)^2 - W(\tau_k)W(t_k) + W(t_{k+1})W(\tau_k) .
\end{equation}
Some slight rearranging yields the following:
\begin{equation}
\frac{W(t_m)^2}{2} +\sum_{k=0}^{m-1} \frac{1}{2}W(t_k)^2 -\frac{1}{2}W(t_{k+1})^2  + W(\tau_k)(W(t_{k+1}) - W(t_k)).
\end{equation}

Note that $\sum_{k=0}^{m-1} \frac{1}{2}W(t_k)^2 -\frac{1}{2}W(t_{k+1})^2$ is a telescoping sum, so we have
\begin{equation}
\frac{W(t_m)^2}{2} +(\frac{1}{2}W(t_0)^2 -\frac{1}{2}W(t_{m})^2)  + \sum_{k=0}^{m-1} W(\tau_k)(W(t_{k+1}) - W(t_k)).
\end{equation}
As you noted, we have standard Brownian motion, so $W(t_0)^2 = 0$, and we obtain the left hand side:
\begin{equation}
\text{RHS } = \frac{1}{2}W(t_0)^2 + \sum_{k=0}^{m-1} W(\tau_k)(W(t_{k+1}) - W(t_k)) = \sum_{k=0}^{m-1} W(\tau_k)(W(t_{k+1}) - W(t_k)) = \text{LHS}.
\end{equation}

Title: Stochastic Integration: Riemann Sums


答:
Here is a partial answer for alternating groups $A_n,n\geq5,n\neq6$.

$A_n$ can be expressed as $A_n/\{e\}$. We take the subgroup $S$ of permutations that preserve $\{1...n-3\}$, i.e. $\{\sigma(1)...\sigma(n-3)\}=\{1...n-3\}$. Note:

$A_n=A_n'$, since $A_n'\trianglelefteq A_n$ and $A_n'$ is nontrivial.

$S$ is maximal in $A_n$ (conjectured with GAP)

Then:

Let $A_n$ act on left cosets $gS$ in $A_n$ by left multiplication. Then $\text{Stab}(eS)=S$ is maximal and the action $\phi:A_n\rightarrow\text{Sym}(gS)$ is transitive.

$S\triangleright\langle(n-2,n-1,n)\rangle$, this subgroup is soluble and conjugates of it generate $A_n$ (since 3-cycles generate $A_n$). This requires $n\geq5$.

Clearly, $A_n\triangleright\text{ker}(\phi)=\{e\}$

Title: Possible Converse to Iwasawa Lemma

Body:
After looking at how Iwasawa's Lemma proves the simplicity of $PSL_n(q)$ and $PSp_{2n}(q)$ groups, I wondered how powerful this lemma is. As such, I thought about this possible converse.

For reference, Iwasawa's Lemma:

For group $G$,

$G$ has a primitive group action $\phi:G\rightarrow\text{Sym}(X)$ on set $X$ (transitive, and $\text{Stab}(x)$ is maximal subgroup of $G$)

$G=G'$ (commutator subgroup)

There is a soluble normal subgroup $A$ of $\text{Stab}(x)$ such that $G=\langle gAg^{-1}|g\in G\rangle$.

Then $G/\text{ker}(\phi)$ is simple.

A possible converse can be postulated:

For every simple group $G$, $G$ can be expressed as $H/N$, where

$H$ has a primitive group action $\phi:H\rightarrow\text{Sym}(X)$ on set $X$ (transitive, and $\text{Stab}(x)$ is maximal subgroup of $H$)

$H=H'$ (commutator subgroup)

There is a soluble normal subgroup $A$ of $\text{Stab}(x)$ such that $H=\langle hAh^{-1}|h\in H\rangle$.

$N=\text{ker}(\phi)$

The overarching question is: Which simple groups can possibly satisfy the converse of Iwasawa's Lemma?

You may assume Classification of Finite Simple Groups.

Here's a possible attempt:

As we know the alternating groups $A_n (n\geq5)$ and cyclic groups of prime order $C_p$ are simple, we can try to look at them first. However, the natural action $A_n$ acting on $\{1...n\}$ does not work because the stabilisers are isomorphic to $A_{n-1}$ with no nontrivial normal subgroups for $n\geq 6$.

On examination, it follows that $G$ must act transitively on $X$ too. However, for $G=C_p$, this means that $|X|=p$ otherwise the action is trivial. I'm not sure how to further analyse this case, but it doesn't seem possible for abelian simple groups.

It was observed that $A_6$ can act on left cosets of $S_4$ (with inclusion map $\sigma\mapsto\sigma$ if $\sigma$ is even and $\sigma(5,6)$ otherwise) in $A_6$ via left multiplication. $A_6$ be expressed as $A_6/\{e\}$ with a maximal subgroup isomorphic to $S_4$ (standard twisted $S_4$). $S_4$ trivially has a soluble normal subgroup $A_4$, containing the 3-cycle $(1,2,3)$, whose conjugates generate $A_6$. This doesn't seem to generalise easily to the other altermating groups.


问:
In how many ways can I select 13 cards from a standard deck of 52 cards so that 5 of those cards are of the same suit?
My professor says the solution is $47 \choose 8$, but I don't understand how and why. Could you please explain the solution or say whether it is correct or not?
Thanks in advance!

