问:
Limit of $a_n=(1-\frac13)^2\cdot(1-\frac16)^2\ldots(1-\frac{1}{\frac{(n)(n+1)}{2}})^2 \; \;\forall n \geq 2$
$$a_n = \left(1-\frac13\right)^2\cdot\left(1-\frac16\right)^2\ldots\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2
 \; \;\forall n \geq 2$$

I have no idea how to solve this, however I will give it a try using Cauchy's second theorem... the following is probably wrong since all terms will depend on $'n'$ after taking $nth$ root

$$\lim_{n\rightarrow\infty} a_n = \left(1-\frac13\right)^2\cdot\left(1-\frac16\right)^2\cdot\ldots\cdot\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2
= \lim_{n\rightarrow\infty}\left[\left(1-\frac13\right)^{2n}\cdot\left(1-\frac16\right)^{2n}\cdot\ldots\cdot\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^{2n}\right]^{\frac1n} = \lim_{n\rightarrow\infty}\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^{2n}=1 $$

答:
Note that:  $1- \frac{2}{n(n+1)}= \frac{(n-1)(n+2)}{n(n+1)}$

So we have :
\begin{align}
& x_n= \left [\frac{1\cdot 4}{2\cdot 3} \cdot \frac{2\cdot 5}{3\cdot 4}\cdot \frac{3\cdot 6}{4\cdot 5} \cdot ...\\ \cdot \frac{(n-2)(n+1)}{(n-1)n} \cdot\frac{(n-1)(n+2)}{n(n+1)} \right]^2 \\
  & = \left (\frac{n+2}{3n} \right)^2 \\
  & = \frac{1}{9}\left (1+\frac{4}{n}+\frac{4}{n^2} \right)
\end{align}
So $\displaystyle \lim_{x \to \infty} x_n= \frac{1}{9}$


问:
Why is the independence of paths for complex integrals not sufficient enough to proof Cauchy's integral theorem?
I have seen multiple proofs of Cauchy's integrals theorem, mostly using the Deformation Invariance theorem or a version of Green's Theorem. But why exacly is this necessary?

From my understanding of the indepence of paths, isn't it obvious that if you take 2 random point z1 and z2 on a closed contour, then the integral from z1 to z2 and back from z2 to z1 cancel out and thus results in zero?

For closed contours consisting of multiple smooth curves, simply taking the sum of the integrals from each start and end point should result in a telescoping sum that again cancels out to zero.

All of this assumes the function is analytic in the domain D. Is there some other condition I am missing? Or is this a circular reasoning? (no pun intended). For reference, my math education is at the level of an undergraduate, with mostly self taught complex analysis.

答: This reasoning is indeed circular, at least most of the time. Generally, Cauchy's integral theorem is used to prove the independence of paths for complex integration, so one can't use the independence of paths to prove Cauchy's integral theorem.</p>

ContentLicense: CC BY-SA 4.0


问:
Showing particular harmonic function is constant
Suppose $u$ is a real valued continuous function on $\overline{\mathbb D}$, harmonic on ${\mathbb D}$\ $\{0\}$ and $u=0$ on $\partial\mathbb D$, show $\mathbb u$ is constant in $\mathbb D$.

I'm going through old exams and I spent quite some time on this one without success.
I think of using log$|z|$ first as an example. But it fails since $u$ is continuous on $\overline{\mathbb D}$. If we are working with holomorphic functions $f$: $\mathbb{C} \to \mathbb{C}$, $0$ is a removable singularity and everything works out. But I don't remember any similar technic to work with harmonic functions. Any help will be appreciated!

答:
Hint: use the maximum principle for harmonic functions.

For the details, it depends a little on what tools you have available. Let me assume that you know the following theorem about removable singularities for subharmonic functions:

Theorem Assume that $\Omega$ is a domain in $\mathbb{C}$ and that $E \subset \Omega$ is polar. If $u$ is an upper bounded subharmonic function on $\Omega \setminus E$, then $u$ extends (uniquely) to a subharmonic function on $\Omega$.

From the theorem, it follows that $u$ is (can be extended to be) subharmonic on $\mathbb{D}$, and so can $-u$. Hence $u$ is in fact harmonic on $\mathbb{D}$, so by the maximum principle $u = 0$ on $\mathbb{D}$.

Just in case, here is a proof of the removable singularities theorem:

Let $\phi$ be a negative subharmonic function such that $\phi = -\infty$ on $E$. (In your particular case, you can take $\phi(z) = \log|z|$.) Let $u_\epsilon = \max \{ u, \epsilon \phi \}$. Then $u_\epsilon$ is subharmonic for every $\epsilon$, since $u_\epsilon = \phi$ near $E$. Furthermore $u_\epsilon$ increases to $u$ on $\Omega \setminus E$ as $\epsilon \to 0^+$, and the function $$v(z) = \limsup_{\zeta \to z} \left(\sup_{\epsilon > 0} u_\epsilon(z)\right)$$ is subharmonic on $\Omega$. By continuity of $u$, $v(z) = u(z)$ on $\Omega \setminus E$, and we have our subharmonic extension of $u$. (In fact, you can check that $v(z) = \limsup_{\zeta \to z} u(\zeta)$ for $z \in E$.)

ContentLicense: CC BY-SA 3.0


问:
Evaluare $I(y)=\int_0^1\frac{1}{y+\cos(x)}dx$. Hence determine $J(y)=\int_0^1\frac{1}{(y+\cos(x))^2}dx$

Consider the following integral:
$$I(y)=\int_0^1\frac{1}{y+\cos(x)}dx $$
with Weierstrasse substitution I showed that

$$ I(y)=\frac{2}{\sqrt{y^2-1}}\arctan\left(\sqrt{\frac{{y-1}}{y+1}}\right). $$

It is this next part that I am not sure about: Hence detertemine
$$J(y)=\int_0^1\frac{1}{(y+\cos(x))^2}.$$

My attempt:

Since $I(y)\pm J(y)$ yields no simplication, the most obvious approach would be to rewrite $I(y)$ as
$$I(y)=\int_0^1\frac{(y+\cos(x))}{(y+\cos(x))^2}=yJ(y)+\underbrace{\int_0^1\frac{cos(x)}{(y+\cos(x))^2}dx}_{K(y)}$$
Applyling Weierstrasse substitituion for $K(y)$:
$$K(y)=2\int_0^1\frac{(1-x^2)}{((y-1)x^2+(y+1))^2}dx=-2\int_0^1\frac{\frac{1}{y-1}((y-1)x^2+(y+1))-\frac{y+1}{y-1}-1}{((y-1)x^2+(y+1))^2}$$
$$=-2\left(\underbrace{\frac{1}{y-1}\int_0^1\frac{1}{(y-1)x^2+(y+1)}dx}_{\text{elementary integral}\to \arctan()}-\underbrace{\frac{2}{y-1}\int_0^1\frac{1}{((y-1)x^2+(y+1))^2}dx}_{\text{evaluated with }
 \tan(u)=\sqrt{\frac{y-1}{y+1}}x}\right)$$
so what follows is quite elmenatary.

However it seems to me that $K(y)$ is by no means simpler than $J(y)$, (i.e. the same approach can be used for $J(y)$ but without having to use the result for $I(y)$) so this is clearly not the point of the question?

答:
As @JackD'Aurizio pointed out, we trying to find:

$$\frac{\partial}{\partial\text{n}}\left(\frac{1}{\text{n}+\cos\left(x\right)}\right)\tag1$$

Using the chain rule, we get:

$$\frac{\partial}{\partial\text{n}}\left(\frac{1}{\text{n}+\cos\left(x\right)}\right)=-\frac{1}{\left(\text{n}+\cos\left(x\right)\right)^2}\cdot\frac{\partial}{\partial\text{n}}\left(\text{n}+\cos\left(x\right)\right)\tag2$$

Differentiate the sum term by term:

$$\frac{\partial}{\partial\text{n}}\left(\frac{1}{\text{n}+\cos\left(x\right)}\right)=-\frac{1}{\left(\text{n}+\cos\left(x\right)\right)^2}\cdot\left(\frac{\partial}{\partial\text{n}}\left(\text{n}\right)+\cos\left(x\right)\cdot\frac{\partial}{\partial\text{n}}\left(1\right)\right)=$$
$$-\frac{1}{\left(\text{n}+\cos\left(x\right)\right)^2}\cdot\left(1+\cos\left(x\right)\cdot0\right)=-\frac{1}{\left(\text{n}+\cos\left(x\right)\right)^2}\tag3$$

So, we have:

$$\int_0^1\frac{1}{\left(\text{n}+\cos\left(x\right)\right)^2}\space\text{d}x=-\int_0^1\frac{\partial}{\partial\text{n}}\left(\frac{1}{\text{n}+\cos\left(x\right)}\right)\space\text{d}x=$$
$$-\frac{\partial}{\partial\text{n}}\left(\int_0^1\frac{1}{\text{n}+\cos\left(x\right)}\space\text{d}x\right)\tag4$$


问:
Limit of sin function
I need some help solving the following limit:

Here is how I started:

$$\lim_{x \to 0^+}\frac{\sin(6x)}{\sqrt{\sin(2x)}} = \lim_{x \to 0^+}\frac{\sin(6x)\sqrt{\sin(2x)}}{{\sin(2x)}}=\lim_{x \to 0^+}\frac{\frac{6 \sin(6x)}{6x}\frac{\sqrt{\sin(2x)}}{x}}{{\frac{2 \sin(2x)}{2x}}}$$

I don't know how to deal with this sqrt...

Thanks! :)

答:
Use the asymptotic equivalence $\sin(x) \sim x$ for $ c\rightarrow 0^{+}$.

Hence you get $$\lim_{x \rightarrow 0^{+}} \frac{6x}{\sqrt{2x}} = \lim_{x \rightarrow 0^{+}} \frac{6\sqrt{x}}{\sqrt{2}} =0 $$

Title: Limit of sin function


问:
find the greatest area of a parallelogram of two vertices lie on $x/a+y/b=1$ and the other two vertices lie on the two axes
find the greatest area of a parallelogram of two vertices lie on $x/a+y/b=1$ and the other two vertices lie on the two axes

答:
Let $A(ka,0)$ and $B(0,kb)$ be the vertices, where $0<k<1$.

By Pythagoras' theorem, the base is $AB=k\sqrt{a^2+b^2}$ and the height is $\dfrac{ab(1-k)}{\sqrt{a^2+b^2}}$.

Can you proceed with that?


问:
Prove that $(1+1/x)^x$ is concave for $x>0$
From the graph it looks like $(1+1/x)^x$ is concave for $x>0$. But in this post, I can only prove that it is concave for $x\ge 1$. It is of interest to see a proof for $x>0$. </p>

答:
Considering the second derivative, it is sufficient to prove that

$$(x+1) \log (1 + 1/x) - 1 \le \frac{1}{\sqrt{x}}$$

We will show that,

$$\log (1 + 1/x) \le \frac{1}{\sqrt{x}}$$

This implies, for $0 \lt x \le 1$, that

$$(x+1) \log (1 + 1/x) - 1 \le \frac{x+1}{\sqrt{x}} - 1 = (\sqrt{x} - 1) + \frac{1}{\sqrt{x}} \le \frac{1}{\sqrt{x}}$$

We can easily show that $f(y) = y - \log(1+y^2)$ is increasing (by considering the derivative), and this implies that $f(y) \ge f(0) = 0$ for $y \ge 0$, and thus $$\log (1 + 1/x) \le \frac{1}{\sqrt{x}}$$


答:
this should be the solution
If you have any question we can discuss in the comments

tou can find the complete formula of d online (distance of a point from a line)

Title: find the greatest area of a parallelogram of two vertices lie on $x/a+y/b=1$ and the other two vertices lie on the two axes


问:
Why does the ODE $xy''+y'+xy=0$ require only one initial condition?
In order to solve second order differential equations with constant coefficients of the form
$$ay''+by'+cy=0$$
we need two initial conditions, for example $y(0)=1$ and $y'(0)=2$.

However, I just crossed the following differential equation with varying coefficients:
$$xy''+y'+xy=0 \\y(0)=1$$ known as the Bessel's Equation which can be solved using power series to give the solution $$y(x)=\sum_{n=0}^{+\infty}\frac{(-1)^n}{4^n(n!)^2}x^{2n}.$$
Why did this equation only require one initial condition? is it because it is not with constant coefficients?

答:
This ODE
$$xy''+y'+xy=0$$ is invariant under $x\to -x$, so it can have two linearly independent
solutions of definite parity. These are $J_0(x)$ and $Y_0(x)$ first one is even but the second one is defined only for $x>0$ and $Y_0(0)=-\infty$ So it cannot have odd prity solution. This aspect is covered by the fact that this ODE has both indicial roots as zrto. Check it by putting $y=x^m$, then equate the coefficient of the lowes power you gwt $m^2=0$. When this happens one solution has logarithmic singularity.
You have found $J_0(x)$ as the solution.

The general solution of the ODE can be written as
$$y(x)=C_1 J_0(x)+ C_2 Y_0(x)$$
Any two conditions other than $x=0$ can determine $C_1$ and $C_2$
for instance $y(a)=c, y(b)=d.$


答:
Your last step has an error.  The $x$ under the square root shouldn't be there.  So your next step should be

$$= \lim_{x\to 0^+} \frac{6}{2}\sqrt{\sin 2x} = 0.$$

Title: Limit of sin function

Body:
I need some help solving the following limit:

Here is how I started:

$$\lim_{x \to 0^+}\frac{\sin(6x)}{\sqrt{\sin(2x)}} = \lim_{x \to 0^+}\frac{\sin(6x)\sqrt{\sin(2x)}}{{\sin(2x)}}=\lim_{x \to 0^+}\frac{\frac{6 \sin(6x)}{6x}\frac{\sqrt{\sin(2x)}}{x}}{{\frac{2 \sin(2x)}{2x}}}$$

I don't know how to deal with this sqrt...

Thanks! :)


问:
How to find recurrence relation for $f(x)=x^3-\alpha=0$ where $\alpha>0$ using newton raphson method?
Alright for $x^3-\alpha=0$ we know that $x=\sqrt[3]\alpha$ is the root of course but unlike defined number it is too complicated to be solved using $x_0=\sqrt\alpha$ or $x_0=1$ for example so what is the idea behind it?

答: If $f(x)=x^3-\alpha$ then the Newton Raphson method would give the recurrence relation $$x_{n+1}=x_n-\frac{x_n^3-\alpha}{3x_n^2}$$


问:
Find $\mathrm{Aut}(G)$, $\mathrm{Inn}(G)$ and $\mathrm{Aut}(G)/\mathrm{Inn}(G)$ for $G = D_4$
Problem

Find $\mathrm{Aut}(G)$, $\mathrm{Inn}(G)$ and $\mathrm{Aut}(G)/\mathrm{Inn}(G)$ for $G = D_4$

My Attempt

I let $D_4 = \{e, x, y, y^2, y^3, xy, xy^2, xy^3\}$

I found that $\mathrm{Inn}(G)$ consists of 4 bijective conjugation functions, namely $\{\phi_e, \phi_x, \phi_y, \phi_{xy}\}$.  For $\mathrm{Aut}(G)$, I found that there are 12 automorphisms.

