问:
How to solve a convection-diffusion equation analytically?
For a 1-d convection equation $u_t + c u_x = 0$ on $\mathbb{R}_+ \times \mathbb{R}$ with a constant $c$, we know the solution is simply $u(x,t)=u_0(x-ct)$ for an initial condition $u_0 := u(x,0)$. However, if we add some diffusion to the RHS of the equation, with a constant $b$, to have
$$
u_t + c u_x = b u_{xx}
$$
then how can we solve this type of convection-diffusion equation analytically?

In my point of view, we may solve the heat equation $u_t = b u_{xx}$ first, by separation of variables, and then add the "transport" features to our solution. It seems requiring some knowledge from Fourier transform, which I am not familiar with, to get the solution of the heat equation, and the form is not fundamental but in a form of convolution. Anyway, I guess we may expect to have a final solution to the convection-diffusion equation in the form like:
$$
u(x,t) = \frac{1}{\sqrt{4 \pi b t}} \int_{-\infty}^{\infty} e^{-\frac{(x-ct-s)^2}{4 b t}} u(s,0) ds
$$
based on Prof. Gilbert Strang's lecture note.

However, I am questioning about the proof in details that the form above is exactly the solution to the convection-diffusion equation given.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Write $\ds{\on{u}\pars{x,t}}$ as
$$
\on{u}\pars{x,t} \equiv
\exp\pars{\alpha x - \beta t}\on{U}\pars{x,t}
$$
where $\ds{\alpha}$ and $\ds{\beta}$ are constants.

Then,
\begin{align}
0 & =
-\beta\exp\pars{\alpha x - \beta t}\on{U}\ +
\exp\pars{\alpha x - \beta t}\on{U}_{t}
\\[2mm] &
+ \alpha c\exp\pars{\alpha x - \beta t}\on{U} +
c\exp\pars{\alpha x - \beta t}\on{U}_{x}
\\[2mm] &
-b\alpha^{2}\exp\pars{\alpha x - \beta t}\on{U} -
2b\alpha\exp\pars{\alpha x - \beta t}\on{U}_{x} \\[2mm] &
-b\exp\pars{\alpha x - \beta t}\on{U}_{xx}
\end{align}
Therefore,
\begin{align}
0 & = \pars{\color{red}{-\beta + \alpha c - b\alpha^{2}}}
\on{U} +
\pars{\color{red}{c - 2b\alpha}}\on{U}_{x} +
\on{U}_{t} - b\on{U}_{xx}
\end{align}
Now, I choose $\ds{\color{red}{-\beta + \alpha c - b\alpha^{2}} = 0}$ and
$\ds{\color{red}{c - 2b\alpha} = 0}$
$\ds{\implies \bbx{\alpha = {c \over 2b}}}$ and
$\ds{\bbx{\beta = {c^{2} \over 4b}}}$

Then,
\begin{align}
\on{u}\pars{x,t} & =
\exp\pars{{c \over 2b}\,x -
{c^{2} \over 4b}\,t}\on{U}\pars{x,t}
\\[2mm] & \mbox{where}\ \on{U}\pars{x,t}\
\mbox{satisfies}\
\bbx{\on{U}_{t} = b\on{U}_{xx}} \\ &
\end{align}


问:
Solve differential equation by integrating factor
I have the differential equation $$2\frac{dy}{dx}+3y=e^{-2x}-5$$

I have determined that this needs solving using the integrating factor method.

My workings out are in the image provided.

Are my workings out correct so far? My main question is, how will I integrate with respect for $x$?

$$\int e^{3x/2} \times e^{-2x}-\frac{5}{2}e^{3x/2}~~dx+c$$

this again is shown at the bottom of my workings out.

after integrating this I will then be able to solve for $y$.

Thank you

My workings out for the question

答:
HINT:
$$\int e^{kx}dx=\frac{1}{k}e^{kx}+c$$
where $k$ is any constant. Ask for more detail if you're still stuck :)

ContentLicense: CC BY-SA 4.0


问:
Greatest Integer function limit problem
I need help solving this problem without redefining the greatest interger function,  is that possible?

$$\lim_{x \, \to \,\frac {1}{2}^-}[3x-\frac {1}{2}]$$

What about this rule

答:
Since we are interested in the behavior of $\left\lfloor 3x-\frac{1}{2}\right\rfloor$ as $x$ tends to $\frac{1}{2}$ from the left, we may assume that $\frac{1}{6}<x<\frac{1}{2}$. Then

$$\frac{3}{6}<3x<\frac{3}{2}$$

so

$$\frac{3}{6}-\frac{1}{2}<3x-\frac{1}{2}<\frac{3}{2}-\frac{1}{2}$$

which reduces to $0<3x-\frac{1}{2}<1$. It follows that $\left\lfloor 3x-\frac{1}{2}\right\rfloor=0$, so

$$\lim_{x\to\frac{1}{2}^-}\left\lfloor 3x-\frac{1}{2}\right\rfloor=0$$

ContentLicense: CC BY-SA 3.0

Title: Greatest Integer function limit problem


问:
$\lim\limits_{n\rightarrow\infty} {\frac{24n-8}{n^2+n+1}} \ne 0.5$
Got stuck in disproving limit, hope you can help/hint me. Disprove $$\lim\limits_{n\rightarrow\infty} {\frac{24n-8}{n^2+n+1}} \ne 0.5$$

My effort :
So I did common factor and got to $\left|{\frac {-n^2+47n-17}{2n^2+2n+2}}\right|$

Now I don't know how to proceed.
I tried to do something and got to  $\left|{\frac {n^2-47n+17}{2n^2+2n+2}}\right|$ but it doesn't help.

答:
Analysis flavored answer: Choose $\varepsilon<\dfrac{5}{362},$ then for all $n>48$ we have $\left| \dfrac{24n-8}{n^2+n+1}-\dfrac{1}{2}\right|>\varepsilon,$ and therefore the limit cannot be $\dfrac{1}{2}.$

 This epsilon works because  $\dfrac{5}{362}=\left(\dfrac{1}{2}-\dfrac{88}{181}\right)$, where $\frac{88}{181}$ is just the result of setting $n=48$ into the sequence term, this is less than $\frac{1}{2},$ and the sequence is monotonically decreasing, so the limit cannot be $\frac{1}{2}$.


问:
Pair of functions with same values and same derivatives at distinct points
Let $f,g:[0,1]\to \mathbb{R}$ be two functions of class $C^1$ such that $f'(x)>0$ and $g'(x)>0$ for all $x\in [0,1]$. Assume that
$$
f(0) = g(0) \qquad \text{and} \qquad f(1) = g(1).
$$
Show that there exist $x,y\in[0,1]$ such that
$$
f(x) = g(y) \qquad \text{and} \qquad f'(x) = g'(y).
$$

I made the following attempts (non of them worked): First, define a function $h:[0,1]\times [0,1]\to \mathbb{R}$ given by
$$
h(x,y) = [f(x)-g(y)]^2 + [f'(x)-g'(y)]^2
$$
and try to show that $h(x,y) = 0$ for some $(x,y)$. Then I defined $k:[0,1]\times [0,1] \to \mathbb{R}$ by
$$
k(x,y) = \frac{1}{2}[f(x)-f(y)]^2
$$
and tried to show that at some point $(x,y)$ the divergence of $k$ is zero, because in this case we have that
$$
0 = \frac{\partial k}{\partial x}(x,y) + \frac{\partial k}{\partial y}(x,y) = (f(x)-g(y))f'(x)-(f(x)-f(y))g'(y) = (f(x)-f(y))(f'(x)-g'(y))
$$
but it didn't work either. May be I'm not following the right way to solve this exercise.

答:
Set $a = f(0) = g(0)$ and $b = f(1) = g(1)$. $f$ and $g$ are strictly increasing from $[0, 1]$ to $[a, b]$ and therefore invertible. If we define the function
$$
 h: [a, b]\to \Bbb R, \, h(t) = f^{-1}(t) - g^{-1}(t)
$$
then $h(a) = h(b) = 0$ and Rolle's theorem shows that for some $c \in (a, b)$
$$
 0 = h'(c) = \frac{1}{f'(f^{-1}(c))} - \frac{1}{g'(g^{-1}(c))} \\
\implies f'(f^{-1}(c)) = g'(g^{-1}(c)) \, .
$$
We can now set
$$
  x = f^{-1}(c) \, , \, y = g^{-1}(c) \, .
$$
$x$ and $y$ are both in the open interval $(0, 1)$ and have the desired properties
$$
 f(x) = g(y) \, , \, f'(x) = g'(y) \, .
$$

Illustration: The graphs of $f$ and $g$ are green. $c$ is the argument on the $y$-axis where the inverse functions have the same slope. $x$ and $y$ are the pre-images of $c$ under $f$ and $g$.

(Created with GeoGebra.)

Remarks:

If $f$ and $g$ are not identical functions then $h$ is not identically zero and we can choose $c\in (a, b)$ as a point where $h$ attains its nonzero minimum or maximum, so that $x - y = h(c) \ne 0$.

This shows that (unless $f=g$) a pair $x, y$ of distinct points in $(0, 1)$ with those properties exists.

It suffices to require that the functions are continuous on $[0, 1]$, and differentiable on $(0, 1)$ with positive derivative. The continuity of the derivative, or differentiability at the endpoints is not needed. (This was noticed by Ari Brodsky in a comment.)


问:
About the eigenvalues and singular values of this matrix
Let be
$T_n=\begin{bmatrix}0&...&0&1&...&1\\ \vdots&&\vdots&\vdots&&\vdots\\0&...&0&1&...&1\\1&...&1&0&...&0\\\vdots&&\vdots&\vdots&&\vdots\\1&...&1&0&...&0 \end{bmatrix}$, proof that the eigenvalues of $T_n$ are $0,-n,n$.

My proof

$p(\lambda)=det(\lambda I-T_n)=det\begin{bmatrix}
\lambda^2-n&-n&...&-n\\-n&\lambda^2-n&...&-n&\\ \vdots&&\ddots&\vdots\\-n&&...&-n
\end{bmatrix}$
Let $D=diag(\lambda^2,...,\lambda^2)$ and $v$ the column vector with $n$ ones, note that
$\begin{bmatrix}
\lambda^2-n&-n&...&-n\\-n&\lambda^2-n&...&-n&\\ \vdots&&\ddots&\vdots\\-n&&...&-n
\end{bmatrix}=D-nvv^T$, we have that $det(D-nvv^T)=det(D)-nv^TAdj(D)v=\displaystyle\prod_{j=1}^{n}\lambda^2-n\displaystyle\sum_{i=1}^{n}\displaystyle\prod_{k\neq i}^{n}\lambda^2=\\=\lambda^{2n}-n(n\lambda^{2(n-1)})=\lambda^{2n}-n^2\lambda^{2(n-1)}=\lambda^{2(n-1)}(\lambda^2-n^2)$

Question

the afirmation $det(A+uv^T)=det(A)+v^TAdj(A)u$ I found in Wikipedia Matrix determinant lemma, but I don't have able to proof it then I try to proof that if $\lambda$ is nonzero then $\begin{bmatrix}
\lambda^2-n&-n&...&-n\\-n&\lambda^2-n&...&-n&\\ \vdots&&\ddots&\vdots\\-n&&...&-n
\end{bmatrix}$ is nonsingular and so I can use that $det(A+uv^T)=det(A)(1+v^TA^{-1}u)$ that is easier to proof, I think the problem is calculate $det\begin{bmatrix}
\lambda^2-n&-n&...&-n\\-n&\lambda^2-n&...&-n&\\ \vdots&&\ddots&\vdots\\-n&&...&-n
\end{bmatrix}$, there exist a alternative way to calculate it?

答:
There are many ways to do this.  Here's one simple way:

$T_n=\begin{bmatrix}\mathbf 0& \mathbf 1_n\mathbf 1_n^T \\ \mathbf 1_n \mathbf 1_n^T&\mathbf 0\end{bmatrix}$

$T_n$ is real symmetric, $\text{rank}\big(T_n\big)=2$ and $\text{trace}\big(T_n\big)=0$

$\implies \lambda_1\in \mathbb R_{\gt 0}\text{ and } \lambda_2 = -\lambda_1\text{ and } \lambda_k = 0$ for $3\leq k\leq 2n$

$2\lambda_1^2 =\lambda_1^2 + \lambda_2^2 + 0+....+0= \Big\Vert T_n\Big\Vert_F^2 = 2\cdot \Big\Vert \mathbf {11}_n^T\Big\Vert_F^2 + 2\cdot \Big\Vert \mathbf 0\Big\Vert_F^2 = 2n^2$

$\implies \lambda_1 =n\text{ and } \lambda_2= -n$ and all other eigenvalues are $0$ as desired.


问:
Multiple player's scenarios in game theory
I am learning game theory in recent lectures and I am curious about the multiple player's scenarios. What if we have three players and two possible strategies. How would the payoff matrix look like? Could you give me an example to interpret the scenario?</p>

答:
In general most $m$ player games are set up by describing the strategy space for each player and their respective payoff functions. Normally only simple normal form games are represented in matrix form, but we can represent a three player normal form game in matrix form. For example we  have three players with $n_1$, $n_2$, $n_3$ strategies each. We would represent this game using $n_3$ different $n_1 \times n_2$ matrices. The payoffs when player 1 plays their second strategy, player 2 plays their first strategy, and player 3 plays their first strategy would be in the second row and first column of the first matrix.

For example here is three players. Player one, two and three each have two strategies, $(A, B)$, $(L, R)$, and $(H, T)$ respectivly.

\begin{array}{|c|c|c|}
\hline
H & L & R \\
\hline
A & (1, 0, 1) & (1, 1, 1) \\ \hline
B & (0, 1, 1) & (1, 1, 3) \\ \hline
\end{array}

\begin{array}{|c|c|c|}
\hline
T & L & R \\
\hline
A & (1, 0, 1) & (1, 2, 0) \\ \hline
B & (0, 0, 1) & (1, 0, 0) \\ \hline
\end{array}


问:
Why is it actually important for $p$ to be prime in the ring of $p$-adic numbers?
In the proof that the set $Z_{[p]}$ of all $p$-adic numbers is a ring, it is not necessary for $p$ to be prime. Is it actually important for $p$ to be prime for the ring of $p$-adic numbers?

答:
For $a=\prod_{p|a} p^{e(p)}$ non-prime we have $$\varprojlim \Bbb{Z}/(a^n)=\varprojlim \Bbb{Z}/(\prod_{p|a}p^{e(p)n})\cong \varprojlim \prod_{p|a}\Bbb{Z}/(p^{e(p)n})$$ $$\cong \prod_{p|a}\varprojlim \Bbb{Z}/(p^{e(p)n})=\prod_{p| a} \Bbb{Z}_p$$
Which is the completion of $\Bbb{Z}$ for the norm $$\| \frac{c}{d} \prod_{p| a} p^{r(p)}\|_a = 2^{-\min_p r(p)}, \qquad \gcd(a,cd)=1$$
This norm is not an absolute value when $a$ has more than one prime divisor.