答:
It is plausible that either your professor misunderstood one detail of the question, or you misheard the question, and thus accidentally omitted a detail. As such, I present what might have been "what was your professor actually giving an answer to":

In the context of a deck of cards, $\binom{47}{8}$ would most naturally be interpreted as the number of ways you could choose 8 cards from the deck after removing 5 specific cards. It is easy to think that the 5 specific cards are part of the "selected" set, and thus the total set is 13, made up of the 5 specified cards plus 8 other cards.

With this, it is plausible that your professor was thinking of "five cards of the same suit" as five specific cards, and it doesn't matter if some of the other cards are of the same suit as those five.

In particular, it would be easy for your professor to inadvertently think it was "how many ways can I select 13 cards from a standard deck of 52 cards so that the 5 'face' cards of one suit are all present". Alternatively, it would be easy for you to miss those specific details.

It is difficult to see other ways to obtain a similar answer. It's possible, however, that your professor mistakenly considered the question to be comparable to similar-sounding questions like how you could seat $x$ people so that a group of $y$ friends are seated together.


问:
I found a curious pattern... any idea of a deductive formula? - [4, 24, 80, 200, 420, 784, 1344]
I found this pattern by taking the absolute value of the difference of both sides of the Cauchy-Schwarz formula with consecutive numbers (I can go into more detail if desired). No worries if this doesn't make sense to you.

I'm looking to find a deductive sequence for this formula (I'm alright if you are able to find a recursive formula too). I can also use a program to generate many more terms of the sequence if you want me to.

Based on intuition, I have a feeling that the answer is going to be a ratio of various factorials.

Let me know if you have any questions... thanks!

答:
It looks to me like being :
$$
a_n = \frac{(n+1)^2((n+1)^2-1)}{3} = \frac{n(n+1)^2(n+2)}{3}$$

starting with $a_1 = 4$.

EDIT : As Gottfried points out below, it is easier to "shift the index" and go with the formula $a_n = \frac{n^2(n^2-1)}{3}$ with $a_2 = 4$ and $a_1 = 0$.

For example : $a_3= \frac{16 \times 15}{3} = \frac{240}{3} = 80$.

Note that you have missed out the term for $n=5$ which is $420$, but all other terms seem to match.

EDIT : Assuming that you have the $420$ in place, so have sufficiently many terms of the sequence to consider, you can use the method of finite differences to check if the terms are polynomial in the index or not.

That is, for a sequence $a_n$ we have the sequence of differences $(\Delta \{a_n\})_k = a_{k+1} - a_k$. Now, if it turns out that $\Delta^t \{a_n\}$ is identically zero, then $a_n$ is in fact a polynomial in $n$ of degree less than $t$.

Once you see this, you can use any $t$ terms of the sequence and try to fit in the coefficients corresponding to the polynomial $a_n  = c_0 + c_1n + ... + c_tn^{t-1}$ to get the coefficients.

For example, considering the sequence of finite differences, we get :
\begin{align}
4 & & 24 & & 80 & & 200 & & 420 & & 784 \\
&\quad \quad 20 & & \quad \quad 56 & & \quad \quad 120 & & \quad \quad 220 & & \quad \quad 364 &  \\
& & 36 & & 64 & & 100 & & 144 && \\
& & & \quad \quad 28 & & \quad \quad \ 36 & & \quad \quad \ 44 &&& \\
& & & & 8 & & 8 & & \\
&&&&&\quad \quad \ \ 0&&&&&
\end{align}
(note : using $1344$ as well one sees that the fifth row gives a sequence of $0$s)

Thus $a_n$ is a quartic in $n$ , and one can use the first four values of the sequence to get the correct expression.

(Alternately, observe that you can hazard a guess at the third row above , which is just the even squares. From there, you can invert the difference operator twice to get the answer).

I have also explained a similar phenomena for powers in this answer, where I showed how one can find powers of numbers by inverting the difference technique.


问:
Doubt about a proof on linear congruence equations
I'm going to copy-paste a proposition and a proof from my textbook.

Proposition

$ax \equiv_m b \land GCD(a,m)=1$ has solutions $\implies$ if $c$ is solution of the congruence $ax \equiv_m 1 \implies c*b $ is the solution of  $ax \equiv_m b$

Proof

$ac \equiv_m 1 \iff ac - mh =1$

$\iff acb = mhb = b$ $\iff a(cb) - b  = m(hb)$
$\iff m|a(bc) - b$

Which is the definition of $a(bc) \equiv_m b$

$\square$

But to me it looks very confusing, looks like it proves the $\Leftarrow $ implication.

I'd have written the proposition in this way:

$ ax \equiv_m 1$ has solutions $\implies ax \equiv_m b$ has $c*b$ as solution $ \land GCD(a,m)=1$.

 Am I wrong?

答:
One too many $\iff$.

$a(cb) - b = m(hb)$ implies $m \mid a(cb) - b$. But $m \mid a(cb) - b$ does not imply $a(cb) - b = m(hb)$. There is no guarantee that $\dfrac{a(cb) - b}m$ is divisible by $b$, since we do not know if $\gcd(m,b) = 1$.

What you actually have is
$$\begin{align}ac \equiv_m 1 &\iff ac - mh =1\\
&\iff acb = mhb = b\\
&\iff a(cb) - b  = m(hb)\\
&\color{red}\implies m\mid a(bc) - b\\
&\implies a(bc) \equiv_m b\end{align}$$

Or one can prove that in general $$u\equiv_m v \implies ut \equiv_m vt$$ which
gets you the same result much faster, and follows simply because if $m \mid (u-v)$, then $m \mid (u-v)t$.


