问:
Solution of a first order PDE
I am trying to understand/develop a numerical code for solving PDEs. I started with a simple first order PDE,
$$
\frac{\partial F(r,\theta)}{\partial r}+\frac{\partial F(r,\theta)}{\partial \theta} = 0
$$
Under the boundary condition that at maximum r $F(r_{max},\theta)=55$ (in my code $r_{max}=3$). Solving the PDF using backward finite difference method I came up with the following solution,
$$
F(r_{i-1},\theta_i) = \frac{(h_r+h_\theta) F(r_{i},\theta_i) - h_r F(r_{i},\theta_{i-1})}{h_\theta}
$$
Here, $h_r$ and $h_\theta$ are the step size for $r$ and $\theta$ respectively. I wrote the following Matlab code for for this solution,

```<code>hr = 0.1;
r = 0:hr:3;
hT = 0.1;
T = 0:hT:2*pi;
F = zeros(length(T), length(r));
F(:,length(r))=10;
for i=length(r):-1:2
    for j=length(T):-1:2
       F(j,i-1) = (((hr+hT)*F(j,i)) - (hr*F(j-1,i)))/hT;
    end
end
surf(F)
</code>```

Based on this I am getting the following surface as solution,

So here are my questions:
1) My solution based on the finite difference shows that having a spherically symmetric boundary condition (no $\theta$ dependence) means that the whole solution is spherically symmetric(?). Is it reflected in the surface plot? I am not sure. May be it is the cartesian coordinate frame that is restricting my imagination to see that this solution actually is spherically symmetric.

2) The solution has weird behavior. For example, if I change the step size in $r$ to a different value, like $h_r = 0.01$ then I get the following surface.

Notice the the scale of $F(r,\theta)$ has shifted by orders of magnitude. This is very odd in my opinion.

3) If I change the resolution of $\theta$ I get the following solution curve,
. Which is odd still because for some reason the structure has shifted and also the scale has changed.

So I am not able to reconcile all these behaviors and explain if this is expected. Unless the solution is highly non-linear so when I change the step size the scale changes (because more non-linear points are included which have extremely high value of the function) but that should be the case for such a simple differential equation, unless of course the PDE admits a solution which actually is this non-linear for the given boundary condition.

Adding a few more cases for reference,
$h_r = 0.1$ and $h_\theta=0.5$

$h_r = 0.5$ and $h_\theta=0.1$

I do not understand how is it possible that by changing the resolution of $r$ has effect on the theta resolution. Or may be
I am looking at it all wrong.

答:
My suggestion is to use a predictor-corrector approach to solve this PDE via the finite difference method. The difficulty in the equation is the lack of symmetry via the finite difference method. The direction in which information is communicated between points is biased by your choice of finite difference scheme: forward or backward. The idea behind a predictor-corrector solver is that you make one iteration with forward difference and one iteration with backward difference. In this particular problem, I believe you need to average two predictor-corrector steps as follows.

Predictor      Corrector   -->   Result
Forward        Backward          Average1
Backward       Forward     -->   Average2
Result  P(next)=0.5*(Average1+Average2)

Here is some example code to get an idea for the equation: dP/dx + dP/dy = C(x,y).

I also suggest you look at the explicit and implicit MacCormack Schemes in  the solution of the Burger's equation. You accidently bumped into the reason why computational fluid dynamics is so difficult. First order derivatives in PDE's are not symmetric, unlike second order derivatives. A predictor-corrector solver is an attempt to add some symmetry into the first order derivatives.

```<code>clear all; clc; close all;

%This script attempts to solve the linear wave Equation with a predictor-corrector solver.

Lx=1;
Ly=1;

%The domain will be M x N
M=100;%points in y-direction
N=100;%points in x-direction

xvec=linspace(0,Lx,N);
yvec=linspace(0,Ly,M);

dx=xvec(2)-xvec(1);
dy=yvec(2)-yvec(1);

[X,Y] = meshgrid(xvec,yvec);

C=@(x,y) -0.5*sqrt((x-0.5).^2+(y-0.5).^2)/sqrt(2);
Cij=C(X,Y);

P=zeros(M,N);

%Apply BC's
P(:,1)=1;%Left Side
P(M,:)=1;%Top Side
P(1,:)=P(2,:);%Bottom Side
P(:,N)=P(:,N-1);%Right Side

interior_x=2:N-1;
interior_y=2:M-1;
%%
const=1/(1/dx+1/dy);
alpha = 1;
Pij1=P;
Pij2=P;
Pij3=P;
Pij4=P;
for nn=1:1:50000
    %Predictor
    Pij1(interior_y,interior_x) = const*(P(interior_y,interior_x+1)/dx + P(interior_y+1,interior_x)/dy - Cij(interior_y,interior_x));
    Pij3(interior_y,interior_x) = const*(P(interior_y,interior_x-1)/dx + P(interior_y-1,interior_x)/dy + Cij(interior_y,interior_x));
    Pij1(1,:)=Pij1(2,:);%Bottom Side
    Pij1(:,N)=Pij1(:,N-1);%Right Side

    Pij3(1,:)=Pij3(2,:);%Bottom Side
    Pij3(:,N)=Pij3(:,N-1);%Right Side
    %Corrector
    Pij2(interior_y,interior_x) = const*(Pij1(interior_y,interior_x-1)/dx + Pij1(interior_y-1,interior_x)/dy + Cij(interior_y,interior_x));
    Pij4(interior_y,interior_x) = const*(Pij3(interior_y,interior_x+1)/dx + Pij3(interior_y+1,interior_x)/dy - Cij(interior_y,interior_x));

    P_next = 0.25*(Pij1 + Pij2 + Pij3 + Pij4);
    P= (1-alpha)*P + alpha*P_next;

    P(1,:)=P(2,:);%Bottom Side
    P(:,N)=P(:,N-1);%Right Side
end
%%
contourf(X,Y,P)
colorbar

%Numerical Derivatives dq/dx
%Second Order
% dq_dx_grid=@(h,q_grid) 1/(2*h)*[-3*q_grid(:,1)+4*q_grid(:,2)-q_grid(:,3),...
%                       q_grid(:,3:end)-q_grid(:,(1:end-2)),...
%                       q_grid(:,end-2)-4*q_grid(:,end-1)+3*q_grid(:,end)];

%Fourth Order
dq_dx_grid=@(h,q_grid) 1/h*[ -25/12*q_grid(:,1) +            4*q_grid(:,2)         - 3*q_grid(:,3)     + 4/3*q_grid(:,4)       - 1/4*q_grid(:,5),...
                                    -q_grid(:,1)/4        - 5/6*q_grid(:,2)       + 3/2*q_grid(:,3)     - 1/2*q_grid(:,4)       +     q_grid(:,5)/12,...
                                     q_grid(:,1:end-4)/12 - 2/3*q_grid(:,2:end-3)                       + 2/3*q_grid(:,4:end-1)     - q_grid(:,5:end)/12,...
                               -1/12*q_grid(:,end-4)      + 1/2*q_grid(:,end-3)   - 3/2*q_grid(:,end-2) + 5/6*q_grid(:,end-1)   + 1/4*q_grid(:,end),...
                                 1/4*q_grid(:,end-4)      - 4/3*q_grid(:,end-3)   +   3*q_grid(:,end-2) -   4*q_grid(:,end-1) + 25/12*q_grid(:,end)];

%Numerical Derivatives dq/dy
% Second Order
% dq_dy_grid=@(h,q_grid) 1/(2*h)*[-3*q_grid(1,:)+4*q_grid(2,:)-q_grid(3,:);...
%                       q_grid(3:end,:)-q_grid((1:end-2),:);...
%                       q_grid(end-2,:)-4*q_grid(end-1,:)+3*q_grid(end,:)];

%Fourth Order
dq_dy_grid=@(h,q_grid) 1/h*[ -25/12*q_grid(1,:) +            4*q_grid(2,:)         - 3*q_grid(3,:)     + 4/3*q_grid(4,:)       - 1/4*q_grid(5,:);...
                                    -q_grid(1,:)/4        - 5/6*q_grid(2,:)       + 3/2*q_grid(3,:)     - 1/2*q_grid(4,:)       +     q_grid(5,:)/12;...
                                     q_grid(1:end-4,:)/12 - 2/3*q_grid(2:end-3,:)                       + 2/3*q_grid(4:end-1,:)     - q_grid(5:end,:)/12;...
                               -1/12*q_grid(end-4,:)      + 1/2*q_grid(end-3,:)   - 3/2*q_grid(end-2,:) + 5/6*q_grid(end-1,:)   + 1/4*q_grid(end,:);...
                                 1/4*q_grid(end-4,:)      - 4/3*q_grid(end-3,:)   +   3*q_grid(end-2,:) -   4*q_grid(end-1,:) + 25/12*q_grid(end,:);];
%%
(sum(sum( (dq_dy_grid(dy,P) + dq_dx_grid(dx,P) - Cij) )))/(N*M)
</code>```

Title: Solution of a first order PDE

Body:
My suggestion is to use a predictor-corrector approach to solve this PDE via the finite difference method. The difficulty in the equation is the lack of symmetry via the finite difference method. The direction in which information is communicated between points is biased by your choice of finite difference scheme: forward or backward. The idea behind a predictor-corrector solver is that you make one iteration with forward difference and one iteration with backward difference. In this particular problem, I believe you need to average two predictor-corrector steps as follows.

Predictor      Corrector   -->   Result
Forward        Backward          Average1
Backward       Forward     -->   Average2
Result  P(next)=0.5*(Average1+Average2)

Here is some example code to get an idea for the equation: dP/dx + dP/dy = C(x,y).

I also suggest you look at the explicit and implicit MacCormack Schemes in  the solution of the Burger's equation. You accidently bumped into the reason why computational fluid dynamics is so difficult. First order derivatives in PDE's are not symmetric, unlike second order derivatives. A predictor-corrector solver is an attempt to add some symmetry into the first order derivatives.

clear all; clc; close all;

%This script attempts to solve the linear wave Equation with a predictor-corrector solver.

Lx=1;
Ly=1;

%The domain will be M x N
M=100;%points in y-direction
N=100;%points in x-direction

xvec=linspace(0,Lx,N);
yvec=linspace(0,Ly,M);

dx=xvec(2)-xvec(1);
dy=yvec(2)-yvec(1);

[X,Y] = meshgrid(xvec,yvec);

C=@(x,y) -0.5*sqrt((x-0.5).^2+(y-0.5).^2)/sqrt(2);
Cij=C(X,Y);

P=zeros(M,N);

%Apply BC's
P(:,1)=1;%Left Side
P(M,:)=1;%Top Side
P(1,:)=P(2,:);%Bottom Side
P(:,N)=P(:,N-1);%Right Side

interior_x=2:N-1;
interior_y=2:M-1;
%%
const=1/(1/dx+1/dy);
alpha = 1;
Pij1=P;
Pij2=P;
Pij3=P;
Pij4=P;
for nn=1:1:50000
    %Predictor
    Pij1(interior_y,interior_x) = const*(P(interior_y,interior_x+1)/dx + P(interior_y+1,interior_x)/dy - Cij(interior_y,interior_x));
    Pij3(interior_y,interior_x) = const*(P(interior_y,interior_x-1)/dx + P(interior_y-1,interior_x)/dy + Cij(interior_y,interior_x));
    Pij1(1,:)=Pij1(2,:);%Bottom Side
    Pij1(:,N)=Pij1(:,N-1);%Right Side

    Pij3(1,:)=Pij3(2,:);%Bottom Side
    Pij3(:,N)=Pij3(:,N-1);%Right Side
    %Corrector
    Pij2(interior_y,interior_x) = const*(Pij1(interior_y,interior_x-1)/dx + Pij1(interior_y-1,interior_x)/dy + Cij(interior_y,interior_x));
    Pij4(interior_y,interior_x) = const*(Pij3(interior_y,interior_x+1)/dx + Pij3(interior_y+1,interior_x)/dy - Cij(interior_y,interior_x));

    P_next = 0.25*(Pij1 + Pij2 + Pij3 + Pij4);
    P= (1-alpha)*P + alpha*P_next;

    P(1,:)=P(2,:);%Bottom Side
    P(:,N)=P(:,N-1);%Right Side
end
%%
contourf(X,Y,P)
colorbar

%Numerical Derivatives dq/dx
%Second Order
% dq_dx_grid=@(h,q_grid) 1/(2*h)*[-3*q_grid(:,1)+4*q_grid(:,2)-q_grid(:,3),...
%                       q_grid(:,3:end)-q_grid(:,(1:end-2)),...
%                       q_grid(:,end-2)-4*q_grid(:,end-1)+3*q_grid(:,end)];

%Fourth Order
dq_dx_grid=@(h,q_grid) 1/h*[ -25/12*q_grid(:,1) +            4*q_grid(:,2)         - 3*q_grid(:,3)     + 4/3*q_grid(:,4)       - 1/4*q_grid(:,5),...
                                    -q_grid(:,1)/4        - 5/6*q_grid(:,2)       + 3/2*q_grid(:,3)     - 1/2*q_grid(:,4)       +     q_grid(:,5)/12,...
                                     q_grid(:,1:end-4)/12 - 2/3*q_grid(:,2:end-3)                       + 2/3*q_grid(:,4:end-1)     - q_grid(:,5:end)/12,...
                               -1/12*q_grid(:,end-4)      + 1/2*q_grid(:,end-3)   - 3/2*q_grid(:,end-2) + 5/6*q_grid(:,end-1)   + 1/4*q_grid(:,end),...
                                 1/4*q_grid(:,end-4)      - 4/3*q_grid(:,end-3)   +   3*q_grid(:,end-2) -   4*q_grid(:,end-1) + 25/12*q_grid(:,end)];

%Numerical Derivatives dq/dy
% Second Order
% dq_dy_grid=@(h,q_grid) 1/(2*h)*[-3*q_grid(1,:)+4*q_grid(2,:)-q_grid(3,:);...
%                       q_grid(3:end,:)-q_grid((1:end-2),:);...
%                       q_grid(end-2,:)-4*q_grid(end-1,:)+3*q_grid(end,:)];

%Fourth Order
dq_dy_grid=@(h,q_grid) 1/h*[ -25/12*q_grid(1,:) +            4*q_grid(2,:)         - 3*q_grid(3,:)     + 4/3*q_grid(4,:)       - 1/4*q_grid(5,:);...
                                    -q_grid(1,:)/4        - 5/6*q_grid(2,:)       + 3/2*q_grid(3,:)     - 1/2*q_grid(4,:)       +     q_grid(5,:)/12;...
                                     q_grid(1:end-4,:)/12 - 2/3*q_grid(2:end-3,:)                       + 2/3*q_grid(4:end-1,:)     - q_grid(5:end,:)/12;...
                               -1/12*q_grid(end-4,:)      + 1/2*q_grid(end-3,:)   - 3/2*q_grid(end-2,:) + 5/6*q_grid(end-1,:)   + 1/4*q_grid(end,:);...
                                 1/4*q_grid(end-4,:)      - 4/3*q_grid(end-3,:)   +   3*q_grid(end-2,:) -   4*q_grid(end-1,:) + 25/12*q_grid(end,:);];
%%
(sum(sum( (dq_dy_grid(dy,P) + dq_dx_grid(dx,P) - Cij) )))/(N*M)


问:
Solving a limit sum involving Cantor sets
Let $P:C\to \mathbb{R}$, where $C$ is the Cantor set and $P$ is continuous. If $C_n$ is the sequence of numerators from the values of endpoints from the defined intervals of iteration $k$

$0, 1, 2, 3, 6, 7, 8, 9, 18, 19, 20, 21, 24, 25, 26, 27, 54, 55, 56, 57, 60, 61, 62, 63, 72, 73, 74, 75, 78, 79, 80, 81, 162, 163, 164, 165, 168, 169, 170, 171, 180, 181, 182, 183, 186, 187, 188, 189, 216, 217, 218, 219, 222, 223, 224, 225, 234, 235, 236, 237, 240, 241, 242, 243,...$

Where $C_1=0$, find the exact value of

$$\lim_{k\to\infty}\frac{\sum\limits_{n=1}^{2^{k+1}} P\left({C_n}/{3^k}\right)}{2^{k+1}}$$

For $P=x^2$ (According to Mathematica, the closed-form is $3/8$).

