答:
The map $\varphi:(\theta,\phi)\rightarrow\Big(f(\theta,\phi),\theta,\phi\Big)$ is not a parametric representation of the spherical surface $r=f(\theta,\phi)$. Rather, $\varphi$ is a parametric representation of the surface $x=f(y,z)$ since parametric surfaces (and curves) are inherently represented in Cartesian form. What you're doing is equivalent to saying $\theta \longrightarrow \Big(\theta^2,\theta\Big)$ is a parametric represention of the polar spiral $r=\theta^2$ when, in reality, $\theta \rightarrow \Big(\theta^2,\theta\Big)$ is parametric representation of the parabola $x=y^2$ while $\theta \rightarrow \Big(\theta^2 \cos(\theta),\theta^2 \sin(\theta)\Big)$ is the spiral $r=\theta^2$.

Let's refer to your example in which you seek to compute the surface area of a sphere.

In spherical coordinates, the equation of a sphere is $r=1$ on the domain $(\theta,\phi)\in[0,2\pi)\times [0,\pi]$. You can represent this parametrically as $$(\phi,\theta) \longrightarrow \Big(\sin(\phi)\cos(\theta),\sin(\phi)\sin(\theta),\cos(\phi)\Big)$$ simply by converting from spherical to cartesian coordinates. However, the image of your function $\varphi:(\phi,\theta)\longrightarrow (1,\phi,\theta)$ on the same domain is the rectangular patch $\{1\}\times [0,2\pi)\times [0,\pi]$ embedded in the vertical plane $x=1$ which has area $2\pi^2$. This is precisely why $$\int_0^{2\pi}\int_0^{\pi}\sqrt{1+\big(f_{\theta}\big)^2+\big(f_{\phi}\big)^2}d\phi d\theta=2\pi^2 \neq 4\pi$$
whenever $f(\phi,\theta)=1$; this integral is calculating the area of the surface $x=1$ on $(y,z)\in [0,2\pi)\times[0,\pi]$ which is the image of your map $\varphi$.

If you want to find the area of the surface given in spherical coordinates by $r=f(\phi,\theta)$ defined on domain  $(\phi,\theta)\in \mathcal{U}$ you will need to express this parametrically as $$\vec{p}(\phi,\theta)=\Big(f(\phi,\theta)\sin(\phi)\cos(\theta),f(\phi,\theta)\sin(\phi)\sin(\theta),f(\phi,\theta)\cos(\phi)\Big)$$ The area of the surface will be $$S=\int \int _{\mathcal{U}}||\vec{p}_\phi \times \vec{p}_\theta||d\phi d\theta$$ If you compute $||\vec{p}_\phi \times \vec{p}_\theta||$ you will surely obtain your desired result.

Title: Surface integrals in spherical coordinates

Body:
If I am given a surface in spherical coordinates $(r,\theta,\varphi)$, such that it is parametrised as:

$$
\begin{align}
r&=r(\theta,\varphi)\\
\theta&=\theta\\
\varphi&=\varphi
\end{align}
$$

What is the area $S$ of such surface? Or more specifically, can you show how to get the result:
$$
S=\int_{0}^{2\pi}\int_{0}^{\pi}\sqrt{r^2+\left(\frac{\partial r}{\partial \theta}\right)^2 + \frac{1}{\sin^2\theta}\left(\frac{\partial r}{\partial \varphi}\right)^2}\;r\sin\theta\;{\rm d}\theta\,{\rm d}\varphi
$$

Some definitions that I am using:

$k$-surface:
Let $k,N\in\mathbb{N}$, $k<N$, $M\subset \mathbb{R}^N$ is called a $k$-surface, if there exists a non-empty open set $E\subset \mathbb{R}^k$ and a map $\varphi:\mathbb{R}^k\to \mathbb{R}^N$, such that: (i) $\varphi(E)=M$, (ii) $\varphi\in C^1(E;\mathbb{R}^N)$, and (iii) the rank of Jacobi matrix of $\varphi$ is equal $k$ everywhere on $E$.
The surface is called simple if $\varphi$ is also injective on $E$ and $\varphi^{-1}$ is continuous of $\varphi(E)$.