The completion of $\Bbb{Q},\|.\|_a$ is isomorphic to $\prod_{p|a} \Bbb{Q}_p$.


问:
Are these rings the same? A detail in the Proj construction
The following detail is needed in the affine-by-affine gluing construction of the Proj of a graded rings. Every single reference that I know of takes the claim for granted.

Let $A$ be a nonnegatively graded ring, and let $f$ and $g$ be two homogeneous elements of positive degree. Then there should be isomorphisms of rings
$$\big((A_f)_0\big)_{g^{\deg f} / f^{\deg g}} \cong (A_{fg})_0 \cong \big((A_g)_0\big)_{f^{\deg g} / g^{\deg f}}$$
Here the subcript $0$ means '$0$-th graded part', while every other subscript denotes localisation at an element.

It is not clear to me at all why this should be true. I do not even have a conjectural map between the various rings, or any idea why $\deg f$ and $\deg g$ should appear.

答:
Let me summarize the discussion from the comments so that this question may be marked as answered.

In the comments, after I suggested looking at an example, you discovered the correct map (in the case where $\deg f=\deg g$) - for an arbitrary element, we can rewrite it as follows: $$\frac{a/f^n}{(g^{\deg f}/f^{\deg g})^m}\mapsto \frac{af^{m\deg g}}{f^ng^{m\deg f}},$$  and then we just multiply by appropriate powers of $\frac{f}{f}$ and $\frac{g}{g}$ to get the bottom to be of the form $(fg)^c$ for some integer $c$ (you can solve this if you want - I don't think it's strictly necessary). The extra degree stuff is perhaps a clerical headache, but it can be dealt with.


问:
Equivalent definitions of Brownian local time
To keep this question simple, we only consider the local time of a standard Brownian motion $B_t$ at zero, we denote it as $L_t^0$. Generally, we define $$\tag{1} L_t^0(B)=\lim_{\varepsilon\to\infty}\dfrac{1}{2\varepsilon}\int_0^t1_{(-\varepsilon,\varepsilon)}(B_s)\,ds. $$
Another definition is given by $$L_t^0(B)= |B_t|-\int_0^t\text{sgn}(B_s)\,dB_s\tag{2}.$$
We can approximate $|B_t|$ pointwise by $\sqrt{\varepsilon+B_t^2}$ by taking $\varepsilon\to0$. Applying Ito formula to $\sqrt{\varepsilon+B_t^2}$ we get, $$\sqrt{\varepsilon+B_t^2}=\sqrt{\varepsilon}+\int_0^t\dfrac{B_s}{\sqrt{\varepsilon+B_s^2}}\,dB_s+\dfrac{1}{2}\int_0^t\dfrac{\varepsilon}{(\varepsilon+B_s^2)^{3/2}}\,ds. $$
If we take the limit, the above formula suggests that $$L_t^0(B)=\lim_{\varepsilon\to0}\dfrac{1}{2}\int_0^t\dfrac{\varepsilon}{(\varepsilon+B_s^2)^{3/2}}\,ds\tag{3}.$$
My question is can we show that by direct computation that equation (3) and (1) are the same?

答:
It may depend on what you mean by "the same". By the occupation time formula for the Brownian local time
$$
\int_0^t{\epsilon\over (\epsilon+B_s^2)^{3/2}}\,ds = \int_{-\infty}^\infty {\epsilon\over(\epsilon+x^2)^{3/2}}L_t^x\,dx=\int_{-\infty}^\infty {1\over (1+u^2)^{3/2}}L^{\sqrt{\epsilon}u}_t\,du.
$$
Because
$$
\int_{-\infty}^\infty {1\over (1+u^2)^{3/2}}\,du=2,
$$
sending $\epsilon$ to $0$ on the right side of  (3) therefore yields $L^0_t$ because of the continuity in $x$ of $L^x_t$.


问:
Pair of orthogonal directions is a submanifold
I would like to ask something i don't understand. In my textbook of manifolds, it says that the subset $M$ of   $\mathbb R P^{n} \times\mathbb R P^{n} $, made from the pairs $(D,D')$ of the orthogonal directions of $\mathbb R^{n+1}-\{0\}$, is a submanifold of the manifold $\mathbb R P^{n} \times\mathbb R P^{n} $.

So, I thought of something, based on what the inner product function that Nick wrote on his answer, but I'm not sure about it. (thinking of $\mathbb R P^{n} $ as $S^{n}$ by identifying the antipodal points). Let $g:\mathbb R P^{n}\times\mathbb R P^{n}\to\mathbb R$, with $f=g\circ \pi$, where $f:S^{n}\times S^{n}\to\mathbb R$ is the inner product function, and $\pi:S^{n}\times S^{n}\to\mathbb R P^{n} \times \mathbb R P^{n} $ is the projection map. Then, $M=g^{−1}(0)$, and because $f$ is a submersion, so is $g$. So $M$ will be a submanifold of $\mathbb R P^{n} \times\mathbb R P^{n} $.

What do you think about it?? Thank you in advance!

答:
Have you learned the theorem about "regular level sets"?

If you have coordinates $(x_0,\dots,x_n)$ and $(y_0,\dots,y_n)$ for two points in $\Bbb{R}^{n+1}$, then the subset $M$ is defined by the equation

$$ x_0 y_0 + x_1y_1 + \cdots + x_n y_n = 0 $$

Let $f$ be the function $f = \sum_i x_i y_i$. Then $M$ is the level set $f=0$. You can check that $0$ is a "regular value" of $f$ (by writing down the gradient of $f$ and seeing that it's non-zero on this set), and the theorem says that the level set is an embedded submanifold. The way I've worded it is for $\Bbb{R}^{n+1} \times \Bbb{R}^{n+1}$, but you can then project to $\Bbb{P}^n \times \Bbb{P}^n$.


问:
distribution of areas of random closed loops on a lattice
Consider a planar square lattice, i.e. ${Z}^2$. I consider the set of the closed paths on this lattice, starting and ending at some fixed base point (imposing a base point is just to avoid considering paths that are identical up to a translation). More specifically, I am interested in the subset of these closed paths that make $2n_1$ steps in the $\pm x$ directions, and $2n_2$ steps in the $\pm y$ directions. Let us call it $\Gamma_{n_1,n_2}$.

Now consider the algebraic area $Area(\gamma)$ of a closed path $\gamma\in\Gamma_{n_1,n_2}$. By algebraic, I mean that the contribution of each lattice plaquette to the area is weighted by the winding number of the path around that plaquette.

Are the moments $\sum_{\gamma\in\Gamma_{n_1,n_2}}(Area(\gamma))^k$   known as a function of $n_1,n_2$ and $k$?  Obviously, the odd moments are zero. Also I believe the asymptotic behavior of the moments for large $n_1,n_2$ is probably easier to obtain. But I am wondering about whether an exact  formula is known for finite $n_1,n_2$.

答:
This question is answered in Epelbaum et al., Ann.Inst.H.Poincare D 3 (2016) 4, 381:
$$\sum_{\gamma\in{\Gamma}_{n_1,n_2}}
\left({\rm Area}\,(\gamma)\right)^{2l}
=\frac{(2(n_1+n_2))!}{n_1!^2n_2!^2}\;{\cal P}_{2l}(n_1,n_2),
$$
where ${\cal P}_{2l}$ is a symmetric polynomial in $n_1,n_2$ of degree
$2l$. The first four of these polynomials read
$${\cal P}_0(n_1,n_2)=1$$
$${\cal P}_2(n_1,n_2)=\frac{n_1n_2}{3}$$
$${\cal P}_4(n_1,n_2)=\frac{n_1 n_2\big({7 n_1n_2}
\!-\!({n_1\!+\!n_2})
\big)}{15}$$
$${\cal P}_6(n_1,n_2)=\frac{n_1 n_2}{21}\big(31 (n_1n_2)^2
-15(n_1n_2)(n_1\!+\!n_2)
+2(n_1+n_2)^2
-({n_1+n_2})
\big)$$
Explicit formulas up to ${\cal P}_{12}$ are given in the above reference. A semi-explicit formula exists for this polynomial, but its evaluation cost becomes prohibitive for large moments.


问:
Probability of defeating a hydra that grows back its heads
Here's the problem statement: "You are trying to save a princess from a dragon who starts with three heads. At each turn, you can either cut off one head, two heads, or no heads; regardless, the dragon will grow back one head unless all of its heads are cut off. For each turn, all outcomes are equally likely. What is the probability you defeat the dragon?"
Let's assume there's no cap on the number of heads the dragon can grow. In that case, we can write down recurrences $$P_3 = P_4/3 + P_3/3 + P_2/3$$
$$P_2 = P_3/3 + P_2/3 + 1/3$$
and many more for $P_4$, $P_5$, etc., where $P_i$ is the probability of defeating the dragon given that there are $i$ heads remaining. I'm struggling to find a way of solving this infinite system of equations, or seeing some kind of symmetry that makes $P_4$ easily expressable in terms of other probabilities. Any suggestions?

答:
Think of it as a random walk: you start at $x=3$, then add a head, remove a head, or stay in the same place with equal probabilities. For simplicity, you can disregard all steps where you stay in the place and think about it as a random walk with equal probabilities to go right and left.

Here comes the punch: A symmetric random walk visits all points with probability 1, so sooner or later, w.p. 1, you will cut all the heads.


问:
Is a Kähler class of the submanifold always deducible from the outer space?
Let $X$ be a compact Kähler manifolds, $Y$ a compact submanifold of $X$, the map $i:Y\to X$ is an inclusion map, then we know $Y$ is a Kähler manifold, and my question is:

For any Kähler class $\omega_Y$ of $Y$, is there always a Kähler class $\omega$ of $X$, such that the restriction of $\omega$ to the submanifold $Y$ is $\omega_Y$?

or equivalently,

Let $C_X,C_Y$ be the Kähler cones of $X,Y$ respectively, then for the map $i:Y\to X$, is the pull back $i^*:C_X\to C_Y$ a surjective map?

this question is important because if we add the condition that $H^2(Y,\mathcal O)=0$, then we can alway pick a Kähler class of $X$, such that the restriction to $Y$, denoted as $\omega_Y$, satisfies: $\omega_Y\in H^2(Y,\mathbb Q)$.

I guess it is true, but I find it hard to prove it, can someone help me? or give a counter-example? Any comment is welcome, thanks!

答: What about this? Take $Q=\Bbb P^1\times \Bbb P^1$, the usual quadric in $\Bbb P^3$. The Kähler form on $\Bbb P^3$ restricts to $\pi_1^*\omega + \pi_2^*\omega$, where $\omega$ is the usual Kähler form on $\Bbb P^1$ and $\pi_j$ are the obvious projections. Take instead the Kähler form $2\pi_1^*\omega + \pi_2^*\omega$ on $Q$. It cannot be the restriction of a Kähler form on $\Bbb P^3$ (since every Kähler form is cohomologous to a positive scalar multiple of the usual one).


问:
Write the solution as a Fourier Series. $u_{tt}=u_{xx},~ u(t,0)=u(t,\pi)=0,~u(0,x)=1, u_t(0,x)=0$
I'm running into a roadblock on this IBV problem

$$u_{tt}=u_{xx},~ u(t,0)=u(t,\pi)=0,~u(0,x)=1, u_t(0,x)=0$$

Assuming separability:

$$u(t,x)=w(t)v(x) \implies  w^{\prime\prime}(t)=\lambda w(t)~~\mbox{and}~~v^{\prime\prime}(x)=\lambda v(x)$$

If $\lambda=-\omega^2<0$ then
\begin{align}
w(t) &= \sin(\omega t) \ \mbox{or} \ w(t) = \cos(\omega t) \\ v(x) &= \sin(\omega x) \ \mbox{or} \ v(x) = \cos(\omega x)
\end{align}

If $\lambda=0$ then
\begin{align}
w(t) &= 1 \ \mbox{or} \ w(t) = t \\
v(x) &= 1 \ \mbox{or} \ v(x) = x
\end{align}

If $\lambda=\omega^2$ then
\begin{align}
w(t) &= e^{-\omega t} \ \mbox{or} \ w(t) = e^{\omega t} \\
v(x) &= e^{-\omega x} \ \mbox{or} \ v(x) = e^{\omega x}
\end{align}

By initial boundary condition $u(t,0) = u(t,\pi) = 0$, we have $v(0) = v(\pi) = 0 \implies v(x) = \sin(\omega x)$.

Here's where I encounter my roadblock. By the IC $u(0,x)=1$, we have $w(0)=\frac{1}{v(x)}$ (which is not a possible solution for $w(t)$. Am I to conclude in this scenario that the function $u(t,x)$ is not separable and I must seek another strategy for finding the solution?

Further Analysis:
Based on $v(x)=\sin(\omega x)$ in conjunction with $u(0,x)=1\implies \sin(\omega x)=1$ at $w(0)=1$.  Thus, $\omega=n\pi$ and $w_n(t)=\cos(n\pi t)$ and therefore $$u_n(t,x)=\cos(n\pi t)\sin(n\pi x)~~~n=1,2,3,...$$

I think I may have solved my own problem.  If I'm making an error here, please correct me.

This brings me to a new stumbling block.  Translating this solution into a Fourier Series solution:

Logically (in my mind at least) $u(t,x)=\sum_{n=1}^{\infty}u_n(t,x)=\sum_{n=1}^{\infty}b_n \cos{(nt)}\sin(nx)$

Solving for $b_n$ is a little tricky for me here.  I'm using v(x) to find the $b_n$ values in the following integral.  I don't see how to put the result back into the Fourier Sum.

$$b_n=\frac{2}{\pi}\int_0^\pi{\sin(\omega x)\sin(kx)}=2~~~\forall ~\omega=k$$

End Further Analysis

This problem is from Chapter 4.2 of "Introduction to Partial Differential Equations" by Peter Olver.  The section is on Separation of Variables, hence the approach I am trying to use.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{u\pars{x,t} =
\sum_{n = 1}^{\infty}a_{n}\pars{t}\sin\pars{nx}}$ satisfies
$\ds{u\pars{t,0} = u\pars{t,\pi} = 0\,\ \forall t}$.

$\ds{
\sum_{n = 1}^{\infty}\bracks{\ddot{a}_{n} + n^{2}a_{n}\pars{t}}\sin\pars{nx} = 0 \implies
\ddot{a}_{n} + n^{2}a_{n}\pars{t} = 0}$.