For $P={1}/{\left[2(x+2)\right]}$. $\left(\text{According to mathematica, it seems the closed form is} \left(\frac{1}{e}\right)^{-1-2e}\cot(e\pi)\cos(e\pi)\ \right)$

For $P=2^x$ (Doesn't appear to have a closed-form in fact most functions don't give a closed form.)

For $P$ in general (In terms of Integrals, Measure, etc.). This should solve my newly defined average for functions defined on the Cantor set.

答:
Tl;dr: there is a recursive formula for monomials based on their degree given in a 1992 paper.

To start, I gathered some numerical evidence. For $0\le m,n \le 6$ I asked Mathematica to evaluate $\int_{C_m} x^n H^s(dx)$. For each fixed $n$, the partial sums had the formulas
$$
\left\{1,\frac{1}{2},\frac{3^{-2 m}}{8}+\frac{3}{8},\frac{1}{16} 3^{1-2
   m}+\frac{5}{16},\frac{9^{1-m}}{32}-\frac{17\ 9^{-2 m}}{320}+\frac{87}{320},-\frac{17\ 9^{-2
   m}}{128}+\frac{25\ 9^{-m}}{64}+\frac{31}{128},-\frac{17}{512} 9^{1-2 m}+\frac{29\
   9^{1-m}}{512}+\frac{3253\ 9^{-3 m}}{46592}+\frac{10215}{46592}\right\}$$

Taking $\lim_{m\to \infty}$ you just keep the constant terms:

$$\left\{1,\frac{1}{2},\frac{3}{8},\frac{5}{16},\frac{87}{320},\frac{31}{128},\frac{10215}{46592}
   \right\}$$

I didn't recognize this sequence and it was somewhat intractable (RSolve, FindSequenceFunction, oeis were stumped). So I googled the sequence; amazingly I got a hit from a 1992 paper. Unfortunately, the paper was in Mandarin, of which I am wholly illiterate. So I asked a good friend to translate and she graciously agreed. Several months later, she emailed me a working draft. I'll give the main points from that paper and will link my edited version and the original (she asked that her version not be distributed).

One establishes that integrating with the Hausdorff measure is well-defined and agrees with the Riemann sum definition for continuous $f$.

Using a recursion formula for the right endpoints of each subinterval, one obtains the functional relation
$$
\int _F f(x) H^s(dx ) = \int _F f(x) H^s(dx ) = \frac{1}{2}\left(\int _F f\left(\frac{x}{3}\right) H^s(dx )+\int _F f\left(\frac{x}{3}+\frac{2}{3}\right) H^s(dx) \right)
$$

In particular, if we let $\alpha_n =\int_F x^n H^s(dx)$, with $\alpha_0=1$, then

$$\alpha_n = \frac{1}{2(3^n-1)}\sum_{k=1}^{n}\binom{n}{k}2^k\alpha_{n-k};$$

this is immediate from linearity and the binomial theorem. One checks that these values agree with the first few that we computed.

For continuous functions $g$ on $[0,1]$, one argues analogously: fix $\epsilon>0$ and use the Stone-Weierstrass theorem to approximate $g$ uniformly by a polynomial to within $\epsilon$. Then use the recursion to integrate the polynomial.

The translation was tricky at this point, but I think one can unite the Hausdorff measure with a 'singular' measure. The idea is to take a piecewise constant function on $[0,1]\setminus C_m$ and take the supremum to yield a function $h$ that is continuous with $h'(x)=0$ a.e. This induces a measure $\mu_h$, after which
$$
\int _F f(x) H^s(dx) = \int _0^1 f(x) \mu_h(dx)
$$

The rest of the paper talks about self-similar sets (of which the [ternary] Cantor set is one) and generalizes the similarity transformations to different parameters. In particular, there is a section on the von Koch curve that I have omitted.

I tried and failed to replicate some of the calculations you mentioned in your post. In particular, I'd love to see how you got the value of $\left(\frac{1}{e}\right)^{-1-2e}\cot(e\pi)\cos(e\pi)$ for that one function.

The original paper: http://fzgx.xpu.edu.cn/OA/pdfdow.aspx?Sid=199203

A link to my summary: https://docdro.id/mJp25vh

https://en.wikipedia.org/wiki/Hausdorff_dimension

ContentLicense: CC BY-SA 4.0


问:
Linear Differentiation
I have to determine whether there is normal linear differentiation equation $a_2(x)y'' + a_1(x)y' + a_0(x)y = 0$ on $\mathbb{R}$ such that $u_1, u_2 \in C^2(\mathbb{R})$ defined by $u_1(x) = x, u_2(x) = sinx$ at each $x \in \mathbb{R}$ are both solutions to own equation on $\mathbb{R}$.

Any help will be appreciated.

Thanks in advance.

答:
So $u_1$ being a solution gives that for all $x$
$$ 0 = a_2(x) u_1''(x) + a_1(x) u_1'(x) + a_0(x) u_1(x) = a_1(x) + a_0(x)\cdot x $$
hence we must have $a_1(x) = -a_0(x)x$ for all $x \in \mathbb R$, $u_2$ being a solution gives
$$ 0 = a_2(x) u_2''(x) - a_0(x)x u_2'(x) + a_0(x)u_2(x) = -a_2(x)\sin x - a_0(x)x\cos x + a_0(x)\sin(x) $$
hence
$$ -a_2(x)\sin x = a_0(x) (x \cos x - \sin x)$$
which is (for example) fulfilled if $a_0(x) = -\sin x$ and $a_2(x) = x\cos x - \sin x$. By the above, we are forced to let $a_1(x) = x\sin x$, leaving us with the equation
$$ (x\cos x - \sin x)y''(x) + x\sin x \cdot y'(x) - \sin x \cdot y(x) = 0 $$
of which both $u_1$ and $u_2$ are solutions.</p>

ContentLicense: CC BY-SA 3.0

Title: Linear Differentiation


问:
$\int_{0}^\pi\frac{\sin^2(\theta)}{(a-b\cos(\theta))^{3/2}}d\theta$
As stated in the title.
$$\int_{0}^\pi\frac{\sin^2(\theta)}{(a-b\cos(\theta))^{3/2}}d\theta$$
where $a,b$ are real, positive constants, $a \neq b$.

I tried Weierstrass substitution, $\theta=\frac{\pi}{2}-u$, trig. identities etc. but they all seem to yield a fourth order polynomial which seems analytically intractable. Is there a (easy?) way to evaluate the integral?

答:
For the integrand to be real over $[0,\pi]$ we require $a>b$. Then using Byrd & Friedman 291.08 – the "most general form" for integrals containing $\sqrt{a-b\cos\theta}$ – we get after some simplification
$$I=\frac8{(a+b)^{3/2}}\int_0^{K(m)}\operatorname{sn}^2u\operatorname{cd}^2u\,du$$
where the parameter $m=\frac{2b}{a+b}$. In turn, B&F 361.34 solves this as (again, after simplifying)
$$I=\frac{2\sqrt{a+b}}{b^2}((2-m)K(m)-2E(m))$$


问:
Using definition of logs, solve $5\log_7(n)=10$
I am to solve $5\log_7(n)=10$. I arrived at $n=7^5$ while the solution in my textbook says it's $49$

My working:

$$5\log_7(n)=10$$
$$\log_7(n^5)=10$$
$$7^{10}=n^5$$
$$n=7^{10-5}$$
$$n=7^5$$

Where did I go wrong and how can I arrive at 49?

(Meta question, is there a way that I can add comments next to each line of my working to say what I was thinking? I tried using '#' next to each line but that buckled the markup)

答:
What you did wrong is $n=7^{10-5}$ does not follow from $7^{10}=n^5$.

To fix your calculation (as WA Don pointed out in a comment),
$$
7^2=(7^{10})^{1/5}=(n^5)^{1/5}=n
$$

Alternatively,

\begin{align}
5\log_7(n)&=10\\
\log_7(n)&=2&\textrm{(dividing by $5$)}\\
n&=7^2=49&\textrm{($\log_ax=y$ iff $x=a^y$)}
\end{align}