Surface integral of the first kind:
Let $k,N\in\mathbb{N}$, $k<N$, $M\subset \mathbb{R}^N$ is a simple $k$-surface parametrized by the map $\varphi:\mathbb{R}^k\to \mathbb{R}^N$, from the open set $E\subset \mathbb{R}^k$ and $f:\mathbb{R}^N\to\mathbb{R}$ is defined on $M$. The surface integral of the first kind is defined by:
$$
\int_M f\,\mathrm{d}S:=\int_E f(\varphi(t))\sqrt{\det{G(D_\varphi(t))}}\,\mathrm{d}t\,,
$$
if the integral on the right exists in the Lebesgue sense and is finite. Here, $G(A)$ denotes the Gramm matrix made from columns of $A$ and $D_\varphi$ is the Jacobi matrix of the map $\varphi$. The numeric value of:
$$
S_k(M):=\int_M f\,\mathrm{d}S\,,
$$
is called the $k$-dimensional surface area of the $k$-surface $M$.

Motivation for the question

Now these definitions can be used to calculate e.g. the surface of a unit sphere. When one describes the sphere by the map (omitting one longitudinal line): $\varphi: (\eta,\psi)\mapsto(\cos\psi\cos\eta,\cos\psi\sin\eta,\sin\psi)$, where $(\eta,\psi)\in E=(0,2\pi)\times(-\frac{\pi}{2},\frac{\pi}{2})$. The Gramm matrix looks like:
$$
\begin{pmatrix}
\cos^2\psi & 0 \\
0 & 1
\end{pmatrix}
$$

one ends up with the (here $\lambda_2$ denotes the Lebesgue measure):
$$
S_2(M) = \int_E 1\sqrt{\det{G(D_\varphi(\eta,\psi))}}\,\mathrm{d}\lambda_2(\eta,\psi)=\int_0^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos\psi\,\mathrm{d}\psi\,\mathrm{d}{\eta}=4\pi
$$

I did not do anything else but blindly followed the definitions.

The wrong result:

If I do the same approach to the above problem at hand. My map is $\varphi:\,(\theta,\phi)\mapsto (r(\theta,\phi),\theta,\phi)$, the Jacobian is then:
$$
\begin{pmatrix}
\frac{\partial r(\theta,\phi)}{\partial \theta} & \frac{\partial r(\theta,\phi)}{\partial \phi} \\
1 & 0 \\
0 & 1
\end{pmatrix}
$$

The rank of this matrix is 2 as needed for a 2-surface. The Gramm matrix is then:
$$G(D_\varphi)=
\begin{pmatrix}
1+ \left(\frac{\partial r(\theta,\phi)}{\partial \theta}\right)^2 & \frac{\partial r(\theta,\phi)}{\partial \theta}\frac{\partial r(\theta,\phi)}{\partial \phi} \\
\frac{\partial r(\theta,\phi)}{\partial \theta}\frac{\partial r(\theta,\phi)}{\partial \phi} & 1+ \left(\frac{\partial r(\theta,\phi)}{\partial \phi}\right)^2
\end{pmatrix}
$$
with determinant evaluated to:
$$
\det(G(D_\varphi)) = 1 + \left(\frac{\partial r(\theta,\phi)}{\partial \theta}\right)^2 + \left(\frac{\partial r(\theta,\phi)}{\partial \phi}\right)^2
$$

which leads to the surface area:

$$S=\int_{S\subset\mathcal{R}^3}{\rm d}S=\int_{0}^{2\pi}\int_{0}^{\pi}\sqrt{1+\left(\frac{\partial r}{\partial \theta}\right)^2 + \left(\frac{\partial r}{\partial \varphi}\right)^2}\;{\rm d}\theta\,{\rm d}\varphi$$

Which is incorrect. The question is why, and please show the correct way with explanation because as suggested by the example with unit sphere, I did not have to do any transformations and it did work with curvilinear coordinates right away. So the answer provided by Quanto does not address this at all.