问:
solve non linear differential equation: $y'\cdot\alpha+y+\beta\cdot e^{\delta\cdot y}+\theta = 0$
Could somebody help me to solve the non linear differential equation, where $y$ is a function of the time and starts with $y(0)=0$
$$
y'\cdot\alpha+y+\beta\cdot e^{\delta\cdot y}+\theta = 0
$$
It will guess it will involve the Lambert W functions.</p>

答:
your equation is of the form

$ y'= f(y) $ this can be expressed in the form (implicit)

$$ x= \frac{1}{a}\int \frac{dy}{-d-be^{cy}-y} $$

of course hre $ a= \alpha $ $b= \beta $ and $c= \delta $ and $ d= \theta $to simplify terms

i do not know how to integrate the expression i guess the integral can not be obtained exactly


问:
Bound on determinant sum if some of rows and columns are zeros
Let $A$ be positive definite and $B$ be positive semidefinite. From the Minkowski inequality,

$$ \det [A + B] \geq \det [A] + \det [ B ] $$

Now, consider the case the $i$th row and column of $B$ are all zeros. Then, $\det [B] = 0$. Let $B’$ denote $B$ with the $i$th row and column deleted. Then, I believe we now have the following inequality:

$$ \det [ A + B ] \geq \det [ A ] + \det [ B’ ] $$

I am not sure how to prove this, but this seems true. My logic is that the following.

The determinant of a matrix is the volume spanned by the columns of the matrix. So, if we have a row and column of zeros, then adding the matrices together increases the volume of A by no more than the volume of $\det [B’]$. At least, this is my intuition. I’m struggling to prove this hypothesis.

So, any suggestion on proving this?

答:
This is not true. Consider
$$A=\begin{bmatrix}1&\epsilon \\\epsilon&\epsilon\end{bmatrix},\ B=\begin{bmatrix}x&0\\0&0\end{bmatrix}$$
for some $0<\epsilon<1$. The matrix $A$ is positive definite since its diagonal entries and determinant are positive, and $B$ is positive semidefinite if $x\geq 0$. Then
$$\det(A+B)=(1+x)\epsilon-\epsilon^2=\det(A)+x\epsilon = \det(A)+\epsilon\det(B')<\det(A)+\det(B').$$


问:
Partial sum of $\sum_{n=1}^m 10^{m-n} \lfloor 10^{n}x - 10 \lfloor 10^{n-1}x \rfloor \rfloor$ or number of digits after decimal point
I wondered how to calculate the minimum place value in $x$ (Or the number of digits after the decimal point) in math way. For quantity of digits in number (or max place value) used is a $\lfloor\lg(n)\rfloor+1$, and I need a "reverse operation". I have not found solutions to this problem, if you know please write.

I myself have come up with only a few ideas. First idea is find partial sum of $\sum_{n=1}^m 10^{m-n} \lfloor 10^{n}x - 10 \lfloor 10^{n-1}x \rfloor \rfloor$, then equate it to $10^m(x-\lfloor x \rfloor)$ and express $m$ ($m$ is number of digits after decimal point). It turns out something like that where $x \in \Bbb{Q }\ \mathit{with}\ \mathit{terminating}\ \mathit{decimal}$:
$$10^m(x-\lfloor x \rfloor) = \sum_{n=1}^m 10^{n-1} \lfloor 10^{m-n+1}x - 10 \lfloor 10^{m-n}x \rfloor \rfloor $$
For example, for $x = 0.365$ then $10^m(x-\lfloor x \rfloor) = 365$. In turn partial sum gets the digits of each place value, and multiplies it by the place value. Unfortunately I am not very sophisticated in deep mathematics, and I know very little about simplify of series, partial sums, convergence and divergence.

Second idea is this infinity series. She definitely gives the correct logic answer for $x \in \Bbb{Q }\ \mathit{with}\ \mathit{terminating}\ \mathit{decimal}$, but I'm not sure can it be shortened and convergence or divergence this.
$$\sum_{n=1}^\infty \lceil 10^{n-1}x - \lfloor 10^{n-1}x \rfloor \rceil $$
This series is round every exist remainder to $1$ after multiplying by $10^{n-1}$.

Thanks in advance for your help.

答: A decimal terminates iff it is of the form $ \frac{a}{2^b 5^c}$, where $a$ is an integer, and $b,c$ are natural numbers. In addition, it terminates at exactly $ \max(b, c)$ decimal places.


问:
Is there a formula for $1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$?

Given
$$f(n) = 1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$$
I am wondering if there is a straightforward formula to compute $f(n)$ and how it may be derived.