For your meta question:

```<code>\begin{align}
5\log_7(n)&=10\\
\log_7(n)&=2&\textrm{(dividing by $5$)}\\
n&=7^2=49&\textrm{($\log_ax=y$ iff $x=a^y$)}
\end{align}
</code>```

Body:
What you did wrong is $n=7^{10-5}$ does not follow from $7^{10}=n^5$.

To fix your calculation (as WA Don pointed out in a comment),
$$
7^2=(7^{10})^{1/5}=(n^5)^{1/5}=n
$$

Alternatively,

\begin{align}
5\log_7(n)&=10\\
\log_7(n)&=2&\textrm{(dividing by $5$)}\\
n&=7^2=49&\textrm{($\log_ax=y$ iff $x=a^y$)}
\end{align}

For your meta question:

\begin{align}
5\log_7(n)&=10\\
\log_7(n)&=2&\textrm{(dividing by $5$)}\\
n&=7^2=49&\textrm{($\log_ax=y$ iff $x=a^y$)}
\end{align}


问:
Expressing "For every positive integer $a$, there exists an integer $b$ with $|b| < a$ such that $|bx| < a$ for every real number $x$" symbolically
I would like to express

For every positive integer $a$, there exists an integer $b$ with $|b| < a$ such that $|bx| < a$ for every real number $x$

symbolically. My attempt:

For every positive integer $a$

This tells us that $\forall a \in \mathbb{Z}^+$.

there exists an integer $b$

This tells us that $\exists b \in \mathbb{Z}$

with $|b| < a$

This tells us that $P(a,b) : |b| < a$.

such that $|bx| < a$ for every real number $x$

This tells us that $Q(a,b,x) : |bx| < a$ and $\forall x \in \mathbb{R}, Q(a,b,x)$.

Now putting it all together, I think we would have:

$\forall a \in \mathbb{Z}^+, \exists b \in \mathbb{Z}, P(a,b) \implies \forall a \in \mathbb{Z}^+, \exists b \in \mathbb{Z},\forall x \in \mathbb{R}, Q(a,b,x).$

Alternatively, it could also be

$\forall a \in \mathbb{Z}^+, \exists b \in \mathbb{Z},\forall x \in \mathbb{R}, P(a,b) \implies Q(a,b,x)$

It could also be the case that neither of them are correct I suppose. Please let me know which you think is the most correct in this case, and why. Thank you!

答:
Welcome on Math SE!

What you proposed is quite good. Nevertheless, let's get back on it, identify then correct the mistakes.
(NB : There different solutions that can be accepted for such question, I'm going to try the one that is best understandable.)

For every positive integer $a$

You said $\forall a\in\mathbb{Z}_+$. But $\mathbb{Z_+}$ has a "name", it is $\mathbb{N}$. So, let's write: $$\forall a\in\mathbb{N}$$

There exists an integer $b$

I will use this one with the following one and combine them:

There exists an integer $b$ with $\vert b\vert < a$.

Then, if $b\in\mathbb{Z}$ and $\vert b\vert < a$, we necessarily have $b\in]\!]-a,a[\![$. So, we can write it as: $$\exists b\in]\!]-a,a[\![$$

such as for every real number $x$, $\vert bx\vert<a$

You got it : $$\forall x\in\mathbb{R}\, (\vert bx\vert<a)$$

Now, let's put the pieces back together:
$$\forall a\in\mathbb{N}\exists b\in]\!]-a,a[\![ (\forall x\in\mathbb{R}, \vert bx\vert<a)$$

If you have any questions, do not hesitate.


问:
Why is it e and why is it raised to a negative power?
Why is $(1-\frac{1}{m})^{kn} \approx e^{-kn/m}$? I mean, when do we know when $e$ should be raised to a negative exponent? and how do we even know it's e?

Little background info:
We have a bit vector of size $m$, $n$ elements, $k$ hash functions, and $\frac{1}{m}$ is the probability that a specific bit is flipped to $1$ (assuming independence at perfect randomness). The above equation is for a Bloom filter. It tells us the probability that a specific bit in the bit vector (size=$m$), after inserting $n$ element with $k$ hash functions, is still $0$. It makes sense, that it should approach $0$ and that it wouldn't make sense if it was raised to a non-negative power, but I still need to know how mathematicians come up with such conclusions? And how to they even know it is e?

SOURCE: scribe notes from CS 6550 – Design and Analysis of Algorithms, Dana Randall on Bloom Filters.

答:
Let m be a positive integer having nothing to do with your question. By binomial expansion, $$(1-\dfrac1m)^m=1-m(\dfrac1m)+\dfrac{m(m-1)}2(\dfrac1m)^2-\dfrac{m(m-1)(m-2)}6(\dfrac1m)^3+...+(-1)^m\dfrac{m!}{m!}(\dfrac1m)^m$$ Each term may written as $(-1)^k\dfrac{m(m-1)\cdots(m-k+1)}{k!}(\dfrac1m)^k$, so if we let $m\to\infty$, the expression becomes $$1-1+\dfrac12-\dfrac16+...+(-1)^m\dfrac1{m!}$$

Let the j-th term be $t_j$, then $|t_j|$ decreases monotinally and $\lim_{j\to\infty}t_j=0$. Therefore, by the altering series test, $1-1+\dfrac12-\dfrac16+...+(-1)^m\dfrac1{m!}$ converges. In fact, the series formed by $|t_j|$ is $$(1+\dfrac1m)^m=1+m(\dfrac1m)+\dfrac{m(m-1)}2(\dfrac1m)^2+\dfrac{m(m-1)(m-2)}6(\dfrac1m)^3+...+\dfrac{m!}{m!}(\dfrac1m)^m,$$ letting $m\to\infty$, this expression tends to $e$. Notice that $$(1-\dfrac1m)^m(1+\dfrac1m)^m=(1-\dfrac1{m^2})^m=\sum_{j=0}^m(-1)^j\dfrac{m(m-1)\cdots(m-j+1)}{j!}(\dfrac1m)^{j}(\dfrac1m)^{j},$$ which tends to $1$ as $m\to\infty$. Hence $\lim_{t\to\infty}(1-\dfrac1m)^m=\dfrac1e$. By the definition of the limit, this means that for any real number $\varepsilon\gt 0$, there exist a positive integer $N$ such that $|(1-\dfrac1m)^m-\dfrac1e|\lt\varepsilon$ whenever $m\ge N.$ In other words, if $m$ is a sufficiently large (finite) positive integer, we have $(1-\dfrac1m)^m\approx\dfrac1e$. Now that $m$ is finite, we can associate it to your question by letting $m$ be the size of your vector (taken from the background info you provided for your question).

Therefore, $(1-\dfrac1m)^{kn}=((1-\dfrac1m)^m)^{kn/m}\approx (\dfrac1e)^{kn/m}=e^{-kn/m}.$


问:
Shape of locus of point R

AO and BO are two fixed straight rods. PQ is a straight rod such that P and Q slide respectively on AO and BO. At each position of P and Q lines PR and QR are drawn perpendicular to AO and BO respectively. Show that the locus of R is the arc of a circle.

My approach was to try and show that angles subtended at R were equal, but could not proceed very far. Any help would be appreciated.

答:
$OR$ is always the diameter of circumcircle of $\triangle POQ$. Since $OR$ subtends right angles at $P$ and $Q$.

From sine rule, $$OR = \frac{PQ}{\sin \angle AOB} = k \,\text{(constant)}$$

Thus $R$ moves on the circle with center $O$ and radius $k$.

Title: Shape of locus of point R


问:
Convergence to $1$ or $-1$
I’m looking for a recurrent sequence $x_0, x_1,\dots, x_n$ that converges to $1$ if $x_0$ is positive and to $-1$ if $x_0$ is negative. The recurrent relation should be something that can be approximated by a polynomial.

Is there such a sequence in mathematics?

P.S. I’m sorry for the lack of formality.

答:
Although the term can be approximated by a polynomial is not rigorous, here's something you may like:
$$
x_{n+1} = x_{n} + (x_{n} - x_{n}^{3}) e^{-x_{n}^{2}}
$$
which satisfies the limit condition (by the proof with my intuition and python experiment) and the function $f(x) = x + (x - x^{3})e^{-x^{2}}$ can be approximated by $p(x) = x$ when $|x|$ is large.

Title: Convergence to $1$ or $-1$


问:
How to use Nullstellensatz to prove Study’s lemma
Let $k$ be a algebraically closed field and $f,g\in k[x_1,x_2,...,x_n]$
be polynomials. Assume $f$ is irreducible.

I want to prove that $V(f)\subseteq V(g)$ if and only if $f\mid g$ by using Nullstellensatz. One direction seems obvious and for the other direction I really have no idea.

答:
I really have no idea.

You just have to take the ideals which are associated to the sets $V(f)$ and $V(g)$. And the strong version of the Nullstellensatz tells us that
$$I(V(f)) = \sqrt{(f)} \quad \text{and} \quad I(V(g)) = \sqrt{(g)}.$$
Since applying $I(\cdot)$ inverts inclusions we get
$$g \in \sqrt{(g)} \subset \sqrt{(f)},$$
i.e. some power of $g$ is contained in the principal ideal $(f)$, say $f \mid g^k$ for some $k \in \mathbb N$. And since $f$ is irreducible, this implies $f \mid  g$.


问:
Evaluate a sum involving n choose r
Evaluate: $\sum_{k=0}^6 (-1)^k \binom{6}{k}$ where $\binom{n}{r}= \frac{n!}{r!(n-r)!}$.

I'm unsure how to compute the part with $\binom{6}{k}$, it should be something along the lines of $\frac{6!}{k!(6-k)!}$ can we simplify this down?

I think there was an expansion method for factorials I could used in the denominator that was somewhere along the lines of $(6-k)(5-k)(4-k)(3-k)(2-k)(1-k)$ ?

答:
Since the question has been answered a few times, let's answer a more general question of the same form.
$$
\begin{align}
\sum_{j=k}^m(-1)^j\binom{n}{j}
&=\sum_{j=k}^m(-1)^j\left(\binom{n-1}{j}+\binom{n-1}{j-1}\right)\\
&=\sum_{j=k}^m(-1)^j\binom{n-1}{j}+\sum_{j=k-1}^{m-1}(-1)^{j+1}\binom{n-1}{j}\\
&=(-1)^m\binom{n-1}{m}+(-1)^k\binom{n-1}{k-1}
\end{align}
$$
The third equality takes advantage of the term by term cancellation for $j=k$ throught $j=m-1$. This leaves only the $j=m$ term on the left and the $j=k-1$ term on the right.

In the particular case at hand, $k=0$ and $m=n=6$. Since $\displaystyle\binom{5}{6}=\binom{5}{-1}=0$, the sum is $0$.

Title: Evaluate a sum involving n choose r


问:
glass and a card
Take a cylindrical glass of height $L$ and radius $R$. Fill the glass to a height $h$.
When you place a card that completely covers the opening of the glass, you can turn
the glass over without the water running out. This is because the tension force keeps
the water in the glass over the small distance between the card and the glass.

Calculate the pressure differential on the card by the increase in mass above it.

This pressure differential is compensated by an isothermal expansion of the air above
the water. Calculate the expansion of the air. How much does the card drop relative
to the edge of the glass (a height $\delta k$)?

Assume that the tension force can bridge a
maximum distance $\epsilon$. What are the conditions on $h$ and $L$ such that $\delta k$ does not exceed
$\epsilon$?

My attempt:

I tried to use the hydrostatic equilibrium.

$F_{fluid}=-\rho_{fluid} g V_{fluid}$. Because the fluid is in a cilinder and the volume of the fluid doesn't chang by turning the glass: $V_{fluid}=\pi R^{2} h$.

$F_{air in glass}=-P_{air in glass} A_{fluid}=-P_{air in glass} \pi R^{2}$

$F_{air on card}=P_{air on card} \pi R^{2}$

$=> \triangle P= \rho_{fluid}gh$

Now for the second question I'm thinking I maybe solved the first question wrong. I don't really know how to calculated this one.

答:
For your first question, you could also have used Bernouilli's principle.

For your second question, you can use the ideal gas law assuming that the temperature stays constant, we get that $$ P_f \, V_f = P_i \, V_i \iff V_f = \frac{P_i}{P_f} V_i.  $$
Where the $i$ index stands for the situation where the glass is filled with water and the $f$ index stands for the glass turned.

Now, you can fill in everything that you know:

$P_i = p_0 $, atmospheric pressure;

$V_i = \pi R^2 (L-h) $;

$P_f = \Delta p - p_0 $;

$V_f = V_i + \Delta V  = \pi R^2 (L-h) + \pi R^2 \delta k  $.

Therefore $\delta k = \frac{\Delta p (L-h)}{p_0 - \Delta p} $.

To answer your third question, you can just work with the inequality: $\delta k < \epsilon$.

Title: glass and a card


问:
Sin wave with double trough length?
I'm not the best at maths but i'm working in 3D and need to create a formula for a spline object that is similar to the sin formula I already have (1).

50*Sin(t*PI)

but I need the trough to be double the length of the crest (2), I have done it manually for the image below

sin wave double trough reference

Any ideas much appreciated, thank you!

答: One period-$1$ option is $\sin\min\left\{\tfrac{3\{x\}}{2},\,\tfrac{1+3\{x\}}{4}\right\}$, where $\{x\}:=x-\lfloor x\rfloor$. For a period $P>0$, replace $x$ throughout with $x/P$.

Title: Sin wave with double trough length?


问:
Determining if this relation is an equivalence relation
$$R=\left\{(f,g)\,\Bigg|\, \exists c\in\Bbb Z,\forall x\in\Bbb Z, \frac{f(x)}{g(c)}\le 2\right\}$$

I can show that this relation is reflexive by showing that $(f,f)$ is in $R$
and so $f(x)/f(c) \le 2$, for $c=x$

But i can't seem to find a way to show this relation is also symmetric and transitive.

Any help would be appreciated!
Thanks.

答:
As Brian pointed out, $R$ is not an equivalence relation because it is not reflexive.

To see why $R$ is not symmetric either, consider the constant functions where for all $n\in \mathbb{Z},\text{ }f(n)=1$ and $g(n)=3$. Then certainly we know that $(f,g) \in R$; for example, if we choose $c=7$, then for all $x \in \mathbb{Z}, \dfrac{f(x)}{g(c)}=\dfrac{f(x)}{g(7)}=\dfrac{1}{3}\le2$. Yet we have $(g,f) \notin R$ since for all $c\in \mathbb{Z}$, there exists an $x \in \mathbb{Z}$ (for example, $x=9$) such that $\dfrac{g(x)}{f(c)}=\dfrac{g(9)}{f(c)}=\dfrac{3}{1}>2$.

To see why $R$ is not transitive either, consider the constant functions where for all $n\in \mathbb{Z},\text{ }f(n)=4,\text{ }g(n)=2,$ and $h(n)=1$. Then since $\dfrac{4}{2}=\dfrac{2}{1}=2\le2$, we know by a similar argument from above that $(f,g),(g,h) \in R$. Yet we have $(f,h) \notin R$ since no matter what values of $c$ and $x$ are chosen, $\dfrac{f(x)}{h(c)}=\dfrac{4}{1}=4>2$.


问:
How to calculate this standard normal probability and also chisquared?
X is standard normal. I want to know

$$\frac{d}{dz}P(-\sqrt{z} \leq X \leq \sqrt{z})$$

I don't know how to do that. The answer should be the pdf of chi-squared (1) distribution.

答:
Hint:

Write your function as a difference of CDF evaluating at two different points. Then differentiate.