Here is the following link that relates to the problem I am doing.

Based on the solution:

Lemma: If $\alpha$ is an automorphism of group $G$ and $G$ has generators $x$ and $y$ with orders $n$ and $m$, respectively then $\alpha(x)$ and $\alpha(y)$ are also generators for $G$ with orders $n$ and $m$.

Proof: First, let us show that the orders agree.  If $g \in G$ has order $n$, let $a = \alpha(g) \in G$ have order $m$.  Then, $a^m = 1$ but by applying $\alpha^{-1} \in \mathrm{Aut}(G)$, we get $g^m = 1$.  However, $g$ has order $n$, so $n$ must divide $m$.  Similarly, $g^n = 1$ and applying $\alpha$, we note that $a^n = 1$, and so we conclude $n = m$.

Secondly, since any element of $G$ can be written as a product of $x$'s and $y$'s, and $\alpha$ is a surjective homomorphism, it follows that any element of $G$ can also be written as a product of $\alpha(x)$ and $\alpha(y)$, hence they generate $G$.

Using this Lemma (or a similar argument), we note that an automorphism of $D_4$ must send $y$ to $y, xy, x^2y$ or $x^3y$ and $x$ to $x$ or $x^3$.  Any such pairing is possible, thus there are $2 \cdot 4 = 8$ such automorphisms...

(The notations that someone use are different from what I denote.)

The question I have is: Why are there 8 automorphisms?  Shouldn't there be 12 automorphisms?  Here is what I have:

$$e \mapsto e$$
$$x \mapsto \text{ either } \{x,y^2, xy^2\}$$
$$y \mapsto \text{ either } \{y, y^3, xy, xy^3\}$$

Then, there are $1 \cdot 3 \cdot 4 = 12$ automorphisms.

Any advices or comments?

答:
To determine $\mathrm{Inn}(D_4)$,first observe that the complete list of inner automorphisms is
$$\phi_{R_{0}},\phi_{R_{90}},\phi_{R_{180}},\phi_{R_{270}},\phi_{R_{H}},\phi_{R_{V}},\phi_{R_{D}},\phi_{R_{D'}}$$

Here $R_0,R_{90},R_{180},R_{270}$ are rotations of square. H and V denote the reflection of square with horizontal and vertical reflection and D and  D' are denote diagonal reflection.Our job is to determine the no. of repetitions in this list.

Now u can check easily that $\phi_{R_{180}}=\phi_{R_{0}}$ also,$\phi_{R_{270}}=\phi_{R_{90}}$

similarly,$H=R_{180} V$ and $D'=R_{180} D$ so we have $\phi_{R_{H}}=\phi_{R_V}$ and $\phi_{R_D}=\phi_{R_D'}$

there are four inner automorphisms
$$\mathrm{Inn}(D_4)=\{\phi_{R_0},\phi_{R_{90}},\phi_{R_H},\phi_{R_D}\}$$


问:
Find the sum of the digits of the smallest value of $k$ if $(a_0 + a_1 +$ ... $+ a_{k-1})$ is divisible by $2005$.

For a positive integer $k$, we have:- $(1 + x)(1 + 2x)(1 + 3x)$ ... $(1 + kx) = a_0 + a_1x + a_2x^2 + $ ... $ + a_kx^k$. Where $a_0,a_1,$ ... $,a_k$ are the coefficients of the polynomial. Find the sum of the digits of the smallest value of $k$ if $(a_0 + a_1 +$ ... $+ a_{k-1})$ is divisible by $2005$.

What I Tried: I have no idea where to start. I was able to guess, that the expression $(1 + x)(1 + 2x)$ ... $(1 + kx)$ consists of a $(+1)$ term and a $(k!x^k)$ term , and I can guess $a_0 = 1$ , $a_k = k!$ , and this might given an idea that $a_1 = 1!$ , $a_2 = 2!$ and so on, but is this enough.

Suppose, after solving this part, I need to find for which $(k-1)$ I will have that :-
$$\rightarrow 0! + 1! + ... + (k - 1)! = 2005m$$

This is easy, because as far as I can see, $2005 = 5 * 401$ and $401$ is prime , so $k$ has to be $402$ in order to be minimum.

I have a doubt in the $1$st part, can anyone help me?

答:
$(1+x)(1+2x) = 1+3x+2x^2$ disproves your claim that $a_1 = 1!$.

Alternatively, consider $f(1)$. We have:

$$f(1) = (1+1)(1+2)\dots(1+k) = a_0+a_1+\dots+a_k = (k+1)!$$

and we want $(k+1)! - a_k = (k+1)! - k! = k(k!)$ to be divisible by $2005$.

By a similar analysis to yours, the minimum $k$ is $401$.


Body:
For a positive integer $k$, we have:- $(1 + x)(1 + 2x)(1 + 3x)$ ... $(1 + kx) = a_0 + a_1x + a_2x^2 + $ ... $ + a_kx^k$. Where $a_0,a_1,$ ... $,a_k$ are the coefficients of the polynomial. Find the sum of the digits of the smallest value of $k$ if $(a_0 + a_1 +$ ... $+ a_{k-1})$ is divisible by $2005$.

What I Tried: I have no idea where to start. I was able to guess, that the expression $(1 + x)(1 + 2x)$ ... $(1 + kx)$ consists of a $(+1)$ term and a $(k!x^k)$ term , and I can guess $a_0 = 1$ , $a_k = k!$ , and this might given an idea that $a_1 = 1!$ , $a_2 = 2!$ and so on, but is this enough.

Suppose, after solving this part, I need to find for which $(k-1)$ I will have that :-
$$\rightarrow 0! + 1! + ... + (k - 1)! = 2005m$$

This is easy, because as far as I can see, $2005 = 5 * 401$ and $401$ is prime , so $k$ has to be $402$ in order to be minimum.

I have a doubt in the $1$st part, can anyone help me?

Body:
Hint:

$$a_0+a_1+a_2+\dots +a_{k} = f(1) = (k+1)! $$


答:
Just note that if you put $P(x)=(1+x)(1+2x)...(1+kx)$ then the value of $a_{0}+a_{1}+...+a_{k-1}$ is actually equal to $P(1)-a_{k}=P(1)-k!=(k+1)!-k!$ Now if $(k+1)!\equiv k! (mod401)$ for a $400\geqslant k$ (Note: It is obviously divisible by $5$ for $k\geqslant 5$ so it suffices that it's divisible by $401$)
Then because $401$ is a prime and $k!,(k+1)!$ are both relatively prime to it, $k+1\equiv 1 (mod401)$ and hence the least such $k$ is $401$ which is a contradiction so none of the numbers less than $401$ satisfy the condition. $401$ itself clearly satisfies the condition so the final answer is $5$.

Title: Find the sum of the digits of the smallest value of $k$ if $(a_0 + a_1 +$ ... $+ a_{k-1})$ is divisible by $2005$.


问:
Taylor Remainder formula for $\sin(x)$
The general formula for remainder of Taylor polynomial is:
$$R_n(x)=\frac{(x-a)^{n+1}}{(n+1)!}f^{(n+1)}(c)$$
where $c$ is an unknown point between $a$ and $x$.

For $\cos(x)$ the book I am reading says :

$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}+(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\cos(c)$$

But I don't understand why last term has $\cos(c)$. In fact $(n+1)$th derivative of $\sin(x)$ at $x=c$ can be either of $\pm\sin c$,$\pm\cos c$.  but why it considered only $\cos c$ among these four?

Title: Taylor Remainder formula for $\sin(x)$

Body:
Let $f(x) = \sin x$. Then observe that $f^{(n)}(x) = \sin {(x + n\pi /2)}$. So
$ f^{(2n+1)}(x) = \sin {(x + \frac{(2n+1) \pi}{2})} $.
Then it can be easily observed that $f^{(2n+1)}(x) = (-1)^n \cos x$ since $ \sin {(x + \frac{(2n+1) \pi}{2}}){}{}{} = \cos x \sin ({(2n+1)\pi /2}) = (-1)^n \cos x$


问:
Does convergence in measure imply "supports getting close" (made precise in question body)?
In connection to my question here, let $P_{n}(n=1,2, \ldots)$ on $([0,1], \mathcal{B}([0,1]))$, where $\mathcal{B}([0,1])$ is the Borel $\sigma$-algebra on $[0,1]$, be a sequence of measures which converges weakly to the finite measure $P$ (i.e. $P_{n} \Rightarrow P$).

Claim: For all $x \in supp \; P$, and $\forall\;\epsilon>0$, $\exists\;K \equiv K_{x,\epsilon} \in \mathbb{N}$ s.t. $\forall\;k\geq K$, $supp\;P_k \cap (x-\epsilon, x+\epsilon)\neq \emptyset$.

Proof: By $a.e.$ continuity of the distribution function associated with $P$, $F(v) \equiv P[0,v]\;\forall\;v\in[0,1]$, there exist $x_1 \in (x-\epsilon,x)$ and $x_2 \in (x,x+\epsilon)$ such that $x_1, x_2 \in C(F)$. $x \in supp\;P,\;\therefore P((x_1,x_2))=P((x_1,x_2])>0$, where the equality follows from $x_1, x_2 \in C(F)$.  $(x_1,x_2]$ is a continuity set of $P$, $\therefore\;\exists\;K$ s.t. $\forall\;k \geq K, P_k((x_1,x_2])>0$. $\therefore \;supp\;P_k \cap (x_1,x_2] \neq \emptyset, \therefore\;\;supp\;P_k \cap (x - \epsilon, x+ \epsilon) \supseteq \;supp\;P_k \cap (x_1,x_2] \neq \emptyset$.

Please let me know if this is correct and also if this is the correct (tightest?) way to make the notion of "convergence in measure implies supports getting close" precise?

答:
Your proof is correct.

But next examples show that the statement  "weak convergence of measures implies supports are getting closer" implies that we should have "convergence" of supports in the next cases:

a) $\sum_{i=1}^n \frac{1}n \delta_{\frac{i}n}$ converges to $U[0,1]$.

b) $N(0, \frac{1}n) \to \delta_0$.

c) $N(1, \frac{1}n) \to \delta_1$.

If we will say that weak convergence of measure implies supports getting closer then we have to say that

a) $\{ \frac{1}n, \frac2{n}, \ldots, 1\}$ are getting closer to $[0,1]$

b)$\mathbb{R}$ is getting closer to $\{0\}$ and at the same time
c)$\mathbb{R}$ is getting closer to $\{1\}$.

So it's an illustration of corollaries of such a thesis, which are not good enough.


问:
Regular functions on locally closed sets
Let $\mathbb{C}^n$ be endowed with Zariski topology.

Definition 1 Let $X \in \mathbb{C}^n$ closed. A function $f : X \to \mathbb{C}$ is regular on $X$ if there exists a polynomial $F \in \mathbb{C}[x_1, ... x_n]$ such that $f(x) = F(x), \forall x \in X$.

Definition 2 Let $X \in \mathbb{C}^n$ locally closed. A function $f: X \to \mathbb{C}$ is regular on $X$ if there exists an open covering $\left\{ U_i \right\}$ of $X$ and polynomials $a_i, b_i \in \mathbb{C}[x_1, ... , x_n]$ such that $\forall i, b_i$ is not identically zero on $U_i$ and $f|_{U_i} = \frac{a_i}{b_i}$.

Now, we can prove the following result:

If $Z \in \mathbb{C}^n$  is closed, let $h \in \mathbb{C}[x_1, ... , x_n]$ a polynomial not identically zero on $Z$ and $X =Z_h$ (i.e. the subset of elements $x$ of $Z$ such that $h(x) \neq 0$)  then every regular function $f$ on $X$ can be written as $\frac{a}{h^d}$, with $a \in \mathbb{C}[x_1, .. x_n]$ and $d$ a positive integer.

It is argued that this result proves that on a closed set Definition 1 coincides with Definition 2, but I can’t seem to understand why. Any hints?

答:
I’m turning my comment into an answer.

Clearly a regular function in the first definition is regular in the second one too. On the other hand, if $f$ is a regular function on a Zariski-closed set $Z$, then $Z=Z_1$ so any regular function on $Z$ must be some $a/1^d=a$ where $a$ is a polynomial – therefore this function is regular in the sense of definition 1.


答:
Here there are some hints. Let's call the polynomial $p(x) $.

In terms of $a_i$, how do you express $p(1) $?

What is the coefficient of $x^k$ in $p(x) $?

After having solved this part, you are almost done... You noticed that 401 is prime :)

Good luck!


问:
Prove that $\mathbb{R}^2 \setminus \{0\}$ is connected?
Prove that  $\mathbb{R}^2 \setminus \{0\}$ is connected ?

My attempt : take $A= (-\infty, 0)$ and $B= (0,\infty)$

$\mathbb{R} \setminus \{0\}=A\cup B = (-\infty, 0)\cup (0,\infty)$

squaring both side we have

$\mathbb{R}^2 \setminus \{0\}=(A\cup B)^2 = ((-\infty, 0)\cup (0,\infty))^2=(0,\infty) \cup (0, \infty)$

where$(0,\infty) \cap (0, \infty)\neq \emptyset$ since $(-\infty,0)^2= (0,\infty)$ and $(-\infty)^2=\infty$

Therefore $\mathbb{R^2} \setminus \{0\}$ is connected

Is its correct ?

答:
Even better, letting $C$ be a countable set, $\Bbb R^2\setminus C$ is path-connected. One way to see this is as follows:

Let $a=(a_1,a_2),b=(b_1,b_2)\in\Bbb R^2$. Consider $A$ the set of all straight lines passing through $a$, and $B$ the respective set for $b$. Note that we can parametrize the lines of $A$ as $x=a_1$ (the vertical line) and $y-a_2=m(x-a_1), m\in\Bbb R$ (the rest of them). Similarly $B=\{x=b_1,y-b_2=k(x-b_1):k\in\Bbb R\}$.

Now, for each point $c\in C$ there is exactly one line in $A$ passing through $c$ (since a straight line is determined by two points). Since $C$ is countable, but $A$ is not (note we have as many lines in $A$ as real numbers), there is some line in $A$ that never passes through any point of $C$ (so that line lies in $\Bbb R^2\setminus C$). Name that line $r$.

$r$ has a certain slope (it could $\infty$ if it is the vertical line), and there's only one line in $B$ with the same slope (a line is determined by a point and its slope), call it $r'$. Now consider the set $B\setminus\{r'\}$, which is still uncountable, so, as before, there is some line in $B\setminus\{r'\}$ that never passes through any point of $C$ (so that line lies in $\Bbb R^2\setminus C$). Name it $s$.

We have two lines, $r$ and $s$, lying in $\Bbb R^2\setminus C$, one passing through $a$ and the other passing through $b$, and the two of them with different slopes, which implies that $r$ and $s$ intersect each other at some point. This means we have got a path from $a$ to $b$ (or viceversa): we can start at $a$, from there go through $r$ up to its intersection with $s$, and then go through $s$ to end up at $b$.