The only reduction I thought about so far would be:

$$n\cdot 1 + (n - 1)\cdot 2 + (n - 3)\cdot 3 +\cdots $$

which seems symmetrical; for example, an odd and even n

```<code> 5 4 3 2 1        4 3 2 1
*                *
 1 2 3 4 5        1 2 3 4
</code>```

but I'm not sure if and how it may help derive a formula.

答:
I think the initial steps towards a formula look a bit simpler than seen so far here.

Beginning with
$$ H^{(2)}(n) = 1 + (1+2) + (1+2+3) + ... + (1+2+3+4+...+n) $$
we count the number of occurences of the $1$, of the $2$
$$ H^{(2)}(n) = 1\cdot n + 2 \cdot (n-1) + 3\cdot (n-2) + ... + n \cdot 1 $$
and rewrite a bit
$$ \begin{align} H^{(2)}(n) &= 1\cdot ((n+1)-1) + 2 \cdot ((n+1)-2) + 3\cdot ((n+1)-3) + ... + n \cdot ((n+1)-n) \\ \qquad \\
&= (1+2+3+...+n)(n+1) - (1^2+2^2+3^2+...+n^2) \\ \qquad \\
&= {n(n+1) \over 2}(n+1) - {2n^3 + 3n^2 + 1n\over 6} \end{align}$$
The last steps stems from the faulhaber formulae for sums-of-like-powers.

Final step, expanding & collecting this gives the same result as derived by other answerers:
$$ H^{(2)}(n) = {3n(n+1)(n+1)-2n^3 - 3n^2 - 1n \over 6} \\
= {n^3+3n^2+2n \over 6} \\
= {n(n+1)(n+2) \over 3!}  $$
For $n=5$ we get the sum $H^{(2)}(5)=35$


答:
I came with a solution but it is not quite satisfying regarding its efficiency. Notice first that the number of digits of an integer is a regular notion in the sense that $n$ has $m$ digits if and only if $10^m\leqslant n<10^{m+1}$. But this is not the case of the number of digits after the decimal point. Let $f(x)$ be a fonction such that $f(x)$ is the number of digits after the decimal point whenever $x$ is a decimal number, then we would have $f(0)=0$, $f(0.1)=1$ but near these numbers, $f\left(\frac{1}{10^m}\right)=m$ for all $m$ therefore such a function $f$ has to be irregular (in particular it cannot be continuous by pieces at all). Thus don't expect $f$ to be pleasing. Anyway I say that
$$ f(x)=\sum_{m=0}^{+\infty}1_{\{10^m x\notin\mathbb{Z}\}} $$
Now, the trick is that for all $x\in\mathbb{R}$, we have $x\in\mathbb{Z}\iff \cos^2(\pi x)=1\iff \lfloor\cos^2(\pi x)\rfloor=1$ because when $x\notin\mathbb{Z}$, $\cos^2(\pi x)\in [0,1)$ and thus $\lfloor\cos^2(\pi x)\rfloor=0$. Therefore,
$$ f(x)=\sum_{m=0}^{+\infty}\left(1-\lfloor\cos^2(\pi 10^m x)\rfloor\right) $$
fits.

Title: Partial sum of $\sum_{n=1}^m 10^{m-n} \lfloor 10^{n}x - 10 \lfloor 10^{n-1}x \rfloor \rfloor$ or number of digits after decimal point

Body:
I wondered how to calculate the minimum place value in $x$ (Or the number of digits after the decimal point) in math way. For quantity of digits in number (or max place value) used is a $\lfloor\lg(n)\rfloor+1$, and I need a "reverse operation". I have not found solutions to this problem, if you know please write.

I myself have come up with only a few ideas. First idea is find partial sum of $\sum_{n=1}^m 10^{m-n} \lfloor 10^{n}x - 10 \lfloor 10^{n-1}x \rfloor \rfloor$, then equate it to $10^m(x-\lfloor x \rfloor)$ and express $m$ ($m$ is number of digits after decimal point). It turns out something like that where $x \in \Bbb{Q }\ \mathit{with}\ \mathit{terminating}\ \mathit{decimal}$:
$$10^m(x-\lfloor x \rfloor) = \sum_{n=1}^m 10^{n-1} \lfloor 10^{m-n+1}x - 10 \lfloor 10^{m-n}x \rfloor \rfloor $$
For example, for $x = 0.365$ then $10^m(x-\lfloor x \rfloor) = 365$. In turn partial sum gets the digits of each place value, and multiplies it by the place value. Unfortunately I am not very sophisticated in deep mathematics, and I know very little about simplify of series, partial sums, convergence and divergence.

Second idea is this infinity series. She definitely gives the correct logic answer for $x \in \Bbb{Q }\ \mathit{with}\ \mathit{terminating}\ \mathit{decimal}$, but I'm not sure can it be shortened and convergence or divergence this.
$$\sum_{n=1}^\infty \lceil 10^{n-1}x - \lfloor 10^{n-1}x \rfloor \rceil $$
This series is round every exist remainder to $1$ after multiplying by $10^{n-1}$.

Thanks in advance for your help.


问:
Why the terms "unit" and "irreducible"?
I'm trying to understand why in a ring we choose the names unit to an invertible element and irreducible element in this definition

Maybe historical reasons?

For example, I suppose the second definition are named as prime elements because of the analogy to prime numbers.

Thanks in advance.

答:
Firstly, units behave "like $1$", which explains their name.

Historically, prime nuumbers were rather defined as what is here called irreducible. And for integers, "irreducible" and "prime" coincide.