Also, you should know that $F'(z)=f(z)$ where $F$ is the CDF and $f$ the PDF.

Notes.

$$
P(-\sqrt{z}\le X\le \sqrt{z})=P(X\le \sqrt{z})-P(X\le -\sqrt{z})=F(\sqrt{z})-F(-\sqrt{z})
$$

Now use the chain rule.


问:
How to simplify the variance of $\sqrt{1+\frac{4\sum_{i=1}^{n}X_i^2}{n}}$?
Let $\{X_i\}_{i=1}^{n}$ be $n \ $ i.i.d random variables such that $X_i \sim N(\theta,\theta)$ where $\theta > 0$.

Now, I am trying to simplify $$\text{var}\bigg(\sqrt{1+\frac{4\sum_{i=1}^{n}X_i^2}{n}}\bigg).$$ I got this expression in some calculation that I am doing.

Only, the square root is creating problems. I can't get my head around this. Please help.

Thanks!

EDIT : I have to show that $$\text{var}\bigg(\sqrt{1+\frac{4\sum_{i=1}^{n}X_i^2}{n}}\bigg) \to 0$$ as $n \to \infty.$

EDIT 2 : MLE of $\theta$ is $$\hat{\theta}_{MLE} = \frac{-1 + \sqrt{1+\frac{4\sum_{i=1}^{n}X_i^2}{n}}}{2}$$.

I am trying to prove the consistency of $\hat{\theta}_{MLE}$.

答:
I do not know your purpose but surely you can approximate

$$\mathbb{V}\Bigg[\sqrt{1+4\overline{X}_n}\Bigg]$$

using Delta Method.

EDIT: your estimator is

$$T=\sqrt{1+4\frac{\Sigma_i X_i^2}{n}}$$

After usual calculations,

$$\hat{\theta}_{ML}=\frac{T-1}{2}$$

The asymptotic distribution of MLE is known...easy solved problem (as an approximation, obviously)

EDIT: at the end, the problem was different.

Reading the complete original text

Let $X_1,...,X_n$ be $n$ i.i.d. random variables with ditribution $N(\theta;\theta)$ for some unknown $\theta>0$. Compute the MLE of $\theta$ and show it is Consistent

After calculating $\hat{\theta}_{ML}$ that you did correctly let's go back to the Score that is

$$l^*(\theta)=-\frac{n}{2\theta}-\frac{n}{2}+\frac{\Sigma_i X_i^2}{2\theta^2}$$

The derivative of the score is

$$l^{**}(\theta)=\frac{n}{2\theta^2}-\frac{\Sigma_i X_i^2}{\theta^3}$$

Thus the Fischer information is

$$I_n(\theta)=-\mathbb{E}[l^{**}(\theta)]=-\frac{n}{2\theta^2}+\frac{(\theta+\theta^2)n}{\theta^3}=\frac{(1+2\theta)n}{2\theta^2}$$

Concluding:

The asymptotic distribution is the following

$$\hat{\theta}_{ML}\dot{\sim} N\Bigg[\theta;\frac{2\theta^2}{(1+2\theta)n}\Bigg]$$

This means that

$$\lim\limits_{n \to\infty}\mathbb{E}[\hat{\theta}_{ML}]=\theta$$

and

$$\lim\limits_{n \to\infty}\mathbb{V}[\hat{\theta}_{ML}]=0$$

Which is a necessary and sufficient condition for consistency


问:
Factoring of a particular polynomial
It is given that $n$ is an odd integer greater than $3$ but not a multiple of $3$. Prove that $x^3 +x^2 +x$ is a factor of $(x+1)^n -x^n -1$.

I tried to write $x^3+x^2+x=x(x^2+x+1) =x(x-\omega^2)(x-\omega)$ where $\omega$ and $\omega^2$ are cube roots of unity $≠1$. Then I tried to do it by putting values but could not get the answer

答:
Modulo $x$, $(x+1)^n-x^n-1\equiv 1^n-0-1=0$, so $(x+1)^n-x^n-1$ is divisible by $x$.

It remains to be shown that $(x+1)^n-x^n-1$ is divisible by $x^2+x+1$.  Well,

$(\omega+1)^n-\omega^n-1=(-\omega^2)^n-\omega^n-1=-(\omega^{2n}+\omega^n+1)$ for $n$ odd.

Furthermore, for $n$ not a multiple of $3$, $(\omega^{2n}+\omega^n+1)(\omega^{n}-1)=\omega^{3n}-1=0$ and $\omega^n-1\ne0$,

so $\omega^{2n}+\omega^n+1=0$.  Therefore, $\omega$ is a zero of $(x+1)^n-x^n-1$.

Title: Factoring of a particular polynomial


问:
How to determine if a vector is in an affine space?
So the question is, how to show if a vector is in an affine space. For a regular vector space it's easy, while all i have to do is to show if a vector is a linear combination of a span. But I am strugling a bit with an affine space. So let's say I have a vector x and an affine space defined like S = v + U.  Is it okay to just subtract the vector v from x and then just chcek wether the result in in U?

x - v = a and then
a ∈ U?

答: Yes, just check that $x-v$ is in $U$. $v$ is fixed so this is an easy reduction to the standard linear case.


问:
Evaluate $\sin (\cot ^{-1} (1+x)) = \cos (\tan^{-1} (x))$
Let $cot^{-1} (1+x)=a$

Then $x =1-\cot a$

So $$\sin a =\cos (\tan^{1}(1-\cot a))$$

$$a=\frac{\pi}{2}- \tan^{-1} (1-\cot a)$$
$$\cot a =\frac 12$$

So $x=\frac 12$, but given answer is $-\frac 12$. What am I doing wrong?

答:
$\sin (\cot ^{-1} (1+x)) = \cos (\tan^{-1} (x)).$
Let $\cot^{-1} (1+x)=A \implies \cot A =1+x.$ So $LHS=\sin A=
\frac{1}{\sqrt{2+2x+x^2}}$ Let $\tan B=x \implies RHS=\cos B =\frac{1}{\sqrt{1+x^2}}$. Finally $LHS=RHS \implies 2+2x+x^2=1+x^2 \implies x=-1/2.$


答:
This is not possible without restrictions. Indeed, the recurrence that you want must be contracting as it eventually maps any positive to $1$, while a polynomial does the opposite.

You can achieve the desired property with the simple non-polynomial recurrence

$$x_{n+1}=\sqrt[3]{x_n}.$$

For a quasi-linear solution,

$$x_{n+1}=\frac{x_n+\text{sgn}(x_n)}2.$$

But this is of little interest, as

$$x_{n+1}=\text{sgn}(x_n)$$ is enough.

Title: Convergence to $1$ or $-1$

Body:
I’m looking for a recurrent sequence $x_0, x_1,\dots, x_n$ that converges to $1$ if $x_0$ is positive and to $-1$ if $x_0$ is negative. The recurrent relation should be something that can be approximated by a polynomial.

Is there such a sequence in mathematics?

P.S. I’m sorry for the lack of formality.


问:
Circle theorem/triange angle question
I am doing practise papers and there is one question I cannot understand even with the mark scheme. I have added the pictures below:

Question (with added annotations):

Mark scheme:

The question is to find angle x (23 degrees, apparently).

At first I thought that angle OCD might be 69 degrees because of the alternate angle theorem, but I realised that it wasn't on a tangent so that wouldn't work. Other than that I have no idea - could someone please explain (in detail if possible, maths isn't my strength!) the steps to find it? Thanks in advance!

答:
To get the $2x$:

$180-(180-2x)$ is obviously $2x$ so $OCD = 2x$ and $CDO = 2x$ because angles in the same segment are equal (the segment being the semcircle).

Title: Circle theorem/triange angle question


问:
Lipschitz condition normed vector space
Am I right that $g: C^{1}[0,2],||*||_{C^1[0,2]} \rightarrow \mathbb{R}$ with $g(f)=f'(1)$ satisfies a Lipschitz condition?

Cause $|f'(1)-h'(1)|=|g(f)-g(h)|\le \text{max} |f-h|+ \text{max} |f'-h'|$, as the last term here contains already the lipschitz condition.

Title: Lipschitz condition normed vector space

Body:
you are right: $$|g(f) - g(h)| = |f'(1) - h'(1)| \le \max_{[0,2]}|f'(x) - h'(x)| \le$$
$$\le \max_{[0,2]}|f'(x) - h'(x)| + \max_{[0,2]}|f(x) - h(x)| = \|f - g\|_{C^1},$$
then it is Lipschitz continuous with Lipschitz constant $L = 1$.


答:
A possible function for your recursion could be

$$f(x) = \frac 4{\pi}\arctan x$$

The interesting fixpoints here are $x=1$ and $x=-1$.

$f$ can be approximated by its Taylor polynomials but it is unclear what you expect if you replace $f$ by an approximating polynomial. The recursion might not have the same fixpoints and might not converge the way you would like to have it.

Back to $f(x) = \frac 4{\pi}\arctan x$:

Since $f$ is an odd function, you only need to check convergence of $x_{n+1} = f(x_n)$ to $1$ for $x_0>0$.

For $x_0>1$ you have $1< x_n < 2$ and $f$ is contracting on $[1,2]$, hence the recursion converges to the fixpoint $x=1$.

For $0<x_0<1$ you have $0<x_n <1$ and additionally $f(x)>x$ (which you may check for yourself). Hence, $x_n$ is increasing and bounded from above, hence it converges as well to the fixpoint $x=1$.

Title: Convergence to $1$ or $-1$


问:
Number of students in school club : $4$ Circles Venn Diagram Puzzle
There are $200$ students in a class. Each student is a member of at least on of the clubs Eagles, Falcons, Tigers and Pythons.

If each club has $100$ students, then the number of students who are members of all the four clubs is at the most .....

If each club has $120$ students, then the number of students who are members of at the most two clubs  is at the least .....

If each club has $140$ students, then the number of students who are members of at least three clubs is at least ....

I came across this question and I tried to solve it using the $4$ circles Venn diagram but it is a very chaotic way to solve this (Too many variables come into picture and it all got messed up) and even I was only able to solve the first question.

Any way or method to simplify the solution approach for this question? Any help is appreciated.

Thanks!

答:
Denote the clubs by $E,F,T,P$ respectively.

There are many regions produced in Venn diagram such as $E \cap F \cap T$ and $E-T-F-P$. To put maximum students into $E \cap F \cap T\cap P$, we'd like to put zeros in most of them, while keeping the numbers of students only in a single club minimum.

Let the number of students who are only members of club $E$ be $A$ and those present in all clubs be $B$. If all other regions have zero students, then number of students who are only members of other clubs is also $A$.

We get $A+B=100$ and $4A+B=200$. Taking difference, $3A=100$.
This is non-integer. But we may just take $A=33$ and $1$ balance can be put into $(E \cap F)-T-P$ and $(T \cap P)-E-F$ each.

Thus we got maximum for the case of $100$ $(1\pmod 3)$ owing to restriction by divisibility by $3$. Can you now solve for $120$ $(0\pmod 3)$ and $140$ $(2\pmod 3)$?


问:
Integrating $f(x,y)=xy$ over parallelogram
I want to integrate $f(x,y)=xy$ over a parallelogram with vertices $(-2,1)$, $(-1,3)$, $(1,3)$, $(0,1)$. My approach:$$\int \limits_{1}^{3}\int \limits_{\frac{1}{2}(y-5)}^{\frac{1}{2}(y-1)}xy\, \mathrm{d}x\, \mathrm{d}y=-2$$ The solution says that it should be $-\frac{10}{3}$. What did I do wrong?

Body:
We can move $y$ into the outer integral:
$$I=\int_1^3y\int_{(y-5)/2}^{(y-1)/2}x\,dx\,dy$$
Then integrate the first integral:
$$=\frac12\int_1^3y[x^2]_{(y-5)/2}^{(y-1)/2}\,dy$$
$$=\frac18\int_1^3y((y-1)^2-(y-5)^2)\,dy$$
$$=\frac18\int_1^3y(8y-24)\,dy$$
$$=\int_1^3(y^2-3y)\,dy$$
The second follows:
$$=[y^3/3-3/2y^2]_1^3=-\frac{10}3$$


问:
Projection on unit vector lipschitz condition
if i have a vector $ v \in C^{n}$, where we do not want to look at vectors with norm $||v||\le 1$ with an arbitrary norm of this space and I am asked whether the map $ f(x)=\frac{x}{||x||}$ satisfies the lipschitz boundary condition $|| f(x)-f(y)|| \le L ||x-y||$. how do i proof this? Somehow it is clear, that lipschitz constant 1 will be sufficient for this, but I have troubles to show it. maybe one needs a clever inequality. </p>

答:
The statement is in general false taking the constant to be equal to $1$. The best possible constant in general is $2$.

It easy to prove that
\begin{equation}
\left\| \frac{x}{\| x \|} - \frac{y}{\| y \|}  \right\|= \left\| \frac{(x-y) \| y \| + y(\| y \| - \| x \|)}{\|x \| \| y \|} \right\| \leqslant \frac{2 \|x-y\| \| y \|}{\|x \| \| y \|} \leqslant 2 \| x-y \|.
\end{equation}

To show that this is, indeed, the best possible constant consider $\mathbb{R}^2$ with the $L^1$ norm given by
\begin{equation}
\| (x_1, x_2 ) \| = |x_1| + |x_2|
\end{equation}
Let $\mathbf{x} = (1,0)$ and $\mathbf{y} = (1,a)$ with $a >0$. Then it is easy to see that the following hold
\begin{align}
\| \mathbf{x} \| &= 1, \quad \| \mathbf{y} \| = 1+a\\
f(\mathbf{x}) &= (1,0), \quad f(\mathbf{y}) = \left(  \frac{1}{a+1}, \frac{a}{a+1} \right)\\
\mathbf{x} - \mathbf{y} &= (0, -a), \| \mathbf{x} - \mathbf{y} \| = a\\
f(\mathbf{x}) - f(\mathbf{y}) &= \left( \frac{a}{a+1}, -\frac{a}{a+1}  \right),\\
\| f(\mathbf{x}) - f(\mathbf{y}) \| &= \frac{2a}{a+1}.
\end{align}
Finally, we have
\begin{equation}
\lim_{a \to 0} \frac{\| f(\mathbf{x}) - f(\mathbf{y}) \|}{\| \mathbf{x} - \mathbf{y} \|} = 2.
\end{equation}


问:
Are there any number fields that extend $\mathbb C$?
If there are number fields that extend $\mathbb C$, what are some examples?

答: Depends on what you mean as a number, but both the field of rational functions, holomorphic functions, or linear operators on them, are extensions of complex numbers, because they include complex numbers as constants.</p>


问:
Calculus - Rolle question
Regarding:

Let $a_{1},a_{2},a_{3},b_{1},b_{2},b_{3} \in \mathbb{R}$, while $a_{1},a_{2},a_{3}>0$ and $b_{1}<b_{2}<b_{3}$.

Prove that the equation $$\frac{a_{1}}{x-b_{1}}+\frac{a_{2}}{x-b_{2}}+\frac{a_{3}}{x-b_{3}}=0$$

has exactly two distinct solutions in $\mathbb{R}$.

--

So I have tried defining a function $f$ which is equal to the left side of the equation, derivate it, and assume by contradiction that it has more than 2 solution.

Unfortunately with no luck...

If someone knows a way, please only give me a hint so I can solve it by myself.

Thanks a lot!

答:
More generally,
suppose there are $n$
$a_k$ and $b_k$ with
$a_k>0$ and
$0 < b_k < b_{k+1}$
for all $k$.

Let
$f(x)
=\sum_{k=1}^n a_k/(x-b_k)
$.

$f'(x)
=-\sum_{k=1}^n a_k/(x-b_k)^2
\lt 0
$
so $f(x)$ is always decreasing
wherever it is defined.

If $x=b_k+c$
where $c$ is small,
$f(x)>0$
because all the other terms are bounded.
Similarly,
if $x=b_k-c$
where $c$ is small,
$f(x)<0$.

This can be made rigorous by choosing
$c_0= \min(b_{k+1}-b_k)/2$,
getting an upper bound on each
$a_j/(x-b_j)$
for $|x-b_j|<c_0$
with $j\ne k$,
and making
$x-b_k$ small enough.

