问:
About the reducedness in algebraic geometry
We adopt the following definition. A ring is said to be reduced if it has no nonzero nilpotents. A scheme $X$ is reduced if $\mathscr{O}_{X}(\mathrm{U})$ is reduced for every open set $\mathrm{U}$ of $\mathrm{X}$.

In the statement of many propositions in AG,  the reducedness assumption is always added.

Question:
1, What is the convenience  of the reducedness in the theory of algebraic geometry?

2, Can someone tell us  some  propositions, which holds in the reduced space while it does not hold in the nonreduced space?

3, Can we always reduce the problem to the reduction of a nonreduced space $X$ and then  go back in the inverse way to get the corresponding information about  $X$?

Any answer is welcomed. Thanks.

答:
Nilpotent elements are useful in mathematics when studying derivatives, differential operators and tangent spaces.

Let $k$ be a field and let $A:=k[x],B:=k[x,y]$ be polynomial rings with $I:=(y-x)\subseteq B$ the ideal defined by the element $y-x$. Let $J(l):=B/I^{l+1}$ and let
$T^l:A \rightarrow J(l)$ be defined by $T^l(f(x)):=f(y)$. The left $A$-module $J(l)$
is free of rank $l+1$ on the elements $(dx)^i:=(y-x)^i$ for $i=0,..,l$. You can prove this using induction.

Example 0. Let $k$ have characteristic zero. It follows from the binomial theorem and an induction that

$T^l(f(x))=\sum_{i=0}^l \frac{f^{(i)}(x)}{i!}(dx)^i$.

Example 1. Let $f(x):=x^2$. We get

$T^l(f(x)):=f(y)=y^2=(x+dx)^2=x^2+2xdx+(dx)^2=$

$f(x)+ f'(x)dx+\frac{f^{(2)}(x)}{2!}(dx)^2$.

Hence the map $T^l$ is the Taylor expansion of the polynomial $f(x)$. The map
$T^l$ is a "differential operator" of order $l$ from $A$ to $J(l)$. If $k$ is the field of real numbers and $\mathfrak{m}:=(x-a)$ for $a\in k$ we may pass to the fiber

$J(l)(a):= \kappa(\mathfrak{m})\otimes_A J(l) \cong k\{(dx)^i\}$

and as an element of $J(l)(a)$ it follows

$T^l(f(x)) = \sum_{i=0}^l \frac{f^{(i)}(a)}{i!}(dx)^i \in J(l)(a)$.

Hence in this case we get the value of the Taylor series of $f(x)$ of order $l$ at the
real number $a\in k$.

If $z=a+ib$ with $a,b$ real numbers and $b \neq 0$ it follows

$p(x):=(x-z)(x-\overline{z})=x^2-2ax+a^2+b^2 \in k[x]$

is an irreducible polynomial with $k[x]/(p(x))\cong \mathbb{C}$. Choosing an explicit isomorphism $\phi: k[x]/((p(x)) \cong \mathbb{C}$ it follows

$0=\phi(p(x))=p(\phi(x))$

hence $\phi(x)$ equals $z$ or $\overline{z}$. Hence if $\mathfrak{p}=(p(x))$ and we pass to the fiber $\kappa(\mathfrak{p})\otimes J(l):=J(l)(\mathfrak{p})$ we get an isomorphism

$J(l)(\mathfrak{p})\cong \mathbb{C}\{(dx)^i\}$.

If we "Taylor expand" $f(x)$ and pass to the fiber $J(l)(\mathfrak{p})$

we get an element $T^l(f(\phi(x))) \in \mathbb{C}\{(dx)^i\}$, and the value of the Taylor series $T^l(f(\phi(x)))$

depends on the choice of an isomorphism $\kappa(\mathfrak{p})\cong \mathbb{C}$. and the choice of $\phi(x)\in \mathbb{C}$. What does not depend on this choice is if the higher order derivative $\frac{f^{(i)}(x)}{i!}$ is zero or not.

Note: If $f(x) \in \mathbb{R}[x]$ is any polynomial it follows $T^l(f(x))=0$ in the fiber at $x-a$ iff $f^{(i)}(a)=0$ for $i=0,..,l$. This holds iff $f(x)=f_l(x)(x-a)^{l+1}$. Hence $T^l(f(x))=0$ in the fiber at $x-a$ iff $a$ is a zero of $f(x)$ of order $l+1$. If $p(x):=(x-z)(x-\overline{z})$ it follows $T^l(f(x))=0$ in the fiber at $p(x)$ iff $f(x)=f_l(x)p(x)^{l+1}$. Hence $T^l(f(x))=0$ iff $z$ and $\overline{z}$ are zeros of $f(x)$ of multiplicity $l+1$.

This construction globalize to any scheme/differentiable manifold/complex manifold/etc.
Hence nilpotent elements and non-reduced ideals are useful in algebraic geometry, differential geometry and complex analysis. In these fields we study tangent spaces and differential operators.

For any smooth projective curve $C \subseteq \mathbb{P}^n_k$ there is a similar construction of a finite rank vector bundle $\mathcal{J}_C^l$ and a universal differential operator

$T^l: \mathcal{O}_C \rightarrow \mathcal{J}_C^l$ which locally "Taylor expands" sections: For any open subset $U \subseteq C$ we get a differential operator

$T^l(U): \mathcal{O}_C(U) \rightarrow \mathcal{J}_C^l(U)$

and the map $T^l(U)$ Taylor expands the section $s\in \mathcal{O}_C(U)$. If $k$ is the field of real or complex numbers, we may view $s$ as a real valued or complex valued function, and in this case $T^l(U)(s)$ is the Taylor expansion of $s$.

Example 2. If $C$ is a complex holomorphic curve and $\mathcal{O}_C$ is the sheaf of holomorphic functions on $C$, there is a similar construction. There is a finite rank holomorphic vector bundle $\mathcal{J}_C^l$ and a differential operator

$T^l: \mathcal{O}_C \rightarrow \mathcal{J}_C^l$

with the same properties.

Example 3. Complex holomorphic curves and nilpotent ideals. The holomorphic vector bundle $\mathcal{J}_C^l$ may be constructed using the ideal of the diagonal $\mathcal{I} \subseteq \mathcal{O}_{C \times C}$. The ideal $\mathcal{I}$ is a coherent sheaf of ideals, and one may prove that the quotient $\mathcal{O}_{C \times C}/\mathcal{I}^{l+1}$ is (as a left $\mathcal{O}_C$-module) locally trivial of rank $l+1$. The corresponding holomorphic vector bundle is isomorphic to $\mathcal{J}_C^l$. The pair $(C\times C, \mathcal{O}_{C\times C}/\mathcal{I}^{l+1})$ may be viewed as a "locally ringed space" with nilpotent elements in the structure sheaf. The sheaf $\mathcal{J}_C^l$ is supported on the diagonal $\Delta(C) \subseteq C \times C$, and it follows $\mathcal{J}_C^l$ is a sheaf of left and right $\mathcal{O}_C$-modules. The sheaf $\mathcal{J}_C^l$ is locally trivial as left and right $\mathcal{O}_C$-module, but these two structures are not isomorphic in general. If you consider the ideal sheaf $\overline{\mathcal{I}} \subseteq \mathcal{J}_C^l$ and choose any non-zero section $s  \in \overline{\mathcal{I}}(U)$ for $U \subseteq C$ an open set it follows $s^{l+1}=0$. Hence the sheaf of rings $\mathcal{J}_C^{l}$ has nilpotent elements. When we restrict $\mathcal{J}_C^{l}$ to the diagonal $\Delta(C)$
we get a sheaf of left and right $\mathcal{O}_C$-modules, but for local sections $s \in \mathcal{O}_C(U)$ and $\omega \in \mathcal{J}_C^l(U)$ it follows $s\omega \neq \omega s$.

You may have seen in Hartshorne exercise II.5.18 that for a scheme $(X,\mathcal{O}_X)$ there is an "equivalence of categories" between the category of locally free finite rank $\mathcal{O}_X$-modules and the category of finite rank geometric vector bundles, and there is a similar result valid for complex holomorphic manifolds. Given a complex holomorphic manifold $(Y, \mathcal{O})$ where $\mathcal{O}$ is the sheaf of complex holomorphic functions
on $Y$, it follows there is an "equivalence of categories" between the category of locally trivial finite rank $\mathcal{O}$-modules and the category of finite rank holomorphic vector bundles on $Y$.

In a calculus course in one variable you define the derivative $f'(x)$ of a real valued smooth function $f(x)$ using limits. In algebraic geometry we work over arbitrary fields (or rings) and we cannot "take limits". Using non-reduced ideals we can still formally "take derivatives" of sections of sheaves as explained above. In characteristic zero we get as you can observe above the correct Taylor expansion of any polynomial (or rational) function. In characteristic $p>0$ the notion is useful as well, ref. Question 1 in your question.


问:
Finding the likelihood function and Posterior Distribution
Let $\underline{X}=(X_1,...X_n)$  be an i.i.d. random sample from an exponential
distribution, with probability density function given by

$f(x; \theta)=\lambda exp${$-\lambda x$} , x>0 where $\lambda$ is an unknown parameter taking values in $\mathbb{R^+}$

A Derive the likelihood function $L(\lambda; \underline{X})$ and derive the Fisher information $I(\lambda)$ measuring the amount of information that $\underline{X}$ carries about $\lambda$

B Given the Jeffreys prior for $\lambda$ is $\pi_J(\lambda)\propto \lambda^{-1}$, derive the posterior distribution for $\lambda$. Find the mean and variance of $\lambda| \underline{X}$

I know that for A I need to use the product up to n and then for the fisher info it is $\frac{1}{-E(l''(\lambda)}$ but I'm not sure how to do this for my given model.

答:
A: the likelihood is the following:

$$L(\lambda)=\lambda^n e^{-\lambda\Sigma_i X_i}$$

To calculate Fischer information, as you stated, you have to calculate

$$-n\mathbb{E}\Bigg[\frac{\partial^2}{\partial\lambda^2}\log f(x;\lambda)\Bigg]$$

thus simply:

$$\log f=\log\lambda-\lambda x$$

$$\frac{\partial^2}{\partial\lambda^2}\log f=-\frac{1}{\lambda^2}$$

So the Fischer information of the n-tuple $(X_1\dots,X_n)$ is

$$\frac{n}{\lambda^2}$$

B

The posterior is the following

$$\pi(\lambda|\mathbf{x})\propto \pi(\lambda)\cdot p(\mathbf{x}|\lambda)$$

that is

$$\pi(\lambda|\mathbf{x})\propto \lambda^{-1}\times \lambda^n e^{-\lambda\Sigma_i X_i}= \lambda^{n-1}e^{-\lambda\Sigma_i X_i}$$

We immediately recongize the kernel of a $Gamma(n;\Sigma_i X_i)$ thus

$$\mathbb{E}[\lambda|\mathbf{x}]=\frac{n}{\Sigma_i X_i}$$

and

$$\mathbb{V}[\lambda|\mathbf{x}]=\frac{n}{(\Sigma_i X_i)^2}$$

ContentLicense: CC BY-SA 4.0


问:
convexity of $f(x)+f(g(x))$
suppose $f$ is a convex function. i.e.,

for all $0\leq\lambda\leq 1$:
\begin{equation}
f(\lambda x+(1-\lambda)y)\leq \lambda f(x)+(1-\lambda)f(y)
\end{equation}
Consider the following function:
\begin{equation}
h(x)=f(x)+f(g(x))
\end{equation}
what can we say about convexity of $h$ w.r.t $x$?

more specifically, under what condition $h$ is a convex function as well? i.e.,
\begin{equation}
h(\lambda x+(1-\lambda y))\leq \lambda h(x)+(1-\lambda)h(y)
\end{equation}

note that there is no restriction on the convexity of $g$.

答:
If $g: \Bbb R \to \Bbb R$ is a linear function then $h(x) = f(x) + f(g(x))$ is convex for any convex function $f: \Bbb R \to \Bbb R$.

Conversely, linear functions are the only functions $g$ with that property:

Choosing the convex function $f(x) = x$ implies that $h(x) = x + g(x)$ is convex, so $g$ must be convex.

Choosing the convex function $f(x) = -x$ implies that $h(x) = -x - g(x)$ is convex, so $g$ must be concave.

A function which is both convex and concave is necessarily linear.

Title: convexity of $f(x)+f(g(x))$


问:
The Uniform convergence of $\sum\limits_{n=1}^\infty{\frac{\arctan(nx)}{1+(x-n)^{3}}}$ on $x \in [0, 10]$.
I've stumbled upon this problem, while preparing to my exams. The problem is that I cannot understand which $M_n$ should I finally come to, trying to apply the Weierstrass M-test. I also couldn't have found similar-to questions here.
I made several attempts, stating that $\forall n \in\mathbb{N}, \forall x\in E[0,10] : \arctan(nx) \leqslant nx$ or $\arctan(nx) \leqslant \frac{\pi}{2}$, but for some reason nothing works out for me. My best attempt for now was to state that the given $|u_n(x)| \leqslant \frac{\pi}{(n-x)^2}$, but the thing is that it is not true for all x in the given $E$.

Thank you in advance for your help.