So for the general ring theory, where they do not coincide in general, one had to coin at least one new name. Since irreducible was established for (the ring of) polynomials and this well matches the notion that these elements cannot be (nontrivially) split into several factors, the name prime could be used for the other notion. (Alternatively, one would have had to coin a name for if-it-divides-a-product-it-divides-a-factor numbers)

Title: Why the terms "unit" and "irreducible"?


问:
Probability density functions that demonstrate the difference between the weak and strong forms of the law of large numbers
If $X_n$ is a sequence of iid random variables with expected value $\mu$, and the random variable
$$
\overline{X}_n := \frac{1}{n} \sum_{i=1}^n X_i, \tag{1}
$$
then the weak form of the Law of Large Numbers states that for all $\epsilon > 0$,
$$
\lim_{n \to \infty} \Pr \left( \left| \overline{X}_n - \mu \right| > \epsilon \right) = 0, \tag{2}
$$
while the strong form states that
$$
\Pr \left( \lim_{n \to \infty} \overline{X}_n = \mu \right) = 1. \tag{3}
$$

I'm just an idiot physicist, so I have a very pedestrian view of random variables: I think of a random variable as basically being equivalent to its probability density function. (I know that not all random variables have PDFs. But like I said: idiot physicist.)

Is there an explicit example of a sequence $\overline{X}_n$ of continuous random variables with PDFs that satisfies condition (2) but not condition (3)? (It doesn't need to be a running average of the form (1); any sequence of random variables will do.) If so, what does the sequence of PDFs look like?

This would help make the difference between the weak and strong forms very concrete in my mind. I'm more used to thinking about sequences of functions than about sequences of random variables.

The answers to this question are useful, but (as far as I can tell) they both discuss sequences of binary variables, which seem less natural than continuous variables in the setting of the LLN because you need to average the values.

答:
With iid random variables with finite mean, the strong law holds. The only way to get the weak law to hold and the strong law to fail in the context of running averages is to break the iid assumption.

If your question is just "what's a sequence of random variables that converges in probability to a constant but doesn't converge almost surely to a constant", the classic example is the "moving block". To construct it, let your probability space be $[0,1]$ with the Lebesgue measure (so the probability of any subinterval is its length). Then define $g_{j,k} : [0,1] \to \mathbb{R}$ by $g_{j,k}(x)=\begin{cases} 1 & x \in [j2^{-k},(j+1)2^{-k}] \\ 0 & \text{otherwise} \end{cases}$ where $k=1,2,\dots$ and $j=0,1,\dots,2^k-1$. Now enumerate the $g_{j,k}$ into a sequence $X_n$: so $X_1=g_{0,1},X_2=g_{1,1},X_3=g_{0,2},\dots$.

The sequence $X_n$ is now $1$ on an interval that gets smaller and smaller as $n$ increases, but nevertheless for any particular $x \in [0,1]$ there are infinitely many $n$ such that $X_n(x)=1$. (It may help to draw a picture to see why this is). Consequently $X_n$ converges in probability to zero even though $X_n(x)$ does not converge for any $x$.

This example is discrete (the PDFs are a weighted combination of two delta functions), but you can adjust the graph of $g_{j,k}$ (removing any flat regions while still basically maintaining the shape) to make it continuous. Ultimately the PDFs are looking like a unit-weight delta function at zero for large $n$ (which is the case when $X_n$ converges almost surely also, so the PDFs don't reveal this detail).

In general, just inspecting the PDFs (or even CDFs) only really allows you to understand convergence in distribution, not convergence in probability, convergence almost surely, etc. These other modes necessarily involve joint distributions, so that you can make sense of the subtraction $X_n-X$.


问:
Segre classes of subvarieties
I am following 13.2 of "3264 and All That".
I read the definition of "Segre classes of subvarieties" but I couldn't calculate with describing specific examples.

The definition is as following.
Let $C \subset X$ be a subvariety cut out by ideal sheaf $\cal I$,
$\cal S$ be the sheaf of graded ring $\oplus {\cal I}^n/{\cal I}^{n+1}$ and $\pi: E = \mathbb{P}{\cal S} \to C$ be the projectivization,
and $\zeta$ be the line bundle on $E$ associated to ${\cal S}(1)$.
Then the Segre class of $C$ in $X$ is defined as $s_k(C,X) = \pi_{*}(c_1(\zeta^k))$.

I want to calculate this in two cases with $X=\mathbb{P}^3 = {\rm Proj}\ k[x,y,z,w]$.