$\ds{
a_{n}\pars{t} = b_{n}\sin\pars{nt} + c_{n}\cos\pars{nt}}$

$\ds{u\pars{x,t} =
\sum_{n = 1}^{\infty}\bracks{b_{n}\sin\pars{nt} + c_{n}\cos\pars{nt}}\sin\pars{nx}}$

$\ds{0 =
\sum_{n = 1}^{\infty}b_{n}n\sin\pars{nx} \implies b_{n} = 0}$

$\ds{u\pars{x,t} =
\sum_{n = 1}^{\infty}c_{n}\cos\pars{nt}\sin\pars{nx}}$

$\ds{1 =
\sum_{n = 1}^{\infty}c_{n}\sin\pars{nx}}$:
\begin{align}
&\overbrace{\int_{0}^{\pi}\sin\pars{nx}\dd x}
^{\ds{1 - \pars{-1}^{n} \over n}}\ =\
\sum_{m = 1}^{\infty}c_{m}\
\overbrace{\int_{0}^{\pi}\sin\pars{nx}\sin\pars{mx}\dd x}^{\ds{{\pi \over 2}\delta_{nm}}}
\\[5mm] &\
\implies c_{n} = {2 \over \pi}\,
{1 - \pars{-1}^{n} \over n}
\end{align}

$\ds{\bbx{u\pars{t,x} =
{4 \over \pi}\sum_{n = 0}^{\infty}
{\cos\pars{\bracks{2n + 1}t}\sin\pars{\bracks{2n + 1}x} \over 2n + 1}}}$


问:
Infinite sum of square root
I tried demonstrate that $\displaystyle \sum_{k=0}^\infty\sqrt{a_n^2+b_n^2}$ is convergent with $a_n$ and $b_n$ are the fourier coefficients of continuos $f(x)$ with $f(x)\in L_2(-\pi, \pi)$ and $f(\pi)=f(-\pi)$. My attempt to show it is to proof that the sum $$\sum_{k=0}^\infty a_n^2+b_n^2$$ is convergent (using Bessel inequation). But I'm not sure if $$\sum_{k=0}^\infty\sqrt{a_n^2+b_n^2}\leq \sum_{k=0}^\infty a_n^2+b_n^2$$

or is it not necessary?

答:
An example of a Fourier series whose sum is continuous but the series is
not absolutely convergent is

$$ \sum_{n=2}^\infty \frac{\sin(nx)}{n \log n} $$

See e.g. Edwards, "Fourier Series: A Modern Introduction", 7.2.2.

Title: Infinite sum of square root


问:
Is every Borel set a countable union of intervals?
Since the Borel $\sigma$ algebra is generated by all open subsets of $\mathbb{R}$ and all open sets are the countable union of disjoint open intervals, I figured that any Borel set is the countable union of intervals. I also used the fact that if we only use '$\sigma $ algebra operations' on intervals then we get an interval or countable union of intervals.
But I ask if this is actually true?

答:
The sets you are talking about are special cases of what are called $\boldsymbol{\Sigma}^0_3$ sets using the Borel hierarchy notation.

To explain, we start with intervals. Every interval can be written as an intersection of countably many open intervals. Sets obtained this way are also called $G_\delta$ sets, and denoted by $\boldsymbol{\Pi}^0_2$ in the Borel hierarchy. So every interval is a $\boldsymbol{\Pi}^0_2$ set. That means that a set that is a union of countably many intervals is, in particular, a union of countably many $\boldsymbol{\Pi}^0_2$ sets. Such sets are denoted by $\boldsymbol{\Sigma}^0_3$ sets. However, the Borel hierarchy does not stop here, but rather continues up through $\boldsymbol{\Pi}^0_\alpha$ and $\boldsymbol{\Sigma}^0_\alpha$ sets for all countable ordinals $\alpha$. Moreover, it can be shown that every level of this hierarchy is proper. So in conclusion there are Borel sets that are much more complicated than unions of countably many intervals.


问:
If for an open cover there exists a partition of unity subordinated to it, then the cover has an open locally finite refinement.
In the Engelkings book "General Topology" in the chapter of paracompactness there is a lemma related to partitions of unity and locally finite covers.

I am trying to understand the proof of the lemma, but I am blocked at the part when $V_s$ sets are defined. I can't get why $V_s$ are open, why $\mathcal{V}$ is a refinement of $\mathcal{U}$ and why $\mathcal{V}$ is locally finite.

The lemma together with the proof is given below:.

答:
For $s\in S$ let $h_s(x)=f_s(x)-\frac12f(x)$; $h_s$ is continuous (because $f_s$ and $f$ are), so

$$V_s=\{x\in X:h_s(x)>0\}=h_s^{-1}\big[(0,\to)\big]$$

is open. For each $x\in V_s$ we have $\color{red}{f_s(x)>\frac12f(x)}>0$ by the definition of $f$, so $x\in f_s^{-1}\big[(0,1]\big]$. The partition of unity is subordinated to $\mathscr{U}$, so by definition there is a $U\in\mathscr{U}$ such that $V_s\subseteq f_s^{-1}\big[(0,1]\big]\subseteq U$, and $\mathscr{V}$ is a refinement of $\mathscr{U}$.

Finally, let $g=\frac12f$, and let $x\in X$; $f(x)>0$, so $g(x)>0$, and therefore there are a nbhd $U_0$ of $x$ and a finite $S_0\subseteq S$ such that $f_s(y)<g(y)=\frac12f(y)$ for all $y\in U_0$ and $s\in S\setminus S_0$. Let $s\in S\setminus S_0$ and $y\in U_0$; then $f_s(y)<\frac12f(y)$, so $y\notin V_s$ (by the red inequality above). Thus, $U_0\cap V_s=\varnothing$ for all $s\in S\setminus S_0$, so $\{s\in S:U_0\cap V_s\ne\varnothing\}\subseteq S_0$, and $U_0$ is therefore a nbhd of $x$ that meets only finitely many members of $\mathscr{V}$. Since $x\in X$ was arbitrary, this shows that $\mathscr{V}$ is locally finite.


问:
Show identity of subgroup is same as identity of group
Let $H$ be a subgroup of $G$. Let $1_H$ and $1_G$ be the identities of $H$ and $G$, respectively. Show that $1_H=1_G$.

My attempt is: since we know that the identity of a group is unique, and hence $1_H=1_G$.

Is my proof correct?

答: Let $h_1, h_2 \in H$ where $h_2$ is the inverse of $h_1$. Note that this is allowed since $H$ itself is a group. Then $h_1h_2 = 1_H$, but $h_1, h_2$ are both elements in $G$ as well, so $h_1h_2 = 1_G$. Therefore, $1_H = 1_G$.</p>


问:
Binomial coefficient sum over top index
I am trying to evaluate a sum over binomial coefficients which is giving me some problems. Specifically I want to calculate:

$$\sum_{r=0}^{c-1}\binom{r+n}{n}\frac{1}{c-r}$$

My main thought was to convert the fraction here into:

$$\int_0^1 x^{c-r}dx,$$
move the integral out of the sum, alongside the $x^c$ and then attempt to rewrite as some closed form function. I however, cannot see what the generating function should be.

Note, my aim here is to avoid having a sum - some product of binomial coefficients would be ideal but I obviously do not know if this exists!

Any help on summing this would be greatly appreciated.

答:
I could not find a nice closed form; however, I have previously computed the generating function
$$
\begin{align}
f_m(x)
&=\sum_{n=1}^\infty\sum_{k=1}^n\frac{\binom{n-k}{m}}{k}x^n\tag1\\
&=\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{\binom{n-k}{m}}{k}x^n\tag2\\
&=\sum_{k=1}^\infty\frac{x^k}k\sum_{n=0}^\infty\binom{n}{m}x^n\tag3\\
&=\sum_{k=1}^\infty\frac{x^k}k\sum_{n=0}^\infty(-1)^{n-m}\binom{-m-1}{n-m}x^n\tag4\\
&=\sum_{k=1}^\infty\frac{x^{\color{#C00}{k}+\color{#090}{m}}}{\color{#C00}{k}}\color{#00F}{\sum_{n=0}^\infty(-1)^{n}\binom{-m-1}{n}x^n}\tag5\\
&=\frac{\color{#090}{x^m}}{\color{#00F}{(1-x)^{m+1}}}\color{#C00}{\log\left(\frac1{1-x}\right)}\tag6
\end{align}
$$
Explanation:

$(1)$: definition

$(2)$: switch order of summation

$(3)$: substitute $n\mapsto n+k$

$(4)$: negative binomial coefficients

$(5)$: substitute $n\mapsto n+m$

$(6)$: $\sum\limits_{n=0}^\infty(-1)^{n}\binom{-m-1}{n}x^n=\frac1{(1-x)^{n+1}}$ and $\sum\limits_{k=1}^\infty\frac{x^k}k=\log\left(\frac1{1-x}\right)$

We can apply $(6)$ to get
$$
\begin{align}
\sum_{r=0}^{c-1}\binom{r+n}n\frac1{c-r}
&=\sum_{r=1}^c\binom{c-r+n}n\frac1r\tag7\\
&=\sum_{r=1}^{c+n}\binom{c+n-r}n\frac1r\tag8\\
&=\left[x^{c+n}\right]\frac{x^n}{(1-x)^{n+1}}\log\left(\frac1{1-x}\right)\tag9\\[3pt]
&=\left[x^c\right]\frac1{(1-x)^{n+1}}\log\left(\frac1{1-x}\right)\tag{10}
\end{align}
$$
Explanation:

$\phantom{1}{(7)}$: substitute $r\mapsto c-r$

$\phantom{1}{(8)}$: the terms with $r\in[c+1,c+n]$ are $0$

$\phantom{1}{(9)}$: apply $(6)$

$(10)$: $\left[x^{c+n}\right]x^nf(x)=\left[x^c\right]f(x)$

Thus, $(10)$ gives the generating function for the sums.

Title: Binomial coefficient sum over top index


问:
Show that any discrete or continuous random variable $X$ satisfies $E(X) = \int_{0}^\infty P(X > x)dx - \int_{0}^\infty P(X < -x)dx$
Show:

$$E(X) = \int_{0}^\infty P(X > x)dx - \int_{0}^\infty P(X < -x)dx$$

$E[X]$ is the expectation value of the Random Variable $X$

$P$ is the probability

We know

$$E(X) = \int_{-\infty}^\infty xf(x)dx$$
$$f(x)dx = \frac{\mathrm{d}}{\mathrm{d}x}P(X\le x)$$

So,
$$
E(X) =-\int_0^\infty x\,\frac{\mathrm{d}}{\mathrm{d}x}P(X\le x)\,\mathrm{d}x
$$

$$
E(X) =-\int_0^\infty x\,\mathrm{d}P(X\le x)
$$
Using Integration by parts:
$$
\mathrm{E}(X)=-\lim_{x\to\infty}x\,P(X\le x)+\int_0^\infty P(X\le x)\,\mathrm{d}x
$$

However, I do not know how to proceed further. Help would be greatly appreciated.

答:
The following answer is useful if you know a little bit of integration theory as given in measure theory.

Of couse the formula only makes sense if $X$ is integrable, i.e. $\mathbb E|X|<\infty$. The result does not depend on the type of the distribution. Just note that
$$\mathbb P(X>x)=\mathbb E[\mathbf{1}_{X>x}], \  \ \ \text{and} \ \ \  \mathbb P(X<-x)=\mathbb E[\mathbf{1}_{X<-x}]$$
Now we apply Fubini (we are allowed to apply Fubini since $X$ is integrable, check it!)
$$\int^\infty_0 \mathbb E[\mathbf{1}_{X>x}-\mathbf{1}_{X<-x}]\,dx=\mathbb E\left[\int^\infty_0 \mathbf{1}_{X>x}-\mathbf{1}_{X<-x}\,dx\right].$$
Note that
$$\int^\infty_0 \mathbf{1}_{X>x}\,dx=X\mathbf{1}_{X\geq 0},$$
and
$$ \int^\infty_0 \mathbf{1}_{X<-x}\,dx=\int^\infty_0 \mathbf{1}_{-X>x}\,dx=-X\mathbf{1}_{X< 0} $$
So
$$\int^\infty_0 \mathbb E[\mathbf{1}_{X>x}-\mathbf{1}_{X<-x}]\,dx=\mathbb E[X\mathbf{1}_{X\geq 0}+X\mathbf{1}_{X< 0}]=\mathbb E[X\underbrace{(\mathbf{1}_{X\geq 0}+\mathbf{1}_{X< 0})}_{=1}]=\mathbb E[X] $$


问:
Continuous functions that are differentiable in only one point
Quite recently I discussed with a colleague of mine the following question: Is there a function $f: \mathbb R \to \mathbb R$ that differentiable in one point only?

The answer is positive and the example is simple to construct: Let
$$ \varphi: \mathbb R \to \mathbb R, \quad \varphi(x) = \begin{cases} 1, &\quad \text{if } x \in \mathbb Q, \\ 0, &\quad \text{if } x \in \mathbb R \setminus \mathbb Q, \end{cases}$$ be the Dirichlet function. Then the function $f: \mathbb R \to \mathbb R, \ f(x) = x^2 \varphi(x)$ is only differentiable at $0$ since
$$ \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^2 \varphi(h)}{h} = 0.$$
Moreover, it is easy to see that $f$ is discontinuous on $\mathbb R \setminus \{0\}$, so that $f$ is not differentiable there.

But this is a good point which naturally leads to the following question: "Is there a continuous function $f: \mathbb R \to \mathbb R$ which is differentiable at only differentiable in $0$?"

I think if such a function exists, it would be given by some cleverly choosen series at $0$. I thought quite some time about the question but I still do not know the answer.

答:
A much more general result was given by Zahorski:

Zygmunt Zahorski, Sur l'ensemble des points de non-dérivabilité d'une fonction continue, Bulletin de la Société Mathématique de France, Volume 74  (1946), p. 147-178 (link)

It says that for a subset of $\mathbb{R}$ to be the set of points where a continuous function is non-differentiable, it is necessary and sufficient that it be a union of a $G_{\delta\sigma}$ of Lebesgue measure zero and a $G_\delta$ set.

OwnerDisplayName: Adriano


问:
Diagonalising a $2 \times 2$ and $3 \times 3$ matrix

For each of the following matrices $A$, find an invertible matrix $P$ over $C$ such that
  $P^{-1}AP$ is upper triangular:

$$A = \begin{bmatrix}4 & 1\\-1 & 2\end{bmatrix} \quad \text{ and } \quad A = \begin{bmatrix}1 & 1 & 1\\1 & 1 & 1\\0 & 0 & 0\end{bmatrix}.$$

For the first matrix, I calculate the eigenvalue to be $3$, with algebraic multiplicity $2$.
However, I can only find one eigenvector for it, namely $\begin{bmatrix}1\\-1\end{bmatrix}$.

The second has two eigenvalues, $0$ (multiplicity $2$) and $2$ (multiplicity $1$)
But again, for the eigenvalue $0$, I can't find two linearly independent eigenvectors.

So, how do I proceed to find $P$ in these cases?

答:
Here is an alternative approach. Note that you are not required to find the Jordan form of $A$. So, when $A$ is $n\times n$, you only need to find one eigenvector of length $n$.