答: If $n>10$ and $x\in[0,10]$, then$$1+(x-n)^3>1+(10-n)^3=1-(n-10)^3,$$and so$$|1+(x-n)^3|>(n-10)^3-1.$$Furthermore, $\arctan(nx)<\frac\pi2$. Therefore,$$\left|\frac{\arctan(nx)}{1+(x-n)^3}\right|<\frac{\pi/2}{(n-10)^3-1}.$$Since the series$$\sum_{n=11}^\infty\frac{\pi/2}{(n-10)^3-1},$$converges, then your series converges uniformly, by the Weierstrass $M$-test.


问:
Some algebraic inequalities with the binomial theorem.
I am working on proving the following limits.

1), $\lim_{n \to \infty}  \sqrt[n]{n} = 1$

2), If $p >0$ and $\alpha \in \Bbb R$, then $\lim_{n \to \infty} {n^{\alpha}\over{(1+p)^n}} =0$

And I am trying to follow the proof but I am stuck on understanding the following inequalities regarding the binomial theorem.

Put $x_n = \sqrt[n]{n} -1$. then $$n = (1+x_n)^n \ge {n(n-1) \over 2}x_n^2 $$

I understand that the right side of the inequality probably comes from $$1 + nx_n + {n \choose 2}x_n^2 + ・・・+x_n^n$$

and taking all the terms after ${n \choose 2}x_n^2$.

But to know this I think that $x_n^k$ has to be greater than $x_n^2$ for $k > 2$, am I right?
Since we can only say that $x_n > 0$, I'm a little puzzled.

For the second one,$$(1+p)^n > {n \choose k}p^k = {n(n-1)・・・(n-k+1) \over {k!}}p^k > {n^kp^k \over {2^k k!}}$$

is supposedly true when $n > 2k$.

I understand the left three expressions, but how does the right side of the inequality work ?

答:
The first confusion is just a misunderstanding. If you have $A=B+C$ with $B,C$ greater than $0$, then no matter what happens to $B,C$'s relationship, $A>B$ is always true. </p>

<p>For the second one all you need to show is $(n-k+1)\ge \frac{n}{2}$ if $n\ge 2k$. But this should be clear to you via some algebraic manipulations. </p>

ContentLicense: CC BY-SA 3.0

Body:
The first confusion is just a misunderstanding. If you have $A=B+C$ with $B,C$ greater than $0$, then no matter what happens to $B,C$'s relationship, $A>B$ is always true.

For the second one all you need to show is $(n-k+1)\ge \frac{n}{2}$ if $n\ge 2k$. But this should be clear to you via some algebraic manipulations.


问:
Finding a function if you know it's directional derivative
I have the following expression
\begin{equation}
\beta \,\,\cdotp \nabla v(x,y) = - u(x,y)
\end{equation}
where $\beta \,\,\cdotp \nabla$ denotes the directional derivative of $v$ in the direction of $\beta$.

Question: How do I get to an explicit expression for $v$?

Intuitively I think I know what the solution should be. $v$ should be the line integral over $u(x,y)$ from $(x,y)$ to some constant point $(a,b)$, where the line along which $u$ is integrated is tangent to the vector $\beta$. In that case the directional derivative of $v$ in the direction of $\beta$ would be equal to $-u(x,y)$ because a small step in the direction of $\beta$ (or up along the line integral) would decrease the integral by $-u(x,y)$.

I don't know how to write this down though and I've looked for a lot of sources.

Thanks in advance!

答:
The given data do not determine the function $(x,y)\mapsto v(x,y)$ up to an additive constant. Consider the simple example
$$v_x=u(x,y)\ ,$$
which corresponds to $\beta=(-1,0)$. In this case you might find a solution $v_0$ by "integrating $u$ with respect to $x$", but you can add an arbitrary function $y\mapsto g(y)$ to this $v_0$ and obtain another solution $v(x,y):=v_0(x,y)+g(y)$.


问:
Tennis Probability Match problem

Question:

A's chance of winning a set at tennis against B is $P_A=2/3$. Find his chance of winning at least 3 sets in a 5 set match,assuming that all 5 sets are to be played.

My approach:

Probability of A winning $$P_A=2/3$$

Probability of A losing: $$P_\bar A=P_B\ (Probability\ of\ B\ winning)=1/3$$
I cannot understand how to find the Probability by considering the winning and losing of A and B respectively. I tried googling for answers and also tried to search similar problems on mathematics stackexchange but no solution was available. And another interesting thing was that 3 answers were given for this problem but as per my knowledge, i think this question can have only 1 solution. This question is given in Frank Model Test Papers for Mathematics (Unsolved Model Test paper 14 ), question number 8.

Given answer:

$$1/2,\ 9/4,\ 2/3$$
Any help or solution would be appreciated.

答:
All 5 sets will be played if and only if after the first 4 sets the game is tied. So assuming that all sets are played, the winning of player $A$ depends only if she wins the last set. And in this situation the winning of $A$ is equivalent to her winning at least 3 sets. Therefore

$$\mathbb{P}(\text{A wins at least 3 sets} \space | \space \text{All the 5 sets are played}) = \mathbb{P}(\text{A wins the last set}) =  \frac{2}{3}.$$

Title: Tennis Probability Match problem


问:
How to determine isomorphism in simple, connected (regular) graphs?
I am learning about regular graphs and have found that there are only 5 different options for a simple connected 3-regular graph with 8 vertices (source: http://www.mathe2.uni-bayreuth.de/markus/reggraphs.html).

However, when I start plotting option for myself, I find it often very hard to determine whether two graphs are isomorphic or not. Here is an example with the first five graphs depicting the five options given in the link above and the sixth one of my additions:

答: The new option is equivalent to option 1. I found this by looking for cycles in the graph and checking in what way the cycles are attached to one another: you have four 3-cycles; two pairs of the four 3-cycles share an edge and then there are two edges (23 and 67 in Option 1) that connect everything.</p>


问:
Classifying extremums using higher order derivatives
I was reading this from my calculus book:

If $f(x)$  has enough consecutive continuous derivatives at $x_0$ and we have
$n\ge2$ where  $$f'(x_0)=\cdots=f^{(n-1)}(x_0)=0,f^{(n)}(x_0)\neq0$$
Then if $n$ is an even number, $x_0$ is local extremum point( for
$f^{(n)}>0$ it is local minimum and for $f^{(n)}(x_0)<0$ it is local
maximum). If $n$ is an odd number $f$ has not extremum at $x_0$.

Can you please explain it . I don't understand this note. in fact I don't have intuition about $f'(x_0)=\cdots=f^{(n-1)}(x_0)=0,f^{(n)}(x_0)\neq0$ and how it can be connected with local extremum?

答:
Hint

Use Taylor formula and notice that $x \mapsto x^n$ has a local minimum at zero when $n$ is even while it is not the case for $n$ odd.


问:
Simplify $\sqrt n+\frac {1}{\sqrt n}$ for $n=7+4\sqrt3$
If $n=7+4\sqrt3$,then what is the simplified value of
$$\sqrt n+\frac {1}{\sqrt n}$$
I was taking LCM but how to get rid of  $\sqrt n$ in denominator</p>

答:
You could as OP suggests, add 2, and take the square root.

For example, $\dfrac{1}{(7+4\sqrt3)}$ is $(7-4 \sqrt3)$, so the sum of these numbers is $14$.  You add $2$ to it to get $16$, and take the square root.

The new number is $4 = x + \dfrac {1}x$,  which leads to $x = \dfrac{\sqrt{4+2} + \sqrt{4-2}}{2}$ = 1.93185165259.


问:
Derive the likelihood function (;Y) and thus the Maximum likelihood estimator ̂ (Y) for . Show that the MLE is unbiased.
Let $\underline{Y}= (Y_1,..., Y_n)$ be an i.i.d. random sample from a Weibull distribution, with probability density function given by $f(x; \lambda)= \frac{k}{\lambda}(\frac{y}{\lambda})^{k-1}exp$ {$-(\frac{y}{\lambda})^k$} where k > $0$ is a known shape parameter, and λ is an unknown scale parameter taking values in $\mathbb{R^+}$.

Consider the parametrisation $\theta= \lambda^k$

Derive the likelihood function $L(\theta; \underline{Y})$ and thus the Maximum likelihood estimator $\hat{\theta}(\underline{Y})$ for $\theta.$ Show that the MLE is unbiased.

What I know so far

take the sum of the pdf up to n to find the likelihood function.  take the log and differentiate and then set to $0$ and solve for the MLE. If the expectation is 0 then the estimator is unbiased. I know the method but I am unsure of how to actually put it into practice. Any help would be greatly appreciated.

答:
First rewrite the density with the new parametrization

$$f(y|\theta)=\frac{ky^{k-1}}{\theta}e^{-\frac{y^k}{\theta}}$$

Calculate the likelihood

$$L(\theta)\propto \theta^{-n}e^{-\frac{\Sigma_i y_i^k}{\theta}}$$

proceeding in the calculation you find that the score function (derivative of the log likelihood with respect to $\theta$) is

$$l^*=-\frac{n}{\theta}+\frac{1}{\theta^2}\Sigma_i y_i^k$$

And thus

$$T=\hat{\theta}_{ML}=\frac{\Sigma_i y_i^k}{n}$$

To show that $\mathbb{E}[T]=\theta$ let's rewrite the score function in the following way

$$l^*=-\frac{n}{\theta}+\frac{nT}{\theta^2}$$

Now simply remembering that (First Bartlett Identity)

$$\mathbb{E}[l^*]=0$$

you get

$$\frac{n}{\theta}=\frac{n\mathbb{E}[T]}{\theta^2}$$

that is also

$$\mathbb{E}[T]=\theta$$

To calculate its variance, using II Bartlett Identity, that is

$$\mathbb{E}[l^{**}]=-\mathbb{E}[(l^*)^2]$$

This identity leads to

$$\mathbb{V}\Bigg[\frac{nT}{\theta^2}-\frac{n}{\theta}\Bigg]=-\mathbb{E}\Bigg[\frac{n}{\theta^2}-\frac{2nT}{\theta^3}\Bigg]$$

that is

$$\frac{n^2}{\theta^4}\mathbb{V}[T]=\frac{n}{\theta^2}$$

$$\mathbb{V}[T]=\frac{\theta^2}{n}$$

Alternative method to calculate expectation and variance of T

Simply transforming

$$W=Y^k$$

you get that $W\sim Exp\Big(\frac{1}{\theta}\Big)$ thus

$$T\sim Gamma\Big(n;\frac{n}{\theta}\Big)$$

thus immediately you get

$$\mathbb{E}[T]=\frac{n}{\frac{n}{\theta}}=\theta$$

$$\mathbb{V}[T]=\frac{n}{\Big(\frac{n}{\theta}\Big)^2}=\frac{\theta^2}{n}$$


答:
I suspect your graph is isomorphic to option 1.

In these types of examples, I would first consider the cycle types of the graph. For instance, in your graph, we can easily see that we have $3$-cycles $(567), (568), (234), (134)$. And note that $(567)$ and $(234)$ has no adjacent vertices, as well as $(568)$ and $(134)$. Option 1 also has four $3$-cycles with the similar properties. So, we may want to suspect that they are isomorphic.

Finding an isomorphism though a little bit intuitional after suspecting. But in some "symmetric" cases it might be easier. For instance, I would consider a map from option 1 to your graph
$$1 \to 3 ,\ 2 \to 1,\ 3 \to 7,\ 4 \to 6,\ 5 \to 5,\ 6 \to 8,\ 7 \to 2, \ 8 \to 4 $$

Title: How to determine isomorphism in simple, connected (regular) graphs?

Body:
I am learning about regular graphs and have found that there are only 5 different options for a simple connected 3-regular graph with 8 vertices (source: http://www.mathe2.uni-bayreuth.de/markus/reggraphs.html).

However, when I start plotting option for myself, I find it often very hard to determine whether two graphs are isomorphic or not. Here is an example with the first five graphs depicting the five options given in the link above and the sixth one of my additions:


问:
A question about the function $S(n)=\sum_{p\le\sqrt n}\Big\lfloor\frac n p\Big\rfloor$
Let's define the function
$$S(n)=\sum_{p\le\sqrt n}\Big\lfloor\frac n p\Big\rfloor$$
where $\,p\,$ is prime and $\,n\gt4$.

A computation of $\,S(n)\,$ suggests that $\,S(n)=S(n-1)\,$ if and only if $\,n\,$ is a prime.

I ask the reason of this behavior.

Many thanks.

答:
(Fill in the gaps as needed. If you're stuck, show your work and explain what you've tried.)

Hint: Show that if $n$ is prime, then $ \lfloor \frac{n}{a} \rfloor = \lfloor \frac{n-1}{a} \rfloor $ for $ a < n$.

Hint: Show that if $n$ is composite, then show that $ \lfloor \frac{n}{a} \rfloor  \geq \lfloor \frac{n-1}{a} \rfloor $, with strict inequality for certain $a$.

Hence, the result follows.


问:
How was the 2D discrete Laplacian matrix calculated?
If I convolute a 2D signal with the L4 kernel, I get the discrete Laplacian. But how was the L4 kernel calculated?