One is $C \simeq {\rm Proj}\ k[z,w] $ cut out by ${\cal I}=(x,y)$,
another is $C'$ cut out by ${\cal I} '=(x^2,y^2)$.

I considered affines $U_1 = {\rm Spec}\ k[x/w,y/w,z/w]$ and $U_2 = {\rm Spec}\ k[x/z,y/z,w/z]$ in $X$.

Then $C$ is covered by $V_1 = C \cap U_1 \simeq {\rm Spec}\ k[z/w]$ and $V_2 = C \cap U_2 \simeq {\rm Spec}\ k[w/z]$.

${\cal I}(V_1)$ is a module generated by $(x/w,y/w)$ over $k[x/w,y/w,z/w]$.

${\cal S}(V_1)$ is a graded ring generated by $a_1 = x/w \bmod {\cal I}(V_1)^2, b_1=y/w \bmod {\cal I}(V_1)^2$ over $k[z/w]$.

Similarly,
${\cal I}(V_2)$ is a module generated by $(x/z,y/z)$ over $k[x/z,y/z,w/z]$,

${\cal S}(V_2)$ is a graded ring generated by $a_2 = x/w \bmod {\cal I}(V_2)^2, b_2=y/w \bmod {\cal I}(V_2)^2$ over $k[w/z]$.

So I described E as covered by 4 affines:

$V_{1a} ={\rm Spec}\ [z/w,a_1/b_1], V_{1b} = {\rm Spec}\ [z/w,b_1/a_1]$,
$V_{2a} ={\rm Spec}\ [w/z,a_2/b_2], V_{2b} = {\rm Spec}\ [w/z,b_2/a_2]$

Take rational section of $\zeta$ is informally of the form $Aa_1+Bb_1$, that is,

$Aa_1/b_1+B$ on $V_{1a}$,

$A+Bb_1/a_1$ on $V_{1b}$,

$Aa_2/b_2+B$ on $V_{2a}$,

$A+Bb_2/a_2$ on $V_{2b}$,

that vanishes in $a_1/b_1 = -B/A$, a line.

So $c_1(\zeta)$ is the class in $E$ corresponding this line, but I don't see how to pushforward this class to $C$.

In the another case with $C'$, doing similarly, $E'$ is covered by

$V'_{1a} = {\rm Spec}\ k[x/w,y/w,z/w,a_1/b_1]/(x^2/w^2,y^2/w^2)$,

$V'_{1b} = {\rm Spec}\ k[x/w,y/w,z/w,b_1/a_1]/(x^2/w^2,y^2/w^2)$,

$V'_{2a} = {\rm Spec}\ k[x/z,y/z,w/z,a_2/b_2]/(x^2/z^2,y^2/z^2)$,

$V'_{2b} = {\rm Spec}\ k[x/z,y/z,w/z,b_2/a_2]/(x^2/z^2,y^2/z^2)$,

I thought $c_1(\zeta)$ is again the class in $E'$ corresponding a line $a_1/b_1=const.$ in $V'_{1a}$, but I don't see how to pushforward it.

(From the book I know I can use conormal sheaf in the first case and the answer is $1-2\alpha$ where $\alpha$ is the class of a point in $C$, but it is not available in the second case.)

EDIT: I found an argument with non-reduced case in 13.3.5, but I could not follow the argument. There, $S$ is 2 dimensional variety parametrized by $[aa:bb:cc:ab:bc:ca]$ in $\mathbb{P}^5$
, $\cal I$ is the ideal sheaf that cut out $S$,
and T is non-reduced subvariety cut out by ${\cal I}^2$.
Then there says: Blow-up along $T$ is the same as the blow-up along $S$, but with the exceptional divisor doubled, so there is the relation $s_k(T,\mathbb{P}^5) = 2^{k+3}s_k(S,\mathbb{P}^5)$, but I could not find the reason.
Why isn't it simply $s_k(T,\mathbb{P}^5) = 2s_k(S,\mathbb{P}^5)$?

答:
I reviewed the definition of pushforward and I noticed I misunderstood the definition.
While $s(C,X) = \sum_{k\geq 0} π_*(\zeta^k)$, $\pi_*$ preserves dimension but not codimension, so in this case with $\dim(C)=1, \dim(E)=2$, the relation is actually $s_k(C,X) = π_*(\zeta^{k+1})$.

Thus, $\pi_* (ζ) = 1$, $\pi_*( ζ^2) = -2\alpha$.
The latter is not trivial. One way is using the blow-up calculated as following:

Let $j:E\to B$ be blow up of $i:C \to X$.
Let $[E], [H_C], [H]$ be respectively the cycle of the exceptional divisor, proper transform of plane including C, proper transform not including C. For example considering rational function $x/z$, I see rational relation $[E]+[H_C]=[H]$.

Then for example using, Proposition 13.12,

$j_*(1) = [E],$

$j_*(ζ) = -j_*(1)\cdot j_*(1) = -[E]\cdot[E],$

$j_*(ζ^2) = -j_*(1)\cdot j_*(\zeta) = [E]\cdot[E]\cdot[E] $

Using the relation $[H]\cdot[H]\cdot[H] = [H]\cdot[H]\cdot[H_C] = 1$ point, $[H_C]\cdot [H_C]=0$, I got $j_*(ζ^2) = -2$ points.

For the case in last EDIT, since $\dim S=2, \dim\mathbb{P}^5=5$, the relation should be $s_k(C,X) = π_*(\zeta^{k+3})$, that is where additional factor $2^3$ comes from.

EDIT: I still had mistake. In this case, the dimension of corresponding exceptional divisor is $4$, so the relation is $s_k(C,X) = π_*(\zeta^{k+2})$. Another factor $2$ seems to come from the "doubleness" itself of $T$.

Title: Segre classes of subvarieties


问:
Possible minor typo from Dummit's Abstract Algebra
On page 4, 0.2/(4)/(b) of Dummit&Foote's Abstract Algebra (3rd ed.):

(If $a,b\in \mathbb{Z}-\{0\}$, there is a unique positive integer $l$, called the least common multiple of a and b ... satisfying:) if $a\mid m$ and $b\mid m$, then $l\mid m$ (so $l$ is the least such multiple).

May I ask if it is necessary to assert "so $l$ is the least such positive multiple" there? (since there also exist $l\mid m$ such that $m\leqslant 0$)

答:
There's no need to highlight that:

there is a unique positive integer

Here it's clear enough. And as what you said in the problem description, you already comprehended it correctly.