Therefore there is exactly one root between each pair
of $b_k$.

If $x$ is greater than all the $b_k$,
then all the terms are positive so there is no root there.

If $x$ is smaller than all the $b_k$,
then all the terms are negative so there is no root there.

Therefore the only roots are between
the $b_k$
so there are $n-1$ of them.

Title: Calculus - Rolle question


答:
We can simplify this equation as:
$f(x)=a_1(x-b_2)(x-b_3)+a_2(x-b_1)(x-b_3)+a_3(x-b_1)(x-b_2)$
$f(b_1)=a_1(b_1-b_2)(b_1-b_3)>0,$
$f(b_2)=a_2(b_2-b_1)(b_2-b_3)<0$
$f(b_3)=a_3(b_3-b_1)(b_3-b_2)>0.$
So by IVT, there is one root in $(b_1,b_2)$ and other one is in $(b_2,b_3)$. So only two real root of this equation which is a quadratic.

Title: Calculus - Rolle question

Body:
Regarding:

Let $a_{1},a_{2},a_{3},b_{1},b_{2},b_{3} \in \mathbb{R}$, while $a_{1},a_{2},a_{3}>0$ and $b_{1}<b_{2}<b_{3}$.

Prove that the equation $$\frac{a_{1}}{x-b_{1}}+\frac{a_{2}}{x-b_{2}}+\frac{a_{3}}{x-b_{3}}=0$$

has exactly two distinct solutions in $\mathbb{R}$.

--

So I have tried defining a function $f$ which is equal to the left side of the equation, derivate it, and assume by contradiction that it has more than 2 solution.

Unfortunately with no luck...

If someone knows a way, please only give me a hint so I can solve it by myself.

Thanks a lot!


问:
Bound on complex triples satisfying some simple equations.
Let $a,b,c \in \mathbb{C}$ be three complex numbers satifying the following equations :
$$\left\{ \begin{array}{rl} a+b+c &= 3 \\ a^2+b^2+c^2 &=9 \end{array} \right.$$

I want to prove that $\max(|a|, |b|, |c|)\geq 2$.

I can formulate this as an optimisation problem for a multiple real variable function and use Lagrange multipliers, but it would probably be extremely long. The problem seems simple enough that there should be a direct and elegant proof.

So, any help would be very appreciated. Thanks.

答:
We may assume $|a|\geq|b|\geq|c|$. We therefore introduce new variables $u$, $v\in{\mathbb C}$ such that
$$b=a u, c= b v=a u v\qquad\bigl(|u|\leq1, \ |v|\leq 1)\ .$$
We then obtain
$$3=a+b+c=a(1+u+uv)\tag{1}$$
and
$$0=(a+b+c)^2-(a^2+b^2+c^2)=2(ab+ac+bc)=2a^2 u(1+v+uv)\ .$$
The case $u=0$ leads to $a=3>2$. Otherwise we have $1+v+uv=0$, or
$$v=-{1\over 1+u}\ .\tag{2}$$
Since we want $|v|\leq1$ we need $|u+1|\geq1$. It follows that the admissible values for $u$ are in the closed unit disc from which the unit disc centered at $-1$ has been removed. Let $\bar \Omega$ be  this moonshaped region. Given an $u\in\bar\Omega$ we have to calculate $v$ from $(2)$ and then have to verify in $(1)$ that $|1+u+uv|\leq{3\over2}$. This would imply that $|a|\geq2$.

Now
$$1+u+uv={1+u+u^2\over1+u}=:f(u)\ .$$
In order to show that   $|f(u)|\leq{3\over2}$ in $\bar\Omega$ it is sufficient to verify this inequality along the two boundary arcs. Along the arc
$$\gamma_1:\quad t\mapsto u(t):=-1+e^{it}\qquad\left(-{\pi\over3}\leq t\leq{\pi\over3}\right)$$
we have $|1+u(t)|=1$, so that it remains to verify that
$$\bigl|1+u(t)+u^2(t)\bigr|=2\cos t-1\in[0,1]\qquad\left(-{\pi\over3}\leq t\leq{\pi\over3}\right)\ .$$
Along the arc
$$\gamma_2:\quad t\mapsto u(t):=e^{it}\qquad\left(-{2\pi\over3}\leq t\leq{2\pi\over3}\right)$$
we have to study the full complex $$w(t):=f\bigl(u(t)\bigr)={1+e^{it}+e^{2it}\over1+e^{it}}\qquad\left(-{2\pi\over3}\leq t\leq{2\pi\over3}\right)\ .$$
The following figure shows the Mathematica plot of this curve $t\mapsto w(t)$. We see that there is indeed a single point with $|w(t)|={3\over2}$, namely the point $w(0)={3\over2}$. It corresponds to $u=1$, $v=-{1\over2}$, and leads to the triple $(a,b,c):=(2,2,-1)$.


问:
Convergence to $1$ or $-1$
I’m looking for a recurrent sequence $x_0, x_1,\dots, x_n$ that converges to $1$ if $x_0$ is positive and to $-1$ if $x_0$ is negative. The recurrent relation should be something that can be approximated by a polynomial.

Is there such a sequence in mathematics?

P.S. I’m sorry for the lack of formality.

答:
The rational function

$$f(x)={2x\over x^2+1}$$

should do the job. It has fixed points at $x=0$, $1$, and $-1$. It's easy to verify that if $x\gt1$ then $0\lt f(x)\lt1$, and if $0\lt x\lt 1$ then $x\lt f(x)\lt1$, so for any $x_0\gt0$, the sequence defined by $x_{n+1}=f(x_n)$ will converge to $1$. By symmetry (since $f$ is an odd function), any $x_0\lt0$ will give a sequence converging to $-1$.

Title: Convergence to $1$ or $-1$


问:
What is the cardinality of $\Bbb{R}^L$?
By $\Bbb{R}^L$, I mean the set that is interpreted as $\Bbb{R}$ in $L$, Godel's constructible universe. For concreteness, and to avoid definitional questions about $\Bbb{R}$, I'm looking at the set ${\cal P}(\omega)$ as a proxy. I would think it needs to be countable, since only definable subsets of $\omega$ are being considered, but I don't see how Cantor's theorem would fail here, since $L$ is a model of $\sf{ZFC}$. (Actually, that's a lie: this probably means that there is no injection $f:\omega\to{\cal P}(\omega)^L$ in $L$, even if there is one in $V$. I'm still a little hazy on all the details, though.) But the ordinals are in $L$, so $L$ is not itself countable, and there must exist genuinely uncountable elements of $L$. What does the powerset operation (in $L$) do to the cardinality of sets (as viewed in $V$), then?

答:
It is impossible to give a complete answer to this question just in $\sf ZFC$.

For once, it is possible that $V=L$ is true in the model you consider, so in fact $\Bbb R^L=\Bbb R$, so the cardinality is $\aleph_1$.

On the other hand it is possible that the universe is $L[A]$ where $A$ is a set of $\aleph_2$ Cohen reals, in which case $\Bbb R^L$ is still of size $\aleph_1$.

And it is also possible that the universe is $L[G]$ where $G$ is a function which collapses $\omega_1^L$, resulting in having $\Bbb R^L$ as a countable set.

There are other axioms, such as large cardinal axioms (e.g. "$0^\#$ exists") which imply that $\Bbb R^L$ is countable, and one can arrange all sort of crazy tricks, where the result is large or small compared to $\Bbb R$.

One thing is for certain, $|\Bbb R^L|=|\omega_1^L|\in\{\aleph_0,\aleph_1\}$.

Title: What is the cardinality of $\Bbb{R}^L$?


问:
Integration with $\int_0^\infty \frac{\cos (ax)}{x^2}dx$
I am trying to integrate $$\int_0^\infty \frac{\cos (ax)}{x^2}dx$$

I get $$-a\operatorname{Si}(ax)-\frac{\cos(ax)}{x}+C$$ as indefinite integral and when I put limits, I understand how to get $-\frac{{\pi}a}{2}$ for the Sine integral. But I don't understand how to put limits for the second term, it seems to me that term goes to infinity. Any help?

Edit:
Sorry for not being clear,
I am trying to integrate this integral $$D_1=\frac4\pi\int_0^\infty\frac{d\lambda}{\lambda^2}\sin\lambda q_1\sin\lambda q_2\sin\lambda q_3\sin\lambda q_4,\tag{25}$$
which I converted to sum of cosines
$$\sum\ \eta_1\eta_2\eta_3\eta_4\frac{\cos ((\eta_1q_1+\eta_2q_2+\eta_3q_3+\eta_4q_4)\lambda)}{16}, \eta_{1,2,3,4}= \pm1$$

I am supposed to get these results,

For $q_1+q_2>q_3+q_4$ and $q_1+q_4>q_2+q_3$, $$D_1=\frac12(q_2+q_3+q_4-q_1)\tag{30}$$

For $q_1+q_2>q_3+q_4$ and $q_1+q_4<q_2+q_3$, $$D_1=q_4\tag{33}$$

I am trying to reproduce calculations given in the Appendix A of this paper.

答:
Since the arguments of your cosines sum to zero, as NHL points out your problem amounts to evaluating $\int_0^{\infty}\frac{\cos(a x)-1}{x^2}\,dx$, $a\in\mathbb{R}$. This can be done with integration by parts (let $dv=x^{-2}\,dx$) and, using $\int_0^{\infty}\frac{\sin(a x)}{x}\,dx=\frac{\pi}{2}$ for $a>0$, each subintegral evaluates to $-\frac{\pi}{2}|a|$. Thus,
$$
D_1=\frac{1}{4} \bigg(-\left| q_1+q_2-q_3-q_4\right| -\left|
   q_1-q_2+q_3-q_4\right| +\left| q_1+q_2+q_3-q_4\right| -\left|
   q_1-q_2-q_3+q_4\right| +\left| q_1+q_2-q_3+q_4\right| +\left|
   q_1-q_2+q_3+q_4\right| +\left| -q_1+q_2+q_3+q_4\right| -\left|
   q_1+q_2+q_3+q_4\right| \bigg)
$$

In particular, if you can impose conditions on the $q_i$, you can simplify this expression accordingly.


问:
Definite integral for $\zeta(3)$
By making use of Mathematica, I detected the following integral expression for zeta(3):
$$\int_0^1\frac{\log(x)}{1+x}\log\left(\frac{2+x}{1+x}\right)dx=-\frac5{12}\zeta(3).$$
Any proof of it would be gratefully appreciated.

答:
The proof is simple : the antiderivative exists (ask Mathematica).

The result is not so complicated (beside logarithms, there are just a few polylogarithm functions).

Use the bounds and simplify to get
$$\text{Li}_3(-3)-2 \text{Li}_3(3)+i \pi  \left(-\text{Li}_2(-3)+2
   \text{Li}_2(3)+\log ^2(3)\right)-$$ $$\text{Li}_2(-3) \log (3)+\text{Li}_2(3) \log
   (9)+\frac{7 \zeta (3)}{4}-\frac{2 i \pi ^3}{3}-\pi ^2 \log (9)$$

This simplifies nicely (just a litle work - I did it by hand).

If you do not know what it is, ask for the numerical value
$$I=-0.5008570429831642855832242$$ and the inverse symbolic calculator will tell you that $I$ satisfies the following Z-linear combination :
$$ 12 I + 5\,  \zeta(3)=0$$

You could have fun with
$$J(a,b,c,d)=\int \frac{\log (x+a) \log \left(\frac{x+b}{x+c}\right)}{x+d}\,dx$$ which has a closed form antiderivative and get a bunch of identities.

Title: Definite integral for $\zeta(3)$


问:
Find Existing Solutions for $\sin(x)\cos(x)=0$ in the interval $[0,2\pi]$
I've managed to understand how to get to $x = \pi$ and $x = \pi/2$, but I have no idea where to go from here or how to find the interval or quadrant. I'm kinda new to trigonometry so I'm very lost.

答: Note that $2\sin x\cos x=\sin(2x)$. So your question is the same as solving $\sin(2x)=0$ on $[0,2\pi]$, which is the same as solving $\sin z=0$ on $[0,4\pi]$. Once you find $z$, set $x=\frac{z}{2}$.


问:
Complex differentiability of a function from first principles.
I have to show a complex function is differentiable at $0$ from first principles. For example take the function:

$f(z) = \overline{z}^2\cos{z}$.

How could I show $f(z)$ is differentiable at $0$?

答: We wish to prove $\lim_{w\to0}\frac{\overline{w}^2}{w}\cos w$ exists and evaluate it. (@mrsamy suggested this approach, but I've changed their $h$ to $w$ to emphasize the limit is in a complex variable.) The $\cos w$ factor is asymptotic to $1$, so we just want $\lim_{w\to0}\frac{\overline{w}^2}{w}$. Since $\left|\frac{\overline{w}^2}{w}\right|=|w|$, the limit is $0$ by the squeeze theorem.


问:
Why do injective holomorphic functions have nonzero derivative
For some open sets $U$, $V$ in the complex plane, let $f:U\rightarrow V$ be an injective holomorphic function. Then $f'(z) \ne 0$ for $z \in U$.

Now I don't understand the proof, but here it is from my text. My comments are in italics.
Suppose $f(z_0) = 0$ for some $z_0 \in U$.

$f(z) - f(z_0) = a(z - z_0)^k + G(z)$ for all $z$ near $z_0$, with $a \ne 0, k \ge 2.$
Also, $G$ vanishes to order $k+1$ at $z_0$.

I'm not clear on what this "vanishing" thing means. Maybe it means that $G$ can be expressed as a power series of order $k+1$ around $z_0$.

For sufficiently small $w$ we can write $f(z) - f(z_0) - w = F(z) + G(z)$, where $F(z) = a(z - z_0)^k - w$.

I'm not sure why we need to have $w$ small. This equation will work for any $w$.

Since $|G(z)| \lt |F(z)|$ on a small circle centered at $z_0$, and $F$ has at least two zeros inside that circle, Rouche's theorem implies that $f(z) - f(z_0) - w$ has a least two zeros there.

Now I think that $|G(z)| \lt |F(z)|$ can follow simply from the fact that $F$ is a polynomial of degree $k$ while $G$ has degree $k+1$. And the remark about the two zeros can follow from the fact that $F$ must have $k$ zeros in the complex plane. But the first part requires that we consider $z$ only on a small circle. The second part requires that our circle be big enough to capture two zeros. How do we know that we can satisfy both?

Since $f'(z) \ne 0$ for for all $z \ne z_0$ sufficiently close to $z_0$, the roots of $f(z) - f(z_0) - w$ are distinct, so $f$ is not injective - a contradiction.

I think that the derivative is never zero for values of $z$ other than $z_0$ because otherwise we would have a sequence of zeros limiting towards $z_0$ which would cause our function to be constant which is a contradiction. But again we have the same problem - we can only consider a small circle. The roots of $f$ may lie outside this circle.

答:
First comment: Yes. I prefer to write $(z-z_0)^k\cdot(a+(z-z_0)H(z))$ in such proofs.

Second comment: $w$ small is not needed immediately, but we can only use a small circle (as guaranted by openness of $U$) and want to have $(z-z_0)^k=w$ at least once (end hence $k$ times) inside that circle. This is what forces $w$ to be small.

Third comment: We make our circle even smaller (and may revise our choice of $w$) in order to make $|G|<| F|$. By the vanishing order of $G$, we have $G(z)\le c\cdot |z-z_0|^{k+1}$ for some $c$ as long as $z\approx z_0$ (with my notation above, you can take any $c>|H(z_0)|$). Then what we need here for $|G(z)|<|F(z)|$ if $|z-z_0|=r$ is to chose $r\le \frac ac$. The very simple polynomial $F$ has $k$ zeroes in the circle because we choose our $w$ small enough (smaller than $ar^k$ if $r$ is the radius of our small circle).

Fourth comment: The zeroes of a holomorphic function are isolated unless the function is (locally) constant. If $f$ is constant on a small open disk, it is already far from injective.