Wikipedia isn't very clear on this and I haven't found other information.
(https://en.wikipedia.org/wiki/Discrete_Laplace_operator#Implementation_via_operator_discretization)

I've also read this but it's still unclear how the convolution of [1, -2, 1] is L4 transposed.

L4 being this:

```<code> L4 =
  [ 0 -1  0;
   -1  4 -1;
    0 -1  0];
</code>```

EDIT:
Also, how is the L8 produced?

```<code> L8 =
  [-1 -1 -1;
   -1  8 -1;
   -1 -1 -1];
</code>```

答:
Basically the 2D "L4" discrete Laplacian operator is constructed by using 4 surrounding points from a central stencil point. These stencil points are north, south, east and west from the central point.
In space these 5 stencil points are then given by:
$$ (x,y), (x+h,y), (x,y+h), (x-h,y), (x,y-h)$$

Remembering that
$$\nabla^{2} f = f_{xx}+f_{yy} $$
and that a 2D Taylor expansion (up to a point) may be seen as:
$$f(x+\delta x,y+\delta y)=f(x,y)+(f_{x}\delta x + f_{y}\delta y)+ \frac{1}{2}(f_{xx}(\delta x)^{2}+2f_{xy} \delta x \delta y + f_{yy}(\delta y)^{2})+O((\delta x)^{3}+(\delta y)^{3})$$
We can then find that
$$f(x\pm h, y)= f(x,y) \pm f_{x}h+\frac{1}{2}f_{xx}h^{2}+O(h^{3})$$

$$f(x, y\pm h)= f(x,y) \pm f_{y}h+\frac{1}{2}f_{yy}h^{2}+O(h^{3})$$

By summing expressions we may obtain
$$(1) \qquad f(x+h,y)+f(x-h,y)=2f+f_{xx}h^{2}+O(h^{3})$$

$$(2) \qquad f(x,y+h)+f(x,y-h)=2f+f_{yy}h^{2}+O(h^{3})$$
Calculating (1) + (2):
$$ f(x,y+h)+f(x,y-h)+f(x+h,y)+f(x-h,y) = 4f +h^{2}(f_{xx}+f_{yy})$$
Then using our very first equation (Cartesian Laplacian definition) we are certain that we approximated our Laplacian up to 2nd order:
$$\nabla^{2} f = f_{xx}+f_{yy}= \frac{1}{h^{2}} \Big[f(x,y+h)+f(x,y-h)+f(x+h,y)+f(x-h,y) -4f(x,y) \Big]$$

As indeed we want to see this in a discrete 2D space, we may correspond the coefficients of our stencils onto a matrix where $h=1$
$$
    L4 = \frac{1}{h^2}\begin{pmatrix}
    c_{f(x-h,y+h)} & c_{f(x,y+h)} & c_{f(x+h,y+h)}\\
    c_{f(x-h,y)} & c_{f(x,y)} & c_{f(x+h,y)}\\
    c_{f(x-h,y-h)} & c_{f(x,y-h)} & c_{f(x+h,y-h)}\\
    \end{pmatrix} = \begin{pmatrix}
    0 & 1 & 0\\
    1 & -4 & 1\\
    0 & 1 & 0\\
    \end{pmatrix}
$$

One can follow a similar procedure to obtain L8 by using a 9-stencil configuration where the stencils would be
$$  f(x,y), f(x+h,y+h),f(x+h,y),f(x+h,y-h), f(x,y-h), f(x,y+h), f(x-h,y-h),f(x-h,y+h),f(x-h,y) $$


问:
How many triples $(x,y,n)$ are there such that $x^n - y^n = 2^{100}$

How many triples $(x,y) \in \mathbb{N^+}^2$ and $n \gt 1$ are there such that $x^n - y^n = 2^{100}$

I dont know how to start. Any hint will be helpful.

I know the identity $x^n-y^n = (x-y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2}+y^{n-1})$.

I think from here we need some combinatorics to get the rest of answer.

答:
Since the question is being asked rather often and the solution in the comments seems to have a flaw I propose the following answer.

We claim: the equation
$$
a^n-b^n=2^k,\tag1
$$
where $a,b,n,k$ are positive integers ($n>1$), has exactly $\left\lfloor\frac{k-1}2\right\rfloor$ solutions $(a,b)$ for $n=2$ and no solutions for $n>2$.

First of all we note that from
$$
a^n-b^n=(a-b)\sum_{i=0}^{n-1}a^{n-1-i}b^i=2^k,\tag2
$$
it immediately follows that the solution can exist for $n>1$ only if $k>0$.

Now let consider the problem case by case.

For $n=2$ from
$$
a-b=2^r,\quad a+b=2^{k-r}
$$
the only solutions are:
$$a=2^{k-r-1}+2^{r-1};\quad b=2^{k-r-1}-2^{r-1},\quad 0<r<\frac{k}2.\tag3$$

For $n=4$ we would have the equation:
$$
(a^2-b^2)(a^2+b^2)=2^k.
$$

This implies
$$a^2-b^2=2^m\implies a=2^{m-r-1}+2^{r-1};\quad b=2^{m-r-1}-2^{r-1}$$
for some $m$ and $r$.
But the equation
$$
2^{2(m-r-1)}+2^{2(r-1)}=\frac{a^2+b^2}2=2^{k-m-1}
$$
cannot hold in view of $r\ne\frac m2$ (cf. (3)).

For odd $n>1$ we note that equation (2) can hold only if both $a$ and $b$ are even: $a=2A$, $b=2B$. Assume now that equation (1) has an integer solution $(a,b)$ for some $k$. Then there must exist some minimal positive $K$ for which the equation holds. However from
$$
(2A)^n-(2B)^n=2^K\implies A^n-B^n=2^{K-n}
$$
it follows that such minimal $k$ does not exist (since $n<K$).

In general case an arbitrary $n>2$ can be represented as $2^p q$, where $q$ is an odd number. If $p>0$ repeating the process:
$$
2^k=a^n-b^n=(a^{n/2}-b^{n/2})(a^{n/2}+b^{n/2})\implies a^{n/2}-b^{n/2}=2^m;\ (
m>0)$$
one reduces the problem either to $n=4$ (if $q=1$) or $n$ odd (if $q\ne1$). The obtained contradiction finalizes the proof.


答:
This is just  a consequence of Taylor's formula. You can decide if $x_0$ is a local extrema observing how the values of $f$ vary when you move away from $x_0$. Taylor's formula states that

$$
f(x_0+h) = f(x_0) + f'(x_0) h + \cdots  +\frac{f^{(n-1)}(x_0)}{(n-1)!} h^{p-1} + \frac{f^{(n)}(\xi)}{n!} h^n
$$

if $f'(x_0) = \cdots =f^{(n-1)}(x_0) = 0$, the formula becomes

$$
f(x_0+h) -f(x_0) = \frac{f^{(n)}(\xi)}{n!} h^n.
$$

Also, assuming that $f^{(n)}$ is continuous, if $h$ is sufficiently small, the sign of $f^{(n)}(\xi)$ is the same as $f^{(n)}(x_0)$. So,

If $n$ is odd, the sign of $h^n$ changes passing through 0, and so the sign of $f(x_0+h)-f(x_0)$ is not fixed ($x_0$ is not an extrema).

If $n$ is even, $h^n \ge 0$ and so the sign of $f(x_0+h)-f(x_0)$ is (for small $h$) determined by the sign of $f^{(n)}(x_0)$. If it is positive, $f(x_0+h) \ge f(x_0)$ and you get a local minimum and, if it is negative, $f(x_0+h)\leq f(x_0)$ we you have a maximum.

Intuition works mainly for the case $n=2$, relating the sign of $f''(x_0)$ with local convexity/concavity.

Title: Classifying extremums using higher order derivatives

Body:
I was reading this from my calculus book:

If $f(x)$  has enough consecutive continuous derivatives at $x_0$ and we have
$n\ge2$ where  $$f'(x_0)=\cdots=f^{(n-1)}(x_0)=0,f^{(n)}(x_0)\neq0$$
Then if $n$ is an even number, $x_0$ is local extremum point( for
$f^{(n)}>0$ it is local minimum and for $f^{(n)}(x_0)<0$ it is local
maximum). If $n$ is an odd number $f$ has not extremum at $x_0$.

Can you please explain it . I don't understand this note. in fact I don't have intuition about $f'(x_0)=\cdots=f^{(n-1)}(x_0)=0,f^{(n)}(x_0)\neq0$ and how it can be connected with local extremum?


问:
Partial Differential Equation with Travelling Wave Solutions
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a smooth function. Consider the partial differential equation:
$$
\frac{\partial u}{\partial t}+\frac{\partial f(u(x))}{\partial x}+\frac{\partial^{3} u}{\partial x^{3}}=0
$$
I wish to find the solutions of the form $u(x, t)=\phi(x-c t)$ (traveling wave solutions) where $\phi$ is a smooth function. Then I am asked to integrate the obtained ODE, any help would be much appreciated since I am completely lost especially in the substitution process.

答:
We first set $\gamma=x-\alpha t$. Then we obtain the ordinary differential
equation
$$
-\alpha \frac{d \phi}{d \gamma}+f(\phi(\gamma))\frac{d \phi}{d \gamma}+\frac{d^{3} \phi}{d \gamma^{3}}=0.
$$
Integrating once we arrive at
$$
-\alpha \phi+f(\phi(\gamma))+\frac{d^{2} \phi}{d \gamma^{2}}=C_{1}
$$
where $C_{1}$ is a constant of integration. One more integration yields
$$
\frac{1}{2}\left(\frac{d \phi}{d \gamma}\right)^{2}=C_{2}+C_{1} \phi+\frac{\alpha}{2} \phi^{2}-F(\phi(\gamma)), \quad F(\phi)=\int_{0}^{\phi} f(y) d y
$$
Where $C_{2}$ is another constant of integration.


答:
In the given situation there are two cases:

1-If the order of first non-zero derivative at $x=x_0$ is odd, there is no local max/ min. e.g. $f(x)=(x-3)^5$ has no local max/min at $x=3$ as $f^{v}(3)\ne 0.$

2-If the order of first non-zero derivative at $x=x_0$ is even:

(a) there is  local min  if its value at $x=x_0$ is positive  e.g., $f(x)=(x-2)^4$ has local min at $x=2$ as $f^{iv}(2)>0$

(b) there is local max if its value at $x=x_0$ is negative e.g., $f(x)=1-(x-4)^6$ has local max at $x=4$ as $f^{vi}(4)<0.$


问:
Projected gradient descent with momentum
Can we apply momentum to projected gradient descent?  If so, how should we do that?

In the domain I'm working on, momentum greatly speeds up gradient descent.  However, I want to do projected gradient descent, rather than plain-old gradient descent.  In project gradient descent, after each update step, we project $x$ to a convex set $\mathcal{C}$.  If I apply momentum naively to projected gradient descent, things seem to work poorly.  I'm guessing this is because the projection changes the input to the next iteration of gradient descent and the algorithm doesn't expect this.

Is there a suitable update rule for projected gradient descent with momentum?  How should we modify the standard rule for momentum to incorporate the projection operation?

答:
Yes. You simply replace the proximal mapping based update rule in FISTA 1 with the projection update rule. That is, replace

$$x_{k+1} = \text{prox}_{h,t}(x_k - t_k\nabla g(x_k))$$

with

$$x_{k+1} = P_C(x_k-t_k\nabla g(x_k))$$

where $t$ is your step size.

[1] A. Beck, M. Teboulle, "A Fast Iterative Shrinkage-Thresholding Algorithm for Linear Inverse Problems", SIAM Journal on Imaging Sciences, vol. 2, no. 1, pp. 183-202, Jan 2009.


问:
Why morphism from initial object to any other given object is unique?
I am following Bartosz Milewski's category theory for programmers.

I understand the argument that the morphism from void -> any exists in the sense that its can't be proved otherwise. But can't the same argument be used to show that there are infinitely many of void -> ()? How is it unique?

答:
If $\mathcal{C}$ is a category, then an object $C$ is called initial if for all objects $D$, the set (class) $\text{Hom}_\mathcal{C}(C,D)$ contains exactly one morphism. Hence, your claim is true by definition of initial object.

Note that if $C$ and $D$ are initial objects, then $C \cong D$ since there are unique morphisms $C \to D$ and $D \to C$. The compositions $C \to C$ and $D \to D$ must necessarily be identities by uniqueness of such morphisms, so the morphisms $C \to D$ and $D \to C$ are inverse to each other.


问:
Metropolis Hasting Algorithm. Whats going wrong?
Hi I wrote a Langevin Metropolis-Hasting Algorithm on python to evaluate

$$ E[\|X\|^2], $$

where $X\sim \pi$, with

$$\pi(x)= \frac{1}{2\sqrt{2d}}e^{\frac{1}{2}\|x-a\|^2-\frac{1}{2}\|x+a\|^2},~~a_i=\frac{1}{\sqrt{2d}}~\forall i=1,\dots d$$

a Gaussian mixture. I.e I built a Markov Chain for some fixed $h>0$ and $\xi_i\sim N(0,I)$

$$ \tilde{x}_{i+1}:=x_i+ h \nabla \log \pi(x_i) +\sqrt{2 h} \xi_i$$

and setting $x_{i+1}$ according to

\begin{align}
&\mathbb{P}( x_{i+1}=\tilde{x}_{i+1})=\alpha(\tilde{x}_{i+1},x_i)
\\
& \mathbb{P}( x_{i+1}=x_i)=1-\alpha(\tilde{x}_{i+1},x_i).
\end{align}

where the acceptance probability is

$$  \alpha(x,y)=\min(1,\frac{\pi(\tilde{x}_{i+1})\exp\Big( -\frac{1}{4h}\| x_i-\tilde{x}_{i+1}-h\nabla \log \pi(\tilde{x}_{i+1}) \|^2 \Big)}{\pi(x_i)\exp\Big( -\frac{1}{4h}\| \tilde{x}_{i+1}-x_i-h\nabla \log \pi(x_i) \|^2 \Big)}).  $$

I calculate $\nabla \log \pi(x)=-2x$.