Note to the edit:

I have added quite a lot of details since the answer by Quanto, but the root problem is the same - to see by calculation explicitly how one arrives to the correct result and to understand why the my calculation for a sphere works (where i am not using Cartesian coordinates) but fails here.


问:
Trouble understanding the question regarding dual spaces

If $(\alpha_0,\alpha_1,\alpha_2,\dots)$ is an arbitrary sequence of
complex numbers, and if $x$ is an element of $\mathcal{P}$,
$x(t)=\sum\limits^n_{t=0}\xi_i t^i$, write
$y(x)=\sum^n_{i=0}\xi_i\alpha_i$. Prove that $y$ is an element of
$\mathcal{P'}$ and that every element of $\mathcal{P'}$ can be
obtained in this manner by a suitable choice of the $\alpha's$.

Here $\mathcal{P'}$ is a dual space of $\mathcal{P}$. Can you explain what the second question means?

答:
For $i\in \Bbb N_0$ and $t\in \Bbb C$ let $f_i(t)=t^i$ (with the usual convention that $0^0=1$).

The 2nd part  asks you to prove that if $y\in P'$ then there is a sequence $(\alpha_0,...)$ such that $y(x)=\sum_i\xi_i \alpha_i.$ Which should be obvious because  $\{f_i:i\in \Bbb N_0\}$ is a vector-space basis for $P.$ That is, $\alpha_i=y(f_i).$

ContentLicense: CC BY-SA 4.0


问:
Is it possible to solve for x in sin^2(x) + tanx - 3/2 =0
original question: Is it possible to solve for x in   $$\sin^2x + \tan x  + \frac{3}{2} = 0 .$$

please show steps

edit:
sorry for typo mistake, the equation should be
$$\sin^2x + \tan x  - \frac{3}{2} = 0 .$$

答:
Starting from @peterwhy's answer, let $t=\tan(x)$ and we need to solve the cubic equation $$2 t^3-t^2+2 t-3=0$$ but $t=1$ is an obvious solution. So
$$2 t^3-t^2+2 t-3=t-1) \left(2 t^2+t+3\right)$$ and the quadratic does not show any real root.


问:
Edge case with sampling and reconstruction.
I know I had been dabbling around this question before, here and here, but does anyone have in their bag of tricks the most simpliest and concise proof that:

$$\sum_{n=-\infty}^{\infty} (-1)^n \, \operatorname{sinc}(t-n) =  \cos(\pi t) $$

where

$$ \operatorname{sinc}(x) \triangleq \begin{cases}
\frac{\sin(\pi x)}{\pi x} \qquad & x \ne 0 \\
\\
1 & x = 0 \\
\end{cases} $$

and $t\in\mathbb{R}$ and $n\in\mathbb{Z}$ ?

I can show that both sides are an even function in $t$ and that both sides have agreement when $t$ is an integer.  But what is the simplest way to show equality for all real $t$ ?

This is something that I want to put together for us Neanderthal electrical engineers.  (and thank you.)

答:
You should be somewhat careful with how you understand the sum but, assuming that you understand $\sum_{n=-\infty}^\infty a_n$ it as the limit as $N\to\infty$ of $\sum_{-N\le n\le N}(1-\frac{|n|}{N})a_n$ (Cesaro summation, which gives the same result as the usual one when the latter makes sense), you can just write
$$
(-1)^n\rm{sinc}(t-n)=\int_{-1/2}^{1/2}e^{-2\pi i n(x+\frac 12)}e^{2\pi i xt}\,dx
$$
so the Cesaro partial sums become $\int_{-1/2}^{1/2}K_N(x+\frac 12)e^{2\pi i xt}\,dx$ where $K_N(z)=\sum_{-N}^N(1-\frac{|n|}{N}) e^{-2\pi i nz}$ is the Fejer kernel. What you want to know now is that $K_N$ is symmetric, non-negative, $1$-periodic, has total integral $1$ over the period and uniformly tends to $0$ outside an arbitrarily small neighborhood of the integers. So, for large $N$, $K_N(x+\frac 12)$ is a function that is nearly $0$ on $(-\frac 12+\delta,\frac 12-\delta)$ for any fixed $\delta>0$ and has integral nearly $\frac 12$ over each of the intervals $[-\frac 12,-\frac 12+\delta]$ and $[\frac 12-\delta,\frac 12]$. When you integrate anything like that against $e^{2\pi i xt}$ over $[-\frac 12,\frac 12]$, you'll get approximately $\frac 12(e^{-\pi it}+e^{\pi i t})=\cos(\pi t)$.