问:
Show that for any $Z\subseteq Y$, $Z$ is closed in $X$
Be $X$ a topological space and $Y \subseteq X$ closed in $X$. Suppose $Y$, with the subspace-induced topology, is a discrete space. Show that for any $Z\subseteq Y$, $Z$ is closed in $X$.

I wonder if my solution is correct.

If $Y$ is a discrete space so any subset of $Y$ belongs to the $Y$ topology.

(i) Can I say that if $Y$, with the subspace-induced topology, is a discrete space then $X$ is a discrete topological space due to this?

If yes, I can say that if $Y$ is closed then $Y$ is not in the $X$ topology, as $Z$ is a subset of $Y$, it means that $Z$ is not an element of the $X$ topology, so it is closed.

答:
If $T_X$ is the topology on $X$ then the subspace topology on any $Y\subset X$ is $T_Y=\{t\cap Y: t\in T_X\}.$ It follows  that if $Z\subset Y\subset X$ then $Cl_Y(Z)=Y\cap Cl_X (Z).$

If $Z\subset Y$ then $Cl_X(Z)\subset Cl_X(Y)$  so $Cl_X(Z)\cap Cl_X(Y)=Cl_X(Z).$

So if $Z=Cl_Y(Z)$ and $Y=Cl_X(Y)$ then $$Z=Cl_Y(Z)=Y\cap Cl_X(Z)=Cl_X(Y)\cap Cl_X(Z)=Cl_X(Z).$$ In other words, if $Z$ is closed in $Y$ and $Y$ is closed in $X$ then $Z$ is closed in $X.$

Now if $Y$ is discrete then any $Z\subset Y$ is closed in $Y.$

$Y$ could be the empty space, which is a closed discrete subspace of every space $X$ so you cannot say that $X$ must be discrete. If you meant to ask whether $Z$ is discrete, then yes, every subspace of a discrete space is discrete.


问:
Prove that $5^n \equiv 8n^2-4n+1 \mod 64$ using induction
I am trying to prove that $5^n \equiv 8n^2-4n+1 \mod 64 $ using induction. I have seen the post which solves this problem using other methods, but I really need to be able to do it using induction.

Here is what I have so far:

Base step:
$5 \equiv 5 \mod 64$

Then, I assumed that $5^k \equiv 8k^2 -4k+1 \mod 64.$

I want to prove that $5^{k+1} \equiv 8(k+1)^2-4(k+1)+1 \mod 64$.

Note that $8(k+1)^2-4(k+1)+1=8k^2+12k+5$.

Hence,
$5^{k+1} \equiv 5^k *5 \mod 64
\equiv (8k^2-4k+1)*5 \mod 64 \equiv 40k^2 -20k+5 \mod 64.$

I am stuck here.
Any insight would be very useful!!
Thanks

答:
A generalization
suggested by J. W. Tanner's answer.

If $0 < j < n$ then

$\begin{array}\\
(1+m)^n
&=\sum_{k=0}^n \binom{n}{k}m^k\\
&\equiv\sum_{k=0}^j \binom{n}{k}m^k \bmod m^{j+1}\\
\text{if }
&j=2\\
(1+m)^n
&\equiv\sum_{k=0}^2 \binom{n}{k}m^k \bmod m^{3}\\
&\equiv 1+nm+\frac12n(n-1)m^2\bmod m^{3}\\
&\equiv 1+n(m-\frac12m^2)+\frac12n^2m^2\bmod m^{3}\\
\text{if}
&m=4\\
5^n
&\equiv 1-4n+8n^2 \bmod 64\\
\end{array}
$


问:
Is there a relation which is neither symmetric nor antisymmetric?
I've proved that there are relations which are both symmetric and antisymmetric ($\forall a \forall b (aRb \rightarrow (a=b))$) and now I'm trying to prove that there are relations which are neither symmetric nor antisymmetric. I got stuck! Any ideas?</p>

答: To say that a relation $R$ on a set $A$ is not symmetric is equivalent to saying that there exist elements $a$ and $b$ in $A$ such that $aRb$ and $\require{cancel}b\cancel{R}a$. To say that a relation $R$ on a set $A$ is not antisymmetric is equivalent to saying that there exists an element $a\in A$ and an element $b\in A$ such that $a\ne b$, $aRb$, and $bRa.$ Consider the relation $R = \{\ (a,b)\ |\ ab^{2}\ \gt\ 0\}$ on the set of all integers $\mathbb Z$. Since $2\cdot (-1)^{2} = 2\gt 0$, the ordered pair $(2, -1)\in R$. However, since $(-1)\cdot 2^{2} = -4 \not\gt 0$, $(-1, 2)\not\in R$, thus $R$ is not symmetric. Similarly, we can show that $R$ is not antisymmetric by noting that the inequality $ab^{2}\gt0$ will hold for any two positive integers $a$ and $b$.  Thus, there exists a distinct pair of integers $a$ and $b$ such that $aRb$ and $bRa$. Hence, $R$ cannot be antisymmetric.


问:
How to show that a valid inner product on V is defined with the formula $[x, y] = \langle Ax, Ay\rangle $?

Let $A \in L(V,W)$ be an injection and $W$ an inner product space with the inner product $\langle \cdot,\cdot\rangle $. Prove that a valid inner product on $V$ is defined with the formula $[x, y] = \langle Ax, Ay\rangle $

$L(V, W)$ = The set of all linear mappings (linear operators) from V to W

To prove this, if I am correct, I need to show that the four properties of an inner products space apply on this formula:

1. $\langle x, y \rangle  = \overline{\langle y,x\rangle }$

2. $\langle \alpha x, y\rangle  = \alpha\langle x,y\rangle $

3. $\langle x+y,z\rangle  = \langle x,z\rangle  + \langle y,z\rangle $

4. $\langle x,x\rangle \space \ge 0 \space \space \forall x$

4.' $\langle x,x\rangle \space = 0 \Longleftrightarrow x=0$

4.