$\textbf{My question :}$ Does the above look ok ? when I run my code sometimes it works perfectly other times it gets stuck why could this be?

答:
Your proposals and acceptance probability seem correct to me.

The main issue I see here, is that you mention "some" $h>0$. Such parameter usually need to be tuned to make sure that your proposals are efficient. Have you done it ?

Cheers!


问:
How do I prove that $ 10 | 4^{n + 2 }+ 5^{n + 2 }+ 4^{n + 3 }+ 5^{n + 3 }$
I have no idea how to prove this. I don't even have an idea where to start from, could someone drop a few hints?</p>

Body:
Note that
$
4^{n + 2 }+ 5^{n + 2 }+ 4^{n + 3 }+ 5^{n + 3 }
=
5\cdot 4^{n + 2 }+ 6 \cdot 5^{n + 2 }
=
80 \cdot 4^n + 150 \cdot 5^n
$.


问:
Throwing a die 4 times and calculating the number of outcomes
Okay, so this is my problem:
How many outcomes where exactly ONE number occurs exactly TWO times
ex (2, 1, 2, 5) or (3, 6, 6, 1) but NOT (1, 2, 1, 2) or (3, 1, 3, 3)

My solution:
So the first throw has 6 possibilities, then this must be followed by one throw with only ONE possibility, then followed by a throw with 5 possibilities, followed by a throw with 4 possibilities.
Then this gives me: 6 * 1 * 5 * 4 = 120 outcomes

This solution gave me 0 points, so my question is where do I go wrong?

答:
Forget the dice, suppose you just had to pick four numbers from 1 to 6 , and you can repeat numbers. For your case, we would need to firstly pick the number that is going to be appearing twice - there are 6 ways to do it

Now, we need to pick two distinct numbers from the remaining 5 - there's ${5 \choose 2}$ ways of doing that

Now for order - we have a sequence of the type $A A B C$ - hence the total number of permutations of this list is $\frac{4!}{2!}$

Put it all together

$$N = 6 \times {5 \choose 2}\times \frac{4!}{2!}$$

Probability = $\frac{N}{6^4}$


答:
If you mean

$N=5^{n+3}+5^{n+2}+4^{n+3}+4^{n+2}$

Then we have:

$N=5^{n+2}(5+1)+4^{n+2}(4+1)= 5k$

But N is even therefore it must be a multiple of $2\times 5 =10$.

Title: How do I prove that $ 10 | 4^{n + 2 }+ 5^{n + 2 }+ 4^{n + 3 }+ 5^{n + 3 }$

Body: I have no idea how to prove this. I don't even have an idea where to start from, could someone drop a few hints?</p>


问:
Reflection principle and independence results, externally and internally
If I understand it correctly, for $\phi_1, \phi_2, \ldots, \phi_n$ any finite subset of ZFC, with the reflection principle and forcing, we can show, for example, that $\mathrm{ZFC} \vdash \mathrm{Con}(\phi_1,\ldots,\phi_n,\mathrm{\neg CH})$. It is quite convincing to conclude metamathematically that if $\mathrm{ZFC}$ is consistent, $\mathrm{ZFC} + \mathrm{\neg CH}$ is also consistent. But is it possible to show it internally, i.e. that $\mathrm{ZFC} \vdash \mathrm{Con}(\mathrm{ZFC}) \rightarrow \mathrm{Con}(\mathrm{ZFC} + \mathrm{\neg CH})$? Can we obtain this result with a similar method ? My idea would be that, if we have $\mathrm{Con}(ZFC)$, we have a model $V'$, and we can construct, in this model, a model for any subset of $\mathrm{ZFC} + \neg \mathrm{CH}$, and by the compactness theorem, we can conclude $\mathrm{Con}(\mathrm{ZFC} + \neg \mathrm{CH})$. But we assume that if $V'$ thinks that some set is a model of a subset of $\mathrm{ZFC} + \neg \mathrm{CH}$, it is also true in the metatheory. Is it the case?

答:
Yes. If ZFC has a model, it has a countable model, and then you can just force directly over that model.

The whole point of using Reflection is that you can use a meta-theory as weak $\sf PRA$, and that you don't need to worry about forcing over non-transitive models.


问:
Cluster growth by random enumeration
Consider the $\mathbb{Z}^2$-lattice and the $l^p$-Ball subgraph $\Lambda_r^p = (V_r^p, E_r^p)$ with $ V_r^p = \{v\in\mathbb{Z}^2 \colon \|v\|_p\leq r \}$ and $E_r^p = \{e = \{v,w\}\colon v,w \in V_r^p \text{ and } \|v-w\|_1 = 1\}$ ($1\leq p \leq \infty$, though the choice of $p$ should not be that important).

Now, let $(v_1,\ldots, v_R)$ be a random enumeration of $V_r$ (every such enumeration has probability $(R!)^{-1}$), and consider the sets
$$\Gamma_0 = \{(0,0)\}\,,\quad \Gamma_i =
\begin{cases}
\Gamma_{i-1} \cup \{v_i\} & \|v_i-w\|_1 = 1 \text{ for some }w\in \Gamma_{i-1}, \\
\Gamma_{i-1} & \text{else,}
\end{cases}$$
for $i=1,\ldots, R$.
The sets $\Gamma_i$ describe a growing cluster, and I am interested in the probability that $\Gamma_R$ contains a boundary node, i.e. a vertex in
$\partial V_r^p = \{v\in V_r^p \colon \|v-w\|_1 = 1 \text{ for some }w\in\mathbb{Z}^2 \setminus V_r^p\}$.

Question: Do constants $C,c>0$ exist such that
$$\mathbb{P}[\Gamma_R \cap \partial V_r^p \neq \emptyset] \leq C\exp(-cr)\,? $$
Is this probability decaying even faster? I have the feeling that it should decay at least exponentially fast, and that $C,c$ can be chosen independently of $p$.

答:
I think I found the answer myself. Let $v\in \partial V_r^p$ and consider the event $v\in \Gamma_R$. For this to happen, it is necessary that a path $v_{i(1)},\ldots, v_{i(n)}$ of adjacent vertices exists, where $v_{i(1)}$ is adjacent to the origin, $v_{i(n)}=v$ and $i(1)<\ldots<i(n)$. Call such a path $v$-path. The number of such paths of length $n$ can roughly be bounded by $4\cdot 3^{n-1}$ (selfavoiding paths starting at the 4 adjacent vertices to the origin, with $\leq 3$ choices at each crossing). The probability that for a fixed path, $i(1) < \ldots < i(n)$ happens is $(n!)^{-1}$. Therefore we get
\begin{align*}
\mathbb{P}[v\in \Gamma_R] & = & \mathbb{P}[\exists (v\text{-path})] \\
& \leq & \sum_n \sum_{(v\text{-paths}) \text{ of length }n} \frac{1}{n!} \\
& \leq & 4 \sum_{n=r}^\infty \frac{3^{n-1}}{n!}\,.
\end{align*}
The sum starts at $n=r$ because any $v$ on the boundary satisfies $\|v\|_1 \geq r$. This bound decays faster than any exponential
function in $r$, i.e. we get
$$\mathbb{P}[v\in\Gamma] \in \mathcal{O}(\exp(-cr)) $$
for any $c>0$. Because the number of boundary vertices grows linearly in $r$, we get
$$ \mathbb{P}[\Gamma_R \cap \partial V_r^p \neq \emptyset] \in \mathcal{O}(\exp(-cr)) $$
for any $c>0$.

Please let me know in the comments if I made an obvious mistake.

Title: Cluster growth by random enumeration


问:
Solving the heat equation with Neumann boundary conditions
The task is to solve the system:
$$
u_t - u_{xx}=tx \;\;\;\;\;\;\;(x,t)\in(0,L)\times[0,\infty)\\u(x.0)=1 \;\;\;\;\;x\in(0,L)\\u_x(0,t)=u_x(L,t) =0 \;\;\;t>0
$$
I am very stuck and so far I have tried a lot of different approaches:

First I tried to find a solution to the homogenous case using separation of variables. When I did this, I solved the differential equation in terms of $t$ but and the differential equation for $x$ only gave a non trivial solution in the case where I assumed that the ratio between $X$ and $X''$ was negative, but even this didn't help.

I thought a particular solution of $\frac{1}{2}xt^2$ might also be useful, but it doesn't equal 0 at the boundary

I realised I need a periodic function probably using $\sin(\frac{n\pi}{L})$ but I haven't been taught this formula that I keep seeing and I certainly haven't been taught Fourier's theorem which I see being used to calcuate the coefficients in these series.

I was wondering if anyone could help me. I think with a model solution that I can follow well, I will be able to solve some of the other problems from this class.

答:
Now that we have covered the solution in class, I know how to solve it:

So consider  the Neumann problem:
$$u_t-u_{xx}=tx \;\;\;(0,T)\times(0,L)\\
u(0,x)=1 \;\;\;0\leq x\leq L\\
u_x(t,0)=u_x(t,L)=0\;\;\;\;t>0$$

We will use the separation of variables method, first analysing the homogenous equation; suppose the solution has the following form:
$$
u(t,x) = \sum_{k=0}^{\infty}c_k(t)v_k(x)
$$
Substitution into the homogenous equation gives us ODEs to solve:

$$c_k'v_k-v_k''c_k=0\\
\frac{c_k'}{c_k}=\frac{v_k''}{v_k}=\lambda\\
v_{k_x}(0)=v_{k_x}(L)=0
$$

It can be seen pretty easily that if $\lambda \geq 0$, then $v_k(x)$ is either 0 or constant so we focus on the case $\lambda <0$. We shall call $\lambda = -\mu_k^2$

$$v_k''=-\mu_k^2v_k\\
v_k=a\cos(\mu_kx)+b\sin(\mu_kx)
$$
The initial condition gives us that $b=0$, and that $\mu_k = \frac{k\pi}{L}$. For ease of notation, we will call $\omega = \frac{\pi}{L}$. Dropping factor constants for now, we have
$$v_k(x)=\cos(k\omega x)$$
Now we  will solve for $c_k$ using the non-homogenous equation:
$$
u_t-u_{xx} = tx  = \sum_{{k=0}}^{\infty}(c_k'-((k\omega)^2\cos(k\omega  x)))
$$
We want to write the function $x$ as a sum of cosines, in other words, in its Fourier series,

$$
x = \frac{a_0}{2} + \sum_{k=0}^{\infty}a_k\cos(k\omega x)
$$

By using Fourier's Theorem, we have that:

$$
a_0 = \frac{2}{L} \int_{0}^{L}x  dx\\
a_k = \frac{2}{L}\int_0^L x\cos(k\omega x) dx
$$
Calculating the integrals (by parts), we  have

$$
a_0 = L\\
a_{2k} = 0\\
a_{2k-1} = \frac{-4}{L(kw)^2}$$
Now we can equate the two expressions:

$$
u_t - u_{xx} = \sum_{k=0}^{\infty}(c_k'-(k\omega)^2c_k)\cos(k\omega x) = t(\frac{a_0}{2}+\sum_{k=1}^{\infty}a_k\cos(k\omega x))
$$
Equating for each $k$ we have:

$$
c_0' = \frac{tL}{2} \;\;\; c_0(0)=1\\
c_0(t) = \frac{t^2L}{4}+1$$

$$
c_{2k}'-(2k\omega)^2c_{2k} = 0\\
c_{2k}(t) = c_{2k}(0)e^{(2k\omega)^2t}=0
$$

$$
c_k' - (k\omega)^2c_k = \frac{-4t}{L(k\omega)^2}$$
Using the integrating factor method, we can solve this last ODE which is valid for $k$ odd:
$$
c_k(t) = \frac{4t}{L(k\omega)^4}-\frac{4}{L(k\omega)^6}-\frac{4}{L(k\omega)^6}e^{-(k\omega)^2t}$$
We can now write down our final solution:

$$
u(t,x) = \sum_{k=0}^{\infty}c_k(t)v_k(x)\\
=\frac{t^2L}{4}+1 + \sum_{k \;\text{odd}} (\frac{4t}{L(k\omega)^4}-\frac{4}{L(k\omega)^6}-\frac{4}{L(k\omega)^6}e^{-(k\omega)^2t})\cos(k\omega x)
$$
$\square$


问:
Minimum number of coconuts
Three friends namely $A$, $B$ and $C$ collected coconuts with the help of monkey and fell asleep. At night, $A$ woke up and decided to have his share. He divided coconuts into three shares, gave the left out coconut to the monkey and fell asleep. In the same way in order $B$ and $C$ got up and not knowing whether anybody woke up divided the coconuts, giving the left out coconut to the monkey. Early in the morning all of them woke up together, divided the coconuts into equal shares and gave the left out coconut to the monkey.