Technically, we'd need to give a continuous function $\gamma:[0,1]\to \Bbb R^2\setminus C$, such that $\gamma(0)=a$ and $\gamma(1)=b$, but I'll leave that to you; you'll have to define $\gamma$ by parts, using (thus having to name) the point of intersection.


答:
Let $ n\in\mathbb{N} $ :

Taylor-Lagrange's theorem states that, for any $ a,x\in\mathbb{R} $, we have : $$ \sin{x}=\sum_{k=0}^{2n}{\frac{\sin^{\left(k\right)}\left(0\right)}{k!}x^{k}}+R_{2n}\left(x\right) $$

Where $ R_{2n}\left(x\right) $ can be witten as follows $$ R_{2n}\left(x\right)=\frac{x^{2n+1}}{\left(2n+1\right)!}\sin^{\left(2n+1\right)}\left(\xi_{x}\right) $$ for some $ \xi_{x}\in\left(a,x\right) $.

Since, for any $ \left(n,x\right)\in\mathbb{N}\times\mathbb{R} $, $ \sin^{\left(2n\right)}\left(x\right)=\left(-1\right)^{n}\sin{x} $, and $ \sin^{\left(2n+1\right)}\left(x\right)=\left(-1\right)^{n}\cos{x} $, we have the following : $$ \sin{x}=\sum_{k=0}^{n}{\frac{\left(-1\right)^{k}}{\left(2k+1\right)!}x^{2k+1}}+\left(-1\right)^{n}\frac{x^{2n+1}}{\left(2n+1\right)!}\cos{\xi_{x}} $$

Title: Taylor Remainder formula for $\sin(x)$

Body:
The general formula for remainder of Taylor polynomial is:
$$R_n(x)=\frac{(x-a)^{n+1}}{(n+1)!}f^{(n+1)}(c)$$
where $c$ is an unknown point between $a$ and $x$.

For $\cos(x)$ the book I am reading says :

$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}+(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\cos(c)$$

But I don't understand why last term has $\cos(c)$. In fact $(n+1)$th derivative of $\sin(x)$ at $x=c$ can be either of $\pm\sin c$,$\pm\cos c$.  but why it considered only $\cos c$ among these four?


问:
$e^z$ is entire yet has an essential singularity (at $\infty$)
Is there no inconsistency? Or does the property of being entire exclude the point $z=\infty$?

p.s. following up from my previous question limit of $e^z$ at $\infty$

答: The complex plane doesn't contain the point at infinity. (Infinity is *not*a number, not even a complex one.) Entire means holomorphic on $\mathbb{C}$. In fact the only functions that are holomorphic on the Riemann sphere (i.e. on $\mathbb{C}\cup \{\infty\}$) are the constant functions.</p>


问:
what is the meaning of this sentence ?.
Im sorry ,my english is weak

After  reading this solution,some  doubt regarding language  has been arise in my mind given below

what is the meaning of this sentence

Let $A_{ij}$ be the $m \times n$ matrix whose entries are all zero except for the entry $ \{i,j\}$ which is one

My attempt :Im thinking like this , take $ m=2 , n=3$  Now \begin{bmatrix} 0&1&1\\1&0&1\end{bmatrix}

for $m=n=2$\begin{bmatrix} 0&1\\1&0\end{bmatrix}

Is its correct ?

答:
Most likely, they mean that all entries in the matrix are zeros, except for one single entry, given by the pair $(i,j)$. For example, $m=3, n=4, i=2, j=1$ you would have the matrix like this:

$$A=\begin{bmatrix}0&0&0&0\\1&0&0&0\\0&0&0&0\end{bmatrix}$$

Note it is an $m\times n=3\times 4$ matrix, and only the entry $A_{ij}=A_{21}$ is equal to $1$.

Title: what is the meaning of this sentence ?.


问:
Prove it's topology
First let's understand who these topologies are:
\begin{equation}
    \tau_1 = \{\{m \in \mathbb{N} : m < n\} : n \in \mathbb{N}\} \cup \{\mathbb{N}\}
\end{equation}

\begin{equation}
    \tau_1=\{\emptyset, \{0\},\{0,1\},\{0,1,2\}, \{0,1,2,3\}, \cdots, \mathbb{N}\}
\end{equation}

Let's verify that $\tau_1$ is a topology:

In fact, $\emptyset$ is subnext to any set, so $\emptyset \in \tau_1$, plus $\mathbb{N}\in \tau_1$.

We need to show that the finite intersection of the $\tau_1$ elements belongs to $\tau_1$. I can see that yes, for example, $\{0\}\cap \{0,1\}=\{0\}\in \tau_1$, $\{0\}\cap\{0,1\}\cap\{0,1,2\}=\{0\}\in \tau_1$ or $\{0,1\}\cap\{0,1,2\}\in \tau_1$. But I can't generalize. Can you help me?

We need to show that the arbitrary union of $\tau_1$ elements belongs to $\tau_1$. I know it's worth that, for example, $\{0\}\cup \{0,1\}\cup \{0,1,2\} \cup \mathbb{N}=\mathbb{N} \in \tau_1$. But I can't generalize. Can you help me?

Thank you!

答:
Hints:

Given finitely many sets of the form $\{0, \ldots, k_1\}, \ldots, \{0, \ldots, k_n\}$, can you see what the intersection must be? What if you put $k = \min\{k_1, \ldots, k_n\}$. What could it be?

Of course, intersecting further with $\varnothing$ or $\Bbb N$ is easy.

Let $I$ be an arbitrary set and for each $i \in I$, suppose you are given $\{0, \ldots, k_i\}$. If the set $\{k_i\}$ is bounded, can you see deduce the union? (What if you consider $\max_{i \in I} k_i$?) What if the set is not bounded? Do you think the union is then $\Bbb N$?

As before, taking a further union with $\varnothing$ or $\Bbb N$ is easy.

Title: Prove it's topology


问:
K Theory: Book Recommendations
Good people!

So I've been hoping to get into K Theory for a while now, and the book that I have been trying to use (and failing) has been Charles Weibel's book by that very title.

The book itself isn't bad, but it's not for the faint-hearted, and can appear quite dense and challenging to a relative newbie like me. As such, I was wondering if someone might be able to recommend a book or two or three or $n$ to help "ease" my way into Weibel. Something that touches upon the constructions he introduces in the first chapter--Picard groups, topological and algebraic vector bundles, Chern classes, etc.--but being more, well, easy for a newbie to process.

I look forward to your suggestions.

答:
From just as much of a newbie, I'd recommend you these notes by Eric Friedlander to begin with.

I'm personally also fond of this introductive article by Inna Zakharevich.

Coming to... larger books, I'm not really aware of any "easier" reference than Weibel's, so I think I'll leave recommendations to users way more expert than me on the subject.

Title: K Theory: Book Recommendations


答:
Set $U_n=\{m\in\mathbb{N}:m<n\}$ and $U_\infty=\mathbb{N}$. Then $U_0=\emptyset$, so $\emptyset\in\tau_1$.

Consider a finite (nonempty) set of members of $\tau_1$ and write it as
$$
\{U_{n_1},U_{n_2},\dots,U_{n_k}\}
$$
Prove that $\bigcap_{i=1}^k U_{n_i}=U_n$, where $n=\min\{n_1,\dots,n_k\}$.

Suppose you have an arbitrary set of members of $\tau_1$. The set is $\{U_{n_i}:i\in I\}$. Now consider $\sup\{n_i:i\in I\}$.

Title: Prove it's topology

Body:
First let's understand who these topologies are:
\begin{equation}
    \tau_1 = \{\{m \in \mathbb{N} : m < n\} : n \in \mathbb{N}\} \cup \{\mathbb{N}\}
\end{equation}

\begin{equation}
    \tau_1=\{\emptyset, \{0\},\{0,1\},\{0,1,2\}, \{0,1,2,3\}, \cdots, \mathbb{N}\}
\end{equation}

Let's verify that $\tau_1$ is a topology:

In fact, $\emptyset$ is subnext to any set, so $\emptyset \in \tau_1$, plus $\mathbb{N}\in \tau_1$.

We need to show that the finite intersection of the $\tau_1$ elements belongs to $\tau_1$. I can see that yes, for example, $\{0\}\cap \{0,1\}=\{0\}\in \tau_1$, $\{0\}\cap\{0,1\}\cap\{0,1,2\}=\{0\}\in \tau_1$ or $\{0,1\}\cap\{0,1,2\}\in \tau_1$. But I can't generalize. Can you help me?

We need to show that the arbitrary union of $\tau_1$ elements belongs to $\tau_1$. I know it's worth that, for example, $\{0\}\cup \{0,1\}\cup \{0,1,2\} \cup \mathbb{N}=\mathbb{N} \in \tau_1$. But I can't generalize. Can you help me?

Thank you!


问:
Formula relating three coplanar vectors
Let $\mathbf{\vec{u}}$, $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ be Euclidean vectors in $\mathbb{R}^3$ such that:

All of the vectors $\mathbf{\vec{u}}$, $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ are non-zero (i.e., $\mathbf{\vec{0}}\notin\{\mathbf{\vec{u}},\mathbf{\vec{v}},\mathbf{\vec{w}}\}$).

The vectors $\mathbf{\vec{u}}$ and $\mathbf{\vec{v}}$ are non-collinear (i.e., $\mathbf{\vec{u}}\wedge\mathbf{\vec{v}}\neq\mathbf{\vec{0}}$).

The vectors $\mathbf{\vec{u}}$, $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ are coplanar (i.e., [$\mathbf{\vec{u}}$,$\mathbf{\vec{v}}$,$\mathbf{\vec{w}}]=0$).

Then, a fortiori, there exist real numbers $\lambda$ and $\mu$ such that:
$$\mathbf{\vec{w}}=\lambda\mathbf{\vec{u}}+\mu\mathbf{\vec{v}}$$

My question is the following:

"Is there a general formula for $\lambda$ and $\mu$ in terms of the vectors $\mathbf{\vec{u}}$, $\mathbf{\vec{v}}$ and $\mathbf{\vec{w}}$ in $\mathbb{R}^3$ (or, if necessary, in $\mathbb{R}^2$)?"

EDIT:
I know of a more general formula relating four Euclidean vectors $\mathbf{\vec{u}}$, $\mathbf{\vec{v}}$, $\mathbf{\vec{w}}$ and $\mathbf{\vec{x}}$ in $\mathbb{R}^3$, which I state as follows:
$$[\mathbf{\vec{u}},\mathbf{\vec{v}},\mathbf{\vec{w}}]\neq 0 \implies \mathbf{\vec{x}}=\displaystyle\frac{[\mathbf{\vec{x}},\mathbf{\vec{v}},\mathbf{\vec{w}}]}{[\mathbf{\vec{u}},\mathbf{\vec{v}},\mathbf{\vec{w}}]}\mathbf{\vec{u}}+\frac{[\mathbf{\vec{u}},\mathbf{\vec{x}},\mathbf{\vec{w}}]}{[\mathbf{\vec{u}},\mathbf{\vec{v}},\mathbf{\vec{w}}]}\mathbf{\vec{v}}+\frac{[\mathbf{\vec{u}},\mathbf{\vec{v}},\mathbf{\vec{x}}]}{[\mathbf{\vec{u}},\mathbf{\vec{v}},\mathbf{\vec{w}}]}\mathbf{\vec{w}}$$
I ask my question above in search for an analogous formula which relates three coplanar Euclidean vectors, if there is one at all.

答:
Let us take a dot product of $w = \lambda u + \mu v$ with $u$ and with $v$.

Thus we have $$(w, u) = \lambda (u,u) + \mu(u,v)$$
$$(w, v) = \lambda (u,v) + \mu(v,v)$$
Hence we have 2 equations for 2 variables.

Put $A =(u,u) (v,v) - (u,v)^2$, $B = (w,u)(v,v) - (u,v)(w,v)$
and $C=  (u,u) (v,w) - (u,w) (u,v)$.
Hence $\lambda = \frac{B}{A}$ and $\mu = \frac{C}{A}$.

Title: Formula relating three coplanar vectors


问:
How to find the second vertex of a line segment from a given angle and length
If you have the (h,k) coordinates of a point, how to you calculate the coordinates of the other endpoint of a line segment if you have the angle $\alpha$ and length$L$?

答:
comment

Rough sketch of the above

Body:
comment

Rough sketch of the above


问:
Complex numbers and quadratic

If $\alpha,\beta$ are roots of $t^2-2t+5=0,(a+1)/2=\cot\phi$ then show
$$((a+\alpha)^n-(a+\beta)^n)/(\alpha-\beta)=
2^{n-1}\operatorname{cosec}^n(\phi)\sin(n\phi)$$

My try was to put values of sum, difference and $a$.
I got
$$2^n\operatorname{cosec}^n(\phi)(\cos(n\phi) \pm i\sin (n\phi)-\sin^n(\phi))/(\pm4i)$$
I can't go any further, can anyone help please?
I found $\alpha,\beta$ and $a$ from the relation substituted it and used De Moivre

答:
$$t=1\pm2i$$

WLOG $\alpha=1+2i,\beta=?\implies\alpha-\beta=?$

$$a+\alpha=2\cot\phi-1+1+2i=\dfrac{2(\cos\phi+i\sin\phi)}{\sin\phi}$$

Using Review of a proof for De Moivre's theorem using mathematical induction to find $$(a+\alpha)^n$$

Similarly find $$(a+\beta)^n$$

Title: Complex numbers and quadratic


问:
Convergence-probability

Let $X_{1}, X_{2}, \ldots,X_{n}$ be independent, $U(0, 1)$-distributed random variables, and set $Z_{n} = \max({X_{1},X_{2}, \ldots,X_{n}})$
and $V_{n} = \min({X_{1},X_{2}, \ldots,X_{n}})$. Determine the limit distribution of $nV_{n}/Z_{n}$ as $n$ goes to infinity.

I solved it in this way :
$$U = V_{n}/Z_{n}$$
$$P(nU< u ) = P( U < u/n ) = F_u (u/n)$$

but for finding $F_u$, I wonder to know if it is possible to find distribution of $\min X_{k}$ and $\max X_{k}$ seprately and then divided to each other, and if not, how can I find $F_u(u)$?

答:
The key is to calculate the joint distribution of $(V_n, Z_n)$.  For $a<b$, we have
$$
\mathbb{P}(V_n > a, Z_n \leq b) = \mathbb{P}(X_1, \dots, X_n \in (a,b]) = (b-a)^n,
$$
so the joint CDF is given by
$$
F_{(V_n,Z_n)}(x,y) = \mathbb{P}(V_n \leq x, Z_n \leq y) = \mathbb{P}(Z_n \leq y) - \mathbb{P}(V_n > x, Z_n \leq y) = y^n - (y-x)^n.
$$
We can then obtain the joint density by differentiating:
$$
f(x,y) = \partial_y \partial_x F(x,y) = n(n-1)(y-x)^{n-2}.
$$

Now we want to calculate, for fixed $t>0$, the probability $\mathbb{P}(n V_n / Z_n \leq t)$.  This amounts to just integrating the joint density $f_{(V_n, Z_n)}$ over the region $x \leq \frac{t}{n} y$:
$$
\mathbb{P}(n V_n / Z_n \leq t) = \int_0^1 \int_0^{\frac{t}{n} y} n(n-1) (y-x)^{n-2} \,dx\,dy = 1 - \frac{(1-t/n)^n}{1-t/n}
$$
(I omit the details of calculating this integral, but it's nothing fancy).