More specifically, suppose $(\lambda_1,v_1)$ is an eigenpair for $A$. Extend $v_1$ to a basis $\{v_1,\ldots,v_n\}$ of $\mathbb{C}^n$ and put the basis vectors together to form an invertible matrix $P_n$. Then
$$
AP_n=(\lambda_1v_1,\,Av_2,\,\ldots,\,Av_n)
=P_n(\lambda_1e_1,\,P^{-1}Av_2,\,\ldots,\,P^{-1}Av_n)
$$
where $e_1=(1,0,\ldots,0)^T$. In other words,
$$
P_n^{-1}AP_n=\pmatrix{
\lambda_1&\ast&\ast&\cdots&\ast\\
0&\ast&\ast&\cdots&\ast\\
0&\ast&\ast&\cdots&\ast\\
\vdots&\vdots&\vdots&&\vdots\\
0&\ast&\ast&\cdots&\ast\\
}
=\pmatrix{\lambda_1&u^T\\ 0&B}
$$
for some $(n-1)$-vector $u$ and some matrix $B$ of order $n-1$. So, if $(\lambda_2,w)$ is an eigenpair of $B$ and $P_{n-1}$ is an invertible matrix whose first column is $w$, then
$$
\pmatrix{1\\ &P_{n-1}^{-1}}(P_n^{-1}AP_n)\pmatrix{1\\ &P_{n-1}}
=\pmatrix{
\lambda_1&\ast&\ast&\cdots&\ast\\
0&\lambda_2&\ast&\cdots&\ast\\
0&0&\ast&\cdots&\ast\\
\vdots&\vdots&\vdots&&\vdots\\
0&0&\ast&\cdots&\ast\\
}.
$$
Proceed recursively, you get an upper triangular matrix.

For your two $A$s, it happens (for different reasons) that you only need one shot each to get a triangular matrix. Take your second $A$ (which is $3\times3$) as an example. $(1,1,0)^T$ is an eigenvector of $A$ corresponding to the eigenvalue $2$. So, if we put
$$
P_3=\pmatrix{1&\ast&\ast\\ 1&\ast&\ast\\ 0&\ast&\ast}
$$
where the second and third columns are arbitrary (as long as $P_3$ is invertible), we have
$$
P_3^{-1}AP_3 = \pmatrix{2&\ast&\ast\\ 0&0&0\\ 0&0&0}.
$$
(Exercise: For each of the two $A$s in question, why only one shot is enough?)


问:
Tangent vectors as directional derivatives on manifolds
I'm reading Modern Differential Geometry for Physicists by Chris J Isham. On page 75 of the book, the author says:

A tangent vector $v \in T_p M$ can be used as a "directional derivative" on functions on $M$ by defining: $$v(f) := \frac{df(\sigma(t))}{dt} \Big|_{t=0}$$

where $v = [\sigma]$ is the equivalence class of the vector $v$ (all curves tangent at $p$). Coming from multivariable calculus, I am struggling to understand the motivation behind this definition. What is the reasoning and motivation behind this definition? And how does this relate to the concept of tangent vectors and tangent planes in ordinary vector calculus?

答:
To keep things simple, let's stick to $3$-dimensions, so it is like familiar "ordinary vector calculus".

If $\sigma(t) = (x(t),y(t),z(t))$ is a curve, then its tangent vector is $\sigma'(t) = (x'(t),y'(t),z'(t))$. Let $p=\sigma(0)$, and $v = \sigma'(0)$.

By the chain rule from "ordinary vector calculus", the right-hand side of the equation you give above is:

$$ \frac{d}{dt} f(\sigma(t)) = \frac{d}{dt}f(x(t),y(t),z(t)) = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} + \frac{\partial f}{\partial z}\frac{dz}{dt} $$
and then of course evaluate at $t=0$. The left-most term in my equation above, $\left. \frac{d}{dt}f(\sigma(t)) \right|_{t=0}$, is by assumption the vector $v$ acting on the function $f$. On the other hand, the right-most part of my equation above looks like a dot product:

$$ \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} + \frac{\partial f}{\partial z}\frac{dz}{dt} = \nabla f \cdot \sigma' $$

This is also the "ordinary vector calculus" definition of the directional derivative in the direction $v$:

$$ D_v f(p) = \nabla f(p) \cdot v $$


问:
About the eigenvalues and singular values of this matrix
Let be
$T_n=\begin{bmatrix}0&...&0&1&...&1\\ \vdots&&\vdots&\vdots&&\vdots\\0&...&0&1&...&1\\1&...&1&0&...&0\\\vdots&&\vdots&\vdots&&\vdots\\1&...&1&0&...&0 \end{bmatrix}$, proof that the eigenvalues of $T_n$ are $0,-n,n$.

My proof

$p(\lambda)=det(\lambda I-T_n)=det\begin{bmatrix}
\lambda^2-n&-n&...&-n\\-n&\lambda^2-n&...&-n&\\ \vdots&&\ddots&\vdots\\-n&&...&-n
\end{bmatrix}$
Let $D=diag(\lambda^2,...,\lambda^2)$ and $v$ the column vector with $n$ ones, note that
$\begin{bmatrix}
\lambda^2-n&-n&...&-n\\-n&\lambda^2-n&...&-n&\\ \vdots&&\ddots&\vdots\\-n&&...&-n
\end{bmatrix}=D-nvv^T$, we have that $det(D-nvv^T)=det(D)-nv^TAdj(D)v=\displaystyle\prod_{j=1}^{n}\lambda^2-n\displaystyle\sum_{i=1}^{n}\displaystyle\prod_{k\neq i}^{n}\lambda^2=\\=\lambda^{2n}-n(n\lambda^{2(n-1)})=\lambda^{2n}-n^2\lambda^{2(n-1)}=\lambda^{2(n-1)}(\lambda^2-n^2)$

Question

the afirmation $det(A+uv^T)=det(A)+v^TAdj(A)u$ I found in Wikipedia Matrix determinant lemma, but I don't have able to proof it then I try to proof that if $\lambda$ is nonzero then $\begin{bmatrix}
\lambda^2-n&-n&...&-n\\-n&\lambda^2-n&...&-n&\\ \vdots&&\ddots&\vdots\\-n&&...&-n
\end{bmatrix}$ is nonsingular and so I can use that $det(A+uv^T)=det(A)(1+v^TA^{-1}u)$ that is easier to proof, I think the problem is calculate $det\begin{bmatrix}
\lambda^2-n&-n&...&-n\\-n&\lambda^2-n&...&-n&\\ \vdots&&\ddots&\vdots\\-n&&...&-n
\end{bmatrix}$, there exist a alternative way to calculate it?

答:
You can compute the eigenvalues and eigenvectors by the definition.

Let $J_n$ be the $n\times n$ matrix with all ones. Then you can write your matrix as
$$
T_n=\begin{bmatrix} 0 & J_n \\ J_n & 0 \end{bmatrix}
$$
Write a $2n$-vector as $\left[\begin{smallmatrix} x \\ y \end{smallmatrix}\right]$ where $x,y$ are $n$-vectors. Then the eigenvector equation is
$$
T_n\begin{bmatrix} x \\ y \end{bmatrix}
= \begin{bmatrix} J_n y \\ J_n x\end{bmatrix}
= \lambda\begin{bmatrix} x \\ y \end{bmatrix}
$$
Thus we need $J_ny=\lambda x$ and $J_nx=\lambda y$. We easily get
$$
J_n^2x=\lambda^2 x,\qquad J_n^2y=\lambda^2y
$$
Since $J_n^2=nJ_n$ and at least one among $x,y$ is nonzero, we see that $\lambda^2/n$ has to be an eigenvalue of $J_n$.

The eigenvalues of $J_n$ are $0$ with multiplicity $n-1$ and $n$ with multiplicity $1$.

Therefore $\lambda\in\{0,n,-n\}$. It's not difficult to show that each of them is an eigenvalue of $T_n$, by taking suitable vectors $x,y$.

Note that the rank of $T_n$ is $2$, so the zero eigenvalue has geometric multiplicity $2n-2$. Since there are other two eigenvalues, the algebraic multiplicity of $0$ cannot exceed $2n-2$ and so is $2n-2$; $n$ and $-n$ have multiplicity $1$.

Since $T_n$ is a symmetric matrix, its singular values are the (nonnegative) square roots of the eigenvalues of $T_n^2$, hence $0$ (with multiplicity $2n-2$) and $n$, with multiplicity $2$.


问:
Finite cyclic groups are isomorphic to their product with $\Bbb Z$?
I'm currently making a start on group theory and have hit a roadblock with a relatively basic theorem on finite cyclic groups. The specific relation killing me is:
$$\mathbb{Z}_n \times \mathbb{Z}_m \cong \mathbb{Z}_{mn} \Leftrightarrow \text{gcd}(m,n) = 1$$
So, the most straightforward result I see there is
$$\mathbb{Z}_n \times \mathbb{Z} \cong \mathbb{Z}_n$$
For some reason this doesn't sit right with me. Why should a cyclic group be unchanged (up to isomorphism) by a direct product with $\Bbb Z$?

Does anybody have a nice example to ease my mind?

Thanks!

答: $\Bbb Z\times\Bbb Z_n\not\cong\Bbb Z_n$, since the former has a nontrivial free part , whereas the latter is just torsion.


问:
How do I find the Laurent series for $\frac{1}{(1+\cos z)^2}$ around $z_0=\pi$?
What I tried so far:

Using that  $\cos z = \sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n)!}z^{2n}$ , for any $z \in \mathbb C$, then, for $0<|z-\pi|<1$:
$$\begin{aligned}\dfrac{1}{(1+\cos z)^2} &= \dfrac{1}{(1+\cos ((z-\pi)+\pi))^2} \\&= \dfrac{1}{(1-\cos(z-\pi))^2}\\ &=\dfrac{1}{\left(1-\left(1-\dfrac{(z-\pi)^2}{2} + \dfrac{(z-\pi)^4}{24}-\dfrac{(z-\pi)^6}{720}+...\right)\right)^2}\\ &=\dfrac{1}{\left(\dfrac{(z-\pi)^2}{2} - \dfrac{(z-\pi)^4}{24}+\dfrac{(z-\pi)^6}{720}-...\right)^2} \end{aligned}$$
Now I don't know how to proceed (and am I even doing things right? I don't know if this is a rigorous development).

答:
May be useful the identity $$\frac{1}{(1+\cos z)^2}=\frac{1}{4} \sec ^4\left(\frac{z}{2}\right)$$
and at $z=\pi$
$$\sec \left(\frac{z}{2}\right)=-\frac{2}{z-\pi }-\frac{z-\pi }{12}-\frac{7 (z-\pi )^3}{2880}+O\left(z^5\right)$$


问:
Probability of spurious correlation between 2 variables (eg, ratings provided by 2 independent movie-raters) depends on the range of the variables?
Suppose you had two people each producing $1$ rating for a set of $n$ movies (suppose $n$ is small, $\approx 10$). Suppose the ground truth is that their ratings are randomly sampled from a discrete uniform distribution. Are their ratings more likely to be spuriously correlated if the range of possible ratings was just $\{1,2\}$ (e.g., $1$ is bad, $2$ is good) compared to the range being $\{1,2,3,4,5\}$ ($1$ is very bad, $5$ is very good)?

I ran simulations, but I would like to approach the answer mathematically.

Let $X_i$ denote movie rating $i$ by rater $1$, sampled $iid$ where $X_i \sim \text{DiscreteUnif}(1,c)$, where $c = 2$ or $5$. Similarly, let $Y_i$ denote $iid$ movie ratings $i$ by rater $2$, sampled from the same distribution as $X_i$. Our random variable of interest is the following:

$\Large R_c = \frac{\sum_{i=1}^n(X_i - E(X_i))(Y_i - E(Y_i))}{\sqrt{\sum_i(X_i - E(X_i))^2}\sqrt{\sum_i(Y_i - E(Y_i))^2}}$

Is the $P(R_{c=2} > a) > P(R_{c=5} > a)$, for any $a \in (0,1)$? Or are those probabilities equal? How do I begin formulating a solution to this problem?

答:
There's probably a more clever way to do this, but one rather naive but brutally computational is to treat $R_c$ as a function of random variables $X_i$ and $Y_i$ and compute $P(R_c>a)$ using the law of total probability and writing out the entire the sample space of every possible outcome of the variables $(X_i,Y_i)_{i=1}^n \in [c]^n\times[c]^n$ where $[c]=\{1,2,\ldots,c\}$. As we are using a uniform distribution, every outcome is equiprobable with probability $\frac{1}{c^{2n}}$. Without taking advantage of any symmetries, the computation ends up being $O(c^{2n})$. [Also, we have to assume the correlation $R$ can be defined in the case that the sample standard deviations of the $X_i$ or of the $Y_i$ is $0$, which we can just say is $0$.]

From doing this out for some very small examples where $n=3$, the probability of a spurious correlation, in the sense that $R_c>0.5$, increases as the number $c$ of options increases. The probabilities for $c=3$ to $c=10$ are 0.09375, 0.18107, 0.231445, 0.258816, 0.277649, 0.288859, 0.297638, 0.303748, 0.308406.


问:
Intuitive understanding of why the sum of nth roots of unity is $0$
Wikipedia says that it is intuitively obvious that the sum of $n$th roots of unity is $0$.  To me it seems more obvious when considering the fact that $\displaystyle 1+x+x^2+...+x^{n-1}=\frac{x^n-1}{x-1}$ and that $\{\alpha \in \mathbb{C}: \alpha^n=1\}$ is a cyclic group.

For me, it is hard to, intuitively, see this in the case that $n$ is an odd composite natural number.

Here is my reasoning thus far:

Let $n=5$. (I understand this is simple when considering the minimal polynomial).

Since the conjugate of an $n$th root of unity is also an $n$th root of unity, it is easy to see that, when written in trigonometric form, the sum (possible reordering here)

$1+\alpha^1+\alpha^2+\alpha^3+\alpha^4$

$=1+( \cos (72)+i \sin (72))+( \cos (144) + i \sin (144))+( \cos (72)-i \sin (72))+( \cos (144)-i \sin (144))$

$=1+2 \cos (72)+2 \cos (144)$.

Somehow $2 \cos (72)+2 \cos (144)=-1$, but I'm not sure this is obvious, or perhaps I'm being silly here.

What do you think?  Silly?

答: There's another intuitive explanation I haven't seen mentioned yet. If you start at the origin and draw a regular $n$-gon with side length 1, starting by going from $(0,0)$ to $(1,0)$ and then proceeding counterclockwise, then the vector corresponding to the $k$th side is exactly $\zeta^{k-1}$. But since the $n$-gon ends out back at the origin, the sum of all these vectors is exactly $0$. Thus the sum of all $n$th roots of unity must be zero.</p>


问:
How to draw a fixed length curve?
Is it possible to draw a curve with some specified length between two points? I'm considering damped sines like WolframAlpha or Bezier curves.</p>

答: If a specified length and specified endpoints are all you want, there are very many ways to do that. The easiest smooth way to proceed is probably to make it a circular arc ... whose center you'll need to find numerically (unless you happen to have an inverse sinc function available).