What is the minimum number of coconuts they counted?

答:
Let $n$ be the total.

$A$ first takes $\dfrac{n-1}3$ and $1$ goes to the monkey. Left out coconuts is $n-1-\dfrac{n-1}3 = \dfrac{2(n-1)}3$.

After $B$ does the same at night, left out coconuts is $\dfrac23 \cdot \left(\dfrac23 (n-1)-1 \right)$.

After $C$ does the same at night, left out coconuts is $\dfrac23 \cdot \left(\dfrac23 \cdot \left(\dfrac23 (n-1)-1 \right) -1\right)$.

Let $m$ be the number of coconuts left in the morning.

Hence, we have
$$\dfrac23 \cdot \left(\dfrac23 \cdot \left(\dfrac23 (n-1)-1 \right) -1\right) = m \implies n = \dfrac{27m+38}8$$

Let the minimum number of coconuts left in the morning be $m$.

This is an integer only when $$27m \equiv 2 \pmod8 \implies m \equiv 6 \pmod8$$ We also know that $m=3k+1$, since they share in the morning and give one to the monkey. Hence,
$$m \equiv 6 \pmod8 \text{ and } m \equiv 1 \pmod3 \implies m \equiv 22 \pmod{24}$$
Since, we want $n$ to be a minimum, so has to be $m$. Hence, we have $$\color{red}{m=22 \implies n = 79}$$

Title: Minimum number of coconuts


问:
Describing functions with number of intersections

Main question

The graph of a continuous function $f: \mathbb{R} \to \mathbb{R}$ intersects a line $k$ times if and only if the same line intersects the graph of $y=x^{n}$ also $k$ times. Is $f(x)$ necessarily $x^n$? If not, for which $n$ is $f(x)$ necessarily $x^n$?

Further research

I look forward to generalizing $x^n$ to an arbitrary polynomial, or even an arbitrary continuous function (i.e., determining if the number of intersections of all line with a the graph of a function uniquely describes that function).

My current approach

So far, I have been trying to gain more intuition through the following perspective.

We describe a function $f$ with a partition of a plane $P_f$: if the number of intersection of a line $y= ax + b$ with $f$ is $k$, we mark the point $(a,b)$ on $P_f$ with $k$. This way, $P_f$ is partitioned into regions marked $0, 1, 2, \dots$, i.e., of $0, 1, 2, \dots$ intersections.

Note that this leaves out vertical lines. However, as we only consider partitions corresponding to continuous functions, with which vertical lines intersects exactly once, I believe it does not matter.

So, the question translates to

Does $P_f = P_{x^n}$ imply $f = x^n$?

More generally,

Is the map $f \to P_f$ injective?

I also wonder if an arbitrary partition $P^*$ has a corresponding continuous function. Perhaps finding a counterexample to this is simpler than the questions above.

As an example, I solved for partitions $P_{x^{2n}}$ and $P_{x^{2n + 1}}$ for $n \geq 1$. Denote the number the point $(a,b) \in P_f$ is marked with by $\gamma$. Then we get the following forms:

My next steps will be to get more insights on these partitions and learn if there are characteristic features for polynomials in them.

答:
For the case of $y=x^{2n}$, the answer is yes, we can confirm that $f = x^{2n}$. Here's why.

The graph of $y=x^{2n}$ splits the plane into 3 parts: itself $(0)$, the region above it $\text{(I)}$, and the region below it $\text{(II)}$. We denote the graph of $f$ by $\mathcal{C}$.

Suppose a point in $\mathcal{C} \cap \text{(II)}$ exists. We can draw a line through it which does not intersect $(0)$. This yields a contradition: the line cuts $\mathcal{C}$ at least once but never meets $(0)$. Hence all points in $\mathcal{C}$ must be either in $(0)$ or $\text{(I)}$.

Now draw the tangent $l$ of $(0)$ at $(x_0, x_0^{2n})$. There must be a point $(x_1, y_1) \in \mathcal{C} \cap l$. But we know that $(x_1, y_1) \in (0)\cup \text{(I)}$ and $l \cap \text{(I)} = \emptyset$. So $(x_1, y_1)$ must be on $(0)$. The only intersection of $(0)$ and $l$ is $(x_0, x_0^{2n})$; therefore, this point must be $(x_1, y_1)$.

As such, we have proved that every point on $(0)$ must also be on $\mathcal{C}$. Consequently, $f$ must be $x^{2n}$.


问:
Prove that the number of common tangents to two circles exterior to one another is 4
Given two circles outside each other, the maximum number of common tangents to the two circles is 4 according to Wikipedia. How can this be proven?

That's a picture above showing what I'm talking about. How can we prove that no other common tangents can be drawn aside those four?

答:
Hint Let the equation of the tangent be $a x + b y =  c$ with $a^2+b^2=1$, and let $(x_i, y_i), r_i$ be the centers and radius of the circles. The following system must hold
\begin{equation}
\begin{array}\cr
a x_1 + b y_1 - c = \pm r_1\cr
a x_2 + b y_2 - c = \pm r_2\cr
a^2 + b^2 = 1\cr
\end{array}
\end{equation}
The two first equations with unknowns $(a, b, c)$ form a linear system with rank 2 if the centers are different, hence for each of the 4 right hand sides, there is a straight line of solutions $(a, b, c)$. Two points at most on this straight line satisfy $a^2 + b^2 = 1$. It gives at most 8 solutions but these solutions come in identical pairs because  $(-a, -b, -c)$ defines the same line as $(a, b, c)$, hence there are at most 4 tangent lines.

Where does the system come from ?
The equation of a line $L$ in the plane can be written $a x + b y = c$ where $(a, b)\not = (0,0)$. Multiplying $a, b, c$ by a non zero factor does not change the line defined by this equation, so we can as well multiply by $1/\sqrt{a^2+b^2}$ to obtain an equation where finally $a^2 + b^2=1$. If $P = (x_i, y_i)$ is any point in the plane, the distance between $P$ and the line $L$ is
\begin{equation}
d(P, L) = |a x_i + b y_i - c|
\end{equation}
Indeed for any $t\in {\mathbb R}$, the point $(x_i + t a, y_i + t b)$ belongs to the line perpendicular to $L$ passing through $P$, because the vector $(a, b)$ is orthogonal to $L$. This point belongs to $L$ iff
\begin{equation}
a (x_i + t a) + b(y_i+ t b) = c
\end{equation}
which implies $t = -(a x_i + b y_i - c)$. The distance from $P$ to the intersection point is then $|t| = |a x_i + b y_i - c|$.

Thus, the two first equations of the system simply say that the distance of the center of each circle to the line $L$ is $r_i$. This property characterizes the tangent lines of the circles.


问:
Finding an Extremal for a function.
I need to find the extremals for the following function :

$I(y) = \displaystyle \int_{x_0}^{x_1} \dfrac{1 + y^2}{(y')^3} dx$

So, by Euler Lagrange Equations

$I_{y}$ -$d/dx(I_{y'}) = 0$

Now, using this I get :

$\dfrac{2y}{y'} + \dfrac{2y}{3} = \dfrac{4(1+y^2)y''}{(y')^3}$

At, this point I am stuck, Please tell me how should I proceed ?

Thank You.

答:
$I(y) = \displaystyle \int_{x_0}^{x_1} \dfrac{1 + y^2}{(y')^3} dx$
reduced Euler-Lagrange equationd for $I=\int F(x,y,y') dx $ is given as $$F-y'\frac{\partial F}{\partial y'}=C.$$ So here we have
$$\frac{1+y^2}{y'^3}+3y'\frac{1+y^2}{y'^4}=C \implies y'= D(1+y^2)^{1/3}.$$
$$\implies \int \frac{dy}{(1+y^2)^{1/3}}=Dx+E.$$
$$\implies y~_2F_1(1/2,1/3,3/2;y^2)=Dx+E.$$
Here $~_2F_1(a,b;c,z)$ is the Gauss hypergeometric function.

Title: Finding an Extremal for a function.


问:
Some algebraic inequalities with the binomial theorem.
I am working on proving the following limits.

1), $\lim_{n \to \infty}  \sqrt[n]{n} = 1$

2), If $p >0$ and $\alpha \in \Bbb R$, then $\lim_{n \to \infty} {n^{\alpha}\over{(1+p)^n}} =0$

And I am trying to follow the proof but I am stuck on understanding the following inequalities regarding the binomial theorem.

Put $x_n = \sqrt[n]{n} -1$. then $$n = (1+x_n)^n \ge {n(n-1) \over 2}x_n^2 $$

I understand that the right side of the inequality probably comes from $$1 + nx_n + {n \choose 2}x_n^2 + ・・・+x_n^n$$

and taking all the terms after ${n \choose 2}x_n^2$.

But to know this I think that $x_n^k$ has to be greater than $x_n^2$ for $k > 2$, am I right?
Since we can only say that $x_n > 0$, I'm a little puzzled.

For the second one,$$(1+p)^n > {n \choose k}p^k = {n(n-1)・・・(n-k+1) \over {k!}}p^k > {n^kp^k \over {2^k k!}}$$

is supposedly true when $n > 2k$.

I understand the left three expressions, but how does the right side of the inequality work ?

答:
For the first one: since it is true that $\,\sqrt[n] n\ge 1\;\;\forall\,n\in\Bbb N\,$ (why?) , we can write

$$\sqrt[n] n=1+c_n\;,\;\;0\le c_n\in \Bbb R\implies $$

$$\implies n=(1+c_n)^n=\sum_{k=0}^n\binom nkc^k_n\ge\binom n2c_n^2=\frac{n(n-1)}2c_n^2\implies$$

$$\implies 0\le c_n\le\sqrt{\frac2{n-1}}\xrightarrow[n\to\infty]{}0$$

The squeeze theorem then gives

$$c_n\xrightarrow[n\to\infty]{}0\implies \sqrt[n]n=1+c_n\xrightarrow[n\to\infty]{}1$$


问:
Evaluate $\lim_{n\rightarrow \infty} \sqrt[n]{\frac{1^n+2^n+3^n+\cdots +n^n}{1+2!+3!+\cdots +n!}}$
Evaluate the following limit:
$$
\lim_{n\rightarrow \infty} \sqrt[n]{\frac{1^n+2^n+3^n+\cdots +n^n}{1+2!+3!+\cdots +n!}}
$$

I tried to apply Stirling's approximation, but it seems that it doesn't work. Can anyone help?

答:
Let's use the squeeze theorem. First,
$$
\sqrt[n]{\frac{\sum_{k=1}^n k^n}{\sum_{k=1}^n k!}}
\leq
\sqrt[n]{\frac{n\cdot n^n}{n!}}
= n^{1/n} \sqrt[n]{\frac{n^n}{n!}}
\xrightarrow[n\to\infty]{} 1\cdot e = e
$$
by, e.g., Stirling's approximation.

Second,
$$
\sqrt[n]{\frac{\sum_{k=1}^n k^n}{\sum_{k=1}^n k!}}
\geq
\sqrt[n]{\frac{n^n}{n\cdot n!}}
= \frac{1}{n^{1/n}} \sqrt[n]{\frac{n^n}{n!}}
\xrightarrow[n\to\infty]{} e
$$
for the same reason.


问:
Name for matrix operation that inverses off-diagonal elements
Is there any operation that sends matrix
$$
S=\left(\begin{array}{cc} a&b\\
c& d
\end{array}\right) \to S'
=\left(\begin{array}{cc} \:\:\:a&-b\\
-c& \:\:\:d
\end{array}\right)
\quad a,b,c,d \in \mathbb{R}
$$
by setting off-diagonal elements of 2x2 matrix to inverse?
My question is: is there a name (or notation) for this kind of matrix operation?

If I would assume that $b,c \in Im$ (imaginary) then the operation obviously can be defined as conjugate $S'=(S^*)^T$. But all values are real in the given case.

Edit: $S \in SL(2, \mathbb{R}) $

答:
The two matrices $S$ and $S'$ are conjugate, or similar since they have the same characteristic polynomial
$$
t^2-(a+d)t+(ad-bc),
$$
provided that they aren't a scalar multiple of the identity, i.e., provided that $(b,c)\neq (0,0)$.


问:
$\int_{0}^{1} \sin x \cdot \ln x dx$
I would like to evaluate the following integral:

$$\int_{0}^{1} \sin x \cdot \ln x dx$$

I had a pretty decent idea to do it, and that is by turning $\sin x$ an $\ln x$ into their respective power series, turning it into a double summation, and integrating. This yields the following:

$$-\int_{0}^{1}\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{x^{2n+1}(-1)^n(1-x)^{m+1}}{(m+1)(2n+1)!} dx$$

Putting the integral inside the double summation and making use the beta function, we get:

$$ =-\sum_{n=0}^{\infty} \sum_{m=0}^{\infty}\frac{(-1)^n \beta(2n+1,m+1)}{(m+1)(2n+1)!}$$

A slight reindexing gives:

$$\sum_{n \space \text{odd}} \sum_{m \space\geq 1}\frac{\beta(n,m)}{n!m}$$

Notice the $-1$ in the front goes away by cancelling with the $-1$ from letting $n$ run through the odds.