The only non-pedestrian step in this argument is switching from the usual summation to the Cesaro one. You can avoid it but then you'll get the Dirichlet kernel instead and the last passage to the limit will be somewhat less obvious (the kernel will not decay uniformly in the bulk of the interval but instead it will oscillate faster and faster there and you'll end up using something like the Riemann-Lebesgue lemma to show that you need to look only at the (small neighborhoods of) endpoints.


问:
If $\tau$ is a name, is $\{\sigma:\Vdash \sigma=\tau\}$ a proper class?
Fix some (atomless) poset $\mathbb P$. Let $V^\mathbb{P}$ be the class of $\mathbb P$-names and $\tau\in V^\mathbb{P}$. Consider the collection
$$
\{\sigma\in V^\mathbb{P} : \Vdash_\mathbb{P} \sigma=\tau\}
$$
My question is: Can this collection be a proper class? I've always believed the answer to be "yes", but I'm struggling to actually prove this.

If $\tau=\emptyset$, then obviously the only name which is forced to be equal to $\tau$ is $\tau$ itself. Indeed, suppose $\sigma\neq \emptyset$, say $(\vartheta,p)\in \sigma$ for some $p\in\mathbb{P}$. Then $p\Vdash \vartheta \in \sigma$, so $p\Vdash \sigma\neq \tau$ and thus $\not\Vdash \sigma=\tau$.

Also, the above is false if we're taking the Boolean algebra approach to forcing. Indeed, since we have access to the bottom element $0$ saying false, we can consider, for any $x\in V$,
$$
\sigma_x=\{(\check x,0)\}\cup \tau
$$
Then $\{\sigma_x:x\in V\}$ is a proper class, and $\|\sigma_x=\tau\|=1$ for every $x$.

It is also false if we replace $\Vdash$ by $p\Vdash$ for some $p\in\mathbb{P}$. Indeed, if $p,q\in \mathbb{P}$ and $p\perp q$, put $\sigma_x=\{(\check x, q)\}$ for $x\in V$. Then $p\Vdash \sigma_x=\emptyset$, and the $\sigma_x$ are pairwise distinct.

I've thought about arguing that
$$
\forall \tau\in V^\mathbb{P} \exists X_\tau\forall \sigma\in V^\mathbb{P} [\sigma \in X_\tau\iff \Vdash \sigma=\tau]
$$
by $\in$-induction. Say $\tau\in V^\mathbb{P}$ and the statement holds for every $\sigma$ of smaller rank. Put $X=\bigcup\{X_\sigma:\sigma\in\text{dom}(\tau)\}$. Suppose there is a proper class of names which are forced to equal $\tau$. Pick a name $\vartheta$ with $\text{dom}(\vartheta)\setminus X\neq \emptyset$ and $\Vdash \vartheta=\tau$. There is some pair $(\pi,p)\in \tau$ with $\pi\not\in X$. We know that $\not \Vdash \pi=\sigma$ for every $\sigma\in \text{dom}(\tau)$, but this is not enough to derive a contradiction, because we only have $p\Vdash \sigma \in \tau$.

答:
Your class is a proper class if $\mathbb{P}$ is atomless.
Let $p$ and $\sigma$ be such that
$\{\mu: p\Vdash \mu=\sigma\}$ is a proper class.
Now consider $\rho=\{(\sigma,p)\}$. If $p\Vdash \mu=\sigma$ and $\rho':=\{(\mu,p)\}$, then

$p\Vdash \rho=\rho'$ (In fact, $p$ thinks $\rho$ is a singleton whose element is $\sigma$, and

If $q$ is incompatible with $p$, then $q\Vdash \rho=\rho'=\varnothing$.