Title: How to draw a fixed length curve?


问:
Solve differential equation by integrating factor
I have the differential equation $$2\frac{dy}{dx}+3y=e^{-2x}-5$$

I have determined that this needs solving using the integrating factor method.

My workings out are in the image provided.

Are my workings out correct so far? My main question is, how will I integrate with respect for $x$?

$$\int e^{3x/2} \times e^{-2x}-\frac{5}{2}e^{3x/2}~~dx+c$$

this again is shown at the bottom of my workings out.

after integrating this I will then be able to solve for $y$.

Thank you

My workings out for the question

Body:
$$2y'+3y=e^{-2x}-5$$
$$y'+\dfrac 23y=\dfrac 12 (e^{-2x}-5)$$
$$(ye^{3x/2})'=\dfrac {e^{3x/2}}2 (e^{-2x}-5)$$
$$ye^{3x/2}=\int \dfrac {e^{3x/2}}2 (e^{-2x}-5)dx$$
Don't forget the factor $1/2$:
$$ye^{3x/2}= \color {red}{\dfrac 12} \int (e^{-x/2}-5e^{3x/2})dx$$

Now to integrate use this result:
$$\int e^{ax}dx=\dfrac 1 a\int \,\,d(e^{ax})= \dfrac {e^{ax}}{a}+K$$

Edit1:
Integration gives us:
$$ye^{3x/2}= {\dfrac 12} \int (e^{-x/2}-5e^{3x/2})dx$$
$$ye^{3x/2}= {\dfrac 12}(-2e^{-x/2}-\dfrac {10}3e^{3x/2})+K$$
$$ye^{3x/2}= -e^{-x/2}-\dfrac {5}3e^{3x/2}+K$$

$$y=e^{-3x/2}( -e^{-x/2}-\dfrac {5}3e^{3x/2})+Ke^{-3x/2}$$
And finally :

$$\boxed {y= -e^{-2x}-\dfrac {5}3+Ke^{-3x/2}}$$


问:
Is it true that if a non-negative function is Riemann integrable, there must exist an upper sum U(f)?
I know that if the upper sum of f, $U(f)$ equals the lower sum of f on an interval $[a,b]$ then f is Riemann integrable on [a,b] with $U(f)=L(f)= \int_{a}^{b} f(x) \,dx.$

Is the reverse true? That is if f is riemann integrable is it the case that there must exists an U(f) and L(f) with $U(f)=L(f)= \int_{a}^{b} f(x) \,dx.$?

Edit:
My books gives the following:

A bounded function $f$ defined on the interval [a,b] is Riemann integrable if $U(f) = L(f).$ In this case we define $\int_{a}^{b} f$ or $\int_{a}^{b} f(x)$ to be $\int_{a}^{b} f = U(f) = L(f).$

They don't use "if and only if" so i was note sure whether it was the case that if f is riemann integrable then there must exists an U(f) and L(f) with $U(f)=L(f)= \int_{a}^{b} f(x) \,dx.$?

答: By definition $U(f)$ always exist to be the infimum of upper sums over partitions. So any bounded function (which includes Riemann integrable functions) would have an upper sum.


问:
Galois Extensions and $n^{\text{th}}$ Roots
I've been studying for my prelims lately, and this problem has me stuck:

(a) Let $K$ be a field with no abelian Galois extensions. Suppose that $n$ is a positive integer and either $char(K)=0$ or the characteristic is relatively prime to $n$. Prove that every element of $K$ has an $n^\text{th}$ root in $K$.

(b) Is the statement still true if $n$ is not relatively prime to the characteristic of $K$?

For part $(a)$, I was thinking about the polynomial $f(x)=x^n-a$ for a given $a\in K$. If it has a root in $K$, then $a$ has an $n^\text{th}$ root. I wasn't able to make it much further than this, I only know some conditions when the polynomial is irreducible, which doesn't necessarily say anything about the roots. As for part $(b)$, I'm pretty unsure...

答: This is just a special case of Kummer extensions. If $a\not\in K^n$, where the hypothesis apply, then $K(\sqrt[n]{a})/K$ is a Kummer extension. These are all abelian.</p>


问:
If countable fragments of an ultrapower can be realized, then the ultrafilter is $\omega_1$-complete
This is exercise 9 of Steel's notes on ultrapowers, found here. I have managed to progress through it, but I am stuck and I would appreciate any help. So this is the statement:

Let $M\models \mathsf{ZFC}^-+``U \mbox{ is a normal ultrafilter on } \kappa\mbox{''}$, with $M$ transitive. Also suppose that whenever $\pi:N \rightarrow M$ is elementary, and $N$ is countable with $\pi(W) = U$, there is a $\sigma$ such that

commutes. Then $U$ is $\omega_1$-complete.

What I have tried so far is fairly standard. First assume that $\langle A_n: n\in \omega \rangle$ is a sequence of elements of $U$, such that $\bigcap_nA_n=\emptyset$. Then choose some $\theta$ large enough and let $\pi:H \rightarrow H_\theta$ be elementary such that $H$ is countable and transitive, there is some $(N, W) \in H$, such that $\pi((N, W)) = (M, U)$ and also there is some sequence $\langle B_n: n\in \omega \rangle \in H$ such that $\pi(\langle B_n: n\in \omega \rangle) = \langle A_n: n\in \omega \rangle$.

Then $\pi|N:N \rightarrow M$ is elementary and $\pi|N(W) = U$ and so $\operatorname{Ult}(N, W)$ can be realized in $M$, via some $\sigma$. Now my strategy was to use $\bigcap_nB_n=\emptyset$(since $\bigcap_nA_n=\emptyset$), to say that $\operatorname{Ult}(N, W)$ is ill-founded and yet is injected into $M$ which is absurd. But the problem is that $\langle B_n: n\in \omega \rangle$ may not live in $N$ and so the regular argument for ill-foundedness won't go through. I would appreciate any fixes to this argument or hints for other approaches.

Remark. We are checking $\omega_1$-completeness in $V$, yet we take ultrapowers with respect to smaller models. And in this context $\omega_1$-completeness means that every countable sequence of members of $U$ has nonempty intersection.

EDIT I.

I am starting to doubt if this is even correct, because at least in my approach I think I am asking for too much information, which isn't there. This exercise is derived from a lemma which is claimed to be due to Jensen, which states the equivalence of the realizability and $\omega_1$-completeness of the ultrafilter. And in the above notes only the other direction is proven. Since this result is attributed to Jensen, I thought it would be best to ask for a reference if anybody knows of one.

答:
After consulting with Dr.M. Golshani, we came to a proof for this exercise. Continuing from where I have left the proof above, since we have $B_n \in W$(because $A_n \in U$) for every $n \in \omega$, then $\xi \in i^N_W(B_n)$ for every $n$(Because $N \models ``W \mbox{ is normal''}$, as $M \models ``U \mbox{ is normal''}$.)(Also here $\xi$ is such that $\pi(\xi) = \kappa$.)

And so $$\sigma(\xi) \in \sigma(i^N_W(B_n)) = \pi(B_n) = A_n,$$ for every $n$, and so $$\sigma(\xi) \in \bigcap_n A_n,$$
which is a contradiction!

[The whole point here is that members of countable fragments of $U$ have these "typical" elements like $\sigma(\xi)$, just like the proof for the other direction.]


问:
Image of sup/inf under Galois connection between lattices
This is Lemma 2.4 in Pete Clark's notes on commutative algebra.

Given two posets $(X,\le)$ and $(Y,\le)$, an (antitone) Galois connection between $X$ and $Y$ is a pair of maps $F:X\to Y$ and $G:Y\to X$ such that $F$ and $G$ are antitone (= order-reversing) and for all $x\in X$ and $y\in Y$
$$x\le G(y)\iff y\le F(x)\quad\quad(*).$$

Lemma 2.4: Given a Galois connection as above, if $X$ and $Y$ are lattices, then for all $x_1,x_2\in X$

$F(x_1\lor x_2)=F(x_1)\land F(x_2)$

$F(x_1\land x_2)=F(x_1)\lor F(x_2)$

I can prove the first formula (proof below), but I don't see how to show the second one.  Can anyone show how?

Proof of $F(x_1\lor x_2)=F(x_1)\land F(x_2)$: Given $x_1, x_2$, for each $y\in Y$ we have
$$
\begin{align*}
& y\le F(x_1\lor x_2) & \\
\iff\;& x_1\lor x_2\le G(y) &\text{ from (*)} \\
\iff\;& x_1\le G(y)\quad\text{and}\quad x_2\le G(y) & \text{ from definition of }\lor\\
\iff\;& y\le F(x_1)\quad\text{and}\quad y\le F(x_2) & \text{ from (*)} \\
\iff\;& y\le F(x_1)\land F(x_2) & \text{ from definition of }\land \\
\end{align*}
$$
The desired formula follows.

答:
In addition to @amrsa's example, here is another simple counterexample.

Take the Boolean algebra with 2 atoms $X=\{0,a,b,1\}$ with $a$ and $b$ not comparable, $a\lor b=1$, and $a\land b=0$.  Take $Y=\{0,1\}$ with $0\le 1$.

Define two maps $F:X\to Y$ and $G:Y\to X$ by
$$F(0)=1\quad\text{and}\quad F(1)=F(a)=F(b)=0$$
and
$$G(0)=1\quad\text{and}\quad G(1)=0\,.$$
The condition $x\le G(y)\iff y\le F(x)$ is easy to check for each value of $y$, so that $F$ and $G$ form a Galois connection between $X$ and $Y$.

We have $F(a\lor b)=F(1)=0=F(a)\land F(b)$ as expected.  But $F(a\land b)=F(0)=1$ is not equal to $F(a)\lor F(b)=0\lor 0=0$.


问:
How to calculate $\int \frac{dp}{p\sqrt{p^2 - c^2}}$
I'm trying to find a solution by using Euler-Lagrange, but I'm stuck with this integral.

$$\int \frac{dp}{p\sqrt{p^2 - c^2}}$$

I know I have to use a substitution, but I don't know how.

I'm stuck there and I don't even know if this is good.

$$\int \frac{dp}{p^2\sqrt{1 - \frac{c^2}{p^2}}}$$

答:
substitute
$$p=c\frac{t+1}{t-1}$$
integral reduces to
$$=-\frac{1}{c}\int{\frac{1}{\sqrt{t}\left( t+1 \right)}dt}=\frac{-2}{c}{{\tan }^{-1}}\left( \sqrt{t} \right)+C=\frac{-2}{c}{{\tan }^{-1}}\left( \sqrt{\frac{p+c}{p-c}} \right)+C$$


问:
how to calculate the 3D cartesin position of a point based on lat and long?
I want to implement a code, where I get a user's latitude and longitude, and I will position it on a x, y, z cartezian plane, where (0,0,0) is the center of the earth.

My order of magnitude will be in millimeters, but if it is possible to leave it and a variable format, it will help a lot too.