Is this a valid strategy? If there is a closed form for this integral, I think this is a pretty cool summation identity. Can someone help out?

答: Multiple comments have already pointed to a special function $\operatorname{Cin}(x):=\int_0^x\frac{1-\cos t}{t}dt$. Integration by parts with $u:=\ln x,\,v:=1-\cos x$ gives $\int_0^1\sin x\ln xdx=[(1-\cos x)\ln x]_0^1-\operatorname{Cin}(1)$. The surface term is $-\lim_{x\to0^+}(1-\cos x)\ln x=-\tfrac12\lim_{x\to0^+}x^2\ln x=0$, so the original integral is $-\operatorname{Cin}(1)$.

Title: $\int_{0}^{1} \sin x \cdot \ln x dx$


问:
Showing that tensoring with $id_A$ preserves injectivity of $\mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/6\mathbb{Z}$
Let $f:\mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}/6\mathbb{Z}$, $n \mapsto 3n$. Let $A$ be any abelian group, then the map
$$id_A \otimes f: A \otimes \mathbb{Z}/2\mathbb{Z} \to A \otimes \mathbb{Z}/6\mathbb{Z}$$ is injective. It seems intuitive to me, because no matter what I choose for $A$ I cannot make $A \otimes\mathbb{Z}/6\mathbb{Z}$ "smaller than" $A \otimes \mathbb{Z}/2\mathbb{Z}$. However I know that tensoring does not preserve injective maps in general, so I am wondering how one can prove this statement?

答:
This is an answer not using the hints in the comments, but probably boils down to some of them.

Let $a\otimes b\in A\otimes \mathbb{Z}/2\mathbb{Z}$. Assume $0=(id\otimes f)(a\otimes b)=id(a)\otimes f(b)=a\otimes f(b)$.

Then we have that either $f(b)=nz$ with $z\in\mathbb{Z}/6\mathbb{Z}$ and $n\in\mathbb{Z}$ such that $na=0$ (in $A$), or $a=mx$ for $x\in A$ and $m\in\mathbb{Z}$ such that $mf(b)=0\mod 6$.

Note that $f(b)$ can only be $0$ or $3\mod 6$ and in the case $f(b)=0\mod 6$, also $b=0\mod 2$ and $a\otimes b=0$, so let as focus on $f(b)=3\mod 6$ ($b=1\mod 2$).

If $f(b)=nz$ with $z\in\mathbb{Z}/6\mathbb{Z}$ and $n\in\mathbb{Z}$ such that $na=0$. Then $n$ is odd. Thus we have $a\otimes b=a\otimes nb=na\otimes b=0$.

Finally, if $a=mx$ for $x\in A$ and $m\in\mathbb{Z}$ such that $mf(b)=0\mod 6$. then clearly $a\otimes b=0$, because $f(b)=3\mod 6$, so $m$ must be even, and in such case $a\otimes b=mx\otimes b=x\otimes mb=0$.


问:
Dot in conditional probability - explanation of notation
I am just reading a paper (https://www.aclweb.org/anthology/P19-1400.pdf) and I came across this notation that I have never seen before and do not understand. See given below:

What is the meaning of this dot in the conditional probability?

答:
The dot is some (innocuous, but possibly confusing) abuse of notation to show what the probability measure used here is.

Specifically, the KL divergence takes two arguments, both probability measures. Here, the second argument is $p_N^\ast$, which is unambiguous; but the first is the (average) conditional measure with pdf
$$
x\mapsto \frac{1}{|D|}\sum_{(m,c)\in D}p_N(x\mid m,c,E^+)
$$
The $\cdot$ is used to indicate that, in the expression, the first argument "$\frac{1}{|D|}\sum_{(m,c)\in D}p_N(\cdot\mid m,c,E^+)$" is a function of the variable which goes where the $\cdot$ is.


问:
Equivalence of definitions of mutually singular measures
I'm wondering if these two definitions of mutually singular measure are equivalent? I've been trying to prove the equivalence but none of my attempts have been fruitful. Any suggestions would be much appreciated:

Let $\mu$ be a signed finite measure on $\mathcal{F}$ a sigma algebra on a set $\mathcal{X}$, and $\nu$ a finite positive measure on $\mathcal{F}$. Then $\mu \perp \nu$ if there exists a measurable $B$ such that $\mu(B^c) = 0 = \nu(B)$.

The second definition is with the same setup as above, except we now require for any measurable $E$, $\mu(B^c \cap E) = 0 = \nu(B \cap E)$.

答: They are equivalent for positive measures.  The second one is fixed up so that it can be used for signed measures (or complex measures).</p>


问:
A short exact sequence of remainder classes is split iff $(n,m)=1$
Let $m, n \in \mathbb{N}, m, n > 1$ and we have the short exact sequence of $\mathbb{Z}$ modules (as is easily seen)
$0 \rightarrow \mathbb{Z}_{m} \xrightarrow{\alpha} \mathbb{Z}_{mn} \xrightarrow {\beta} \mathbb{Z}_n \rightarrow 0$, where $\alpha (\bar{k}) = \hat{nk}, \beta (\hat{l}) = \bar{\bar{l}}$. I'm trying to prove that it is split iff $(m,n)=1$.

If $(m,n) = 1$ it follows easily that $\mathbb{Z}_{mn} = n\mathbb{Z}_{mn} \oplus m\mathbb{Z}_{mn}$ and I'm done since $ker(\beta) = Im (\alpha) = n \mathbb{Z}_{mn}$. I'm having trouble with the other implication:

Supposing $\mathbb{Z}_{mn} = n\mathbb{Z}_{mn} \oplus X$ for $X$ a subgroup of $\mathbb{Z}_{mn}$, then, writing $X = a\mathbb{Z}_{mn}$ with $0 < a < mn$ and using that $\hat{1} \in \mathbb{Z}_{mn} = n\mathbb{Z}_{mn} + a \mathbb{Z}_{mn}$, I get that $(n,a) = 1$.

So I am left to see that $a = m$, and obviously I should use here that $a\mathbb{Z}_{mn} \cap n\mathbb{Z}_{mn} = \hat{0}$, but I don't know how.
I'm trying to use things like $\hat{m} = \hat{ak} + \hat{nl}$ for some $\hat{k}, \hat{l}$ or like $\widehat{1 + m} =  \hat{ak} + \hat{nl}$ for some other $\hat{k}, \hat{l}$, but these don't seem to help.

答:
$\require{AMScd}$

Suppose $(n,m)=1$. Then it is easily checked that the following is an isomorphism of exact sequences:
\begin{CD}
0@>>>\mathbb{Z}_m @>{\alpha}>> \mathbb Z_{nm} @>{\beta}>> \mathbb Z_{n} @>>> 0\\
& @V{n}VV @V{f}VV @V{\text{id}}VV\\
0@>>>\mathbb{Z}_m @>{p}>> \mathbb Z_{n}\oplus\mathbb Z_m @>{q}>> \mathbb Z_{n} @>>> 0
\end{CD}

Here, $p$ and $q$ are the canonical homomorphisms and $f:\mathbb Z_{nm}\to\mathbb Z_n\oplus\mathbb Z_m:\overline a\mapsto(\overline a,\overline a)$. Coprimeness is used to say $n:\mathbb Z_m\to\mathbb Z_m$ is an isomorphism.

Conversely, if the exact sequence is split, then we need an isomorphism $\mathbb Z_{nm}\cong\mathbb Z_n\oplus\mathbb Z_m$, which means $(n,m)=1$.


问:
Minimization of iron costs: financial mathematics for an high school
Meanwhile I apologize to everyone. I have never treated financial mathematics. A colleague of mine asked me for help but I can't give an answer. I have two exercise. This is the first.

A company operates in the iron sector, and sustains the following fixed costs equal to $ 648 \, €$ for month. A variable cost of $4$ € euros per quintal produced. An additional cost determined by the use of a particular ironworking machinery, equal to half of the square of the quintals produced.

What are the function of the total monthly cost and the function of the average cost?

What is the production at which you have the lowest average cost and what is the total monthly cost at this quantity?

The results are:
$$a) \quad C=0.5q^2+4q+648, \quad C_m=0.5q+\frac{648}q+4$$
$$b)\quad  q=36\, \frac{\text{quintals}}{\text{month}};\quad  C(36)=€\, 1440$$

My observation: with the first informations of the problem I have a parabola. Why? For the rest of the questions I think that there are some derivative to be done. Never studied financial mathematics. Thank you all for your understanding.

答:
Before to understand the exercise I had to study a bit on the internet. Regarding the average cost $C_m(q)$ one must keep in mind that the average cost is equal to the cost per unit of production; while the marginal cost $C_{\text{mar}}$ represents the change in cost due to a unit increase in production which we will denote by the independent variable $q$. We therefore have:

\begin{equation}
    C_m(q)=\frac{C(q)}{q}
\end{equation}

and
\begin{equation}
    C_{\text{mar}}(q)=\frac{dC(q)}{dq}
\end{equation}

So in Economics we use linear, quadratic, cubic, exponential etc. functions for the cost function. The curves of the average cost and marginal cost intersect at the stationary points of the average cost (i.e. at the points where the average cost has zero derivative).
Given the nature of the problem with the indications provided of the exercise we have three parameters. Excluding the circumference which is not a function unless there are appropriate domain restrictions, the simplest case is represented by a parabola.
$$C(q)=0.5q^2+4q+648$$
with average cost equal to
$$C_m(q)=\frac{0.5q^2+4q+648}{q}=0.5q+\frac{648}{q}+4$$
Now for the marginal cost it is necessary to use the derivative, equal to
$$C_{\text{mar}}=\frac{dC(q)}{dq}=q+4$$
Intersecting the two functions unit or average cost $C_m(q)=(0.5q^{2}+4x+648)/q$, branch of nonequilateral hyperbola, and marginal cost we have the intersection point $(36, 40)$, so $q=36\, \text{quintals}/\text{month}$ and $C(36)=1440$ euros.


问:
How to determine limits of a semi-infinite volume of a cone?
I'm struggling to find the limits need for this equation to give me the answer the question requires.
$z=\sqrt{x^2+y^2}$ and $x>=0$ for the integral $\int\int\int(xe^{-z}) dV$.

I've got that z is between 0 and infinity and theta between 0 and $\pi$ and R between 0 and z but when that is integrated i can't get 4, which is what the question requires the answer to be.

Can anyone tell me where i've gone wrong?

答: try limits of r bounded by z and 0. limits of z bounded by zero and infinity. And your limit for theta should be between pi/2 and -pi/2. The limit for theta is because the cone is between the posotive y-axis and the negative y-axis. so theta is pi/2 at the posotive y-axis and theta is negative pi/2 when it goes anti-clockwise from the posotive x-axis.</p>


问:
Every manifold has unique orientation in base ring Z/2
How is it that every manifold has a unique $\mathbb{Z}/2$ orientation? Came across the more general conversation here and here (they're linked).

I'm familiar with Hatcher's Algebraic Topology which has the local orientability defintion: A local $R$ orientation of $X$ at $x$ is a choice of a generator of the $R$-module $H_n(X,X−x)$.

答:
An orientation, as defined by Hatcher, is a function $\mu$ assigning to each $x \in M $ a generator $\mu_x$ of $H_n(M,M \setminus \{x\};R) \approx R$ which satisfies a suitable local consistency condition.

$\mathbb Z_2$ has only one generator, thus there is only one function $\mu$ and the consistency condition is automatically satisfied. If $R$ has more than one generator, then the local consistency condition gives a serious restriction on the function $\mu$.


问:
If $p$ is a polynomial approximating a continuous function $h$, why can we assume $\left\|p\right\|_\infty\le\left\|h\right\|_\infty$?
Let $K\subseteq\mathbb R$ be compact and $h:K\to\mathbb R$ be continuous and $\varepsilon>0$. By the Stone-Weierstrass theorem, there is a polynomial $p:K\to\mathbb R$ with $\left\|h-p\right\|_\infty<\varepsilon$.

Why can we choose $p$ such that $\left\|p\right\|_\infty\le\left\|h\right\|_\infty$?

I'm not sure whether it's really crucial that $p$ is a polynomial. In general, if $K$ is a compact Hausdorff space, $\mathcal C$ is a subalgebra of $C(K)$ with $1\in\mathcal C$ and $f:K\to\mathbb R$ is continuous with $c:=\left\|f\right\|_\infty\ne0$, then there is a $\tilde g\in\mathcal C$ with $$\left\|\frac fc-\tilde g\right\|_\infty<\frac\varepsilon2.$$ Now $g:=c\tilde g\in\mathcal C$ and $\left\|f-g\right\|_\infty<\varepsilon$. But only if $\left\|\tilde g\right\|\le1$, we obtain $\left\|g\right\|_\infty\le c=\left\|f\right\|_\infty$ ...