Hence $\{r:r\Vdash\rho=\rho'\}$ contains a dense subset, namely $$\{r: r\le p \lor r\perp p\}.$$
Thus $1\Vdash \rho=\rho'$.


问:
Find Area of shaded region.
In the given figure $AB$ = $9cm$, $AD$ = $BC$ = $10cm$, $DB$ = $AC$ = $17cm$. If area of the shaded
region is ${m\over n}$, where ($m, n$) = $1$, then evaluate m + n .

My Work :-

Well initially I had ideas of joining $D$ and $C$ and then maybe using something related to similar triangles. Well then I had another idea of maybe finding the height of the trapezium $ABCD$ and then maybe find height of the shaded triangle and similar ones. However whenever I am approaching either ways, I am just getting stuck in variables.

So I need an advice as to how to approach this problem.

答:
Let the intersection point of $AC,BD$ be denoted by $O$. Let $V$ be the point on $AB$ such that $OV\perp AB$. Since $\triangle DAB\cong\triangle CBA(SSS),\triangle AOB$ is isosceles and $AV=BV=4.5$.

Then $\tan\angle OAB=\frac{OV}{AV}$ which gives the required area $\Delta=OV\times AV=4.5^2\tan\angle OAB$.

Let $X$ be the point on $AC$ such that $BX\perp AC$. Then $BX^2=9^2-AX^2=10^2-(17-AX)^2$ giving $AX=\frac{135}{17}$.

Now, $\tan\angle OAB=\frac{BX}{AX}$. Can you complete?

$\tan\angle OAB=0.5\bar3$ giving $\Delta=10.8=\frac{54}5.$

Title: Find Area of shaded region.


问:
Landmarks of subjects of mathematics
In order to have a good view of the whole mathematical landscape one might want to know a deep theorem from the main subjects (I think my view is too narrow so I want to extend it).

For example in natural number theory it is good to know quadratic reciprocity and in linear algebra it's good to know the Cayley-Hamilton theorem (to give two examples).

So, what is one (per post) deep and representative theorem of each subject that one can spend a couple of months or so to learn about? (In Combinatorics, Graph theory, Real Analysis, Logic, Differential Geometry, etc.)

答:
Differential Geometry: the Gauss-Bonnet theorem.

I took a one-semester intro course on differential geometry class and we got to this towards the end of the semester, so I feel that a couple of months is an appropriate time frame for this theorem.

ContentLicense: CC BY-SA 3.0

Title: Landmarks of subjects of mathematics


问:
Critical points of nonlinear second-order ODE
I want to find the critical points of the autonomous nonlinear ODE $y'' = \sin(y) - y'y$. I know these are the points where $0 = y'' = \sin(y) -y'y$.

I'm unsure if I need to solve this system outright. Other than $y = 0$ I'm having a hard time visualizing any other critical points. Any help?

答:
If we rewrite the equation as a system:
$$
y'=z,\qquad z'= \sin y - yz,
$$
where $z=y'$, then it becomes clear that the equilibrium points are given by
$$
\left\{\begin{array}{lll}
z&=&0\\
\sin y - yz&=&0
\end{array}\right.
$$
or
$$
\left\{\begin{array}{lll}
y'&=&0\\
\sin y &=&0.
\end{array}\right.
$$
Hence, the equilibrium points are $y=\pi k$, $k\in\mathbb Z$.


问:
Compute $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^4}} \cdot \int_0^1 \frac{x^2\mathrm{d}x}{\sqrt{1-x^4}}$

Compute$$\int_0^1 \dfrac{\mathrm{d}x}{\sqrt{1-x^4}} \cdot \int_0^1 \dfrac{x^2\mathrm{d}x}{\sqrt{1-x^4}}$$.

The primitive function of each one seems not elementary function.

I tried like this
$$
\int_0^1 \dfrac{\mathrm{d}x}{\sqrt{1-x^4}} + \int_0^1 \dfrac{x^2\mathrm{d}x}{\sqrt{1-x^4}}=\int_0^1\dfrac{\mathrm{d}(x-\frac{1}{x})}{\sqrt{\frac{1}{x^2}-x^2}},
$$
but it still does not work.