I thought of something like:

```<code>Lo = longitude and La = latitude
So:
x = rcos (La) cos (Lo)
y = rcos (La) sin (Lo)
z = rsin (La)
</code>```

can someone tell me if this is right? and how do I have these rapids in order of magnitude of millimeters?

答:
The formulae you used here are assuming that the Earth is a sphere. Keep in mind that your Earth-centered x,y,z values can be as much as 20 km off the real values with this approximation.

Those Earth-centered x,y,z coordinates also depend on the elevation of your points. So if you do not have that information, you won't be able to compute the exact coordinates, let alone with millimeter accuracy. Moreover, precision and accuracy aren't the same thing, you could calculate and express values with millimeter precision, but if the accuracy of your points is only, say, a few meters, then that kind of precision is overkill. To put that in context, millimeter accuracy is used mainly in geodesy and seismology, but for more general applications, perhaps you're not dealing with those accuracies, and don't need that much precision in your final values. A general rule of thumb is to express your values with a precision no more than 1 order of magnitude smaller than your accuracy, otherwise, it could be misleading.

If you are willing to assume that the Earth is an ellipsoid of revolution, you can use the following formulae:

$$X = \left(N\left(\phi\right)+h\right)\text{cos }\phi\text{ cos }\lambda$$
$$Y = \left(N\left(\phi\right)+h\right)\text{cos }\phi\text{ sin }\lambda$$
$$Z = \left(\frac{b^2}{a^2}N\left(\phi\right)+h\right)\text{ sin }\phi$$
$$\text{Where:}$$
$$N\left(\phi\right)=\frac{a}{\sqrt{1-e^2\text{ sin}^2\phi}}$$
$$e^2=1-\frac{b^2}{a^2}$$
$$a=6378137\text{ m (in WGS84)}$$
$$b=6356752.314\text{ m (in WGS84)}$$
($\phi, \lambda, h$ are respectively latitudes, longitudes and ellipsoidal heights)

The ellipsoid is within about 100 meters from the geoid (essentially mean sea level), so it is a better approximation that the sphere. To get the best accuracy, you would need to add the geoid height at each of your points to get actual ellipsoidal heights, and use that in the formulae. This would ensure accuracy and consistency across the whole globe, but if you are mainly concerned with relative accuracy in a specific region, then you may not have to consider this.


问:
Uniqueness of adjoint functors up to isomorphism
Suppose we are given functors $F:\mathcal{C}\rightarrow\mathcal{D}$ and $G,G':\mathcal{D}\rightarrow\mathcal{C}$ such that $G$ and $G'$ are both right adjoint to $F$. To show that $G$ and $G'$ are isomorphic, we wish to come up with a natural isomorphism between the two functors. This is how Awodey does it:

For any $D\in\mathcal{D}$ and any $C\in\mathcal{C}$, we have that $Hom(C,GD)\cong Hom(FC,D)\cong Hom(C,G'D)$ simply using the hom-set definition of adjunctions. By the Yoneda principle, this means that $GD\cong G'D$ for all $D\in\mathcal{D}$.

Awodey then states that

But this isomorphism is natural in $D$,  again by adjointness.

I dont' quite understand how the naturality follows from adjointness. Any help is appreciated.

答:
I'm not sure that I answer precisely on your question, but I want to share my idea.

Let's view the category $C$ as a full subcategory of its presheaves (image of the Yoneda embedding $C\to\mathbf{Set}^{C^{op}}$).

Adjunction gives you a natural isomorphism $\hom_C(C,G(D))\to \hom_C(C,G'(D))$, which is actually a natural isomorphism between the following two functors:
$$
X=(\hom_C\circ(I_{C^{op}}\times G))\colon C^{op}\times D\to\mathbf{Set};
$$
$$
Y=(\hom_C\circ(I_{C^{op}}\times G'))\colon C^{op}\times D\to\mathbf{Set},
$$
which are objects in the category $\mathbf{Set}^{C^{op}\times D}$. Consider an exponential functor:
$$
E\colon\mathbf{Set}^{C^{op}\times D}\to(\mathbf{Set}^{C^{op}})^D
$$
So we have two isomorphic functors $E(X),E(Y)\colon D\to\mathbf{Set}^{C^{op}}$. It's easy to check that their images are in $C$, and constriction on $C$ yields $E(X)|_C=G, E(Y)|_{C}=G'$. Thus, the natural isomorphism between $E(X)$ and $E(Y)$ induce a natural isomorphism between $G$ and $G'$.


问:
A bag containing counters each with a different currency value
A bag contains 10 counters labelled 1 to 10. The first 6 are worth £1 each, the next 2 are worth £5 each, the next one is worth £10 and the final one is worth £50. You draw a counter from the bag at random. Its value being X pounds.
Let $\Omega= \lbrace1,2,\dots,10\rbrace$.

The questions are as follows:

Write down the function corresponding to a random variable X, i.e, for each $\omega \in \Omega$, write down $X(\omega)$

What subset of $\Omega$ does the event $\lbrace X \leq 5 \rbrace$ correspond to? Find $\mathbb{P}\lbrace X \leq 5 \rbrace$

Write down the probability mass function p of X, i.e. write down in a table all values X can take and their corresponding probabilities.

Use the probability mass function to find $\mathbb{P} (X \geq 10)$

So for question one I have no idea what to do, the notation confuses me honestly and I think maybe that's why I'm stuck.

For question 2. I believe the subset would be $A = \lbrace 1,2,3,4,5,6,7,8 \rbrace$ and I believe the probability is 0.8.

Again for number 3 I'm unsure what to do, do I simply write a table with $1,2,\dots,10$ in one row and then underneath for the probability just simply put $\frac{1}{10}$ for each counter. If this is the case what do I put in the tables as the titles? For the probability one would it be $\mathbb{P}(X=x)$ ?

Finally for number 4 is it just 0.2.

On an unrelated note, I'm struggling with interpreting notation while reading my lecturers notes for probability, will this improve over time just with practice. Or is there anywhere else I can look for additional help on mathematical notation. Thanks.

答:
Your answers to the second and fourth questions are correct.

The elements $\omega \in \Omega$ are the counters $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$.  The values $X(\omega)$ are the values of the counters.  Hence,
\begin{align*}
X(\omega) & = 1~\text{pound}, ~\text{if}~1 \leq \omega \leq 6\\
X(\omega) & = 5~\text{pounds}, ~\text{if}~7 \leq \omega \leq 8\\
X(\omega) & = 10~\text{pounds}, ~\text{if}~\omega = 9\\
X(\omega) & = 50~\text{pounds}, ~\text{if}~\omega = 10
\end{align*}

As for the probability mass function, we want to find the probability that the value of the counter is equal to a particular value for each possible value of $X$.
\begin{align*}
\Pr(X = 1~\text{pound}) & = 0.6\\
\Pr(X = 5~\text{pounds}) & = 0.2\\
\Pr(X = 10~\text{pounds}) & = 0.1\\
\Pr(X = 50~\text{pounds}) & = 0.1
\end{align*}


问:
We have $\pi(R) = \operatorname{End}_\mathbb{C}(V)$ is simple?

Let $R$ be a $\mathbb{C}$-algebra and $(\pi, V )$ a finite dimensional simple $R$-module. Show that

(a) The ring $\pi(R) \subset \operatorname{End}_{\mathbb{C}}(V)$ is simple.

(b) We have $\pi(R) = \operatorname{End}_\mathbb{C}(V)$.

So I want to show that $\pi(R)$ is Artinian, and then show that its Jacobson radical is $0$, so then I can conclude it is semisimple. I'm not sure how to show it is Artinian, and where do I go after I show it is semisimple. Maybe work with the finite dimensionality of the endomorphism ring?

答:
Let us put $S=\pi(R)$ and $J$ its Jacobson radical. Note that $V$ is a simple $S$-module.

$S$ is semi-simple. Indeed, $S$ is artinian since its left ideals are complex subspaces of $S\subset\mathrm{End}_\Bbb{C}(V)$, which has finite dimension, and thus satisfy the descending chain condition. $S$ is also Jacobson semi-simple since by Nakayama's lemma and finite generation of $V$, $JV$ is a proper $S$-submodule of $V$ and therefore zero by simplicity of $V$ (*). It follows that $J=0$ and thus that $S$ is semi-simple.

(*) Alternatively, and more to the point, $JV=0$ by simplicity of $V$ as an $S$-module and by definition of the Jacobson radical as those elements in $S$ killing all simple $S$-modules.

$S$ is simple. By the Artin-Wedderburn theorem and the fact that there are no nontrivial finite dimensional complex division algebras we have $S\simeq\prod_{i=1}^rM_{n_i}(\Bbb{C})$. In fact, $r=1$ and $S\simeq M_m(\Bbb{C})$. Indeed, if there were two factors or more, the element $e$ of $S$ corresponding to $(I_{n_1},0,\dots,0)$ would define a nontrivial central idempotent of $S$, and thus two nontrivial $S$-submodules $\mathrm{Ker}(e)$ and $\mathrm{Im}(e)$, contradicting the simplicity of $V$ as an $S$-module.

$S=\mathrm{End}_\Bbb{C}(V)$. Recall that $M_m(\Bbb{C})$ has up to isomorphism a single simple module which is isomorphic to $\Bbb{C}^m$. We conclude that $V$ has dimension $\mathrm{dim}(V)=m$ and $S$ dimension $\mathrm{dim}(V)^2$. Therefore $S=\mathrm{End}_\Bbb{C}(V)$.


问:
Right continuity property of the cumulative distribution function?
I have been studying probability from Sheldon Ross' A First Course in Probability. And the following property of cumulative distribution function was stated: CDFs are right continuous. And the following example was presented:

I am not totally sure, as to why when we do the probability of P(X $\le$ c) where c is some constant, we basically have that additionally $$\frac{1}{n}$$ term. Can someone help me understand it? I specifically want to know what is the physical interpretation behind having that term and why we need it to rigorously prove P(X $\le$ c).

答:
The reason is that you want to find the probability of $P(X<c)$ whereas $F(x)$ only gives you the probability of $P(X\leq c)$. Take part a) for example, we know that
$$\bigcup_n\left\{X\leq 3-\frac{1}{n}\right\} = \{X<3\}$$
and the sets on the LHS is nested such that
$$\left\{X\leq 3-\frac{1}{n}\right\}\subseteq \left\{X\leq 3-\frac{1}{n+1}\right\}$$
for each $n\in\mathbb{N}$. Thus, by continuity from below of the probability measure, we have
$$P(\{X<3\}) = P\left(\bigcup_n\left\{X\leq 3-\frac{1}{n}\right\}\right) = \lim_n P\left(\left\{X\leq 3-\frac{1}{n}\right\}\right).$$
Now, you have a good handle on the RHS of the above equation by simply using $F$.


答: When a book uses the phrasing, "We say that an A has property P if condition C holds", they mean "exactly if" or "if and only if".  Just using the word "if" is a universal shorthand for mathematical definitions.</p>

Title: Is it true that if a non-negative function is Riemann integrable, there must exist an upper sum U(f)?

Body:
I know that if the upper sum of f, $U(f)$ equals the lower sum of f on an interval $[a,b]$ then f is Riemann integrable on [a,b] with $U(f)=L(f)= \int_{a}^{b} f(x) \,dx.$

Is the reverse true? That is if f is riemann integrable is it the case that there must exists an U(f) and L(f) with $U(f)=L(f)= \int_{a}^{b} f(x) \,dx.$?

Edit:
My books gives the following:

A bounded function $f$ defined on the interval [a,b] is Riemann integrable if $U(f) = L(f).$ In this case we define $\int_{a}^{b} f$ or $\int_{a}^{b} f(x)$ to be $\int_{a}^{b} f = U(f) = L(f).$

They don't use "if and only if" so i was note sure whether it was the case that if f is riemann integrable then there must exists an U(f) and L(f) with $U(f)=L(f)= \int_{a}^{b} f(x) \,dx.$?


问:
Position of squares on a checkerboard from a sequence of numbers
I want to position squares on a checkerboard in such a way that they fill the square they are on.

How do I calculate the position of one of those orange squares such that $f(i) = x,y$ coordinate, $i$ being the number of the orange square starting from $0$.

Also, the size of the checkerboard is not definite, I need the pattern to repeat for any size of the checkerboard horizontally and vertically.

With a checkerboard height of 5 instead of 4.

答:
You have to play with how many times you have reached $5.$ They way to do this is with division, so taking ceils is the right approach. The row is just $\left \lfloor \frac{i}{5}\right \rfloor,$ the column depends on which row are we. It is $5\times \left \lfloor \frac{\left \lfloor \frac{i}{5}\right \rfloor}{5}\right \rfloor +i\pmod 5.$ This gives you
$$f(i)=\left (5\times \left \lfloor \frac{\left \lfloor \frac{i}{5}\right \rfloor}{5}\right \rfloor +\left(i\pmod 5)\right )\, ,\,
\left \lfloor \frac{i}{5}\right \rfloor\right ).$$
here the columns and rows are indexed starting at $0.$

The code in sage looks like:

```<code>A = [[0 for i in range(0,10)] for j in range(0,10)]
for i in range(0,50):
    x = i//5
    y = 5*(x//5) + i%5
    A[x][y]=i+1
for i in range(0,10):
    print(A[i])
</code>```

Edit:
Your new formula will have to have a period of $2$ that alternates it.
The code for your new pattern looks like

```<code>  A = [[0 for i in range(0,20)] for j in range(0,20)]
    for i in range(0,25*8):
        x = i//10
        tal = (i%10)//5
        retal = (i//50)%2
        y = 5*retal + 5*tal+ i%10
        A[x][y]=i+1
    for i in range(0,20):
        print(A[i])
</code>```

Edit 2:
For any height and even width, the following does what you want

```<code>h = 4
w = 8
A = [[0 for i in range(0,5*w)] for j in range(0,5*h)]
for i in range(0,(h*w*25)/2):
    ww = ceil(w/2)*5
    x = i//ww
    tal = (i%ww)//5
    retal = (i//(ww*5))%2
    y = 5*retal + 5*tal+i%ww
    if x<h*5 and y<5*w:
        A[x][y]=1+i
for i in range(0,5*h):
    print(A[i])
</code>```

The explanation is the same as before, so I am not sure what is unclear or why you keep changing the size instead of generalizing it. Why is the odd width case different? How to fix it?

Body:
You have to play with how many times you have reached $5.$ They way to do this is with division, so taking ceils is the right approach. The row is just $\left \lfloor \frac{i}{5}\right \rfloor,$ the column depends on which row are we. It is $5\times \left \lfloor \frac{\left \lfloor \frac{i}{5}\right \rfloor}{5}\right \rfloor +i\pmod 5.$ This gives you
$$f(i)=\left (5\times \left \lfloor \frac{\left \lfloor \frac{i}{5}\right \rfloor}{5}\right \rfloor +\left(i\pmod 5)\right )\, ,\,
\left \lfloor \frac{i}{5}\right \rfloor\right ).$$
here the columns and rows are indexed starting at $0.$

The code in sage looks like:

A = [[0 for i in range(0,10)] for j in range(0,10)]
for i in range(0,50):
    x = i//5
    y = 5*(x//5) + i%5
    A[x][y]=i+1
for i in range(0,10):
    print(A[i])

Edit:
Your new formula will have to have a period of $2$ that alternates it.
The code for your new pattern looks like

  A = [[0 for i in range(0,20)] for j in range(0,20)]
    for i in range(0,25*8):
        x = i//10
        tal = (i%10)//5
        retal = (i//50)%2
        y = 5*retal + 5*tal+ i%10
        A[x][y]=i+1
    for i in range(0,20):
        print(A[i])

Edit 2:
For any height and even width, the following does what you want

h = 4
w = 8
A = [[0 for i in range(0,5*w)] for j in range(0,5*h)]
for i in range(0,(h*w*25)/2):
    ww = ceil(w/2)*5
    x = i//ww
    tal = (i%ww)//5
    retal = (i//(ww*5))%2
    y = 5*retal + 5*tal+i%ww
    if x<h*5 and y<5*w:
        A[x][y]=1+i
for i in range(0,5*h):
    print(A[i])

The explanation is the same as before, so I am not sure what is unclear or why you keep changing the size instead of generalizing it. Why is the odd width case different? How to fix it?


问:
Does a linear time system imply that for a zero input, a zero output is given?
My question is exactly that. The definition of a linear (discrete) time system is, as with most definitions of linearity, that a linear combination of signals given as input must correspond to the same combination of individual outputs. I.e:

$$T(\alpha_1\overline{x_1} + \alpha_2\overline{x_2}) = \alpha_1T(\overline{x_1}) + \alpha_2T(\overline{x_2})$$

Does this imply that given a zero signal, the output must also be zero?

As background, I was given some signals and responses of a system and a task to determine the linearity. If alphas are set to zero, the signals may be whatever, and naturally multiplying the responses by zero yields a zero response. But by combining vectors such that the input is zero, alphas are most likely non-zero. So additionally I'm wondering, is there a way to say something meaningful about the system based on those few signal-response pairs.

答:
Proof:

Let $T:U\to V$

$T(\alpha \cdot 0_U)=T(0 \cdot u)=0\cdot T(u)=0_V.$


问:
Solving a differential equation in distribution theory
Let $$a_n\frac{d^ny(t)}{dt^n} + a_{n-1}\frac{d^{n-1}y(t)}{dt^{n-1}} + \dots +a_0y(t) = \delta(t)$$What's the general solution for $y(t)$? I don't know how to interpret and solve this equation in the sense of distribution theory. Certainly $\delta(t)$ is not a function, so it can't be evaluated at point. I'm looking for a rigorous solution which gives general solution. Here is a specific example which is solved by md2perpe but doesn't seem can be generalized to the general case.

答:
Especially if we want to behave as though there were "methods" beyond "tricks", we might want to see how to solve the differential equation in a way that genuinely uses distributions. (I'm not going to say anything here that's not available in some form in the references linked-to by Wikipedia's page on the Malgrange-Ehrenpreis theorem, though the "more elementary arguments" that were published in the Math Monthly do not seem to be widely known.)

Let the equation be $P(i\partial)u=\delta$, where $\partial=d/dx$, and $P$ is a complex polynomial. When $P$ has no real roots, using Fourier transform and simply "dividing" suffices. The opposite case is more interesting: suppose $P(x)=0$ has $n$ distinct real roots $a_1,\ldots,a_n$. We still use Fourier transform, but cannot naively "divide".