答:
Assuming $K$ is not empty and $h\ne 0.$

Let $e=\min (\|h\|/2, \varepsilon /2).$

For $x\in K$ let $\bar h(x)=h(x)$ if $|h(x)|< \|h\|-e\,;$ let $\bar h(x)=\|h\|-e$ if $h(x)\ge \|h\|-e\,;$ let $h(x)=e-\|h\|$ if $h(x)\le e-\|h\|.$

Now $\bar h$ is continuous. So take $p$ with $\|p-\bar h\|<e.$


问:
probability density functions
Suppose $Y$ is a random variable pdf $f(y)=ky , y=3/n,6/n,9/n...,3n/n$

Find the value of the constant $k$ and write down $Y$'s cdf.

Find simple general expressions for $EY, \text{Var} \,Y, P(Y=3/2)$ and $P(Y>3/2)$

For the case $n=10$, evaluate $EY, \text{Var}\,Y,P(Y=3/2)$ and $P(Y>3/2)$, then write down $Y$'s pdf and cdf.

Repeat (3) for the limiting case as $n$ tends to infinity.
HINT: The random variable in (4) has cdf equal to the limit of the cdf (1) as $n\to \infty$.

Wondering if it's possible to consider $Y$'s cdf by cases, like when $y=3/n,y=6/n...$ and sum up to get the value of k.

Could anyone help? thanks.

答:
Indeed. For the function $f$ to be a valid PDF one must have that $f(y)\geq 0$ for all possible $y$-values and that
$$
\sum_{y}f(y)=1\tag{1}
$$
where the sum is taken over the possible values of $y$. Note that the possible values $$\frac3n, \frac6n,\ldots,\frac{3n}n$$ are of the form
$\frac{3i}{n},$ where $i=1,2,\ldots,n$, and hence $(1)$ becomes
$$
\sum_{i=1}^n k\frac{3i}{n}=1.
$$
Using that $\sum_{i=1}^n i = \frac{(n+1)n}{2}$ we obtain that
$$
1=\frac{3k}{n}\sum_{i=1}^ni=\frac{3k}{n}\frac{n(n+1)}{2}=k\frac32(n+1)
$$
exactly as proposed by @Argha.

To find the CDF of $Y$ we note that $P(Y=y)=f(y)=ky$ for $y=\frac{3i}{n}$ for some $i=1,\ldots,n$. The CDF is by definition given by
$$
F(y)=P(Y\leq y)=\sum_{y'\leq y}f(y')
$$
so you're simply summing $f(y')$ for the values $y'$ that are less than $y$. Hence $F(y)$ becomes a step function that jumps exactly at the values $\frac{3i}n$ for $i=1,\ldots,n$.

Title: probability density functions


问:
$S=\{x\in X: \|x\|=1\}$ cannot be covered by a finite family of closed balls in X
I am working with an exercise and have som questions.
The exercise is

Let X be an infinite dimensional normed vector space  over $\mathbb{K}$, where $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C}$

Show that one cannot cover the unit sphere $S=\{x\in X: \|x\|=1\}$ with a finite family of closed balls in $X$ such that none of the balls contains 0.

I am thinking to show that S is not covered by a finite family of closed balls in X such that none of the balls contains 0. To show this I am thinking that I have to show $S\not\subset \cup \{x\in X| d(x,p)\leq r\}$ and to show this I am thinking to take a $x\in S$ and show that $x \notin \cup \{x\in X| d(x,p)\leq r\}$.

I note that for $x\in S$ I have that $\|x\|=1$. And to show that $x\notin \cup \{x\in X| d(x,p)\leq r\}$ I am thinking that I have to show that $d(x,p)>r$ . And to show this I am thinking that I somehow have to use Riesz lemma, but to be able to use this lemma I have to have that $\cup \{x\in X| d(x,p)\leq r\}$ is a closed proper subspace of X. It is obviously closed but I don't know how to show that it is a proper subspace of X.
But after I have argued that it is a proper subspace of X then I am thinking that I have from Riesz lemma that there exist a $x\in X$ with $\|x\|=1$ such that $\|x-y\|\geq a$ for all $y\in
\cup \{x\in X| d(x,p)\leq r\}$, where $0<a<1$. Since we have that $p\in \cup \{x\in X| d(x,p)\leq r\}$ then we have that $\|x-p\| \geq a$, and what we want to show is that $d(x,p)>r$ i.e. $\|x-p\|>r$. So dont now how I get from $\|x-p\| \geq a$ to $\|x-p\|>r$

答:
In the union $\cup\{x \in X, d(x,p) \leq r \}$, what are the values of $p$? I assume that $p$ and $r$ run over a finite set. Let me denote the closed ball around $p$ with radius $r$ by $B(p,r) = \{x \in X, d(x,p) \leq r \}$ and assume by contradiction that $S \subset \cup_{i=1}^n B(p_i,r_i)$. Since the balls are centered on $S$ and cannot contain $0$, we have $||p_i|| = 1$ and $0 < r_i < 1$.

Now, let $e_1$ be a unit vector. Then, $e_1 \in B(p_{i_1},r_{i_1})$ for some $1 \leq i_1  \leq n$. Riesz's lemma now finds an $e_2 \in S$, s.t. for $y \in lin\{e_1,p_{i_1}\} = W_1$ we have $||y-e_2|| \geq r_{i_1}$. In particular, $e_2 \notin B(p_{i_1},r_{i_1})$.
Iterating this construction, we find closed subspaces $W_k = lin\{e_1,p_{i_1},\ldots,e_k,p_{i_k}\}$, $e_{k+1} \notin W_k \cup \cup_{j=1}^k B(p_{i_j},r_{i_j})$ and $i_r \neq i_s$ for $r \neq s$. Since $X$ is infinite dimensional, $W_k$ are always proper subspaces of $X$. But then we obtain a contraction when $k=n$.


问:
How do I prove that $ 10 | 4^{n + 2 }+ 5^{n + 2 }+ 4^{n + 3 }+ 5^{n + 3 }$
I have no idea how to prove this. I don't even have an idea where to start from, could someone drop a few hints?</p>

Body:
Hint: $\ ({a,b\!+\!1})(\color{#c00}{a\!+\!1,b})\mid a^j(\color{#c00}{a\!+\!1}) + \color{#c00}{b^k}(b\!+\!1)\,$ if $\,j,\color{#c00}{k>0}$.

OP is case $\ a,b = 4,5,\ j,k = n\!+\!2$


问:
Why are elements of $GF(p^n)$ represented as polynomials mod some irreducible polynomial, while elements of $GF(p)$ are integers mod $p$?
My background is in computer engineering, not pure math. I've been studying the application of Finite Fields. Wikipedia on Finite field arithmetic says:

GF(p), where p is a prime number, is simply the ring of integers modulo p.

...

Elements of GF($p^n$) may be represented as polynomials of degree strictly less than n over GF(p).

I would really appreciate a simple, accessible explanation as to why this is the case? What fundamental property of Finite Fields makes it that the elements of $GF(p)$ and $GF(p^n)$ behave so differently and yet are both considered Finite Fields?

答:
The field $GF(p)$, $p$ prime, is a so-called prime field and given by the integers modulo $p$.

The field $GF(p^n)$ is a field extension of $GF(p)$; i.e., it contains $GF(p)$ as subfield.

The practical way to construct $GF(p^n)$ is to choose a polynomial $f(x)$ in $GF(p)[x]$ of degree $n$, which is irreducible over $GF(p)$.
Then the field $GF(p^n)$ is given by the quotient ring $GF(p)[x]$ modulo the ideal $\langle f(x)\rangle$; practically, the computations are modulo the polynomial $f(x)$.

There are other ways to consider $GF(p^n)$, e.g. as the zero-set of a polynomial - but this would be more abstract and not suitable for practical computations.


问:
A question with infinity
I am a sophomore in high school and my math teacher did a very short lesson on infinity, here's how it went:  (Try to solve each part yourself the first two are easy)

Part 1

You have an inf. number of boxes each labeled like so

```<code>[1][2][3][4].....
</code>```

and that there is a person in each one of these boxes.  But you have another set of five boxes like so

```<code>[1][2][3][4][5]
</code>```

You want to only have one set of boxes what do you do?  You must be able to tell each person what box they should goto next.

Spoiler

Each person is told to move down five boxes.  Or n+5 is you new box, and the five people from the other boxes are able to move into the first five boxes or just n+0.

Part 2

Well you actually have two of these inf. box sets, but you really only want one.  What now? (same rules apply)

Each person from set one is told to go to box 2n and each person from set two is told to box 2n-1

Part 3

Now you have an inf. number of these inf. box sets arranged like so

```<code>1: [1][2][3][4][5]...
2: [1][2][3][4][5]...
3: [1][2][3][4][5]...
4: [1][2][3][4][5]...
5: [1][2][3][4][5]...
</code>```

and you want to fit them all into one inf. box set, how do you do it.  She refused to tell us the answer and no one in my class figured it out, but I think I just did (correct me if I'm wrong, I'm also looking for alternate solutions).

How I went about solving this problem

Step 1

I assumed that there had to be some sort of pattern so I tried a few

Step 2

I picked the following patter.

1: [1][3][6][10][15]...

2: [2][5][9][14][]...

3: [4][8][13][][]...

4: [7][12][][][]...

5: [11][][][][]...

The numbers in the box represent the new box of each person

Step 3

I started to create a formula for the people to find there new room based of there current room number and box set number.  I found the following sequence in for people in the first row, 1 3 6 10 15.  Or +2 +3 +4 +5.  Which told me that it was quadratic so the equation had to be ax^2 + bx + c = new room number

Step 4

I found three equation by plugging in room numbers and their correct new room numbers to find these three equations:

(1^2)a + 1b + c = 1 or a + b + c = 1

(2^2)a + 2b + c = 3 or 4a + 2b + c = 3

(3^2)a + 3b + c = 6 or 9a + 3b + c = 6

Step 5

After solving this system I got the following equation 1/2x^2 + 1/2x + 0 = new room number where x = current room number.  This of course only tell the people in the boxes on the bottom where to go. Now I needed to find how to place the rest of the people.  The first thing I noticed was that people on the same diagonal as seen bellow were very much related to each other when you look at the rooms they will end up in.

[x-4][][][][]

[][x-3][][][]

[][][x-2][][]

[][][][x-1][]

[][][][][x]

The solution

Working this knowledge into my equation was fairly easy (I was able to do it while going for a run so I won't explain how I got my equation).  The equation is:

 (1/2(r+c-1)^2 + 1/2(r+c-1) + 0)-r+1 = new room number | r = row | c = column

Now my question to all of you is

Is my equation correct?

Are there any alternate solutions to this problem?

How could you fit a 3D array of boxes (inf boxes of people in all direction) into a 1D line of boxes?

Can you find a way to fit a n dimensional array of boxes into a 1D line of boxes?

For any of you into JAVA here's a a short snippet of code I made to prove my answer

```<code>    package main;

public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int f = 2;
        int r = 1;
        int c = 1;
        while(true){
            double x = (.5*(r+c-1)*(r+c-1) + .5*(r+c-1) + 0)-r+1;
            System.out.print(x +", ");
            c++;
            r--;
            if(r < 1){
                r = f;
                f++;
                c = 1;
            }
        }
    }
}
</code>```

答:
The following answer is not yet complete solution, but I think it might be an idea how to approach the problem yourself.

I recall the Collatz-problem in the "syracuse"-statement and look at the inverse transformation which allows to construct an infinite tree with infinitely many infinitely long "boxes" where each odd natural number occurs exactly one time (well, whether all odd natural numbers occur at least once is really open, but it would not spoil the resulting infinite indexing-system).

The "syracuse"-statement of the Collatz-transformation is with an odd a and b and a natural $A \ge 1$ is $$ b = {3a+1 \over 2^A} $$ where the value of A and that of b are of course uniquely determined by the value of a. But there are a lot (actually infinitely many) of different a which lead to the same b, so if we invert the transformation,
$$ b = {a 2^A - 1 \over 3} $$ we have for each a by infinitely many A's infinitely many different b, so we could write infinitely many indexed b as $$ b_{A} = {a 2^A - 1 \over 3} $$ and now each of the infinitely many $b_A$ can be treated similarly to get $$c_{A,B} = {b_A 2^B - 1 \over 3} $$
This scheme implements an infinite long index (the infinite number of boxes) and each box is of infinite length. One need now to show, that the occuring values are unique, which I think is not too difficult (it is not solving Collatz, because that (unsolved) conjecture is that the values are not only unique but also including all odd positive natural numbers which makes it so difficult).

(I have also another statement from the same problem for the infinitude of infinite bins, maybe I can reproduce it here later as an even easier-to-see representation, I'm short of time at the moment, sorry)

Title: A question with infinity


答:
Integrating by parts
$$\int \log x \sin x \, dx=-\cos x \log x-\int\left(- \frac{\cos x}{x}\right)\,dx=$$
$$=-\cos x \log x+\text{Ci}(x)+C$$
The requested improper integral is
$$\int_0^1 \log x \sin x \, dx=\text{Ci}(1)-\underset{x\to 0}{\text{lim}}(\text{Ci}(x)-\log x \cos x)=^*\text{Ci}(1)-\gamma$$
where $\gamma$ is the Euler-Mascheroni constant.