Finally, taking $n \to \infty$ we find
$$
\lim_{n \to \infty} \mathbb{P}(n V_n / Z_n \leq t) = 1-e^{-t},
$$
which we recognize as the CDF of the exponential distribution with parameter $1$.

Title: Convergence-probability


问:
prove convergence of $|u_m|^{p-2} u_m \to |u|^{p-2} u$ in $L^{p'}(U)$
Assume $u_m\to u$ in $L^p(U)$ such that $p>2$.

Prove that $|u_m|^{p-2}u_m \to |u|^{p-2}u$ in $L^{p'}(U)$ where $1/p + 1/p' = 1$.

I did the following: $\||u_m|^{p-2}u_m -|u|^{p-2}u\|_{p'} \le \||u_m|^{p-2}u_m -|u_m|^{p-2}u\|_{p'} + \||u_m|^{p-2}u -|u|^{p-2}u\|_{p'}$

The first term on the RHS can be proved to converge to zero, since $u_m$ converges in $L^p$ , hence the norm $\|u_m\|_p\le M$ . For the second term we apply Holder. There should be a term like this:

$$\int||u_m|^{p-2} - |u|^{p-2}|^{\frac{p}{p-2}}dx$$

I have no idea on how to prove that the term above also converges to zero.For $p-2<1$ we have

$$\int||u_m|^{p-2} - |u|^{p-2}|^{\frac{p}{p-2}}dx \le C\|u_m- u\|_p^p \to 0$$

What about $p>3$?I think the convergence result may not hold?

答:
Hi I hope you are having a merry christmas!

First it's useful to observe that $p>2 \implies p>p^{\prime}$ and that $||f||_{p^{\prime}}<||f||_p$. Thus $||u_m-u||_{p^{\prime}}<||u_m-u||_{p}\rightarrow 0$, regarding the second half maybe something like using the inequality $|a^p-b^p|\leq |a-b|^p$ when $a,b>0$ and  $0<p<1$. Now first $p^{\prime}=\frac{p}{p-1}$ so $\int_{U}||u_m|^{p-2}-|u|^{p-2}|^{p^{\prime}}=\int_{U}||u_m|^{p-2}-|u|^{p-2}|^{\frac{p}{p-1}}\leq\int_{U}||u_m|-|u||^{\frac{p(p-2)}{p-1}}\leq\int_{U}||u_m|-|u||^{p^{\prime}}=|||u_m|-|u|||_{p^{\prime}}$
and using the reverse triangel inequality $||a|-|b||\leq |a-b|$ we can get $|||u_m|-|u|||_p^p\leq ||u_m-u||_p^p\rightarrow 0$

I think this argument Is valid.


问:
Convolution of tight measures is tight
Let $E$ be a Banach space (think of $E=\mathbb R^d$, if it's easier to understand the following for you), $(\mu_n)_{n\in\mathbb N}$ and $(\nu_n)_{n\in\mathbb N}$ be tight sequences of finite nonnegative measures on $\mathcal B(E)$.

I would like to conclude that the sequence $(\mu_n\ast\nu_n)_{n\in\mathbb N}$ of convolutions is tight as well.

Let $\varepsilon>0$. By the tightness assumption, there is a compact $K\subseteq E$ with $$\max\left(\sup_{n\in\mathbb N}\mu_n(K^c),\sup_{n\in\mathbb N}\nu_n(K^c)\right)<\varepsilon\tag1.$$

Now, $K+K$ is clearly compact and \begin{equation}\begin{split}(\mu_n\ast\nu_n)(K+K)&=(\mu_n\otimes\nu_n)(\{(x,y)\in E^2:x+y\in K+K\})\\&\ge(\mu_n\otimes\nu_n)(K\times K)=\mu_n(K)\nu_n(K)\end{split}\tag2\end{equation} for all $n\in\mathbb N$.

If I assume that $\mu_n,\nu_n$ are probability measures for all $n\in\mathbb N$, then $(2)$ yields $$(\mu_n\ast\nu_n)(K+K)\ge(1-\varepsilon)^2\tag3$$ and hence we obtain the claim.

But how can we show the claim in general?

答:
As I wrote in the comments, the claim is not true in general. To see this, let $\delta_x$ be the Dirac measure at $x$, meaning that $\delta_x(A) = 1$ if $x \in A$ and $\delta_x(A) = 0$ otherwise. Then define $\mu_n = n^{-1} \cdot \delta_n$ and $\nu_n = n \cdot \delta_0$.

Then choosing $K = \{0\}$, we see that $(\nu_n)_n$ is a tight sequence. Likewise, for each $N$, we can choose $K = \{1,\dots,N\}$ to get $\sup_n \mu_n(K^c) \leq 1/N$ for all $n \in \Bbb{N}$, showing that also $(\mu_n)_n$ is a tight sequence.

But $\mu_n \ast \nu_n = \delta_n$, which easily implies that $(\mu_n \ast \nu_n)_n$ is not a tight sequence.

Probably the answer is positive if one in addition assumes that $(|\mu_n|)_n, (|\nu_n|)_n$ are both uniformly bounded.

If one assumes the measures to be uniformly bounded, meaning that there is $C > 0$ with $\mu_n(E) \leq C$ and $\nu_n (E) \leq C$ for all $n$, then the claim is true.

To see this, simply note that what you already shows implies
\begin{align*}
(\mu_n \ast \nu_n)((K + K)^c)
& = (\mu_n \ast \nu_n)(E) - (\mu_n \ast \nu_n)(K + K) \\
& \leq \mu_n(E) \nu_n(E) - \mu_n(K) \nu_n(K) \\
& =    \mu_n(E) \cdot (\nu_n(E) - \nu_n(K)) + \nu_n(K) (\mu_n(E) - \mu_n(K)) \\
& \leq C \epsilon + C \epsilon .
\end{align*}

Title: Convolution of tight measures is tight


问:
Does a tiling being monohedral and monogonal imply that it is isohedral or isogonal?
I’m working through Grünbaum and Shepard’s Tilings & Patterns and I have been unable to make much progress on this problem.

Does there exist a tiling that is monohedral and monogonal, but is neither isohedral nor isogonal?

I suspect there is not a tiling that is both monohedral and monogonal but neither isohedral nor isogonal based on all the monohedral and monogonal tilings I’ve been able to find. However, I’m unable to formulate any sort of proof as to why.

The way I’ve been attempting to prove it is by starting with a vertex and its incident edges and seeing when it is possible to “tile” the plane with it while producing a monohedral tiling. For example, I’ve found that every possible configuration of three edges that meet together at a single vertex can “tile” the plane monohedrally (as well as isohedrally and isogonally) by forming hexagons. However, I’m not sure this is a feasible route especially since I’m unsure if there exist k-valent tilings in which k is not 3, 4, or 6.

Something I think may be useful but am not sure how to use is that due to the monogolality of the tiling all angles and edges of the prototile must meet at every vertex and therefore be present in every vertex-incident edge “tile”.

Edit Definitions:

Monohedral: every tile in the tiling is the same.

Monogonal: every vertex and its incident edges (the edges connected to the vertex) forms a figure that is congruent to every other vertex and its incident edges.

Isohedral: every tile can be mapped to any other tile using only symmetries of the tiling (imagine tracing out the tiling on a separate sheet of paper and then being able to line up any tile on your paper to any tile on the original and still have the tracing line up with the original everywhere).

Isogonal: every vertex can be mapped to any other vertex of the tiling using only symmetries of the tiling (like isohedral but with vertices).

答:
Let $\alpha_4 = 0. a_1 a_2 a_3 \ldots_4 $ be an irrational number in base 4 subject to $|a_i - a_{i+1} | = 1$.

(There are many ways to construct this, like starting from any irrational number in base 4 to determine the "height of the peaks".)

Consider unit squares laid out in the plane, where the lower left corner is $(m + \frac{1}{4} a_{|n| + 1 } , n )$, for integer $m, n$.

This is essentially rows of unit squares stacked upon each other, and shifted left/right by $1/4$ from the row above/below it, though in a aperiodic way.

It is monohedral as all tiles are unit squares.

It is monogonal as each vertex is a T with incident edges of length $1$ (for the leg of the T), and lengths $3/4, 1/4$ (for the arms of the T).  We allow for reflections of the T.

It is neither isohedral nor isogonal as the only symmetries of the tiling are left/right translations due to the aperiodic nature.


问:
Example of a map which is not closed but every closed ball maps into closed set
I'm trying to find a counter-example for the following statement:
"Consider $X$ and $Y$ be any two metric spaces. If $f: X\rightarrow Y $ is a map such that $f$ maps closed balls in $X$ to closed Sets in $Y$ then $f$ is a closed map "
I thought about inclusion maps , projection maps, constant maps, but I was unable to think counter example from there. Please help me on how to think of any counter example? I am a beginner in topology and metric spaces.

答: You can take, for instance, $\arctan\colon\Bbb R\longrightarrow\Bbb R$, with $\Bbb R$ endowed with the usual topology. It is not a closed map, since $\arctan(\Bbb R)=\left(-\frac\pi2,\frac\pi2\right)$. But it maps closed balls into closed sets.


问:
torsion of abelian variety commutes with base change?
Let $X/k$ be a variety and $A/k$ be an abelian variety. Set $\mathcal{A} = A \times_k X$ and $\bar{X} = X \times_k \bar{k}$.

Is it true that $\mathcal{A}[\ell^n] = A[\ell^n] \times_k X$? And what is  $\mathcal{A}[\ell^n](\bar{X})$?

答:
Assuming that by $\mathcal A[\ell^n]$ you mean the subgroup scheme of $\mathcal A$ consisting of the (scheme-theoretic) kernel of multiplication by $\ell^n$, then yes, it is true that $\mathcal A[\ell^n] = A[\ell^n]\times_k X$.  This follows from the description of $\mathcal A[\ell^n]$ as the fibre product of the map
$\ell^n:\mathcal A \to \mathcal A$ and the map $X \to \mathcal A^n$ giving the identity section (and the analogous description of $A[\ell^n]$).

Also (assuming that $\overline{X}$ is reduced or that $\ell$ is prime to the characteristic of $k$), $\mathcal A[\ell^n](\overline{X})$ will be a product of copies of
$A[\ell^n](\overline{k})$, one copy for each connected component of $\overline{X}$.


答:
Let $ n\in\mathbb{N}^{*} $.

\begin{aligned}\int_{0}^{1}{\left(\left\lfloor\frac{n}{x}\right\rfloor -n\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm{d}x}&=\int_{1}^{+\infty}{\frac{\left\lfloor nx\right\rfloor - n\left\lfloor x\right\rfloor}{x^{2}}\,\mathrm{d}x}\\ &=\sum_{p=1}^{+\infty}{\int_{p}^{p+1}{\frac{\left\lfloor nx\right\rfloor - n\left\lfloor x\right\rfloor}{x^{2}}\,\mathrm{d}x}}\\ &=\sum_{p=1}^{+\infty}{\int_{p}^{p+1}{\frac{\left\lfloor nx\right\rfloor - np}{x^{2}}\,\mathrm{d}x}}\\ &=\sum_{p=1}^{+\infty}{\sum_{\ell = np}^{n\left(p+1\right)-1}{\int_{\frac{\ell}{n}}^{\frac{\ell+1}{n}}{\frac{\left\lfloor nx\right\rfloor - np}{x^{2}}\,\mathrm{d}x}}}\\ &=\sum_{p=1}^{+\infty}{\sum_{\ell = np}^{n\left(p+1\right)-1}{\int_{\frac{\ell}{n}}^{\frac{\ell+1}{n}}{\frac{\ell - np}{x^{2}}\,\mathrm{d}x}}}\\ &=n\sum_{p=1}^{+\infty}{\sum_{\ell = np}^{n\left(p+1\right)-1}{\frac{\ell - np}{\ell\left(\ell +1\right)}}}\\ &=n\sum_{p=1}^{+\infty}{\left(H_{n\left(p+1\right)}-H_{np}-np\left(\frac{1}{np}-\frac{1}{n\left(p+1\right)}\right)\right)}\\ &=n\sum_{p=1}^{+\infty}{\left(H_{n\left(p+1\right)}-H_{np}-\frac{1}{p+1}\right)}\\ &=\lim_{k\to +\infty}{n\left(H_{nk}-H_{n}-H_{k}+1\right)}\\ \int_{0}^{1}{\left(\left\lfloor\frac{n}{x}\right\rfloor -n\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm{d}x}&=n\left(\ln{n}-H_{n}+1\right)\end{aligned}

So what you did is correct.

Title: Calculate $\int_0^1 \left\lfloor \frac{2020}{x} \right\rfloor - 2020\left\lfloor\frac1x\right\rfloor\,dx$

Body:
Calculate the definite integral $$\int_0^1 \left\lfloor \frac{2020}{x} \right\rfloor - 2020\left\lfloor\frac1x\right\rfloor\,dx$$

My Attempt:

Let's make a start. Write $[2020/x]=n\ge 2020$ then ${{2020}\over {n+1}}\le 1/x < {{2020}\over n}$ and ${n\over {2020}}\le 1/x <{{n+1}\over {2020}}$. Hence $[1/x]=[n/2020]$. For the integral we find

$$\sum_{n=2020}^\infty \int_{2020/(n+1)}^{2020/n} (n-2020[n/2020]) dx=2020\sum_{2020}^\infty {1\over {n(n+1)}} (n-2020[n/2020]) $$
In order to get $[n/2020] = m$ we need $m \leq n/2020 < m+1$ or $2020m \leq n \leq 2020m + 2019$. Now write:
$$2020\sum_{n=2020}^\infty \frac{1}{n(n+1)} (n-2020\lfloor n/2020\rfloor) = 2020\sum_{m=1}^\infty \sum_{n=2020m}^{2020m+2019} \frac{1}{n(n+1)} (n-2020m)$$$$ = 2020\sum_{m=1}^\infty \left[\left(\sum_{n=2020m}^{2020m+2019} \frac{1}{n+1}\right) - 2020m\left(\frac{1}{2020m} - \frac{1}{2020(m+1)}\right)\right]$$$$ = 2020\sum_{m=1}^\infty \left[\left(\sum_{n=2020m}^{2020m+2019} \frac{1}{n+1}\right) - \frac{1}{m+1}\right]$$Now an analysis of partial sums...
$$\sum_{m=1}^N \left[\left(\sum_{n=2020m}^{2020m+2019} \frac{1}{n+1}\right) - \frac{1}{m+1}\right] = \sum_{k=2021}^{2020N+2020} \frac{1}{k} - \sum_{k=2}^{N+1} \frac1k$$$$ = H_{2020N+2020} - H_{N+1} - H_{2020} + 1 \to \log(2020) - H_{2020} + 1$$So the answer is $\boxed{2020\left(\log(2020) - H_{2020} + 1\right)}$.
However, I still wonder whether my answer is correct.
Please help on this problem, thank you.