Supposing $u$ is at least a tempered distribution, take the Fourier transform: $P(x)\widehat{u}=1$. We would like to divide, but the functions $1/(x-a_j)$ are not locally $L^1$, so are not immediately (tempered) distributions. But we might recall that $xv=1$ for distribution $u$ implies that $v$ is a constant multiple of the principal_value integral against $1/x$. At least qualitatively and heuristically, $1/P(x)=\sum_j c_j/(x-a_j)$ for suitable constants $c_j$, so it is reasonable/plausible that the tempered distribution $v$ such that $P(x)\cdot v=1$ is the corresponding linear combination of principal value integrals. (And this is true, but does take a little proof, which will vary depending on one's set-up.)

We also know that the Fourier transform of PV $1/x$ is (up to a constant) the sign function. Fourier transform converts translation to multiplication by exponentials. Thus, the (inverse) Fourier transform is (a constant multiple of) $\sum_j c_j\cdot ({\mathrm sign}(x))\cdot e^{ia_jx}$.

To check that this is correct requires a little more info about the coefficients $c_j$, for example, that $c_j=1/P'(a_j)$. This uses the distinctness of the $a_j$.


问:
Set of fixed points is convex.
Problem: Suppose that $f:\mathbb{R}^n \to \mathbb{R}^n$ has Lipschitz constant $1$. Is it true that the set of fixed points of $f$ is convex?

My attempt: I managed to show that if $x,y$ are two fixed points of $f$ and $z  = (1-\lambda)x+\lambda y$ for $\lambda \in (0,1)$, then $||f(z)-x||=||z-x||$ and $||f(z)-y||=||z-y||$. But I'm not quite sure what to do after this. Is there any hint or suggestion?

Title: Set of fixed points is convex.

Body:
Let $x_0,x_1\in\mathbb R^n$, $x_0\ne x_1$.

Assume $f(x_0)=x_0$ and $f(x_1)=x_1$, and $x_t=tx_1+(1-t)x_0$, $t\in (0,1)$. We need to show that $f(x_t)=x_t$.

Assume that $f(x_t)\ne x_t$.

Then
$$
\|x_1-x_0\|=\|f(x_1)-f(x_0)\|\le \|f(x_t)-f(x_0)\|+\|f(x_t)-f(x_1)\|\\ \le \|x_t-x_0\|+\|x_t-x_1\|
=t\|x_1-x_0\|+(1-t)\|x_1-x_0\|=\|x_1-x_0\|.
$$
Hence
$$
\|f(x_1)-f(x_0)\|= \|f(x_t)-f(x_0)\|+\|f(x_t)-f(x_1)\|   \tag{1}
$$

Meanwhile
$$
\|f(x_t)-f(x_0)\|\le\|x_t-x_0\|=t\|x_1-x_0\|. \tag{2}
$$
Similarly
$$
\|f(x_t)-f(x_1)\|\le (1-t)\|x_1-x_0\|. \tag{3}
$$
Adding (2) and (3) we obtain
$$
\|f(x_t)-f(x_0)\|+\|f(x_t)-f(x_1)\| \le t\|x_1-x_0\|+(1-t)\|x_1-x_0\|=\|x_1-x_0\|
$$
and hence
$$
\|x_1-x_0\|=\|f(x_1)-f(x_0)\|\le
\|f(x_t)-f(x_0)\|+\|f(x_t)-f(x_1)\| \le \|x_1-x_0\|
$$
which implies that
$$
\|x_1-x_0\|=\|f(x_t)-f(x_0)\|+\|f(x_t)-f(x_1)\|=\|f(x_t)-x_0\|+\|f(x_t)-x_1\|
$$
and we obtain that (2) and (3) are equalities.

Lemma. If $\,a,b\in\mathbb R^n\,$ and $\,\|a+b\|=\|a\|+\|b\|$, then $a=\lambda c$ and $b=\mu c$, for some $c\in\mathbb R^n$ and $\lambda,\mu\ge 0$. In particular, if
$\|a-b\|=\|a-c\|+\|c-b\|$, then $c=sa+(1-s)b$, for some $s\in [0,1]$.

This Lemma implies that
$$
f(x_t)=sx_1+(1-s)x_0, \quad\text{for some $s\in [0,1]$}
$$
and due to (2) or (3), $s=t$, and hence $f(x_t)=x_t$.


问:
What is this group $O(2)/\mathbb{Z}_2$?
I'm talking from a physics student perspective. What is the properties of this group?. Is it isomorphic to something more known in the physics community?.

Background of the question:

I found this group while playing around with the properties of $O(2)$ group of ortogonal matrices. At first sight, I thought that $O(2)/\mathbb{Z}_2$ was isomorphic to $SO(2)$, but I think it's wrong.

答:
Interestingly, the group $PO(2, \Bbb R)$ is isomorphic to $O(2, \Bbb R)$.

To see this, let us write $r(\theta)$ for $\begin{pmatrix}\cos \theta & \sin\theta \\-\sin\theta&\cos\theta\end{pmatrix}$ and write $t(\theta)$ for $\begin{pmatrix}\cos \theta & \sin\theta \\\sin\theta&-\cos\theta\end{pmatrix}$.

We then have: \begin{eqnarray}r(\theta)r(\theta') &=& r(\theta + \theta'),\\r(\theta)t(\theta') &=& t(\theta' - \theta),\\t(\theta)r(\theta') &=& t(\theta + \theta'),\\t(\theta)t(\theta') &=& r(\theta' - \theta).\end{eqnarray}

Now define a map $\phi:O(2, \Bbb R) \rightarrow O(2, \Bbb R)$ sending $r(\theta)$ to $r(2\theta)$ and $t(\theta)$ to $t(2\theta)$. From the above formulas, it is clear that $\phi$ is a surjective (topological) group homomorphism.

The kernel of $\phi$ is clearly the subgroup $\{\pm I_2\}$, thus $\phi$ induces an isomorphism from $PO(2, \Bbb R)$ to $O(2, \Bbb R)$.

Title: What is this group $O(2)/\mathbb{Z}_2$?


问:
Is it true that for all $k>1$, $ x^k > x \log x $ for sufficiently large $x$?
I was doing some problems in Big-O notation, and I got intrigued by this question as it was never really addressed. If we plot $x^2$, it's obviously bigger than $x\log x$ for all $x > 0$. However, I started plotting graphs for smaller and smaller values of the exponent, and saw that the crossing point between $x^k$ and $x\log x$ got further and further away. I wondered if maybe there was some particular constant which was the smallest value for $k$ such that the two graphs would cross. But then I found out that even $x^{1.1} >x\log x$ when $x> 3.43 \times 10^{+15}$. This has lead me to believe that for any $k>1$, $x^k$ will eventually overtake $x\log x$, but I think it would be beyond my capabilities to prove it, and I don't know where to start. Does anyone know the answer to this, and is the proof understandable for an undergrad math student?

Edit: Also the functions cross at two points, and I noticed that the first crossing point converges to $e$ as $k$ goes to $1$. What's the explanation for this?

答:
Yes, it is  a basic result in asymptotic analysis that  for any $\alpha,\beta >0$,
$$\ln^\alpha x=_{\infty}o\bigl(x^\beta\bigr). $$


问:
Finding the area bounded by a curve, a line, and the x-axis
I am to find the area R bounded by the kurve $x=y^2+2$, $y=x-4$ and $y=0$.

I found the points of intersection by first setting the lines equal eaxh other and used the quadratic formula:
\begin{align*}
    y^2+2=y+4
    \\y^2-y-2=0
    \\
    \\y_{1,2}=\frac{(-b) +/- \sqrt{{b^2}-4ac}}{2a}=\frac{-(-1)+/-\sqrt{{(-1)^2}-4*1*(-2)}}{2}=\frac{1 +/- \sqrt{9}}{2}=\frac{1+/-3}{2}\\
    \\y_1 = 2 \\y_2=-1
\end{align*}

I am using $y_1=2$ and $y_2=0$ as boundaries because $y=0=x$ is given. Then I sliced horizontally:
\begin{align*}
A=\int_0^2{(y+4)-(y^2+2)}dy\\
A=\int_0^2{2+y-y^2}dy\\
A=[2y-\frac{y^2}{2}-\frac{y^3}{3}]_0^2\\
A=[(2*2)-\frac{2^2}{2}-\frac{2^3}{3}]-[(2*0)-\frac{0^2}{2}-\frac{0^3}{3}]\\
A=4-2-\frac{8}{3}\\
A=-\frac{2}{3}
\end{align*}

I get a negative area, what am I doing wrong here?\
If I slice it vertically, I use the boundaries $x_1=2$ and $x_2=6$ and get the integral
\begin{align*}
A=\int_2^6{(x-4)-(\sqrt{x-2})}dx\\
A=\int_2^6{1-\frac{1}{2\sqrt{x-2}}}\\
A=[1-\frac{1}{2\sqrt{x-2}}]\\
A=(1-\frac{1}{2\sqrt{4}})-(1-\frac{1}{2\sqrt{0}})\\
A=(1-\frac{1}{4})-(1-undefined)
\end{align*}

What is the proper way to solve this?

Body:
You made a mistake in the sign when integrating
\begin{align*}
A=\int_0^2{(y+4)-(y^2+2)}dy\\
A=\int_0^2{2+y-y^2}dy\\
A=[2y\color{red}{+}\frac{y^2}{2}-\frac{y^3}{3}]_0^2\\
A=[(2*2)\color{red}{+}\frac{2^2}{2}-\frac{2^3}{3}]-[(2*0)\color{red}{+}\frac{0^2}{2}-\frac{0^3}{3}]\\
A=4\color{red}{+}2-\frac{8}{3}\\
A=\frac{10}{3}
\end{align*}

Integrating wrt to $x$ we have $$A=\int_{2}^{4}\sqrt{x-2}dx+\int_{4}^{6}(\sqrt{x-2}-(x-4))dx$$
$$=\int_{2}^{6}\sqrt{x-2}dx+\int_{4}^{6}(4-x)dx$$
$$=\frac{2}{3}{(x-2)^{\frac{3}{2}}}\big|_{x=2}^{x=6}+\big[4x-\frac{x^2}{2}\big]_{x=4}^{x=6}$$
$$=\frac{16}{3}+\big[6-8]=\frac{16}{3}-2=\frac{10}{3}$$


问:
A question about basis linear space
See $\mathbb{R}$ as a linear space over $\mathbb{Q}$. $B\subset\mathbb{R}$ is a basis with positive reals.

Can $B$ be a group where multiplication is standard?

答:
No.

Suppose it is otherwise. Then we have $\sqrt{2}=\sum_{j=1}^n \alpha_jb_j$ for some $\alpha_j\in {\bf Q}\setminus\{0\},b_j\in B$ with $0<\lvert b_1\rvert<\ldots<\lvert b_n\rvert$. Then also
$$2=\left(\sum_{j=1}^n \alpha_jb_j\right)^2=\sum_{i,j}\alpha_i\alpha_jb_ib_j$$
and similarly
$$4=\sum_{i,j,k,l}\alpha_i\alpha_j\alpha_k\alpha_lb_ib_jb_kb_l$$
Summing things up, we get that
$$0=\sum_{i,j,k,l}\alpha_i\alpha_j\alpha_k\alpha_lb_ib_jb_kb_l-\sum_{i,j=1}2\alpha_i\alpha_jb_ib_j$$
Now we have the following facts:

If $\lvert b_n\rvert>1$, then the coefficient by $b_n^4$ is $\alpha_n^4$, but it must be zero, which is a contradiction (because $b_n^4$ is the largest element of $B$ in the expression on right hand side).

If $\lvert b_1\rvert<1$, then likewise the coefficient of $b_1^4$ is $\alpha_1^4$, again yielding a contradiction.

Therefore, we must have $\lvert b_1\rvert=\lvert b_n\rvert=1$, so $n=1$ and $\sqrt 2$ is a rational number, which is a contradiction, so we're done.

Title: A question about basis linear space


问:
Proof that in the family of sets $A_t = (\frac{t}{2} ; \frac{t+1}{2})$ for $t \in (0;1)$, $\bigcup A_{t} = (0;1)$, $\bigcap A_{t} = \{\frac{1}{2}\}$
As in the title, I have to proof that for the family of sets given as: $A_t = (\frac{t}{2} ; \frac{t+1}{2})$ for $t \in (0;1)$:

$\bigcup\limits_{t \in (0;1)} A_{t} = (0;1)$

$\bigcap\limits_{t \in (0;1)} A_{t} = \{\frac{1}{2}\}$

I don't know how to do the proof form right to left. I mean proving that for every $x$ there is $t$ that represents a set that includes $x$. Usually I would use Archimedes' Axiom but out here I have no natural numbers. How then prove that for x close to 0 there is a set with a boundary that is even closer to 0? I need to do it "in the way of set theory" - I mean that I can not use boundary-value analysis nor function analysis.

答:
I suggest to prove the first equality with double inclusions:

since for every $t\in(0,1)$ one has $0<\frac t2<\frac{t+1}2<1$, it follows that $A_t\subset(0,1)$ for all such $t$, hence $\bigcup_{t\in(0,1)}A_t\subseteq(0,1)$.

Conversely take any $x\in(0,1)$ and search for some $t\in(0,1)$ such that $x\in A_t$, that is $\frac t2<x<\frac{t+1}2$; you'll see that $t=2x-\frac12$ will work. Hence $x\in\bigcup_{t\in(0,1)}A_t$ so you conclude that $(0,1)\subseteq
\bigcup_{t\in(0,1)}A_t$ and we are done with the first one.

Second one: we want to study $\bigcap_{t\in(0,1)}A_t$, that is we search for the elements $x$ belonging to $A_t$ for all $t\in(0,1)$:
\begin{align*}
x\in A_t\;\forall t\in(0,1)\;
&\Longleftrightarrow\frac t2<x<\frac{t+1}2\;\;\forall t\in(0,1)\\
\end{align*}
you can see that $x=1/2$ is satisfies this condition, that is
$$
\left\{\frac12\right\}\subseteq\bigcap_{t\in(0,1)}A_t$$
Moreover for every $x<1/2$ or $x>1/2$ you can find some $t\in(0,1)$ such that the last two equality are not simultaneously satisfied, from which the last inclusion is indeed an equality.


问:
Open, discontinuous function from $\mathbb{R}$ to $\mathbb{R}$
I am looking for some examples of open functions from $\mathbb{R}$ (or $\mathbb{R}^n$) to $\mathbb{R}$ that are not continuous. I know that the classic example for an open, discontinuous function is mapping the points on the unit circle to the interval $[0, 2\pi)$, but the domain is not reals. Maybe there is a way to compose this function with another one to get a function that is from $\mathbb{R}$ to $\mathbb{R}$.

For clarity, a function $f:\mathbb{R}^k \rightarrow \mathbb{R}^n$ is open if for each $A \subset \mathbb{R}^k$ with $A$ open, $f(A)$ is open. In other words, the image of every open subset is open.

I know that if the preimage of every open set is open, then the function is continuous. So I thought a discontinuous function whose inverse is continuous might work. However, I couldn't find such a function.