问:
How to find if an extreme point is degenerate algebraically?
Is there a method to algebraically determine if an extreme point is degenerate?  Can someone please explain to me.

Here's my example:
$$\max x + 2y $$

subject to,
\begin{align}
x - 4y &\leqslant 4 \\
-2x +y &\leqslant 2 \\
-3x +4y &\leqslant 12 \\
2x + y &\leqslant 8 \\
x, y &\geqslant 0
\end{align}
And you find your extreme points $(0,0), (4,0), (20/11, 48/11), (4/5, 18/5)$ and $(0,2)$.
I know the optimal point is $(20/11, 48/11)$, but how can I determine if any of these points are degenerate without the graphs? thank you.

答:
Substitue those points in the inequalities and see how many are active. If exactly one pair of them is active, then it is not degenerate. For example $(0,0)$ only make $x \ge 0$ and $y \ge 0$ active. It is not degenerate.

Otherwise, if you can find more than two constraints. Verify if you can find two sets of two-constraints that are linearly independent.
For example $(4,0)$ makes $x-4y \le 4$, $2x+y \le 8$ and $y \ge 0$ active. We can verify that $x-4y=4, 2x+y=8$ gives the solution uniquely. Also $x-4y=4$ and $y=0$ uniquely as well. It is degenerate.


问:
Prove that there exists an uniformly continuous $g$ such that $f = g$ a.e
Suppose $f \in L^{\infty}(\mathbb{R})$, $f_{h}(x) = f(x+h)$, and
$$
\lim_{h \rightarrow 0}||f_{h} - f||_{\infty} = 0
$$
Prove that there exists a uniformly continuous function $g$ on $\mathbb{R}$ such that $f = g$ a.e

Firstly, I used the theorem such that if $f$ is Borel measurable and integrable, then there exists continuous function $g$ having compact support. However, I can't show $f$ is integrable, and Borel measurable.

And I wonder about a notation $||f_{h} - f||_{\infty}$ means $||f(x+h)-f(x)||_{\infty}$.

Can anybody help?

答:
I stumbled upon a similar question in Exercise 8.4 of Folland's "Real Analysis".

As I struggled to find my mistake of mixing up convergence with respect to the uniform norm and $\lVert \cdot \rVert_\infty$, I believe it is worth to point out a few remarks. I will make use of Julián Aguirre's notation:

Let $h>0$. $f^h$ is a uniformly continuous function. Indeed: given $x,y \in \mathbb{R}$;

$$
\lvert f^h(x)-f^h(y)\rvert
\leq
\lVert f \rVert_\infty
\int
\lvert
\chi_{(x-h,x+h)}
-
\chi_{(y-h,y+h)}
\rvert
d m.
$$
One may use the DCT to prove that
$$
\lim_{y \to x}
\int
\lvert
\chi_{(x-h,x+h)}
-
\chi_{(y-h,y+h)}
\rvert
d m
=
0.
$$
Due to translation-invariance of the Lebesgue measure, we obtain uniform continuity.

If $\{ h_n\}_n$ is a sequence of positive numbers such that $\lim_n h_n = 0$, then $\lim_n \lVert f^{h_n} -f \rVert_\infty = 0$. To see that, first prove that $(f^{h_n})_n$ is a Cauchy sequence in $L^\infty$ and $f^{h_n} \xrightarrow[n]{} f$ a.e.

Up until now, I did not use the hypothesis $\lim_h \lVert f_h - f \rVert_\infty = 0$! This hypothesis is only necessary to prove that $\lim_h \sup \lvert f_h - f\rvert=0$.

To check out that difference, one could work out the details of approaching Heaviside's function $H\colon \mathbb{R} \to \{0,1\}$ with the means $H^h$: the sequence one obtains is a sequence of uniformly continuous functions which converges in $L^\infty$ yet does not convege with respect to the uniform norm. (hence the non-continuous limit)


问:
Selecting points of intersection given a point and equations of lines

I have an $x$-axis and a $y$-axis both beginning from the origin. I also have $n$ straight lines with known equation ($n$=4 in the diagram). If all intercepts of the lines to the axes are positive, then at most I have ${n \choose 2} + 2n$ points of intersection.

The question is, of these points of intersection, how do I determine the ones that form the boundary of the region that contains a point $M$ with known coordinates? In the diagram they are coloured red. I am interested in knowing how these points can be selected from all points.

答:
For every intersection point $L_i \cap L_j$ and every other line $L_k$, check if $L_k$ separates $L_i \cap L_j$ from $M$. If no line separates $L_i \cap L_j$ from $M$, then $L_i \cap L_j$ is one of the adjacent points.

The check for separation is done by the sign test outlined in the other answer. If $L_k$ has the equation $ax+by-c = 0$, we can check the signs of $ax+by-c$ at $M$ and at $L_i \cap L_j$. The line $L_k$ separates the two points if we get different signs.

In terms of efficiency: this takes $\Theta(n^3)$ operations, which might be the best we can do if we only need to check this for one point $M$.

If we're going to be doing this check for lots of points, we can probably get this down to an $\Theta(\log n)$ search. "Probably" because I'm not sure which guarantees we can make about balance; but definitely we can promise $O(n)$. There's only a few different possible answers: one for each region. On average, a single sign test distinguishes between about half the regions. You just need to build a binary tree that plans out which signs to check in which order.

Start with all $\binom n2 + 2n$ intersection points. Pick a line $L_i$ such that the two groups "points with a nonnegative sign test" and "points with a nonpositive sign test" are as equal in size as possible. Then repeat with each group, recursively. At a level where no lines distinguish the remaining points, we know that we have the corners of a region, and that's a leaf of our tree.


问:
when $cl_{\tau} (A^c) = A^c$?
A subset $S$ of $ (X, \tau) $ is said to be  preopen  if $ S \subset int_{\tau} (cl_{\tau} (S))$.
The family of all   preopen subsets of $ (X, \tau) $ is denoted by $ PO(X, \tau)$.
The complement of a  preopen set in $ (X, \tau) $ is called preclosed.

let $A \subset X$ be both preopen and preclosed subset in $ (X, \tau) $. Is $cl_{\tau} (A^c) = A^c$?

Do space $X$ or $A$ require a specific condition to $cl_{\tau} (A^c) = A^c$?

答: No to the first question, take $A=\Bbb Q$ in the reals (standard top.) Then $A$ is pre-open and pre-closed,but the closure of $A^\complement$ is $\Bbb R$, not $A^\complement$.

Title: when $cl_{\tau} (A^c) = A^c$?


问:
Integrate: $\int_0^1 \frac{1}{\sqrt[3]{x^2 - x^3}}dx$
How to integrate using Residue theorem.
$$\int_0^1 \frac{1}{\sqrt[3]{x^2 - x^3}}dx$$

How do I choose my branch-cut particularly? I was reading this article on wikiepdia and I think it is related. What I don't understand is

"The cut of $z^{3/4}$ is therefore $(−∞, 0]$ and the cut of $(3−z)^{1/4}$ is $(−∞, 3]$. It is easy to see that the cut of the product of the two, i.e. $f(z)$, is $[0, 3]$". What is this product? Is it $(-\infty, 3]\setminus (-\infty, 0] \cup {0}$ ?

Why is $0\le\arg((3-z)^{(1/4)}) \le 2\pi$? the start of angle is taken at negative $x$ axis (counterclockwise) while $0\le\arg(z^{(3/4)}) \le 2\pi$ is taken along positive axis (counterclockwise)?

Also in my integral, I think I don't have residue outside the contour, do I? Thanks for your help in advance!!

ADDED::

My given hint says to use this contour but this quite different from the one in Wikipedia?

答:
In a case like this, you have to contend with a residue at infinity.  You can see this from the contour in your hint: as the radius of the circular contour $R \to \infty$, the integral about that contour approaches

$$i R \int_0^{2 \pi} d\phi \, \frac{e^{i \phi}}{\left(R^2 e^{i 2 \phi}-R^3 e^{i 3 \phi}\right)^{1/3}} \sim i 2 \pi (-1)^{-1/3}$$

However, you have to subtract out that dumbbell piece which excludes the branch points from the interior.  This is where things get tricky.  To this effect, we define

$$z^{-2/3} = e^{-(2/3) \log{z}}$$

such that $\arg{z} \in [-\pi,\pi)$.  This definition is a result of the original branch cut of this factor being $(-\infty,0]$.  The reaosn the branch is defined this way is because the argument of the log is negative real along the branch cut.  Further define

$$(1-z)^{-1/3} = e^{-(1/3) \log{(1-z)}}$$

such that $\arg{(1-z)} \in [0,2\pi)$.  This definition is a result of the original branch cut of this factor being $(-\infty,1]$.   The reason the branch is defined like this is because, along the branch, $1-z$ is positive real along the branch cut.

To summarize, on the lines above and below the real axis, $z=x \in [0,1]$ and therefore $\arg{z} = 0$.  On the line above the real axis, however, $\arg{(1-z)} = 2 \pi$.  Therefore above the real axis,  $z^{-2/3} (1-z)^{-1/3} =  x^{-2/3} (1-x)^{-1/3} e^{-i 2 \pi/3}$  Below the real axis,  $z^{-2/3} (1-z)^{-1/3} =  x^{-2/3} (1-x)^{-1/3}$ because there, $\arg{(1-z)} = 0$.

Further, it should be clear that the integrals about the small circular arcs of radius $\epsilon$ around the branch points vanish as $\epsilon^{1/3}$.

Therefore, we may write

$$\left ( 1-e^{-i 2 \pi/3}\right) \int_0^1 dx \: x^{-2/3} (1-x)^{-1/3} = i 2 \pi (-1)^{-1/3}$$

Because that residue was calculated from the $1-z$ term, then $-1=e^{i \pi}$ and we have

$$\int_0^1 dx \: x^{-2/3} (1-x)^{-1/3} = i 2 \pi \frac{e^{-i \pi/3}}{1-e^{-i 2 \pi/3}} = \frac{\pi}{\sin{(\pi/3)}} = \frac{2 \pi}{\sqrt{3}}$$


问:
Evaluate $\sin (\cot ^{-1} (1+x)) = \cos (\tan^{-1} (x))$
Let $cot^{-1} (1+x)=a$

Then $x =1-\cot a$

So $$\sin a =\cos (\tan^{1}(1-\cot a))$$

$$a=\frac{\pi}{2}- \tan^{-1} (1-\cot a)$$
$$\cot a =\frac 12$$

So $x=\frac 12$, but given answer is $-\frac 12$. What am I doing wrong?

答:
$\sin(\cot^{-1}(1+x))=\cos\left(\dfrac\pi2-\cot^{-1}(1+x)\right)=\cos(\tan^{-1}(1+x))$

We need $$\tan^{-1}(1+x)=2n\pi\pm\tan^{-1}x$$  where $n$ is any integer

Considering the '-' sign,   $$\tan^{-1}(1+x)+\tan^{-1}x=2n\pi$$

Using Principal values, $$-\pi<\tan^{-1}(1+x)+\tan^{-1}x<\pi\implies n=0$$

Consequently, $$\tan^{-1}(1+x)=-\tan^{-1}x=\tan^{-1}(-x)$$

$$\iff1+x=-x\iff x=?$$

Check for the '+' sign


问:
What does this notation mean. (Asymptotic approximation, from Hardy's Number Theory textbook)
From An Introduction to the Theory of Numbers, Hardy, 6th ed, on Page 9, it says,

We may also observe that $f$ ~ $\phi$ is equivalent to $f = \phi + o(\phi)$ or to

$f= \phi{\{1 + o(1)\}}$

In these circumstances we say that $f$ and $\phi$ are asymptotically equivalent...

So what does $f= \phi{\{1 + o(1)\}}$ mean? I am especially confused about the brackets.

Thanks!

答: It is a weird and non-standard choice of brackets, but clearly means nothing more than $φ(1+o(1)) = φ+φ·o(1) = φ+o(φ)$.


问:
How to find the area of triangle ABC?
How to find the area of triangle ABC if BP is an angle bisector?

Title: How to find the area of triangle ABC?

Body:
$$AB=x;\;BC=y$$
$$\begin{cases}
\frac{x}{1} =\frac{y}{2}\to y=2x\\
x^2+y^2=9\\
\end{cases}
$$
$$x^2+(2x)^2=9$$
$$x=\frac{3}{\sqrt 5};\;y=\frac{6}{\sqrt{5}}$$
$$Area=\frac{1}{2}xy=\frac{9}{5}$$


问:
Determine whether language L is regular
Let $L\subseteq \{0, 1\}^*$ be the set of natural numbers in binary notation, which together compose the infinite arithmetic progression with first term 4 and common difference 3. Is it true that $L$ is regular?

I tried to tackle this down using $a^{-1}L$ but failed to accomplish much.

Any help is very much appreciated.

答: Essentially you are looking for all of the binary numbers greater than 1 that have a remainder of 1 modulo 3.  If you think about how appending a 0 or 1 to the end of a binary number changes its modulus modulo 3, you shouldn't have any trouble building a FSM for this language.</p>

Title: Determine whether language L is regular


问:
Finiteness and Lower Semi-Continuity of an Functional.
Assume I have a probability density $\rho$ on $\mathbb{R}^n$ with finite second moment

$$ \int_{\mathbb{R}^n}\|x\|^2\rho(x) dx<C. $$

I'm now interested in the following functional

$$ F(\rho):=\int_{\mathbb{R}^n} f(x) \rho(x)dx, $$

for some $f\in C(\mathbb{R}^n)$ which is bounded from below : $f(x)\geq c$ for some $c\in\mathbb{R}$.

Two questions, under what conditions on $f$ do we need for

$F$ finite?

$F$ lower semi-continuous?

Is $f$ Lipschitz enough or Im guessing it WILL depend on $\rho$?

EDIT : maybe Theorem $2.38$ of Functions of bounded variation and free discontinuity problems by Luigi Ambrosio could help?

答:
For $F$ to be finite you need that $\frac{f(x)}{\|x\|^2+1}$ is bounded.

This is clearly sufficient.

Lets check that it is necessary:

Suppose you have a sequence $x_n$ with $\frac{f(x_n)}{\|x_n\|^2+1}≥ 4^n$. By continuity of $f$ these $x_n$ will eventually escape every compactum, so you may assume $\|x_n- x_m\|>1$ for $n\neq m$ and $\|x_n\|>1$ for all $n$. Around each $x_n$ you have a ball on which $f(x)≥4^n\|x_n\|^2$ holds, wlog this ball has radius $<1$ so that all these balls around $x_n$ are disjoint.

On a given such ball look at some bump-function $\rho_n$ supported on the ball so that $\int \rho_n = \frac{2^{-n}}{\|x_n\|^2+1}$. It is clear that $\rho:=\sum_n \rho_n$ remains smooth, positive and
$$\int \rho<\infty, \quad \int\|x\|^2\rho ≤\sum_n 2^{-n}=1$$
By rescaling $\rho$ you may make it into a probability measure, it is then of the class you are considering. Now
$$F(\rho) =\sum_n \int f\, \rho_n ≥\sum_n 4^n\|x_n\|^2\int \rho_n = \sum_n 2^n \frac{\|x_n\|^2}{\|x_n\|^2+1}=\infty$$

Note that if $f$ is Lipschitz this condition is satisfied.

Now about lower-semi-continuity:

Fix some $\rho$ and consider a net $\rho_\alpha\to\rho$ weakly. Let $\psi_n$ be the characteristic function of the ball of radius $n$. By weak convergence you've got $\int \psi_n\rho_\alpha \to \int\psi_n \rho$ as $\alpha\to\infty$. But you've also got that
$$\int \psi_n \rho\to 1$$
as $n\to\infty$. This means that for every $\epsilon$ you've got an $n(\epsilon)$ so that the mass of $\rho$ outside of the  ball of radius $n$ is smaller than $\epsilon$. Additionally the two equations mean you've got an $A(\epsilon)$ so that for all $\alpha>A$ the mass of $\rho_\alpha$ outside of the ball of radius $n(\epsilon)$ is smaller than $2\epsilon$.

Now $f\cdot \psi_n$ is bounded and hence
$$\int f \psi_n\, \rho_\alpha\to \int f\psi_n \,\rho$$
Make use of this, for $n=n(\epsilon)$ and $\alpha ≥ A(\epsilon)$ you've got:
$$\int f \rho_\alpha = \int f \psi_n\,\rho_\alpha + \int f(1-\psi_n)\rho_\alpha ≥ \int f \psi_n \rho_\alpha + 2c \epsilon$$
the term on the right converges, as $\alpha\to\infty$, towards
$$\int f\psi_n\, \rho+ 2c\epsilon$$
where $\epsilon$ is as small as you please although $n$ will get larger with smaller $\epsilon$. But as $n$ goes to infinity the integral $\int f\psi_n\,\rho$ converges to $\int f\rho$ (here you again use that $f$ is bounded below and something like Beppo-Levi).

All these words amount to:
$$\limsup_\alpha \int f\,\rho_\alpha ≥ \int f\rho$$

So $F$ is lower semi-continuous if $f$ is bounded below (finiteness not necessary).


问:
Combining multiple probability density functions.
I'm trying so hard to clear out my simple confusion but I'm getting nowhere near to it :(

Let's say, the PDF for weight measurement of all adults (age $\ge$ 18) is given $\large p_{adults}$, using this, I can tell what's the probability that an adult person would weigh between a Kgs to b Kgs...

And the PDF for weight measurement of all non-adults (age $\lt $ 18) is given $\large p_{non-adults}$, and using this, I can tell what's the probability that a non-adult person would weigh between c Kgs to d Kgs...

Since all humans is either an adult or not... can't I find a PDF that would represent the weight measurement of all humans? Or in other words, what would be a single PDF such that can cover all the cases of $\large p_{adults}(x)$ and $\large p_{non-adults}(x)$?

If this is possible to find then how?? If not then why not??

I am thinking that the answer should be yes, and that single PDF might be represented by some operation between $\large p_{adults}(x)$ and $\large p_{non-adults}(x)$, like addition or multiplication, or maybe convolution or something like that... what would that be?

答:
Use the law of total probability. Let $F_A$ and $F_C$ be the cumulative distribution functions for the weights of adults and children, respectively. If the proportion of adults in the population is $p$, then
\begin{align*}
\mathbb P(X\leq x)&=\mathbb P(X\leq x\mid\text{Adult})\cdot\mathbb P(\text{Adult})+\mathbb P(X\leq x\mid\text{Child})\cdot\mathbb P(\text{Child})\\
&=pF_A(x)+(1-p)F_C(x).
\end{align*}
Differentiating implies that the density is given by $f(x)=pf_A(x)+(1-p)f_C(x)$.

This is a relatively simple example of mixture distributions.


问:
Prove that the stopping time is finite

Let $Y_0 = 0$ and $(Y_n)_{n \in \mathbb{N}_0}$ be an iid sequence of random variables with zero mean and attaining exactly values in $\{1, 0, -1\}$ with positive probability. Consider the filtration $F^Y$ generated by $Y = (Y_n)_{n \in \mathbb{N}_0}$ . Define the process $X = (X_n)_{n \in \mathbb{N}_0}$ via $X_n = \sum_{k=0}^n Y_k$ for all $n \in \mathbb{N}_0$. Moreover, define $\tau = \inf \{n \in \mathbb{N}:X_n = 1\}$, which is a stopping time corresponding to $F^Y$. Is $\tau$ finite?

This is an exercise I need to solve. Intuitionally, I think it is not true. I found a similar topic here (Martingale: Show $p\{T<+\infty \}=1$.), but the definition of $\tau$ in that topic is different, and that difference enables us to conclude from the equation $()−(+1)=(−1)−()$ and $(−)=()=1$ that $u(i) = 1$ for all i. In this situation, we only have $u(1) = 1$, so we cannot use the same reasoning to conclude that. Can anyone give me some hint on how to solve this?

答:
Let $p=P[Y_n=1] = P[Y_n=-1]$ (equal because $E[Y_n]=0$), and $r=P[Y_n=0]$ for $n\ge 1$. Clearly $2p+r=1$. Let $b:=P[\tau<\infty]$. By conditioning on the first step of the walk you see that
$$
b = p+rb+pb^2.
$$
That is, because $1-r = 2p$, $b$ solves the quadratic
$$
pb^2-2pb+p=0.
$$

Title: Prove that the stopping time is finite


问:
Determine the cardinality of the set $E$
Determine the cardinality of the set:

$$E=\{f:\mathbb{N}\longrightarrow\mathbb{N} \mid\forall n\in\mathbb{N} \ , f(n)=f(n+1)+f(n+2)\}$$

What can we say about the decreasing functions $f:\mathbb{N}\longrightarrow\mathbb{N}$?

I've tried looking for examples of something similar in some books, but haven't found anything about it.

答: The set $E$ is empty as for any $n \in \Bbb N$, consider $f(n) \in \Bbb N$. Then by construction we must have infinitely many distinct natural numbers $f(n+1),f(n+2),\dots$ such that each of them are $< f(n)$, which is a contradiction as $[1,f(n)) \cap \Bbb N $ is a finite set!


问:
Let $\exp( i u X_{t}+\frac{1}{2}u^{2}t)$ be a local martingale for any $u$, show that $X_{t}$ is a Brownian motion
Let $X$ be an $\mathcal{F}_{t},\; t\geq 0$, adapted a.s. continuous process with $X_{0}=0$ and    $M^{u,X}_{t}:=\exp( i u X_{t}+\frac{1}{2}u^{2}t)$ be a local martingale for any $u$, show that $X_{t}$ is a $\mathcal{F}_{t}$-Brownian motion.

My idea:

We need to show that $\operatorname{cov}(X_{t},X_{s})=t \land s$ and that $X$ is a Gaussian process.

Let $0 \leq t_{0} < ... < t_{n} <\infty$. I know that for $T > 0, \; u \in \mathbb R$ the process $M_{t\land T}^{u,X}=\exp( i u X_{t\land T}+\frac{1}{2}u^{2}(t\land T))$ is a local martingale too. Furthermore, $\lvert M^{u,X}_{t\land T}\rvert\leq \exp(\frac{1}{2}u^{2}T)< \infty$.

Therefore, $(M_{t\land T}^{u,X})_{t\geq 0}$ is a martingale. By Doob's optional sampling Theorem, we obtain for $t < T$,  $$E[M_{t}^{u,X}]=E[M_{t\land T}^{u,X}]=E[M_{0\land T}^{u,X}]=E[M_{0}^{u,X}]=1$$

This would then imply that $$ E[\exp(iuX_{t})]=\exp(-\frac{u^{2}t}{2})$$ which is the characteristic function of a normal random variable $\mathcal{N}(0,t)$ evaluated at $u$. But I am struggling to do the same for the random vector $(X_{t_{0}},...,X_{t_{n}})$ and $0 \leq t_{0} < ... < t_{n}< \infty$. I think I need to use induction and the tower property to reduce it all to the random variable $X_{t_{0}}$, but that seems less efficient.

On the issue of showing $ \operatorname{cov}(X_{t},X_{s})= t\land s$, I am completely stuck. Maybe I can set $u:=\log(E[X_{t}X_{s}])$ and by assumption: $ M^{u,X}_{r}:=\exp( i \log(E[X_{t}X_{s}]) X_{r}+\frac{1}{2}\log(2E[X_{t}X_{s}])r)$ is a local martingale

We then have:

$$ E[\exp( i \log(E[X_{t}X_{s}]) X_{r}]=\exp(-\frac{\log(E[X_{t}X_{s}])r}{2})$$

Then

$$ E[\exp( i \log(E[X_{t}X_{s}]) X_{r}]=E[X_{t}X_{s}]^{-\frac{r}{2}}$$

I do not see how I get to $t\land s $ from here. Any ideas?

Or is something simply missing in the question's assumptions?

答:
You've made a good start. An other approach to the covariance and the fact that $X$ is a Gaussian process would be to use the martingale property of $M^{u,X}$ and $M^{v,X}$ to
compute $E[\exp(iuX_s+ivX_t)]$ for $0<s<t$. This will show that $X_u$ and $X_v$ are jointly Gaussian with the anticipated covariance. Similar reasoning handles the joint law of $X_{t_1}, X_{t_2},\ldots , X_{t_n}$.


问:
If $F$ and $R$ are subspaces of vector space $E$, then $F \cap R \neq \varnothing$
I need to prove the following:

let $F \cap R$ intersection of vector subspaces $F$ and $R$ of vector space $E$, then $F \cap R \neq \emptyset$

Thanks in advance!

答: Hint: What vector must every vector space (or subspace, which is in its own right a vector space) contain? </p>


问:
Calculate rotation matrix that flips the frame up
Let's start with the standard basis frame: [1, 0, 0], [0, 1, 0], and [0, 0, 1].

Imagine this frame goes through some arbitrary rotation R, with one constraint, that we know that after the R is applied to the frame, the z-axis becomes [0, 0, -1].

I want to calculate a rotation that "flips" the [0, 0, -1] back to the [0, 0, 1] vector with the smallest angle possible. Note: Inverting the R is NOT correct, because we want the smallest angle, and inverting R would always completely erase the rotation to give us the identity matrix.

The top answer here: Calculate Rotation Matrix to align Vector A to Vector B in 3d? does not apply, because I am describing the edge case that Jur van den Berg says is not supported.

Geometrically, if we start with the orientation corresponding with the z-axis at [0, 0, -1], I think to "flip" the frame we need to swap the corresponding x and y axises, and simply multiply [0, 0, -1] by -1 to get [0, 0, 1]. You can understand what I'm saying if you use your hand to form a frame by right hand rule, then 'flip' your thumb to negate the z-axis, and then 'flip' it back. How do I calculate this 'flip' relative rotation for any arbitrary orientation?

答:
To expand a bit on tpofofn's answer, there is an infinite number of such transforms, described by a 180° rotation around an unix axis vector $\hat{a}$,
$$\hat{a} = \left [ \begin{matrix} \cos\varphi \\ \sin\varphi \\ 0 \end{matrix} \right ]$$

In other words, all rotation matrixes of form
$$\mathbf{R} = \left [ \begin{matrix}
\cos\varphi & \sin\varphi & 0 \\
\sin\varphi & -\cos\varphi & 0 \\
0 & 0 & -1 \\
\end{matrix} \right ]$$
will work. $\varphi$ only determines the direction of the rotation axis on the $x y$ plane.

By OP's own metric, all these rotation matrices are equally preferable, because they rotate by angle $\theta = 180°$; only the axis around which we rotate changes.


问:
Prove that a strictly decreasing function from $f:\Bbb R \to \Bbb R$ is one-to-one
I would like to prove that a strictly decreasing function from $f:\Bbb R \to \Bbb R$ is one-to-one.

We want to show that show that $f(a) = f(b)$ implies $ a = b$ for all $a, b \in \Bbb R$.

One proof I saw online was as follows (although I did the same proof using contrapositive technique), but I just want to get better understanding as to why he did the proof as follows:

Proof:

Since the function is strictly decreasing, it means that if  $ x \lt y \implies f(x) \gt f(y)$. To proof that it's one-to-one function, we need to prove that if $f(a)=f(b) \implies a=b$.

Let $f(a) = f(b)$.

Case 1:  Consider when $a \lt b$, then this implies that $f(a) \gt f(b)$ since $f(x)$ is strictly decreasing. This implies that $f(a) \ne f(b) \therefore a\ge b $.

Case 2: Consider when $a \gt b$, then this implies that $f(a) \lt f(b)$ since $f(x)$ is strictly decreasing. This implies that $f(a) \ne f(b) \therefore a = b $.

Questions:

It seems the proof that was used in the question is proof by cases, was not it?

Why it was assumed, in Case 1, that $f(a) = f(b)$ although what is given in the question is that $f(x)$ is strictly decreasing?

Why it was concluded ,in Case 1, that since $f(a) \ne f(b) \therefore a\ge b $?

I assume that it was finally concluded that $\therefore a = b $ is because no other scenarios left as to why $f(a) = f(b)$ except by equality of $a$ and $b$.

答:
This proof really overcomplicates it. Rather, prove the contrapositive:

$$x \neq y \implies f(x) \neq f(y)$$

This is then easy, because either $x < y$ or $y < x$ and thus either $f(x) < f(y)$ or $f(y) < f(x)$. Hence, $f(x) \neq f(y)$ when $x\neq y$.


问:
Find the parameters in $\int_{-\pi/2}^{\pi/2}f(x)\,dx\approx w_1f(x_1)+w_2f(x_2)$ such that the formula will be exact for $\sin(2x),\sin(x),\cos(x),1$

Consider the following integration rule on $[-\pi/2,\pi/2]$: $$\int_{-\pi/2}^{\pi/2}f(x)\,dx\approx w_1f(x_1)+w_2f(x_2).$$ Find the parameters such that the formula will be exact for $\sin(2x),\sin(x),\cos(x),1$.

I've done the following system of equations

\begin{align}
&\int_{-\pi/2}^{\pi/2}1\,dx=\pi=w_1+w_2\tag{1}\\
&\int_{-\pi/2}^{\pi/2}\cos(x)\,dx=2=w_1\cos(x_1)+w_2\cos(x_2)\tag{2}\\
&\int_{-\pi/2}^{\pi/2}\sin(x)\,dx=0=w_1\sin(x_1)+w_2\sin(x_2)\tag{3}\\
&\int_{-\pi/2}^{\pi/2}\sin(2x)\,dx=0=w_1\sin(2x_1)+w_2\sin(2x_2)\tag{4}
\end{align}

I've tried to solve the system but didn't get anywhere.

Body:
A start for a solution:

Expand $(4)$
$$
2w_1\sin(x_1)\cos(x1) = -2w_2\sin(x_2)\cos(x_2)
\tag{$4'$}$$
Rewrite $(3)$
$$
w_2\sin(x_2) = -w_1\sin(x_1)
\tag{$3'$}$$
Substitute $(3')$ into $(4'$)
$$
2w_1\sin(x_1)\cos(x_1) = 2w_1\sin(x_1)\cos(x_2)
$$
Then reduce to three cases:

(A) $w_1 = 0$;

(B) $\sin(x_1)=0$ so $x_1 = 0$;

(C) $\cos(x_1) = \cos(x_2)$ so $x_1=-x_2$.

Try to continue...

I get:
$$
w_1=\frac{\pi}{2}\\
w_2=\frac{\pi}{2}\\
x_1=\arccos\frac{2}{\pi}\\
x_2=-\arccos\frac{2}{\pi}
$$


问:
$|\operatorname{Aut}(K/F)| \leq [K:F]_s$ holds in general?
Let $K/F$ be finite extension. $$|\operatorname{Aut}(K/F)| \leq [K:F]$$ is a fundamental equation, and equality holds if the extension is Galois.

My question is; more strict inequality,
$$|\operatorname{Aut}(K/F)| \leq [K:F]_s$$ holds in general? $[K:F]_s$ is separable degree of $K/F$. My text seems to be using this without assumption that $K/F$ is separable.