问:
Calculate padding on coordinates bounding box
I'm working on a GIS application and I can't seem to warp my head around this problem.
I have bounding box defined as 4 coordinates pair (A, B, C, D in the picture) and a center and I want to calculate a new bounding box (A', B', C', D' in the picture) defined as the original bounding box, plus a padding on all sides. The padding value is defined as a percentage of the longest side of the original bounding box P.

What I'm trying to achieve is: given A, B, C, D, CENTER and a percentage value P, calculate A', B', C', D'.

Any ideas?

Cheers!

Here is a picture of this for reference:

答:
You can erect a local Cartesian coordinate system if the area is small.  Small here is relative to the radius of the earth, about $4000$ miles ($6400$ km).  You can use the center as the origin.  The origin has latitude $\phi$ and longitude $\lambda$ and local coordinates $(0,0)$.  Distances North/South come from the fact that one degree of latitude corresponds to a distance of $\frac {R_e \pi}{180}$.  Distances East/West come from the fact that one degree of longitude corresponds to a distance of $\frac {R_e \pi}{180}\cos(\phi)$.  The earth radius varies, but for a spherical earth you choose one value and use it consistently.  Most often I have seen the equatorial radius used, $R_e=6378 km \approx 3963 mi$.

The dimensions of the box are then $\frac {R_e \pi}{180}\Delta \phi$, North/South where $\Delta \phi$ is the difference in latitude between the top and the bottom in degrees and $\frac {R_e \pi}{180}\cos \phi \Delta \lambda$ East/West.  You can find the linear distance you want to pad and convert back to the change in latitude/longitude


问:
$f(z+1)-f(z)$ is a polynomial map, then $f(z)$ is a polynomial map
Let $f: \mathbb{Z}\rightarrow\mathbb{Z}$ be a map of sets, if the map $f(z+1)-f(z)$ can be written as some polynomial of $z$ with $\mathbb{Q}$-coefficients, then $f(z)$ can be also written as a polynomial of $z$ with $\mathbb{Q}$-coefficients.

How to prove this?

Thanks in advance.

答:
(The statement is true even for $ f: \mathbb{Z} \rightarrow \mathbb{Q} $, which should make sense because we could multiply by the LCM of the denominator.)

Set up the difference table where $ D_0 (z) = f(z) $ and $ D_{i+1} (z) = D_{i} (z+1) - D_{i} (z)$ for $ i \geq 0$.

We are given that $D_i(z)$ are rational for $i\geq 1$.

Claim: $$ f(x) = f(1) + \sum_{i=1}^{n+1} \frac{ D_i (1) }{i!} \times f_i (x),$$ where $f_i(x) = (x-1) \times (x-2) \times \ldots \times (x-i)$ are polynomials.

Proof: Refer to Method of Differences for polynomial interpolation.

Hence, $ f(x)$ is indeed a rational polynomial.

As an alternative but equivalent interpretation,

(Fill in the gaps, if any. If you're stuck, show your work and explain what you have tried.)

Define $ f_i (x) = (x-1) (x-2) \ldots (x-i)$ for $ i \geq 1$, which is a polynomial of degree $i$.

Observe that $f_{i+1} (x+1) - f_{i+1} (x) = i f_{i} (x)$ for $ i \geq 1$.

Observe that $ f_i (1) = 0 $ for $ i \geq 1$.

Claim: There exists a representation
$$f(x+1) - f(x) = \sum a_i f_i (x),$$

where $a_i$ are rational numbers.

Sketch of Proof: Work from the highest degree on the LHS downwards, to obtain such a representation.

Note: The RHS are polynomials of degree $i$ which are linearly independent, so the representation is unique.

Corollary: $f(x+1) - f(x) = \sum a_i f_i (x) = \sum \frac{a_i}{i} \left[ f_i (x+1) - f_i (x) \right]$, so we can obtain the representation $ f(x) = \left[ \sum \frac{a_i}{i} f_i (x) \right] + C$, for some constant $C$.

Substituting $ x = 1$, we get $ C = f(1)$.

This gives us our rational polynomial representation.


问:
Why isn't $x^\frac{-1}{2}$ equal to $-\sqrt{x}$
A fraction multiplied by $-1$ can be written in different ways:

$\frac{-a}{b}$

$\frac{a}{-b}$

$-\frac{a}{b}$

So $x^\frac{-1}{2}$ can also be written as $x^\frac{1}{-2}$ then why can't we take the whole $-2$ down and turn it into $-\sqrt{\bullet}$ and say:

$x^\frac{1}{-2}$ = $-\sqrt{x}$

答:
$x^{-y}$ is $\frac1{x^y}$, not $- x^y$.

Added: $\sqrt[-2]{x}$ is the positive real number $u$ such that $u^{-2}=x$, so it's $u=\frac1{\sqrt x}$.


问:
Sequences in which, one value of $K$ work for all $ε$
We say a sequence $(x_n)$ of real numbers converges to $x\in\mathbb{R}$ "if $\forall\epsilon>0$ $\exists$ $K\in\mathbb{N}$ such that, $|x_n-x|<\epsilon$ for all $n≥K$"

My question : Are there any non constant sequences where unique value of $K$ works for all value of $\epsilon$? Is there any special name given to such sequences?

答:
Let $x_1=0$ and $x_n = 1$ for $n \ne 1$. Clearly $(x_n) \to 1$ and choosing $K = 2$ we have $ |x_n - 1 | < \epsilon $ whenever $n \geq K$

If $K$ is fixed then the sequence $(x_n)$ must have finite number of distinct terms. To see this suppose for all $ \epsilon > 0$ a unique $K$ works, let $n$ be fixed number then $|x_n - x | < \epsilon $ for all $ \epsilon > 0$. We claim that $x_n = x$ for all $n \geq K$ , otherwise there would be fixed $ \epsilon$ below which the equality doesn't hold which certainly not possible since choosing $ \epsilon /2$ instead of $ \epsilon$ will be a contradiction.


问:
Round off error in modified Newton-Raphson
Newton's method for approximating roots of $f(x)=0$ is given by $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}, \quad n=0,1,2,\ldots,$$ where $x_n$ denotes the $n$th approximation of the root, and some sufficiently good initial approximation $x_0$ is required for convergence.

In general, this converges quadratically, but if the root, say $x^*$, satisfies $f'(x^*)=0$, the convergence is only linear. I'm aware that quadratic convergence can be restored by using the scheme $$x_{n+1}=x_n-\frac{F(x_n)}{F'(x_n)},$$ where $F(x_n)=f(x_n)/f'(x_n)$.

After substituting the definition of $F(x_n)$, it can be seen that $$x_{n+1}=x_n-\frac{f(x_n)f'(x_n)}{f'(x_n)f'(x_n)-f(x_n)f''(x_n)}.$$

I'm told that this raises another issue, namely that the denominator involves computing the difference of two nearly equal quantities and can lead to large rounding errors.

Questions

Why are the quantities on the denominator nearly equal?

How do we get around this issue?

答:
Since both $f(x)\to0$ and $f'(x)\to0$, it is clear both terms on the denominator tend to $0$ quickly. Furthermore, if $f''(x)\to c\ne0$, then by Taylor's theorem one may use the estimations:

$$f(x)\sim\frac c2(x-x^\star)^2,\quad f'(x)\sim c(x-x^\star)$$

and the terms in the denominator are

$$f'(x)f'(x)\sim c^2(x-x^\star)^2,\quad f(x)f''(x)\sim\frac{c^2}2(x-x^\star)^2$$

and as you can see, they only differ by a factor of $2$ asymptotically. To try to avoid cancellation error, you can rewrite the expression as

$$x-\frac{f(x)}{f'(x)}\left(1-\frac{f(x)}{f'(x)}\frac{f''(x)}{f'(x)}\right)^{-1}$$

where one can see that

$$\frac{f(x)}{f'(x)}\frac{f''(x)}{f'(x)}\sim\frac12$$

$$\left(1-\frac{f(x)}{f'(x)}\frac{f''(x)}{f'(x)}\right)^{-1}\sim2$$

and that evaluations are more numerically stable since you get tricky divisions out of the way before other operations.

Some notable remarks:

If it is known ahead of time that $f^{(n)}(x^\star)$ is $0$ for all $n<N$ and non-zero for $n=N$, then you can replace the term multiplying $f(x)/f'(x)$ by $N$. The proof is the same Taylor argument as above.

Since it can be expected that the term multiplying $f(x)/f'(x)$ tends to a constant except in extreme examples (such as $\exp(-x^{-2})$, which has $f^{(n)}(0)=0$ for all $n$), it suffices to replace the term by the last numerically stable value you were able to compute, if you were to have issues evaluating it.

If $f$ is known to be smooth, you can round the term multiplying $f(x)/f'(x)$ to the nearest integer.

The above form is also advantageous if you can evaluate $f(x)/f'(x)$ and $f'(x)/f''(x)$ faster, or as fast as, or more numerically stable than $f(x),f'(x),f''(x)$ individually.


问:
Set Theory Notation Crises
For those who are familiar with the following notation, could you explain it in plain English because I picked up a set theory textbook but the book assumes the reader is familiar with the notation without giving a formal explanation anywhere.

1) $$\{x:\mathscr{P}x\}$$
2) $$\{\mathscr{a}_x: \mathscr{P}x\}$$
3) $$\left(\bigcup_{i \in I} \mathscr{a}_i\right)^{c}$$

4) $$x \in \{y: y>2\}$$
5) (distributive laws) $$B \cap \left(\bigcup_{i \in I} \mathscr{a}_i\right)$$
6) $$\{x: x \not\in x\}$$

The Px is like a P that is curly and looks like old english writing or might be a greek and same with a_i. a is like a fat curly a which I'm thinking might stand for "for all." Any input with explanation would be great or any reference would work as well. The book I'm reading is 'Set Theory An Introduction' by Robert Vaught.

答:
1) and 2) are explained in page 8. ${\scr P} x$ indicates that $x$ satisfies some predicate or some property (for example consider $x$ satisfies $\scr P$ if and only if $x$ is a positive integer then $2,23$ does not satisfy $\scr P$ and therefore is not an element of $\{x:{\scr P} x\}$), and $a _x$ is a function of $x$, so you can just substitute that by $f(x)$ for some fixed $f$ defined whenever $x$ satisfies $\scr P$.

6) Is an instance of 1) it is therefore a set formed out of all the sets $x$ so that $x$ is not a member of itself, it is generally used to show a paradox which arises when one allows himself to form sets $\{x:{\scr P} x\}$ without any restriction on the predicate $\scr P$.

$\bigcup_{i\in I} a_i$ Is the set formed by all elements $x$ such that $x\in a_i$ for at least one $i\in I$, here the $a_i$ are said to form a family of sets indexed on the set $I$, so 5), becomes just the intersection of this family with $B$, that is, elements that lie both in $B$ and in some $a_i$, the (distributive law) refers to the property
$$ B\cap (\bigcup_{i\in I} a_i) = \bigcup_{i\in I} (B\cap a_i)$$
Now given a set $a$, $a^c$  usualy refers to the complement of the set (which is seen to be contained in some ommited larger set), thus 3) is defined if all of the $a_i$ lie inside the same set $A$ in which case it expands to
$$(\bigcup_{i\in I}a_i)^c = A\setminus (\bigcup_{i\in I} a_i) $$

4) Is, again, an instance of 1) where by saying that $x\in \{y:y>2\}$ we mean $x>2$ (note that in order for this to make sense $x$ must be a real number)

Title: Set Theory Notation Crises


问:
What is the asymptotic complexity of $\log x / \log(1+1/x)$?
I have encountered the function $$f(x) = \frac{\log x}{\log(1+\frac{1}{x})}$$ in the context of computational complexity and am wondering whether it is possible to simplify it. In particular, I am wondering if
$$\Theta(f) = x \log x$$
which seems plausible from looking at its graph. If so, how to prove it - and if not, is there a more succinct way to represent $\Theta(f)$?

答:
Since $\log(1+u)\sim u$ when $u\to 0$, we have:

$$\frac{\log x}{\log(1+1/x)}\sim\frac{\log x}{1/x}=x\log x$$

when $x\to\infty$.


问:
about topological manifold
A fundamental theorem in Topology asserts that if $U \subset \mathbb{R}^{n}$ and $V \subset \mathbb{R}^{m}$ are homeomorphic, then $m=n$.

(i)- for above shape , try to write a description in terms of familiar topological spaces.

(ii)- Prove that a sphere with a hair (above shape), is not a topological manifold.

for (ii) we have :A connected manifold has a unique dimension $n$, and every point of $X$ then has an open neighbourhood homeomorphic to the open unit  ball  $\mathbb D^n\subset \mathbb R^n$.

However in the pictured $X$ the points different from $q$  on the hair have an open neigbourhood homeomorphic to $\mathbb D^1$ , whereas the points different from $q$  on the sphere have an open neigbourhood homeomorphic to $\mathbb D^2$.

Since $X$ is connected this proves that it is not a manifold, since it cannot have a unique dimension.so above shape is not a topological manifold.

how we can answer first question ? also we know that above shape is homeomorphic to sphere and sphere is topological manifold but above shape is not a topological manifold .so we find two homeomorphic space such that one of them is is not a topological manifold and other is a topological manifold. is this true ?

答:
Call the space $X$. You can write the space as a glueing of a sphere $\mathbb{S}^2$ and a half open interval

$$ X = \mathbb{S}^2 \coprod_{q = 0} [0,1)$$

In other words, $X$ is the pushout of the maps $\{\ast\} \rightarrow \mathbb{S}^2$ given by $\ast \mapsto q$ and $\{\ast\} \rightarrow [0,1)$ given by $\ast \mapsto 0$.

Title: about topological manifold


Title: Sequences in which, one value of $K$ work for all $ε$

Body:
We say a sequence $(x_n)$ of real numbers converges to $x\in\mathbb{R}$ "if $\forall\epsilon>0$ $\exists$ $K\in\mathbb{N}$ such that, $|x_n-x|<\epsilon$ for all $n≥K$"

My question : Are there any non constant sequences where unique value of $K$ works for all value of $\epsilon$? Is there any special name given to such sequences?

Body: $$|x_n-x|<\epsilon\enspace\forall\epsilon\in\mathbb{R}_{>0}\implies x_n=x$$


问:
Geometry problem - prove that incentre of two triangles coincide

$AB$ and $AC$ are tangent to the circle at $B$ and $C$ respectively. Let $OA$ and $BC$ intersect at M. Let $E$ be an arbitrary point on the circle. Extend $EM$ to meet the circle at F.

(1) Prove that $\Delta AEF$ and $\Delta ABC$ have a common incentre.

(2) Prove that $\Delta AEF$ and $\Delta ABC$ have a common excentre.

I verified (1) on Geogebra and one of the interesting things I found was that $\Delta AEM \sim \Delta AOF$, but I have no idea how to prove it (or will it even be useful).

My Question

I want to know how to solve (1). To do it I tried to show that $OA$ is the angle bisector of $\angle MAF$ first. However, although I was quite sure that it is true, I couldn't come up with a proof of it. Anyone can give me a hint?