$^*$ MacLaurin expansions are

$\text{Ci}(x)=\log x+\gamma +O\left(x^2\right)$

$\log x\cos x=\log x+O\left(x^2\right)$

Title: $\int_{0}^{1} \sin x \cdot \ln x dx$

Body:
I would like to evaluate the following integral:

$$\int_{0}^{1} \sin x \cdot \ln x dx$$

I had a pretty decent idea to do it, and that is by turning $\sin x$ an $\ln x$ into their respective power series, turning it into a double summation, and integrating. This yields the following:

$$-\int_{0}^{1}\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{x^{2n+1}(-1)^n(1-x)^{m+1}}{(m+1)(2n+1)!} dx$$

Putting the integral inside the double summation and making use the beta function, we get:

$$ =-\sum_{n=0}^{\infty} \sum_{m=0}^{\infty}\frac{(-1)^n \beta(2n+1,m+1)}{(m+1)(2n+1)!}$$

A slight reindexing gives:

$$\sum_{n \space \text{odd}} \sum_{m \space\geq 1}\frac{\beta(n,m)}{n!m}$$

Notice the $-1$ in the front goes away by cancelling with the $-1$ from letting $n$ run through the odds.

Is this a valid strategy? If there is a closed form for this integral, I think this is a pretty cool summation identity. Can someone help out?


问:
Let $p=\lim_{x\rightarrow 0^{+}}{\frac{1}{x}}$. Is $f(x)=x$ continuous on $[0,p]$?
I finished my introductory calculus class and just had this question in my mind when I was preparing for the exam (Engineering Math). My lecturer has never gone through this type of question when he was teaching the concept of limit, but he just told us that what you evaluate from the limit expression is 'very close' to the actual value of the expression. (For example, in his words, $p=\lim_{x\rightarrow 0^{+}}{\frac{1}{x}}=+\infty\neq\frac{1}{0}$). According to this, $f(x)=x$ should be continuous on $[0,p]\implies[0,+\infty]$. But in reality, $f(x)=x$ is continuous on $[0,+\infty)$ but not $[0,+\infty]$. Is there something to do with the 'very close' thing to answer this type of question?

答: For any topological space $X$ (so in particular also for $[0,\infty]$ if you know how to view it as a topological space), the identity $X\to X$ is continuous.


问:
Evaluate the integral $\int \frac{\sec x}{\sqrt{3+\tan x}}dx$
Evaluate the following integral.$$\int \frac{\sec x}{\sqrt{3+\tan x}}dx$$

On putting $t=\tan x$ I am getting $$\int\frac{1}{\sqrt{(t^2+1)(t+3)}}dt$$.

How should i proceed from here.

答:
The other answers have already pointed out that there is no elementary solution and that elliptic integrals are the "simplest" approach. So we will use elliptic integrals; after the substitution $t=\tan x$ we can apply Byrd and Friedman 239.00. (Cf. here.)
$$\int\frac1{\sqrt{(t^2+1)(t+3)}}\,dt=10^{-1/4}F\left(\cos^{-1}\frac{\sqrt{10}-3-t}{\sqrt{10}+3+t},m=\frac{\sqrt{10}+3}{2\sqrt{10}}\right)+C$$
Then just substitute back $t=\tan x$ to get the final result.


问:
The Uniqueness of a Fixed Point
Let $D(0,r) = \{ x \in \mathbb{R}^n | \  \rVert x \rVert \le r \}$ and $f: D(0,r) \to \mathbb{R}^n$ be a map with $a) \rVert f(x) - f(y) \rVert \le 1/3 \rVert x - y \rVert, \ \ \ b) \rVert f(0) \rVert \le (2/3)r$.

I want to determine if there is a unique fixed point of $f$.

I think all I need to do is show that $f$ maps $\text{cl}(D(0,r))$ into itself. Then $f$ obviously satisfies the hypotheses of the contraction mapping principle, but I'm having difficulty showing $f$ maps $\text{cl}(D(0,r))$ into itself. Can someone give me a hint? Thanks!

EDIT: I just came up with a way of showing it.
$$
\rVert f(x) \rVert \le \rVert f(x) - f(0) \rVert + \rVert f(0) \rVert \le \frac13 \rVert x \rVert + \frac23r \le r
$$
So $f$ maps $D(0,r)$ to itself. So by the contraction mapping principle, $f$ has a unique fixed point.
Can someone verify my proof?

答:
First note that $\text{cl}(D(0,r)) = D(0, r)$ because according to your definition, $D(0,r) = \{ x \in \mathbb{R}^n \mid \;   \rVert x \rVert \le r \}$ is already closed.

Now if $x \in D(0, r)$ then $\Vert x \Vert \le r$ and therefore
$$
 \Vert f(x) \Vert \le \Vert f(x) - f(0) \Vert + \Vert f(0) \Vert \\
\le \frac 1 3 \Vert x - 0 \Vert + \frac 23 r \le r
$$
so that $f(x) \in D(0, r)$.

Title: The Uniqueness of a Fixed Point


问:
Square Root Trick
$\sqrt{\mathbf{X}}
\approx \frac{\mathbf{X}+\mathbf{Y}}{2 \sqrt{\mathbf{Y}}}$

And $|\sqrt{\mathbf{X}} -\frac{\mathbf{X}+\mathbf{Y}}{2 \sqrt{\mathbf{Y}}}| \le 10^{-1}$

$Y=$ The Nearest number have root

For example :

$\sqrt{23}=\frac{23+25}{2 \sqrt{25}}=4.8$ ,
$\sqrt{50}=\frac{50+49}{2 \sqrt{49}}=7.071$

Is this  true above approximate?

答:
With $Y=n^2$, $n^2-n<X\le n^2+n$, you can directly apply the binomial identities to get
$$
\left|\sqrt{X}-\frac{n^2+X}{2n}\right|=\frac{(\sqrt{X}-n)^2}{2n}=\frac{(X-n^2)^2}{2n(\sqrt{X}+n)^2}\\
\le\frac{n^2}{2n(X+n^2+2n\sqrt{X})}\le\frac{n}{2((n^2-n)+n^2+2n(n-1))}=\frac1{2(4n-3)}
$$
For $n\ge 2$ this is indeed smaller than $\frac1{10}$.

Note that this approximation is the first step of the Babylonian-Heron-Newton method for the square root $w_{k+1}=\frac12(w_k+\frac{X}{w_k})$ starting from $w_0=n$.

Title: Square Root Trick


问:
Prove that the number of common tangents to two circles exterior to one another is 4
Given two circles outside each other, the maximum number of common tangents to the two circles is 4 according to Wikipedia. How can this be proven?

That's a picture above showing what I'm talking about. How can we prove that no other common tangents can be drawn aside those four?

答: I believe this can be made more rigorous, but here is an intuitive argument. Imagine a tangent to one of the circles meeting it at a point. As you rotate that circle once around its centre the two ends of the tangent will each pass through the other circle. Each end of the tangent will have a point of contact as it enters the other circle and as it leaves.</p>


问:
Sum with integer coefficients
Suppose the series
$$
\sum_{n=0}^\infty a_nx^n
$$
converges for $\left\vert x\right\vert>1$ and $a_n\in\mathbb Z$. Is it true that $a_n=0$ for large enough $n$? I think this should be true, but can't seem to prove it.

答:
Yes.

The fact that the series converges for a $x$ such that $|x|>1$ implies that $$\lim_{n \rightarrow 0} a_n x^n = 0, \quad \quad  \text{so} \quad \quad a_n = o \left( x^{-n}\right), \quad \quad  \text{so} \quad \quad \lim_{n \rightarrow 0} a_n  = 0$$

Now you can conclude.

Title: Sum with integer coefficients


答: It is true. If the series converges for a certain $x$ satisfying $|x|>1$, you must have $\lim_{n\to \infty} a_n x^n=0$. Since $x^n$ does not converge to zero, you must have $\lim a_n = 0$. If the terms of $a_n$ are integers, this can only happen if $a_n = 0$ for $n > n_0$.

Title: Sum with integer coefficients

Body:
Suppose the series
$$
\sum_{n=0}^\infty a_nx^n
$$
converges for $\left\vert x\right\vert>1$ and $a_n\in\mathbb Z$. Is it true that $a_n=0$ for large enough $n$? I think this should be true, but can't seem to prove it.


问:
Uniform convergence of a sequence of functions on real line.
Let the sequence $\{f_n(x)\}$ be given by $$f_n(x)=\cos\frac{(n+1)x}{n}.$$
Then, it is easy to see that $$\lim_{n\to\infty}f_n(x)=\cos x:=f(x).$$
Is the above  convergence uniform over $\mathbb{R}$. I tried to obtain an upper bound on $|f_n(x)-f(x)|$ which is
$$|f_n(x)-f(x)|=\bigg|\cos\frac{(n+1)x}{n}-\cos x\bigg|$$ $$=2\bigg|\sin \frac{(n+1)x+nx}{2n}\bigg|\ \bigg|\sin \frac{x}{2n}\bigg|.$$
How to proceed further? Please help.

答:
The convergence is not uniform. For example consider the sequence $x_n=n\pi$. Then,
\begin{align}
\bigl|f_n(x_n)-f(x_n)|&=\bigl|\cos(n+1)\pi-\cos n\pi\bigr|\\
&=\bigl|\cos n \pi\cos \pi-\cos n\pi\bigr|\\
&=2|\cos n \pi|=2
\end{align}
If the convergence was uniform, then for $\epsilon=1$ we would be able to find $N$ such that
$$\sup_{x\in \mathbb{R}}\bigl|f_n(x)-f(x)|\leq 1$$
for every $n\geq N$. But this contradicts the fact that $|f_n(x_n)-f(x_n)|=2>1$ for all $n$.


问:
First order Autoregressive model
How do I solve this? How can I obtain the lag-one autocorrelation coefficients just from the data??

Following are $10$ years of observation of annual streamflows in millions of cubic meters:

$$\begin{array}{|c|cl|}\hline\hline
\text{Year}&1&2&3&4&5&6&7&8&\\\hline\text{Discharge}&145.78&95.43&116.66&96.12&122.09&175.02&101.98&146.14\\\hline
\text{Year}&9&10\\\hline\text{Discharge}&126.01&132.73
\end{array}$$

Use the above data to estimate the parameters of an  first order Autoregressive model and the parameters of an first order Moving Average Model.

答:
For AR(1)[i.e. first order Auto-regressive model] $U_t=\alpha U_{t-1}+\epsilon_t$ we know that

$\hat \alpha=r_1$=the auto-correlation of lag $1$

$\,\,\,\,\,=$The product moment correlation coefficient obtained from $(x_1,x_2),(x_2,x_3),\dots(x_{n-1},x_n)$

Here $n=10$.

Title: First order Autoregressive model


问:
On the probability of the existence of path from one corner to its opposite
Consider a $n\times n$ grid, whose nodes are randomly colored black and white (with probability $p$ and $1-p$ respectively). Let $A$ be the event that there exists a path of all black nodes connecting the top-left corner and the bottom-right corner, and $B$ be the event that there exists a path of all black nodes connecting the bottom-left corner and the top-right corner.

Now please prove that $P(B|A)\ge P(B) $

It can be understood intuitively that $A$ and $B$ are positively correlated, however, there is possibly no closed form of these probabilities (see this problem). Therefore pure calculation might not be able to solve this problem.

答:
Your question is an elementary consequence to the fact that the site percolation model (as you defined it) satisfies the Fortuin-Kasteleyn-Ginibre inequality for increasing events.

In your model, say that two colorings $\sigma$ and $\tau$ satisfy
$$\sigma \leq \tau$$
if and only if
$$\sigma(v) \leq \tau(v) \text{ for all gridpoints }v\,,$$
where we say that 'white < black'.
This defines a partial order on colorings. Call an event $A$ increasing if $\sigma \in A$ and $\sigma \leq \tau$ implies $\tau \in A$. The events $A$ and $B$ of diagonal black crossings you described are increasing events w.r.t. this partial order.

The FKG-inequality states that

$$P(A\cap B) \geq P(A) P(B) $$
for increasing events $A$ and $B$, or equivalently
$$ P(B\mid A) \geq P(B)\,.$$

To see that this is true, consider all colorings in $\sigma \in A$
and define the following Markov chain:

For a given $\tau \in A$, choose a vertex $v$ randomly

Recolor $\tau(v)$ white. If this new configuration is in
$A$, accept it with probability $1-p$, otherwise recolor it black.

The stationary distribution of this Markov chain is exactly $P(\cdot \mid A)$. Call this Markov chain $(\tau^t)$, with $tau^0$ the all-black coloring. Now, define the chain $\sigma^t$ via the process

For a given $\sigma$, choose a vertex $v$ randomly

Recolor $\sigma(v)$ white with probability $1-p$, otherwise black.

with $\sigma^0$ the al-white coloring. The stationary distribution of this Markov chain obviously is $P(\cdot)$. Now, we have $\sigma^0\leq \tau^0$, and the Markov chains can be coupled to ensure $\sigma^t \leq \tau^t$ for all $t\geq 0$ (simply choose the same vertex in Step 1, and base the choice in Step 2 on the same random number generated uniformly in $[0,1]$). As a consequence,
we get
$$P(B) = \lim_{t\to\infty} P(\sigma^t\in B) \leq \lim_{t\to\infty} P(\tau^t\in B) = P(B\mid A)\,.$$