If there are counter examples, it is also appreciated. Thank you for your help.

答:
Lemma. Let $E/k$ be separable, and let $\sigma: k\rightarrow k^a$. Then $\sigma$ extends to an embedding of $E$ in $k^a$, and there are $[E:k]$ distinct extensions of $\sigma$. The number of such distinct monomorphism is $\lt [E:k]$ if $E/k$ is not separable.

Proof. We will prove the lemma by induction. Let $\alpha\in E$, then the minimal polynomial $\text{Irr}(\alpha,k)$ has no multiple roots. Any embedding of $k(\alpha)$ sends $\alpha$ to a root of $\text{Irr}(\alpha,k)$, so there are $\text{deg}(\text{Irr}(\alpha,k))=[k(\alpha):k]$ distinct extensions. For any $\beta\in k(\alpha)$, since $\text{Irr}(\beta,k(\alpha))\mid \text{Irr}(\beta,k)$, it follows from our assumption that $\text{Irr}(\beta,k(\alpha))$ is separable. By induction hypothesis, the number of distinct extensions of a given embedding of $k(\alpha)$ on $E$ is $[E:k(\alpha)]$. Hence the number of distinct extensions of $\sigma$ on $E$ is $[E:k(\alpha)]$. The second assertion is proven in essentially the same way.

Let $K$ be an extension of $F$. Given any $\alpha\in K$, let its minimal polynomial over $F$ be $f(x)$. Let the roots of $f(x)$ be $\alpha_1=\alpha, \alpha_2,\cdots, \alpha_
r$, then $f(x)=\prod_{j=1}^r (x-\alpha_j)^{m_j},$ and for each $\alpha_j$ there exists an isomorphism $\sigma: k(\alpha)\rightarrow k(\alpha_j)$, which may be extended to an automorphism of $K^a$. Then all $m_j$ are equal.

If $\text{char}F=0$, any irreducible polynomial in $F[x]$ is separable, so $[K:F]=[K:F]_s$. We assume that $\text{char}F=p\gt 0$. If $f(x)$ is inseparable over $F$, its formal derivative $f'(x)$ shares a common root with $f(x)$. But this is impossible unless $f'$ is identically $0$, and thus $p\mid m_j$ i.e. $f(x)=g(x^p)$ for some $g(x)\in F[x]$. Proceeding inductively, we have $f(x)=h(x^{p^m})$ for some separable polynomial $h(x)\in F[x]$. Then $h$ is irreducible in $F[x]$ and hence is the minimal polynomial of $\alpha^{p^m},$ and $\text{deg}(g)=r=[K:F]_s$. Therefore by our lemma above, given $\sigma: F\to F^a$, the number of distinct monomorphism extending $\sigma$ is $[K:F]_s$. Hence $|Aut(K/F)|\le [K:F]_s$.

We remark that if $K/F$ is normal, any monomorphism $\sigma: K\to F^a$ is an automorphism on $K$, so $|Aut(K/F)|= [K:F]_s$.


问:
Shortest distance between two vertices
In the graph, we have to find the shortest distance between Newark and Cape May. I have tried using the Dijkstra algorithm but I can't seem to figure out how it would work. According to my working, the distance is coming out to be 175 whereas it should be 85.

答:
Djikstra's algorithm gives Newark-Woodbridge-Camden-Cape May with a distance of $165$.

I suspect that your error was to permanently box Atlantic City (on $130$) before you had boxed Camden (on $80$).

Title: Shortest distance between two vertices


问:
Hypothesis testing, is my calculation correct?
Is one able to say that the average weight of candy bars is at least 18.2g?

$n = 10$

$\bar{x} = 18.57$

$\sigma = 0.2$

level of significance $\alpha = 5.71\%$

$$H_{0} = \mu \leq \mu_{0}$$
$$H_{1} = \mu > \mu_{0}$$

(I'm not sure whether this is correct. I've heard that the thing you're trying to prove should always be $H_{1}$. The thing we're trying to prove is $\mu \geq \mu_{0}$, but then i heard that the equal sign should never be in $H_{1}$)

Assuming my $H_{0}$ and $H_{1}$ are correct:

Critical value of $\alpha$ : $-1,579$

$$ z=\frac{\bar{x} - \mu_{0}}{\frac{\sigma}{\sqrt{n}}}=\frac{18.57-18.2}{\frac{0.2}{\sqrt{10}}}= 5.85$$

Since z-value > critical value, reject $H_{0}$

答:
First, note that the null hypothesis or $H_0$ is the statement that we try to reject. If we succeed in rejecting $H_0$ then we can accept the alternative hypothesis, or $H_1$. So we always formulate our work such that we can (maybe) reject the case that we are not trying to show!

A small but important point is that we do not assume that $H_0$ and $H_1$ are both correct. Rather, we start by assuming that $H_0$ is correct, and we will see if it leads us to a conclusion that is probable or not probable (and our criteria for deciding what probabilities to accept is the level of significance). Edit: I may have misunderstood your sentence there. If you meant to assume that you have formulated $H_0$ and $H_1$ correctly, then I agree with you. (Thanks @5201314 for pointing it out)

Now, your calculation of z-value is correct. But there is a problem with the critical value that you identified. I see you have checked the tables (or used a software) to find the critical value corresponding to $\alpha = 5.71\%$. This is a mistake in this exercise. Here, you should find the critical value corresponding to $100\%-\alpha$ (or $1-\alpha$ if it is written as a proportion). This is because of the $H_0$ of this problem: we are assuming that the mean is less than or equal to $\mu_0$. You might say that the improbable values that violate our assumption are then very large values. The figure below shows the idea.

Finally, your conclusion is correct, because it just happens that the z-value is still larger than the correct critical value, but in an exam, your answer would probably lose marks.


问:
Conditions for number of solutions for a given system of equations
I need to find the conditions for $a$ and $b$ such that
$$3x-2y+z=b$$
$$5x-8y+9z=3$$
$$2x+y+az=-1$$
has (i) unique solution, (ii) no solution (iii) infinitely many solutions

Here's what I did. Firstly, I put the situation into an augmented matrix form, and tried reducing to row echelon form to get the rank. Let C be the augmented matrix $C=[A:B]$ for the system of equations $AX=B$

$$ C= \left[
\begin{array}{ccc|c}
  3&-2&1&b\\
  5&-8&9&3\\
  2&1& a & -1\\
\end{array}
\right] $$

The operations that I used, in order, are:

$$R_1 \rightarrow R_1-R_2$$

$$R_1\rightarrow \frac{-R1}{2}$$

$$R_2\rightarrow R_2-5R_1$$

$$R_3\rightarrow R_3-2R_1$$

$$R_3 \rightarrow R_3-R_2$$
Reducing using these operations to row echelon form, I got:

$$ C= \left[
\begin{array}{ccc|c}
  1&-3&4&\frac{3-b}{2}\\
  0&7&-11&3+5(\frac{3-b}{2})\\
  0&0& a+3 & \frac{7b-29}{2}\\
\end{array}
\right] $$

I know, for unique solution, the ranks of A and C must be equal, and they should be equal to 3 (the number of unknowns), i.e.,

$$\rho(A)=\rho(C)=3$$

$($rank of the matrix A is being denoted by $\rho(A)$$)$

$\implies a+3\neq0$ and, $b \in \mathbb{R}$

This is the correct answer.

However, for no solution, I know that $\rho(A)\neq\rho(C)$

That is possible when, $a+3=0$ and $\frac{7b-29}{2} \neq 0$ $\implies b \neq \frac{29}{7}$

But, the answer says the conditions are $a = -3 $ and $b \neq \frac{1}{3}$ for the no solution part. What is my mistake here? I don't know where I went wrong, any help would be nice. If someone can point out the error in my steps, it would be very helpful

答: It seemed to be a calculation error. I have now found that the previous row-echelon form that I got was incorrect</p>


答: $85$ does not make sense: from Cape May you could only reach Camden and Atlantic City.

Title: Shortest distance between two vertices

Body: In the graph, we have to find the shortest distance between Newark and Cape May. I have tried using the Dijkstra algorithm but I can't seem to figure out how it would work. According to my working, the distance is coming out to be 175 whereas it should be 85.


问:
Solve $e^{2z} - 2e^z + 2= 0$
So I've started by looking at

\begin{align} e^{z} =  x:\\
x^2 - 2x + 2 = 0
\end{align}

Whose solutions should be: \begin{align}\ 1+i, 1-i \end{align}

Then I did the following:

\begin{align}
e^z=e^{x+iy}=e^x(\cos y+i\sin y)&=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}=\frac{1}{2}+\frac{i \cdot 1}{2}\\
e^x= {\sqrt{2}} \quad \text{and}\quad y&=\frac{\pi}{4}+2\pi k\quad k \in \Bbb{Z}\\
x&= \sqrt{2} \quad \text{and} \quad y=\frac{\pi}{4}\\
\text{Also,} \quad e^x(\cos y+\sin y)&=\cos(\frac{7\pi}{4})+i\sin\frac{7\pi}{4})=\frac{1}{2}-\frac{i \cdot 1}{2}\\
e^x= \sqrt{2} \quad \text{and}\quad y&=\frac{7\pi}{4}+2\pi k\quad k \in \Bbb{Z}\\
x&= \sqrt{2} \quad \text{and} \quad y=\frac{7\pi}{4}\\
\end{align}

Are the solutions the following? \begin{align} \sqrt{2} + \frac{\pi}{4} \quad \text{and} \quad \sqrt{2} + \frac{7\pi}{4} \end{align}

Forgive the shoddy formatting as this is my first ever post here. Thanks.

答: Yes, the solutions of $x^2-2x+2=0$ are $1+i$ and $1-i$. So, the solutions of $e^{2z}-2e^z+2=0$ are all those numbers $z$ such that $e^z=1+i$ or that $e^z=1-i$. But$$1+i=\sqrt2\left(\frac1{\sqrt2}+\frac i{\sqrt2}\right)=\sqrt2e^{\pi i/4}. $$Therefore$$e^z=1+i\iff z=\log\sqrt2+\frac{\pi i}4+2\pi in,$$for some $n\in\Bbb Z$. The case of the equality $e^z=1-i$ is similar.

Title: Solve $e^{2z} - 2e^z + 2= 0$


问:
Convergence of improper unbounded integral

Show that $\displaystyle \int_{0}^{1}  \frac{1}{x^2 + \sqrt{x}} dx$ converges.

My method : This is a type of improper integral where the function becomes unbounded at the lowe limit of integration and both the limits of integration are finite. I took the given function as $f(x)$ being integrated from $0$  to $1$ and another function $g(x) =  \frac{1}{\sqrt{x}}$ by taking $\frac{1}{\sqrt{x}}$ common from $f(x)$ and observing that at $x=0$ the convergence and divergence of the function $f(x) $ was solely dependent by this $\frac{1}{\sqrt{x}}$

I was taught a limit test to find the convergence or the divergence. So it says that when the function is of the unbounded type with finite bounds to integrate  :

Find $\lim_{x \to 0 } \frac{f(x)}{g(x)}$ which I get as 1 in this question. Since the limit exists and is not equal to zero, whatever is the behaviour of $g(x)$, that is the behaviour of $f(x) $ also. Therefore by using the test integral I find that $g(x)$ diverges therfore $f(x) $ should also diverges. But the answer says that it converges. How is this possible ?

Test integral used :

$\displaystyle \int_{a}^{\infty}  \frac{1}{x^p} dx$ where $a>1$;
If p $\leq$ 1 then it diverges.
If p $>$ 1 than it converges.

I know that in the test integral $a>1$ is given but then my prof used it irrespective of it in other examples of the same type (that is in sums of unbounded type where the limits of integration were $0$ to $3$ and $0$ to $4\pi$ in the two examples he gave) . And I also do not understand the $\infty$ limit in the integration of test integral as in the given question I need to integrate only from $0$ to $1$

I know I have asked two questions but since they're related please don't close my answer.

答:
Hint

$$\frac{1}{x^2+\sqrt x}\leq \frac{1}{\sqrt x}.$$


问:
Continuity of a function on C[0,1]
Consider the space $C[0,1]$( space of all complex valued continuous functions on $[0,1]$) with the sup norm $\|{\cdot}\|_\infty$.

Now define the function $\Lambda$ on $C[0,1]$ as $\Lambda(f)=f(x_0)$ where $x_0$ is the point where $\sup_{x\in[0,1]}|f(x)|$ is attained .
Will $\Lambda$ be a continuous function on $(C[0,1], \|{\cdot}\|_\infty)$?

答: So basically, $\Lambda(f)=||f||_{\infty}$, then $\Lambda$ is clearly a sublinear map and also note that $\Lambda(f)=||f||_\infty \implies |\Lambda(f)-\Lambda(g)|=|||f||_{\infty}-||g||_\infty|$ thus it is continuous!

Title: Continuity of a function on C[0,1]


问:
primes of the form $p = x^2 + ny^2$: primitive element for the Hilbert class field of $\mathbb{Q}(\sqrt{-n})$ is a real algebraic integer?
The following proposition is found in $\S 5$ of Primes of the form $p = x^2 + ny^2$: Fermat, Class Field Theory, and Complex Multiplication by David A. Cox:

Proposition 5.29. Let $K$ be an imaginary quadratic field and let $L$ be a finite extension of $K$ which is Galois over $\mathbb{Q}$. Then

(i) There is a real algebraic integer $\alpha$ such that $L = K(\alpha)$.

(ii) Let $\alpha$ be as in (i) and $f(x) \in \mathbb{Z}[x]$ denote its minimal polynomial. If $p$ is a prime not dividing the discriminant of $f(x)$, then
$$
p\text{ splits completely in } L \Leftrightarrow
\cases{(d_K/p) = 1 \text{ and } f(x) \equiv 0\bmod{p}\cr
\text{has an integer solution.}
}
$$
My trouble concerns (i). In his proof Cox uses the fact that $[L \cap \mathbb{R} : \mathbb{Q}] = [L : K] < \infty$ to conclude there exists $\alpha \in \mathscr{O}_L \cap \mathbb{R}$ such that $L \cap \mathbb{R} = \mathbb{Q}(\alpha)$.

The Primitive Element Theorem guarantees the existence of $\alpha \in L \cap \mathbb{R}$ such that $L \cap \mathbb{R} = \mathbb{Q}(\alpha)$, but I do not understand how we are able to select $\alpha$ so that $\alpha \in \mathscr{O}_L$. Perhaps I am missing something obvious?

答:
I was missing something obvious. Following the suggestion of @franzlemmermeyer, I found the following proposition in Chapter 2 of J. S. Milne's course notes for Algebraic Number Theory:

Let $A$ be an integral domain and $L$ be a field containing $A$. Then we have

Proposition 2.6. Let $K$ be the field of fractions of $A$ and let $L$ be a field containing $K$. If $\alpha \in L$ is algebraic over $K$, then there exists a $d \in A$ such that $d\alpha$ is integral over $A$.

Taking $A = \mathbb{Z}$, $K = \mathbb{Q}$, and $L \cap \mathbb{R} = \mathbb{Q}(\alpha)$ the field containing $K$ allows us to select $\alpha \in \mathscr{O}_L \cap \mathbb{R}$.