答:
First of all, note that $I_{ABC}\equiv$ Midpoint of the arc BC.
That's mainly because this incenter shall be on $AO$ and on the line that connects $B$ to the midpoint of the arc $BC$ (because it is the bisector of the angle $ABC$)
So we just have to prove that the incenter of $AEF$ lies on the midpoint of arc $BC$. For proving this note that $AE.AF=AC^2$ and thus if we name the intersection of $AF$ and the circle as $T$, $AT.AF=AC^2$ and so $AT=AE$. This shows (in the same time) that the midpoint of the arc $BC$ lies on the bisector of the angles $AFE$ and $FAE$ and hence we are done.


问:
Does $\,x>0\,$ hint that $\,x\in\mathbb R\,$?

Does $x>0$ suggest that $x\in\mathbb R$?

For numbers not in $\,\mathbb R\,$ (e.g. in $\mathbb C\setminus \mathbb R$), their sizes can't be compared.

So can I omit "$\,x\in\mathbb R\,$" and just write $\,x>0\,$?

Thank you.

答:
My favorite would be "$x$ is a positive real number". Simple, clear, unambiguous, and no strange symbols to decipher.

It's a bit long, maybe, but who cares -- the point of writing is clarity, not brevity.


问:
Check whether the equation $3x^3+5y^5+z^3+4z=13$ defines the function $z(x,y)$ at a point $A(1,1,1)$

Consider the equation
$$
3x^3+5y^5+z^3+4z=13
$$
Check whether the equation defines the function $z(x,y)$ at a point $A(1,1,1)$

I don't understand what am I supposed to in this problem. Can someone give me a clue as to how to start?

答:
$$3 x^3+5 y^5+z^3+4 z=13\tag{1}$$
plugging $x=1,y=1$ we get
$$z^3+4 z+8=13\to z^3+4 z-5=0\to (z-1) \left(z^2+z+5\right)=0$$
which has the only real solution $z=1$

therefore $(1)$ defines implicitly a function at $(1,1,1)$


问:
Is there an archive or a collection of quadratic reciprocity proofs?
After Pythagorean theorem the quadratic reciprocity has the largest amount of proofs, I heard there are more than 160 proofs of quadratic reciprocity. I have seen a few of these proofs but I was wondering has anyone collected these proofs, or at least a generous amount of them?</p>

答:
Yes ;-)

Franz Lemmermeyer did: https://www.mathi.uni-heidelberg.de/~flemmermeyer/qrg_proofs.html


答:
Continuing on from my other answer -

Property 1:

The quantity $x^{-a}$ is one such that $x^a x^{-a}=1$. This gives us
$$x^{-a}=\frac{1}{x^a}$$
Property 2:

The quantity $x^{1/a}$ is one such that ${\left(x^{1/a}\right)}^a=x$. If we define the operation $\sqrt[a]{}$ as the inverse of taking something to the exponent $a$, we can write explicitly that $x^{1/a}=\sqrt[a]{x}$.

With this in mind, let's compare the expressions $x^{-(1/2)}$ and $x^{1/(-2)}$. Using the first property mentioned, we can write
$$x^{-(1/2)}=\frac{1}{x^{1/2}}=\frac{1}{\sqrt{x}}$$
On the other hand, taking a look at our other expression and using the second property,
$$\left(x^{1/(-2)}\right)^{-2}=x$$
Which, by our first property means
$$\frac{1}{\left(x^{1/(-2)}\right)^{2}}=x$$
So taking the square root of both sides,
$$\frac{1}{x^{1/(-2)}}=\sqrt{x}$$
Meaning
$$x^{1/(-2)}=\frac{1}{\sqrt{x}}$$
Thankfully we reach the desired conclusion that $$x^{-(1/2)}=x^{1/(-2)}=x^{\frac{-1}{2}}=x^{\frac{1}{-2}}.$$

Title: Why isn't $x^\frac{-1}{2}$ equal to $-\sqrt{x}$

Body:
A fraction multiplied by $-1$ can be written in different ways:

$\frac{-a}{b}$

$\frac{a}{-b}$

$-\frac{a}{b}$

So $x^\frac{-1}{2}$ can also be written as $x^\frac{1}{-2}$ then why can't we take the whole $-2$ down and turn it into $-\sqrt{\bullet}$ and say:

$x^\frac{1}{-2}$ = $-\sqrt{x}$


问:
Countability of $\mathbb{Q}$, function notation
Proof

Let $\mathbb{Q} = \{\frac{a}{b}: a,b \in \mathbb{Z}, b \neq 0\}$

Define a function $f$ given by $f(a,b)=2^a3^b \in \mathbb{N}$

Then, by the fundamental theorem of arithmetic, each natural number can be rewritten as the unique product of primes. This implies that each element in the image is unique and a natural number. Thus, $\mathbb{N}$ is countable. $\square$

So here's my question. Is the proper function notation $f:\mathbb{Q} \to \mathbb{N}$ or is it $f:\mathbb{Z} \to \mathbb{N}$? Because, while each $a$ and $b$ are derived from $\frac{a}{b} \in \mathbb{Q},$ each $a$ and $b$ are integer values.

Ultimately, I'm just asking a question regarding proper notation. Pay no attention to the details of my proof. I'm just asking what the proper notation is.

答:
If you want to prove that $\mathbb{Q}$ is countable, then you define your function as being $f:\mathbb{Q}\rightarrow \mathbb{N}$.

But actually this function can't be defined as $f:\mathbb{Z}\rightarrow \mathbb{N}$ even if you consider $f(a,b)$ since it is then defined on $\mathbb{N}^2$...

But if you want to prove that $\mathbb{Q}$ is countable, I suggest you rather show that there exists a bijection between $\mathbb{Q}$ and $\mathbb{Z}\times \mathbb{N}^*$, which is countable since it is the product of two countable spaces.


问:
What is the simplest way to do this combinatorics problem?
Question: Bob is about to hang his eight shirts in the wardrobe. He has 4 different styles of shirt, two identical ones of each particular style. How many different arrangements are possible if no two identical shirts are next to each other?

Here are the solutions:

In this case, we take the first case of all possible combinations: $\frac{8!}{2^4}$
But we see that we counted the possibilities with at least one pair together
so we have to subtract $\frac{7!}{2^3}\cdot \binom{4}{1}$ if we tie the two offending shirts together.

^This I don't get - why and what is being double-counted?

However, we see now that we double subtracted ones that have two pairs thus we must add the chance of two pairs so we must add $\frac{6!}{2^2}\cdot \binom{4}{2}$ and so on

Thus the final answer is equal to $864$

Is there a better way that doesn't involve BOTH adding AND subtracting?

答:
We denote with $a,b,c,d$ the four different types of shirts. We are looking for words of length $8$ built with all letters from the multiset $\{a,a,b,b,c,c,d,d\}$ which do not contain the bad words in $\{aa,bb,cc,dd\}$.

This answer is based upon a generating function of generalized Laguerre polynomials
\begin{align*}
  L_k^{(\alpha)}(t)=\sum_{i=0}^k(-1)^k\binom{k+\alpha}{k-i}\frac{t^i}{i!}
  \end{align*}

The Laguerre polynomials have some remarkable combinatorial properties and one of them is eminently suited to answer problems of this kind. This is nicely presented in Counting words with Laguerre series by Jair Taylor. We find in section $3$ of this paper:

Theorem: If $m_1,\ldots,m_k,n_1,\ldots,n_k$ are non-negative integers, and $p_{m,n}(t)$ are polynomials defined by
\begin{align*}
  \sum_{n=0}^\infty p_{m,n}(t)x^n=\exp\left(\frac{t\left(x-x^m\right)}{1-x^m}\right)
  \end{align*}
then the total number of $k$-ary words that use the letter $i$ exactly $n_i$ times and do not contain the subwords $i^{m_i}$ is
\begin{align*}
   \int_0^\infty e^{-t}\prod_{j=1}^k p_{m_j,n_j}(t)\,dt\tag{1}
  \end{align*}

Here we consider a $4$-ary alphabet $\{a,b,c,d\}$ and words built with all letters from the multiset $\{a,a,b,b,c,c,d,d\}$ and the bad words $\{aa,bb,cc,dd\}$.

Following the theorem above we have $m_j=2, n_j=2, 1\leq j\leq 4$ and the wanted number of valid words is according to (1)
\begin{align*}
 \int_0^\infty e^{-t}\left(p_{2,2}(t)\right)^4\,dt\tag{2}
\end{align*}

We obtain with some help of Wolfram Alpha

\begin{align*}
p_{2,2}(t)&=[x^2]\exp\left(\frac{t\left(x-x^2\right)}{1-x^2}\right)\\
&=[x^2]\left(1+tx+\frac{1}{2}t(t-2)x+\cdots\right)\\
  &=\frac{1}{2}t(t-2)
\end{align*}

According to (2) and recalling $\int_0^{\infty}e^{-t}t^{n}\,dt=\frac{1}{n!}$ we obtain as wanted number

\begin{align*}
&\int_0^\infty e^{-t}\left(\frac{1}{2}t(t-2)\right)^4dt\\
&\qquad=\frac{1}{16}\int_{0}^{\infty}e^{-t}t^4\left(t^4-8t^3+24t^2-32t+16\right)dt\\
&\qquad=\frac{1}{16}\left(8!-8\cdot7!+24\cdot 6!-32\cdot 5!+16\cdot 4!\right)\\
&\,\,\qquad\color{blue}{=864}
\end{align*}
in accordance with OPs statement.


问:
Invert Trigonometric System of Equations
In my research, I have a system of trig equations that I would like to invert. I know $\alpha$ and $\beta$ ($\theta$ is a constant) in the following system.

$$\cos \alpha  = \cos \theta \cos x + \sin \theta \sin x \cos y$$
$$\tan \beta = \frac{\cos x +\cos \theta \cos \alpha}{\sin \theta \sin x \sin y} $$

I would like to find an inversion in the form:
$$\text{trig}(x) =\; ...$$
$$\text{trig}(y) = \;...$$

How can this be done?

答: Not an answer, just illustrating what I said in my comment:


问:
Find the coefficient of ${x^9}$ in the expansion of $ (1 + x)( 1 + x^2)( 1 + x^3)..(1 + x^{100})$
Find the coefficient of ${x^9}$ in the expansion of $\left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^3}} \right)..\left( {1 + {x^{100}}} \right)$.
The official answer is 8.

How do I find the general term,

Dividing the above equation by $(1-x)$ is not generating the required result.

答:
An "x^9" occurs in this expression exactly when a sequence of powers of $x$s add up to a total exponent of $9$. For instance, the first and eighth terms contribute $x$ and $x^8$, for a total of $x^9$ (where in all other terms you multiply by $1$).