答:
Define an equivalence relation $\sim$ on $\Bbb R$ by $x\sim y$ iff $x-y\in\Bbb Q$. For each $x\in\Bbb R$ the $\sim$-class of $x$ is $[x]=x+\Bbb Q=\{x+q:q\in\Bbb Q\}$, which is clearly dense in $\Bbb R$. Let $\mathscr{C}=\{[x]:x\in\Bbb R\}$, the partition of $\Bbb R$ into $\sim$-classes; each member of $\mathscr{C}$ is countable, so $|\mathscr{C}|=|\Bbb R|$, and there is a bijection $\varphi:\mathscr{C}\to\Bbb R$. Define

$$f:\Bbb R\to\Bbb R:x\mapsto\varphi([x])\,.$$

Let $U$ be any non-empty open set in $\Bbb R$, and let $y\in\Bbb R$; there is a $C\in\mathscr{C}$ such that $\varphi(C)=y$. $C$ is dense in $\Bbb R$, so there is an $x\in U\cap C$, and $$f(x)=\varphi([x])=\varphi(C)=y\,,$$ so $f[U]=\Bbb R$. That is, $f[U]=\Bbb R$ for each non-empty open $U\subseteq\Bbb R$, so $f$ is open.

Clearly, however, $f$ is not continuous at any point: for each $x\in\Bbb R$ and $C\in\mathscr{C}$ there is a sequence in $C$ converging to $x$, so that continuity of $f$ at $x$ would require that $f(x)=\varphi(C)$ for each $C\in\mathscr{C}$, i.e., that $f(x)=y$ for each $y\in\Bbb R$.


问:
Cancelling items in product over two indices
$$ \int_0^1 x^n (1-x)^m dx$$

Do parts:

$$ I_{n.m} = \frac{1}{n+1}\left( \left[x^{n+1} (1-x)^m \right]_0^1 + m \int_0^1 x^{n+1} (1-x)^{m-1}\right)$$

Or,

$$ I_{n,m}= \frac{m}{n+1} \left(I_{n+1,m-1} \right)$$

Product both sides over m from $1 \to k$ (refer here)

$$ \prod_{m=1}^k I_{n,m} = \prod_{m=1}^k \frac{m}{n+1} I_{n+1,m-1}$$

Or,
$$ \prod_{m=1}^k I_{n,m} = m! \prod_{m=1}^k \frac{1}{n+1} I_{n+1,m-1} \tag{1}$$

Product both sides over $n$ from $1 \to j$,

$$ \prod_{n=1}^j\prod_{m=1}^k I_{n,m} = \prod_{n=1}^jk! \prod_{m=1}^k \frac{1}{n+1} I_{n+1,m-1} \tag{1}$$

Or,

$$ \prod_{n,m=1}^{j,k} I_{n,m} = \frac{(k!)^j}{(j!)^m} \prod_{n,m=1}^{j,k} I_{n+1,m-1}$$

The left is summed over the domain $n \times m= (1,j) \times (1,k)$ and the right is over the domain $n \times m = (2,j+1) \times (0,k-1)$. I need advice on how to cancel like terms, the cancellation part is really tripping me over.

答:
We consider $I_{n,m}=\int_{0}^1x^n(1-x)^m\,dx$ and the recurrence relation:
\begin{align*}
I_{n,m}&=\frac{m}{n+1}I_{n+1,m-1}\qquad\qquad n,m\geq 1\\
I_{n,0}&=I_{0,n}=\frac{1}{n+1}\qquad\qquad\quad n\geq 0\tag{1}
\end{align*}

Let $j,k$ be positive integers. We obtain
\begin{align*}
\prod_{n=1}^j\prod_{m=1}^k I_{n,m}&=\prod_{n=1}^j\prod_{m=1}^k \left(\frac{m}{n+1}I_{n+1,m-1}\right)\\
&=\prod_{n=1}^j\frac{k!}{(n+1)^k}\prod_{m=1}^k I_{n+1,m-1}\\
&=\frac{\left(k!\right)^j}{\left((j+1)!\right)^k}\prod_{n=1}^j\prod_{m=1}^k I_{n+1,m-1}\\
&=\frac{\left(k!\right)^j}{\left((j+1)!\right)^k}\prod_{n=2}^{j+1}\prod_{m=0}^{k-1} I_{n,m}\tag{2}\\
&=\frac{\left(k!\right)^j}{\left((j+1)!\right)^k}\left(\prod_{m=0}^{k-1}I_{j+1,m}\right)\prod_{n=2}^{j}\prod_{m=0}^{k-1} I_{n,m}\tag{3}\\
&=\frac{\left(k!\right)^j}{\left((j+1)!\right)^k}\left(\prod_{m=0}^{k-1}I_{j+1,m}\right)
\left(\prod_{n=2}^j I_{n,0}\right)\prod_{n=2}^{j}\prod_{m=1}^{k-1} I_{n,m}\tag{4}\\
&=\frac{\left(k!\right)^j}{\left((j+1)!\right)^k}\left(\prod_{m=0}^{k-1}I_{j+1,m}\right)
\left(\prod_{n=2}^j \frac{1}{n+1}\right)\prod_{n=2}^{j}\prod_{m=1}^{k-1} I_{n,m}\tag{5}\\
&\,\,\color{blue}{=\frac{2\left(k!\right)^j}{\left((j+1)!\right)^{k+1}}\left(\prod_{m=0}^{k-1}I_{j+1,m}\right)
\prod_{n=2}^{j}\prod_{m=1}^{k-1} I_{n,m}}\tag{6}\\
\end{align*}

Comment:

In (2) we shift the indices $n$ and $m$ to derive the representation $I_{n,m}$ as preparation for cancellation.

In (3) we separate the factors $I_{j+1,.}$.

In (4) we separate the factors $I_{.,0}$.

In (5) we use the identity (1).

In (6) we use the identity $\prod_{n=2}^j\frac{1}{n+1}=\frac{2}{(j+1)!}$.

On the other hand we have
\begin{align*}
\color{blue}{\prod_{n=1}^j\prod_{m=1}^k I_{n,m}}&=\left(\prod_{n=1}^j I_{n,k}\right)\prod_{n=1}^j\prod_{m=1}^{k-1} I_{n,m}\tag{7}\\
&=\left(\prod_{n=1}^j I_{n,k}\right)\left(\prod_{m=1}^{k-1} I_{1,m}\right)\prod_{n=2}^j\prod_{m=1}^{k-1} I_{n,m}\tag{8}\\
&\,\,\color{blue}{=\left(\prod_{n=1}^j I_{n,k}\right)\left(\prod_{m=1}^{k-1}\frac{1}{m^2+3m+2}\right)\prod_{n=2}^j\prod_{m=1}^{k-1} I_{n,m}}\tag{9}
\end{align*}

Comment:

In (7) we separate factors $I_{n,k}$ to get the upper index $j$ in the product as in (6).

In (8) we separate factors $I_{1,m}$ to start with $n=2$ as in (6).

In (9) we use the identity $I_{1,m}=\frac{1}{m^2+3m+2}$ valid for $m\geq 1$.

We can now cancel the double product in (6) and (9) and obtain
\begin{align*}
\color{blue}{\frac{2\left(k!\right)^j}{\left((j+1)!\right)^{k+1}}\prod_{m=0}^{k-1}I_{j+1,m}
=\prod_{m=1}^{k-1}\frac{1}{m^2+3m+2}\prod_{n=1}^j I_{n,k}}
\end{align*}


问:
How do I find the Laurent series for $\frac{1}{(1+\cos z)^2}$ around $z_0=\pi$?
What I tried so far:

Using that  $\cos z = \sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n)!}z^{2n}$ , for any $z \in \mathbb C$, then, for $0<|z-\pi|<1$:
$$\begin{aligned}\dfrac{1}{(1+\cos z)^2} &= \dfrac{1}{(1+\cos ((z-\pi)+\pi))^2} \\&= \dfrac{1}{(1-\cos(z-\pi))^2}\\ &=\dfrac{1}{\left(1-\left(1-\dfrac{(z-\pi)^2}{2} + \dfrac{(z-\pi)^4}{24}-\dfrac{(z-\pi)^6}{720}+...\right)\right)^2}\\ &=\dfrac{1}{\left(\dfrac{(z-\pi)^2}{2} - \dfrac{(z-\pi)^4}{24}+\dfrac{(z-\pi)^6}{720}-...\right)^2} \end{aligned}$$
Now I don't know how to proceed (and am I even doing things right? I don't know if this is a rigorous development).

答: If, as you wrote in the comments, you're after a proof of the fact $\operatorname{res}_{z=\pi}\frac1{(1+\cos z)^2}$, using the Laurent series, note that if$$\frac1{(1+\cos z)^2}=\sum_{n=-\infty}^\infty a_n(z-\pi)^n,$$then\begin{align}\sum_{n=-\infty}^\infty(-1)^na_n(z-\pi)^n&=\sum_{n=-\infty}^\infty a_n\bigl((2\pi-z)-\pi\bigr)^n\\&=\frac1{(1+\cos(2\pi-z))^2}\\&=\frac1{(1+\cos z)^2}\\&=\sum_{n=-\infty}^\infty a_n(z-\pi)^n\end{align}and therefore, if $n$ is odd, then $a_n=0$. In particular, $a_{-1}=0$.


问:
Prove $M(\alpha(-s)) = \alpha(s)$ for all s in $(-\epsilon,\epsilon)$, where M: $\mathbb{R}^2 \rightarrow \mathbb{R}^2$
Let $\epsilon>0$ and $\alpha:(-\epsilon,\epsilon) \rightarrow \mathbb{R}^2$ be a regular plane curve parametrized by arc-length. Suppose that $k(s) = k(-s)$ for all $s \in (-\epsilon,\epsilon)$.
Prove that $M(\alpha(-s)) = \alpha(s)$ for all $s$ in $(-\epsilon,\epsilon)$, where $M: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is reflection in the plane about the normal line of $\alpha$ at $s = 0$.

I know I have to use the uniqueness part from the Fundamental Theorem of Curves in the plane. But I don't know how to start it. Does someone have a little tip????

答: Hint: To keep track of things, let $\beta(s) = M\alpha(-s)$. Then note that $\beta$ is also parametrized by arc-length, and $\kappa_\beta(s) = \kappa_\alpha(s)$ (because $s \mapsto -s$ changes the sign of the curvature once, and $M$ being a reflection changes this sign back to the original $+$). We also have $\beta(0) = \alpha(0)$ (clear), $T_\beta(s) = T_\alpha(s)$ (why?), and thus $N_\beta(0) = N_\alpha(0)$. Hence...?


问:
What does the phrase "limit exists" graphically mean?
I know that "limit exists" means the left-hand limit and the right-hand limit of a function are equal.

However, What does the phrase "limit exists" graphically mean?

What I understand is, "limit exists" means:

the function is 'almost' continuous with a 'hole' on the curve

How can my above statement be reworded using 'mathematical' language?

答:
I would say, graphically speaking, "the limit of $f$ exists at $a$ if $f$ is continuous and defined on at least every point close to $a$."

I say "at least", because we can have definition and continuity at $a$, which is sufficient for the limit existing, but not necessary.


问:
How to Find a Finite-Difference Matrix
I have read several websites trying to explain finite-differential equations, but I haven't been able to find one that explains how it's put into the matrix form.

$f(x) = -\frac{d^2u}{dx^2}$ where $u(0) = 0$ and $u(1) = 0$ becomes

$\begin{bmatrix}
2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
0 & -1 & 2 & -1 & 0 \\
0 & 0 & -1 & 2 & -1 \\
0 & 0 & 0 & -1 & 2
\end{bmatrix}$

multiplied by one column of values between $u_1$ to $u_5$ is equal to

$h^2$$\begin{bmatrix}
f(h) \\
f(2h) \\
f(3h) \\
f(4h) \\
f(5h)
\end{bmatrix}$

Looking at the equation that causes this matrix, it confuses me.  The difference equation is
$$-u_{j+1}+2u_j-u_{j-1}=h^2f(jh)$$

The initial term, when $j=1$, should make it so that the equation is:

$$u_2+2u_1-u_0$$

Would this not cause the first row to become $(-1, 2, 0, 0, 0)$ because $u_0$ has been defined as 0?  I know that my thinking is wrong, since the book tells me so, but I don't understand how the first and last row is determined.

On a related note, how does the matrix equation if the boundaries are changed?  For an example, if $u_0 = 1$ and $u_1 = 2$ on the original equation.  Will the answer become

$\begin{bmatrix}
2 & 0 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
0 & -1 & 2 & -1 & 0 \\
0 & 0 & -1 & 2 & -1 \\
0 & 0 & 0 & -1 & 2
\end{bmatrix}$

答:
Just adding to the answer, summarizing Jonathan.

\begin{eqnarray}
\begin{pmatrix}
2 & 0 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
0 & -1 & 2 & -1 & 0 \\
0 & 0 & -1 & 2 & -1 \\
0 & 0 & 0 & -1 & 2
\end{pmatrix}
\begin{pmatrix}
u_1 \\
u_2 \\
u_3 \\
u_4 \\
u_5
\end{pmatrix}
=
\begin{pmatrix}
f(h) \\
f(2h) \\
f(3h) \\
f(4h) \\
f(5h) \\
\end{pmatrix}
\end{eqnarray}

Only $ u_1 $ to $u_5$ goes in the matrix due the Boundary Condition $u_0=u_6=0$, that explains why your logic about the first and last rows were wrong. Just remember the multiplication logic for matrix/array.

Title: How to Find a Finite-Difference Matrix


问:
Why are we interested in total variation and how does it tie into its integral definition?
I am studying measure theory and had a few questions. In the text I am using, it defines total variation of a measure $\nu$ as

$$|\nu| = \nu^+ + \nu^-$$
My first question is why are we interested in this quantity to begin with? Secondly, the text claims that the above implies $\nu$ is of the form $\nu(E) = \int_E f d|\nu|$, for any measurable set $E$. I cannot see how this integral definition follows the total variation definition. Lastly, the text claims the inequality $|\nu(E)| \leq |\nu|(E)$ is obvious, but I cannot quite see how. One last question I have is one of notation. I see notation such as $g(\frac{d\nu}{d\mu})$, what does this mean exactly?

Thank you.

答:
(1) Why are we interested in the total variation?

Firstly, because of (2) and (3), below.  We can often prove things about signed measures using this.

(2) Why is $\nu(E) = \int_E f d|\nu|$?

Hahn decomposition.

(3) Why is $|\nu(E)| \leq |\nu|(E)$?

Use your solution to (2).

(4) I doubt you often see $g(\frac{d\nu}{d\mu})$, that is a function $g$ evaluated on a Radon-Nikodym derivative.  But you do often see $g\;\frac{d\nu}{d\mu}$, that is a function $g$ multiplied by a Radon-Nikodym derivative.