$ [x, x] = \langle Ax, Ax\rangle , Ax \in W$, and since $W$ is an inner product space, $\langle Ax, Ax\rangle \space \ge 0$ implies $[x, x] \ge 0$.

4.'

Since A is an injection: $Ax = 0 \implies x = 0$ and since $W$ is an inner product space and $Ax \in W \implies \langle  Ax, Ax \rangle  = 0 \implies Ax = 0 \implies [x,x] = 0 \Longleftrightarrow x=0$

3.

$[x+y,z] = \langle A(x+y), Az\rangle  = A$ is linear $= \langle Ax + Ay, Az\rangle = Ax, Ay, Az \in W$ and $W$ is an inner product space $= \langle Ax, Az\rangle  + \langle Ay, Az\rangle  = [x,z] + [y,z]$

2.

$[\alpha x, y] = \langle A(\alpha x), Ay\rangle  = A$ is linear $ = \langle \alpha (Ax), Ay \rangle  = W$ is an inner product space $ = \alpha \langle Ax, Ay\rangle  = \alpha [x, y]$

1.

$[x, y] = \langle  Ax, Ay \rangle  = W$ is an inner product space $= \overline{\langle Ay, Ax\rangle } = \overline{[y, x]}$

But this seems to me a little to easy, did I maybe conclude something that can't be concluded so easily or maybe is my approach to prove this completely wrong?

答: Yes, that's it all. So called 'routine verification'. $A$ embeds $V$ into $W$, and $[-,-]$ is just the inherited inner product from $W$, along this embedding.


问:
How to to find the sol. By pascal's roule and mathematical induction
\begin{equation}
\begin{array}{l}
\text { - For } n \geqslant 2 \text { , prove that } \\
\left(\begin{array}{c}
2 \\
2
\end{array}\right)+\left(\begin{array}{c}
3 \\
2
\end{array}\right)+\left(\begin{array}{c}
4 \\
2
\end{array}\right)+\ldots+\left(\begin{array}{c}
n \\
2
\end{array}\right)=\left(\begin{array}{c}
n+1 \\
3
\end{array}\right)
\end{array}
\end{equation}

答:
Generalizing Tuvasbien's answer,
since
$\binom{p}{k}
=\binom{p+1}{k+1}-\binom{p}{k+1}
$,
if $k \ge 2$,

$\sum_{p=2}^n\binom{p}{k}
=\sum_{p=2}^n\left(\binom{p+1}{k+1}-\binom{p}{k+1}\right)
=\binom{n+1}{k+1}-\binom{2}{k+1}
=\binom{n+1}{k+1}
$.

Your result is $k = 2$.


问:
There are 6 apples in a basket. Two of them are red and four are green.
a) What's the probability of selecting a red apple when choosing at random?

This one is easy enough for me it's just $P(R) = \cfrac{2}{6}$ where R is the event we pick a red apple

b) What's the probability that, if one apple is randomly chosen per day and then eaten and not replaced, red apples are chosen on the first two days and green apples are chosen on the last four days.

I'm a little confused on this one, please let me know if my methodlogy/reasoning is correct.

There are $\cfrac{6!}{4!2!}=15$ ways/permutations to pick and eat the apples when we don't care about any constraints

Red apples being eaten on the first two days and green apples being eaten the last four days is ONE specific permutation of the $15$ permutations. So I believe that the answer is the Probability $= \cfrac{1}{15}$ Do I have b right? Please help me note any flaws in my thinking

答:
The order in which we need to pick the apples is $\mathrm{RRGGGG}$. The first time we pick apples, probability of picking $\mathrm{R}$ is $\frac 26 = \frac 13$. After that, we're left with 1 red apple and 4 green apples. The proabbility of picking a red apple on the second day is then $\frac 15$. After this, since all the apples left are green apples, the probability of picking a green apple is $1$ on all following days.

The final probability is then $P = \frac {1}{3} \times \frac{1}{5} = \frac{1}{15}$


答: You are absolutely right. When using counting to solve probability problems, always remember you should treat both the denominator and the numerator in the same way, i.e. both labeled(distinguishable) counting or both unlabeled(undistinguishable) counting.</p>

Title: There are 6 apples in a basket. Two of them are red and four are green.

Body:
a) What's the probability of selecting a red apple when choosing at random?

This one is easy enough for me it's just $P(R) = \cfrac{2}{6}$ where R is the event we pick a red apple

b) What's the probability that, if one apple is randomly chosen per day and then eaten and not replaced, red apples are chosen on the first two days and green apples are chosen on the last four days.

I'm a little confused on this one, please let me know if my methodlogy/reasoning is correct.

There are $\cfrac{6!}{4!2!}=15$ ways/permutations to pick and eat the apples when we don't care about any constraints

Red apples being eaten on the first two days and green apples being eaten the last four days is ONE specific permutation of the $15$ permutations. So I believe that the answer is the Probability $= \cfrac{1}{15}$ Do I have b right? Please help me note any flaws in my thinking