The sequences we care about

start at 1

are increasing

Have no two exponents the same

So now you can just start writing them down:

```<code>9
1,8
2,7
3,6
4,5
1,2,6
1,3,5
2,3,4
</code>```

...
and there are eight of those.

Body:
An "x^9" occurs in this expression exactly when a sequence of powers of $x$s add up to a total exponent of $9$. For instance, the first and eighth terms contribute $x$ and $x^8$, for a total of $x^9$ (where in all other terms you multiply by $1$).

The sequences we care about

start at 1

are increasing

Have no two exponents the same

So now you can just start writing them down:

9
1,8
2,7
3,6
4,5
1,2,6
1,3,5
2,3,4

...
and there are eight of those.


答:
Hint:

$$\begin{align}9=9+0\\=8+1\\=7+2\\=6+3\\=6+2+1\\=5+4\\=5+3+1\\=4+3+2\end{align}$$

We don't have to worry about 4 summands, since $1+2+3+4>9$. There is no known closed form for the general term.

Title: Find the coefficient of ${x^9}$ in the expansion of $ (1 + x)( 1 + x^2)( 1 + x^3)..(1 + x^{100})$

Body:
Find the coefficient of ${x^9}$ in the expansion of $\left( {1 + x} \right)\left( {1 + {x^2}} \right)\left( {1 + {x^3}} \right)..\left( {1 + {x^{100}}} \right)$.
The official answer is 8.

How do I find the general term,

Dividing the above equation by $(1-x)$ is not generating the required result.


问:
Range of a function
Find the range of

$$f(x)=\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}$$ where $a, b, c$ are distinct real numbers such $a\neq b\neq c\neq a$.

答:
$$f(x)=\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}$$

$$f(a)= \dfrac{(a-b)(a-c)}{(a-b)(a-c)}=1$$

$$f(b)= \frac{(b-c)(b-a)}{(b-c)(b-a)}=1$$

$$f(c)= \frac{(c-a)(c-b)}{(c-a)(c-b)}=1$$

Now what is $f'(x)?$, $f'(x)=\dfrac{2x-(a+b)}{(c-a)(c-b)}+\dfrac{2x-(b+c)}{(a-b)(a-c)}+\dfrac{2x-(c+a)}{(b-c)(b-a)}$. Now watch the house of cards fall. Then you have a constant function, and you are aware of that constant.

Title: Range of a function


问:
Can a sum of an elementary and non-elementary definite integral be elementary or evaluated without special functions?
Hi I'm asking this in context of this:

Suppose we have non-elementary integral $\chi(x) = \int K(x)dx$ and we want to find

$$\int^b_aK(x)dx$$

Now lets say there is some function $F(x)$  which has elementary integral $\int F(x)$

such that $K(x) + F(x) $ has elementary integral so we can do this :

$$\int^b_aK(x)dx = \int^b_a(K(x)+0)dx$$

$$ = \int^b_a(K(x)+ F(x) - F(x))dx$$

$$=\int^b_aK(x)+ F(x)dx + \int^b_a-F(x)dx$$

Maybe $\int^b_a K(x) + F(x)dx$ is not elementary but lets say I dilute this condition to there exist some definite integral property transformation or substitution which allows us to arrive at numerical value for e.g just like $\int \ln(\sin(x))dx$ which is non-elementary according to WolframAlpha but it gives into some computable value for appropriate limits like

$\int^{\pi/2}_0 \ln(\sin(x))dx$

So my end goal goal is converting $\int^b_a K(x) + F(x)dx$ into some function like that atleast.

答:
I'm not familiar with the deeper details of definite integration. I only want to show that one of the assumptions in your question is wrong and that there are at least some cases (maybe single cases) where a definite non-elementary integral can be evaluated without special functions.

1.) The set of all elementary functions is closed regarding addition (It is an algebraically closed field.). $\int K(x)+F(x)\ dx=\int K(x)\ dx+\int F(x)\ dx$ cannot be elementary therefore. Your prerequisite "such that $K(x)+F(x)$ has elementary integral" is wrong therefore.

2.) Definite integrals can be calculated also by other methods than indefinite integration.

Lichtblau, D.: Symbolic definite integration: methods and open issues. ACM Communications in Computer Algebra 45 (2011) (1/2) 1-16

Raab, C. G.: Definite Integration in Differential Fields. PhD thesis Johannes Kepler University Linz, Austria, 2012

Algorithms for symbolic definite integration?

Davenport, G.: An Exploration of Three Related Parametric Definite Integrals. 2016

Davenport, G.: The Difficulties of Definite Integration

A non-elementary definite integral and therefore a sum of an elementary and a non-elementary definite integral can therefore be evaluated in some cases without special functions.

Nyblom, M. A.: On the evaluation of a definite integral involving nested square root functions. Rocky Mountain J. Math. 37 (2007) 4 1301-1304

3.) One could ask: Is there a theory of integration in elementary terms for definite integrals? But it seems there is no such theory.

4.) The value of a definite integral is a number. One could ask if this number can be an elementary number. But that's a completely different mathematical problem.


问:
Calculating a limit to infinity on e

$$\lim_{x\to0^+}(1+3x)^\frac{-2}{3x^3}$$

Okay so I got to the point where I have
$$\lim _{t\to \infty }\left(\left(1+\frac{1}{t}\right)^{t}\right)^{-18t^{2}}$$

Now, my instructors don't allow us to use L'Hopital and we haven't proven anything about situations such that
$$\lim _{x\to \infty }\left(f\left(x\right)^{g\left(x\right)}\right)$$

so is my only option is to use squeeze rule? since I know the expression is basically $e^{-18t^{2}}$
and we know that $2 < e < 3$..
is there a simpler way to solve this? if so then how? if not then how do I prove $2^{-18t^{2}} < e^{-18t^{2}} < 3^{-18t^{2}}$?

答:
IDC :)

We did learn AOL of pow in lecture 14, so this is valied:

$\lim _{x\to \infty }\left(f\left(x\right)^{g\left(x\right)}\right)$

so why not just define $g(x) = {-18t^{2}}\space $ and use AOL to get $\frac{1}{e^\infty} = 0$ ?

Title: Calculating a limit to infinity on e


问:
K Theory: Book Recommendations
Good people!

So I've been hoping to get into K Theory for a while now, and the book that I have been trying to use (and failing) has been Charles Weibel's book by that very title.

The book itself isn't bad, but it's not for the faint-hearted, and can appear quite dense and challenging to a relative newbie like me. As such, I was wondering if someone might be able to recommend a book or two or three or $n$ to help "ease" my way into Weibel. Something that touches upon the constructions he introduces in the first chapter--Picard groups, topological and algebraic vector bundles, Chern classes, etc.--but being more, well, easy for a newbie to process.

I look forward to your suggestions.

答:
Personally, I like:
An Algebraic Introduction to K-Theory , Bruce A. Magurn

https://www.cambridge.org/an/academic/subjects/mathematics/algebra/algebraic-introduction-k-theory?format=HB&isbn=9780521800785

THis is suitable for beginners, with lot of interesting sources of $K_0$ and $K_1$, and classical theorems (such as excision exact sequences)

Title: K Theory: Book Recommendations


问:
Evaluate $\lim_{r \to \infty} \iint_{D_r} \frac{1}{1+x^2+y^2} dx dy$
Evaluate
$$\lim_{r \to \infty} \iint_{D_r} \frac{1}{1+x^2+y^2} dx dy$$
where $D_r=\{(x,y) \in \mathbb{R}^2 \ \text{s.t.} \ 0 \leq y \leq 1, \ x \geq 0, \ x^2+y^2 \leq r^2 \}$.

My try is the following: I use a change of variable in polar coordinates with pole in the origin, hence
$$D_r \mapsto E_r=\{(\rho, \theta) \in \mathbb{R}^2 \ \text{s.t.} \ 0 \leq \rho \sin \theta \leq 1, \rho \cos \theta \geq 0, \rho^2 \leq r^2, \rho \geq 0, 0 \leq \theta <2\pi\}$$
$$\iff \left\{(\rho, \theta) \in \mathbb{R}^2 \ \text{s.t.} \ 0 \leq \rho \leq \frac{1}{\sin \theta}, 0 \leq \theta \leq \frac{\pi}{2}, 0 \leq \rho \leq r\right\}$$
And
$$\lim_{r \to \infty} \iint_{D_r} \frac{1}{1+x^2+y^2} dx dy=\lim_{r \to \infty}\iint_{E_r} \frac{\rho}{1+\rho^2} d \rho d \theta$$
Since $\rho$ has two upper bounds I discuss $0 \leq \rho \leq \min\left(r, \frac{1}{\sin \theta}\right)$. Now, being interested in what happens when $r \to \infty$, I assume that $r>\frac{1}{\sin \theta}$ and so I get that $\min\left(r, \frac{1}{\sin \theta}\right)=\frac{1}{\sin \theta}$; hence
$$\lim_{r \to \infty}\iint_{E_r} \frac{\rho}{1+\rho^2} d \rho d \theta=\lim_{r \to \infty} \int_0^{\frac{\pi}{2}} \left(\int_0^{\min\left(r,\frac{1}{\sin\theta}\right)} \frac{\rho}{1+\rho^2}d\rho\right)d\theta$$
$$=\lim_{r \to \infty} \int_0^{\frac{\pi}{2}} \left(\int_0^{\frac{1}{\sin\theta}} \frac{\rho}{1+\rho^2}d\rho\right)d\theta$$
At this point both the set is independent of $r$ and so
$$\lim_{r \to \infty} \int_0^{\frac{\pi}{2}} \left(\int_0^{\frac{1}{\sin\theta}} \frac{\rho}{1+\rho^2}d\rho\right)d\theta=\int_0^{\frac{\pi}{2}} \left(\int_0^{\frac{1}{\sin\theta}} \frac{\rho}{1+\rho^2}d\rho\right)d\theta=\int_0^{\frac{\pi}{2}} \frac{1}{2}\log\left(1+\frac{1}{\sin^2 \theta}\right)d\theta$$
$$=\frac{\pi}{2} \log(1+\sqrt{2})$$
The result is correct, however I'm not sure when I declare that since $r \to \infty$ I can assume $r>\frac{1}{\sin \theta}$; the doubt comes from the fact that $\theta \in \left[0,\frac{\pi}{2}\right]$ and so when $\theta$ gets really close to $0$ the function $\frac{1}{\sin \theta}$ is unbounded. So I feel like there is some kind of "battle" between $r$ and $\frac{1}{\sin \theta}$ going to infinity that makes impossible to understand which one is actually bigger for discussing the minimum.

Am I missing something and there is no problem for $\frac{1}{\sin \theta}$ when $\theta \to 0^+$ (and if there isn't, why it isn't? Why can I ignore the fact that it tends to infinity for small values of $\theta$ in the domain of integration?) or I was just lucky and my approach was wrong because I can't make that assumption? Thanks.

Edit: Added a term $\frac{1}{2}$ that I forgot.

答:
This is another way, which avoids the problem you are pointing out. For $r>1$,

$$I_r:=\iint_{D_r} \frac{1}{1+x^2+y^2} dx dy=\int_{0}^1\left(\int_{0}^{\sqrt{r^2-y^2}}\frac{dx}{1+x^2+y^2}\right)\,dy
=\int_{0}^1\frac{\arctan\left(\frac{\sqrt{r^2-y^2}}{\sqrt{1+y^2}}\right)}{\sqrt{1+y^2}}\,dy.$$
Now for $y\in[0,1]$ and $r>1$, we have that
$$0\leq \frac{\pi}{2}-\arctan\left(\frac{\sqrt{r^2-y^2}}{\sqrt{1+y^2}}\right)=\arctan\left(\frac{\sqrt{1+y^2}}{\sqrt{r^2-y^2}}\right)\leq
\frac{\sqrt{2}}{\sqrt{r^2-1}}.$$
Hence, as $r\to \infty$,
$$0\leq \frac{\pi}{2}\int_0^1\frac{dy}{\sqrt{1+y^2}}-I_r\leq
\frac{\sqrt{2}}{\sqrt{r^2-1}}\int_0^1\frac{dy}{\sqrt{1+y^2}}\leq \frac{\sqrt{2}}{\sqrt{r^2-1}}\to 0,$$
and we may conclude that
$$\lim_{r\to\infty}I_r=\frac{\pi}{2}\int_0^1\frac{dy}{\sqrt{1+y^2}}=\frac{\pi}{2}\, \log(1+\sqrt{2}).$$


答:
What you have defined as "$\mathbb{Q}$" is actually just the set of positive rational numbers (assuming that $0\notin \mathbb{N}$).

Expanding on the comment by projectilemotion:

Your function "$f(a,b)=2^{a}\cdot 3^{b}$" is not a function on $\mathbb{Q}$ because, for instance $\frac{1}{2}$ and $\frac{2}{4}$ are the same rational number, but $f(1,2)=18$ while $f(2,4) = 324$.

So as noted in the comments by projectilemotion, your function is from $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$

Title: Countability of $\mathbb{Q}$, function notation

Body:
Proof

Let $\mathbb{Q} = \{\frac{a}{b}: a,b \in \mathbb{Z}, b \neq 0\}$

Define a function $f$ given by $f(a,b)=2^a3^b \in \mathbb{N}$

Then, by the fundamental theorem of arithmetic, each natural number can be rewritten as the unique product of primes. This implies that each element in the image is unique and a natural number. Thus, $\mathbb{N}$ is countable. $\square$

So here's my question. Is the proper function notation $f:\mathbb{Q} \to \mathbb{N}$ or is it $f:\mathbb{Z} \to \mathbb{N}$? Because, while each $a$ and $b$ are derived from $\frac{a}{b} \in \mathbb{Q},$ each $a$ and $b$ are integer values.

Ultimately, I'm just asking a question regarding proper notation. Pay no attention to the details of my proof. I'm just asking what the proper notation is.


问:
Why does knowing that the MLE of the mean = the arithmetic mean of the sample uniquely determine the Normal sampling distribution?
Currently reading through E.T Jayne on Bayesian inference and in one chapter dedicated to the Normal distribution he argues that Gauss proved the following:

Given a sample from a population distribution with unknown mean, our MLE for the mean is the arithmetic mean of our sample if and only if the sampling distribution of the mean is Normal

I’m really struggling to see the intuition behind this, can anyone help out? Thanks!

答:
You're referring to an argument Jaynes attributes to Gauss in Sec. 7.3. The log-likelihood of $n+1$ observations $\{x_i|0\le i\le n\}$ when the estimator is unbiased is of the form $\sum_i\ln f(x_i|\theta)$, which we want to maximise when $\theta=\bar{x}$.

Jaynes actually adds an assumption, which as @paulinho notes is inapplicable for the Bernoulli distribution, namely that the likelihood depends only on the $\theta-x_i$. If this seems like sleight of hand, bear in mind the point of his Chapter 7 is only to motivate why sampling errors would be Gaussian. (In particular, sampling means being approximately Gaussian for large samples is a completely unrelated issue!) It is not an "everything is Gaussian" argument. (In fact, Sec. 7.12 explores why sometimes even errors would be non-Gaussian.) The assumption we're adding is reasonable for sampling errors.

Compare $\sum_i(\bar{x}-x_i)=0$ to $\left.\sum_i\frac1f\frac{\partial f}{\partial\theta}\right|_{\theta=\bar{x}}=0$. For $\frac1f\frac{\partial f}{\partial\theta}$ to be proportional to $\theta-x$ (which is equivalent to $\ln f$ being quadratic in $\theta-x$ and peaked at $x=\theta$, making $\theta$ the mode of a Gaussian distribution) is clearly sufficient, and the case $x_i=-\tfrac1nx_n$ for all $i<n$, when $n\ge1$, proves necessity.

As for an intuition rather than a proof, note that if $\frac1f\frac{\partial f}{\partial\theta}$ is nonlinear you should be able to tweak initially constant $x_i$, without changing $\bar{x}$, so that $\frac1f\frac{\partial f}{\partial\theta}$ changes in value. The above example shows you can; see Jaynes for the full proof.


问:
Prove $f:P(\mathbb{N})^2\rightarrow P(\mathbb{N})^2$ , is defined by $f(\langle A,B\rangle) = \langle A,A \Delta B \rangle$ is an bijection.
Prove $f:P(\mathbb{N})^2\rightarrow P(\mathbb{N})^2$ , is defined by $f(\langle A,B\rangle) = \langle A,A \Delta B \rangle$ is an bijection.

($A\Delta B $ is a Symmetric difference of A, B)

Attempt:

I tried to find the inverse function.

I tried to use a Symmetric difference in the outcome as the inverse function.

$(A\Delta B) \Delta A$

I'm having trouble finding a bijection in groups. I am used to finding bijections in regular functions.

答:
To give some intuition: we have two sets, and the function $f$ returns the first set $A$, followed by the set whose elements are the elements that are exclusively in one of the sets ($A \Delta B$). Clearly, we can reconstruct the first set $A$, but can we reconstruct the second set $B$?

The answer is yes! We can notice that the intersection of $A$ and $A \Delta B$ (call it $I$) is exactly the elements which are not in $B$, so if we remove these elements from $A$, then we will be left with exactly $A \cap B$. Furthermore, if we remove $I$ from $A \Delta B$, we recover the elements which are only in $B$, but not in $A$ (why?). So we have both the set of elements that are in $B$ and not in $A$, as well as the set of elements in $B$ and in $A$. Clearly, if we take the union we will recover $B$.

Putting our procedure into symbols, the inverse function will be
$$f^{-1}(\langle A, B\rangle) = \langle A, (A \setminus (A \cap B)) \cup (B \setminus (A \cap B)) \rangle$$
(Here $\setminus$ represents the set difference.) But as lisyarus suggests, this expression can be written more compactly: can you see what it is? Hint: it should look familiar...

Remark: If you are at all interested in boolean logic and bits, this statement is the equivalent of saying for two bits $a$ and $b$, the function $f$ which takes $(a, b)$ to $(a, a \oplus b)$ is invertible (and is its own inverse). Here $\oplus$ represents the XOR operation between the bits.


问:
Find the closed-form solution to a integral with the floor function
I have the following integral:

$$\int_0^1k^{\left\lfloor\frac{1}{x}\right\rfloor}dx$$

My question is, is there a nice closed form of this integral?

I have not any idea where to start.

Maybe I could use the substitution $u=\frac{1}{x}$, so I get:

$$\int_0^1k^{\left\lfloor\frac{1}{x}\right\rfloor}dx=-\int_\infty^1\frac{k^{\left\lfloor u\right\rfloor}}{u^2}du$$

答:
Assignment:

Find a closed-form solution to the following integral:

$$\mathscr{S}_\epsilon:=\int_0^1\epsilon^{\left\lfloor\frac{1}{x}\right\rfloor}\space\text{d}x\tag1$$

Where $\left\lfloor x\right\rfloor$ is the Floor function.

Well, first let $\text{s}:=\frac{1}{x}$ this implies that:

$$\mathscr{S}_\epsilon=-\int_\infty^1\frac{\epsilon^{\left\lfloor\text{s}\right\rfloor}}{\text{s}^2}\space\text{ds}=\int_1^\infty\frac{\epsilon^{\left\lfloor\text{s}\right\rfloor}}{\text{s}^2}\space\text{ds}\tag2$$

It is now not hard to see (because $\lfloor\text{s}\rfloor$ is constant on all the sub-intervals that we are integrating over) that we can write:

$$\mathscr{S}_\epsilon=\sum_{\text{n}\ge1}\int_\text{n}^{\text{n}+1}\frac{\epsilon^{\left\lfloor\text{s}\right\rfloor}}{\text{s}^2}\space\text{ds}=\sum_{\text{n}\ge1}\int_\text{n}^{\text{n}+1}\frac{\epsilon^\text{n}}{\text{s}^2}\space\text{ds}=\sum_{\text{n}\ge1}\epsilon^\text{n}\int_\text{n}^{\text{n}+1}\frac{1}{\text{s}^2}\space\text{ds}=$$
$$\sum_{\text{n}\ge1}\epsilon^\text{n}\cdot\left[-\frac{1}{\text{s}}\right]_\text{n}^{\text{n}+1}=\sum_{\text{n}\ge1}\epsilon^\text{n}\cdot\left[\frac{1}{\text{s}}\right]_{\text{n}+1}^\text{n}=\sum_{\text{n}\ge1}\epsilon^\text{n}\cdot\left(\frac{1}{\text{n}}-\frac{1}{\text{n}+1}\right)\tag3$$

Because if $\text{n}\le\text{s}<\text{n}+1$ we have $\lfloor\text{s}\rfloor=\text{n}$.

Now, I let you prove that when $\left|\epsilon\right|\le1$ the equality following holds:

$$\sum_{\text{n}\ge1}\epsilon^\text{n}\cdot\left(\frac{1}{\text{n}}-\frac{1}{\text{n}+1}\right)=\frac{\epsilon+\ln\left(1-\epsilon\right)-\epsilon\ln\left(1-\epsilon\right)}{\epsilon}\tag4$$

And you can do that using Taylor- or/and Maclaurin series.


问:
$f(x)=(x-2)(x-7)(x-9)(x-12)$ how many roots is there to $f'(x)=0$ and $f''(x)=0$? without taking derivative.

if $f(x)=(x-2)(x-7)(x-9)(x-12)$ how many roots is there to $f'(x)=0$ and $f''(x)=0$? without taking derivative.

What I know from Rolle's theorem that between every 2 roots of the function theres atleast 1 root of the derivative, so for example $f'(x)=0$ has atleast 3 roots.
 I'm struggling to get how im supposed to know its Exactly 3 roots and not atleast without differentiating since that's what the question is asking for and I feel like im missing something.
 any help is really appreciated.

答: $f(2)= f(7)=f(9)=f(12)=0$. By mean value theorem there exists  atleast one $c$ in $(2,7)$ ,one $c'$ in  $(7,9) $and one $ c''$ in $(9,12)$ such that $ f'(c)=f'(c')=f(c'')=0$. Also degree of $f'(x) $ is $3$ which implies $f'(x)$ has atmost  3 real roots. Therefore number of real roots of $ f'(x)$ is $3$. Let the roots of $f'(x) $ be $a,b,d$. As the roots of $f'(x)$  lie in distinct intervals( i.e $(2,7), (7,9), (9,12)$) so they are distinct. As $f'(a)= f'(b)= f'(d)=0$, again using mean value theorem and proceeding similarly we get $f''(x) $ has 2 real roots.


问:
Calculating the residues
What are the residues at $s=0,\pm 1$ for the following function
$$I=\frac{1}{e^{2\pi is}-1}\frac{e^{isu}}{s^2(s^2-1)}?$$

Now it appears to me that, two pieces, $e^{2\pi is}$ and $s^2(s^2-1)$, diverge in the denominator. What are the residues then?

答:
Since you asked only for the residues and not how to get them:
$$\text{res}\left(I,0\right)=-\frac{i \left(3 u^2-6 \pi  u+2 \pi ^2-6\right)}{12 \pi }$$
$$\text{res}\left(I,1\right)=\frac{e^{i u} (2 u+5 i-2 \pi )}{8 \pi }$$
$$\text{res}\left(I,-1\right)=\frac{e^{-i u} (-2 u+5 i+2 \pi )}{8 \pi }$$

To get the first residue, we note that $s=0$ is a $4$th order pole of the function $I(s)$, then we must compute
$$\text{res} (I,0)={\frac {1}{3!}\lim _{s\to 0}{\frac {d^3}{ds^3}}}\left((s-0)^{4}I(s)\right)$$
which is quite "time consuming" :) (I did it with Mathematica)
$$\frac{1}{3!}\underset{s\to 0}{\text{lim}}\left(\frac{\left(-\frac{i e^{i s u} u^3}{-1+e^{2 i \pi  s}}+\frac{6 e^{i u s+2 i \pi  s} i \pi  u^2}{\left(-1+e^{2 i \pi  s}\right)^2}+3 e^{i s u} \left(\frac{4 e^{2 i \pi  s} \pi ^2}{\left(-1+e^{2 i \pi  s}\right)^2}-\frac{8 e^{4 i \pi  s} \pi ^2}{\left(-1+e^{2 i \pi  s}\right)^3}\right) i u+e^{i s u} \left(\frac{48 i e^{6 i \pi  s} \pi ^3}{\left(-1+e^{2 i \pi  s}\right)^4}-\frac{48 i e^{4 i \pi  s} \pi ^3}{\left(-1+e^{2 i \pi  s}\right)^3}+\frac{8 i e^{2 i \pi  s} \pi ^3}{\left(-1+e^{2 i \pi  s}\right)^2}\right)\right) s^2}{s^2-1}+3 \left(-\frac{e^{i s u} u^2}{-1+e^{2 i \pi  s}}+\frac{4 e^{i u s+2 i \pi  s} \pi  u}{\left(-1+e^{2 i \pi  s}\right)^2}+e^{i s u} \left(\frac{4 e^{2 i \pi  s} \pi ^2}{\left(-1+e^{2 i \pi  s}\right)^2}-\frac{8 e^{4 i \pi  s} \pi ^2}{\left(-1+e^{2 i \pi  s}\right)^3}\right)\right) \left(\frac{2 s}{s^2-1}-\frac{2 s^3}{\left(s^2-1\right)^2}\right)+\frac{e^{i s u} \left(\left(\frac{24 s}{\left(s^2-1\right)^3}-\frac{48 s^3}{\left(s^2-1\right)^4}\right) s^2+6 \left(\frac{8 s^2}{\left(s^2-1\right)^3}-\frac{2}{\left(s^2-1\right)^2}\right) s-\frac{12 s}{\left(s^2-1\right)^2}\right)}{-1+e^{2 i \pi  s}}+3 \left(-\frac{2 i e^{i u s+2 i \pi  s} \pi }{\left(-1+e^{2 i \pi  s}\right)^2}+\frac{i e^{i s u} u}{-1+e^{2 i \pi  s}}\right) \left(\left(\frac{8 s^2}{\left(s^2-1\right)^3}-\frac{2}{\left(s^2-1\right)^2}\right) s^2-\frac{8 s^2}{\left(s^2-1\right)^2}+\frac{2}{s^2-1}\right)\right)=\frac{i \left(-3 u^2+6 \pi  u-2 \pi ^2+6\right)}{12 \pi }$$

$s=1;\;s=-1$ are a double poles
$$\text{res}\left(I,1\right)=\underset{s\to 1}{\text{lim}}\frac{dI}{ds}(s-1)^2$$
and
$$\text{res}\left(I,-1\right)=\underset{s\to 1}{\text{lim}}\frac{dI}{ds}(s+1)^2$$
It is useful L'Hopital rule in the last two limits

Title: Calculating the residues


答:
Suppose $ \left|k\right|<1 $, then substituting $ \left\lbrace\begin{matrix}y=\frac{1}{x}\ \ \ \\ \mathrm{d}x =-\frac{\mathrm{d}y}{y^{2}}\end{matrix}\right. $, we get :

\begin{aligned} \int_{0}^{1}{k^{\left\lfloor\frac{1}{x}\right\rfloor}\,\mathrm{d}x}&=\int_{1}^{+\infty}{\frac{k^{\left\lfloor y\right\rfloor}}{y^{2}}\,\mathrm{d}y}\\ &=\sum_{n=1}^{+\infty}{\int_{n}^{n+1}{\frac{k^{\left\lfloor x\right\rfloor}}{x^{2}}\,\mathrm{d}x}}\\ &=\sum_{n=1}^{+\infty}{\int_{n}^{n+1}{\frac{k^{n}}{x^{2}}\,\mathrm{d}x}}\\ &=\sum_{n=1}^{+\infty}{k^{n}\left(\frac{1}{n}-\frac{1}{n+1}\right)}\\ &=\sum_{n=1}^{+\infty}{\left(\frac{k^{n-1}}{n}-\frac{k^{n}}{n+1}\right)}+\left(1-\frac{1}{k}\right)\sum_{n=1}^{+\infty}{\frac{k^{n}}{n}}\\ \int_{0}^{1}{k^{\left\lfloor\frac{1}{x}\right\rfloor}\,\mathrm{d}x}&=1-\left(1-\frac{1}{k}\right)\ln{\left(1-k\right)} \end{aligned}

We were able to split the sum in the fifth line, because we know that $ \sum\limits_{n\geq 1}{\left(\frac{k^{n-1}}{n}-\frac{k^{n}}{n+1}\right)} $ converges (telescopic series), and that $ \sum\limits_{n\geq 1}{\frac{k^{n}}{n}} $ does also converge (taylor series of the function $ f:\left(-1,1\right)\rightarrow\mathbb{R},\ x\mapsto -\ln{\left(1-x\right)} $ at $ x=k $).

Title: Find the closed-form solution to a integral with the floor function

Body:
I have the following integral:

$$\int_0^1k^{\left\lfloor\frac{1}{x}\right\rfloor}dx$$

My question is, is there a nice closed form of this integral?

I have not any idea where to start.

Maybe I could use the substitution $u=\frac{1}{x}$, so I get:

$$\int_0^1k^{\left\lfloor\frac{1}{x}\right\rfloor}dx=-\int_\infty^1\frac{k^{\left\lfloor u\right\rfloor}}{u^2}du$$


问:
How can I describe a triangle or tetrahedron with barycentric coordinates?
How would I, for example, describe a triangle given by the vertices $(2,1),(4,2),(2.5,2)$ with barycentric coordinates?

My guess would be $(a:b:1-a-b)$ where $0\leq a,b\leq 0.75$ and $a+b\leq 0.75$ since $A_\Delta=0.75$.

Is that true?

答:
Here is how I would describe it -

If we consider masses $(\alpha, \beta, \gamma)$ placed at the vertices of the $\triangle ABC, A(2,1), B(4,2), C(2.5,2) \,$ respectively then all points in the triangle can be described as -

$(2 \alpha + 4 \beta + 2.5 \gamma, \alpha + 2 \beta + 2 \gamma)$

$0 \leq \alpha, \beta, \gamma \leq 1, \alpha + \beta + \gamma = 1$

or substituting $(\gamma = 1 - \alpha - \beta)$, we can represent all points in the triangle as $(2.5 - 0.5 \alpha + 1.5 \beta, - \alpha +2 ), 0 \leq \alpha, \beta \leq 1, \alpha + \beta \leq 1$.

Vice versa, we can find barycentric coordinates $(\alpha, \beta, 1 - \alpha - \beta)$ of any given point in the triangle using the above.

As you mentioned in your working, for any point $P$, the ratio of area can also be used to find its barycentric coordinates.

$\alpha = \frac{\triangle PBC}{\triangle ABC}, \beta = \frac{\triangle PCA}{\triangle ABC}, \gamma = \frac{\triangle PAB}{\triangle ABC} \,$ or $\, (1 - \alpha - \beta)$


问:
Integral identity with Markov kernel
Let $K:E\times\mathcal{G}\rightarrow\left[0,+\infty\right]$ be a positive Markov kernel. The $n$-th power of the $K$ is defined iteratively as
$$K^0(x,A)=\delta_{x}(A)$$
where $\delta_x(A)$ is the Dirac measure and
$$
K^{n} = K^{n-1}\circ K
$$
where $\circ$ denotes the kernel composition
$$
(K^{n-1}\circ K)(x,A) = \int_{E}K^{n-1}(x,dx^{\prime})\,K(x^{\prime},A)
$$
In a book by Revuz it is said that the following chain of identity holds
\begin{eqnarray}
K^{n}(x,f) &=& \int_{E}K^n(x,dx^{\prime})f(x^{\prime}) \\
&=& K(x,K^{n-1} f) \\
&=& \int_{E}K(x,dx^{\prime})K^{n-1}f(x^{\prime})\\
&=& \int_{E}K(x,dx^{\prime})\int_{E}K^{n-1}(x^{\prime},dx^{\prime\prime})f(x^{\prime\prime}) \\
&=&K^{n-1}(x,Kf) \\
&=&\int_{E}K^{n-1}(x,dx^{\prime})Kf(x^{\prime}) \\
&=& \int_{E}K^{n-1}(x,dx^{\prime})\int_E K(x^{\prime},dx^{\prime\prime})f(x^{\prime\prime})
\end{eqnarray}
From the first to the fourth equation there are no problems, just definitional equations, but how from the fourth are we going to the fifth? I suspect that behind there is an application of the Fubini theorem that I cannot see.

答:
It's not Fubini's Theorem: the key is "$K$ commutes with $K^{n - 1}$".  (Basically, this is analogous to the identity $a \cdot a^{n - 1} = a^{n - 1} \cdot a$.)

Claim: for all $n \in \mathbb{N} \cup \{0\}$ and all test functions $f$,
\begin{align*}
K^{n}(x,Kf) = K(x,K^{n}f) = K^{n + 1}(x,f).
\end{align*}

Proof by induction: when $n = 0$, this follows from the fact that $K^{0}(x,g) = g(x)$.

Assume the identity holds when $n = m$.  In the case $n = m + 1$, we write
\begin{align*}
K^{m + 1}(x,Kf) &= \int_{E} K(x,dx') K^{m}(x',Kf)
\end{align*}
Here we used the definition of $K^{m + 1}$ in terms of $K$ and $K^{m}$, nothing more.  Now we are in a position to use the inductive hypothesis:
\begin{align*}
K^{m + 1}(x,Kf) &= \int_{E} K(x,dx') K^{m}(x',Kf) \\
&= \int_{E} K(x,dx') K^{m + 1}(x',f) \\
&= K(x,K^{m+1}f).
\end{align*}
Finally, observe that, by definition,
\begin{align*}
\int_{E} K(x,dx') K^{m + 1}(x',f) = K^{m + 2}(x,f).
\end{align*}
Putting it all together, we get $K^{m+1}(x,Kf) = K(x,K^{m+1}f) = K^{(m+1)+1}(x,f)$.

Finally, let me say this is an argument against this kernel notation in favor of transition matrices.  Denote by $\mathcal{F}$ the space of test functions and let us define $P : \mathcal{F} \to \mathcal{F}$ as follows:
\begin{equation*}
(Pf)(x) = K(x,f).
\end{equation*}
Notice this is a linear operator --- a matrix, if you will.  (In the context of Markov chains, $P$ is called the transition matrix.)  With this notation, the above identities reduce to the familiar one: $P \circ P^{n} = P^{n} \circ P = P^{n + 1}$, where $\{P^{n}\}$ are the "matrix powers" defined through composition of operators.

Title: Integral identity with Markov kernel