Actually, I can compute it by Gamma and Beta function. But I wonder is there other approach? Appreciate any help!

Body:
Use $\beta$ function: https://en.wikipedia.org/wiki/Beta_function

Let $x=\sin^{1/2} t$, then
$$I_1=\int_{0}^{1} \frac{dx}{\sqrt{1-x^4}}=\frac{1}{2} \int_{0}^{\pi/2} \sin^{-1/2}t ~ dt=\frac{1}{4}\frac{\Gamma(1/4)\Gamma(1/2)}{\Gamma(3/4)}$$
Next, $$I_2=\int_{0}^{1} \frac{x^2dx}{\sqrt{1-x^4}}=\frac{1}{2} \int_{0}^{\pi/2} \sin^{1/2}t~ dt=\frac{1}{4}\frac{\Gamma(1/4)\Gamma(1/2)}{\Gamma(3/4)}=\frac{1}{4}\frac{\Gamma(3/4)\Gamma(1/2)}{\Gamma(5/4)}$$
Then$$I_1.I_2=\frac{1}{16}\frac{\Gamma(1/4)\Gamma(1/2)^2}{\Gamma(5/4)}=\frac{\pi}{4}$$


问:
How does $\frac{1}{2^n-1}+\frac{2^n-2}{2^n-1}\left(\frac{1}{2^{n-2}}\right)^2 = \frac{1}{2^{n-1}}$??
No matter what I try I cannot simplify it. However, even symbolab and wolfram alpha do not simplify this. Is it just me or is the question faulty? It comes from this:

Eight players play in the quarter-finals at Wimbledon. The winner of the quarter-finals plays again in the semi-final to decide who enters the finals.

a) Assume all 8 players have an equally likely chance of winning. Show that the probability that baby two players winning is $\frac{1}{4}$

b) What is the probability that two particular people will play each other if the tournament begins with 16 players?

c) What is the probability if the tournament starts with $2^n$ players?

The worked solutions provided for parts a) and b) are $\frac{1}{7}+\frac{6}{7}(\frac{1}{2})^3 = \frac{1}{4}$. I understand the $\frac{1}{7}+\frac{6}{7}$ but not the half, and especially to the power of 3???

Similarly for b) $\frac{1}{15}+\frac{14}{15}(\frac{1}{4})^2 = \frac{1}{8}$. This is even more confusing! Why is it to the power of 2 now, and why is it not a half but a quarter?

For part c),it is easy to reason the formula $\frac{1}{2^n-1}+\frac{2^n-2}{2^n-1}\left(\frac{1}{2^{n-2}}\right)^2$ (or at least the part excluding the brakets). I do not understand how the part inteh brackets is derived and simplified (as per the title)

答:
Firstly, let us evolve a simple way of computing the probability of any particular pair meeting in an $2^n$ player knockout tournament.

It should be obvious that there will be a total of $(2^n-1)$ matches (because in each match one player loses, and $(2^n-1)$ players must lose)

Also, number of pairs possible $= \dbinom{2^n}{2}$,

Thus required probability $=\dfrac{2^n-1}{\dbinom{2^n}{2}}$, which simplifies to $\dfrac{1}{2^{n-1}}$

So far as the title is concerned, there is obviously
some typographical error. The answers tally, I'll try to figure out the logic used, but idk why such a tortuous method needs to be used at all !

Added explanation re book formula:

We know that in any round, the number of players is in the form $2^n$ and the Pr of a particular pair meeting is $\frac 1 {2^n-1}$

Let $2^n = x$, and assuming they have not met in this round, then for them to meet in the next round, we need
P(not meet in this round)$\times$ P(both win this round) $\times$P(meet in next round)

$=  (1 - \frac 1 {x-1})\cdot\frac 1 4 \cdot\frac 1 {\frac {x}2-1}$ which simplifies to $\frac 1 {2(x-1)}$

ie half the probability of this round

Thus the formula for a tournament of $2^n$ people will be

$\left(\frac 1 {2^n -1}\right)\left(1 + \frac 1 2 + \frac 1 4 +.. \;to\;
 n\; terms\right)$ = $\frac {1 }{2^n-1}\cdot  \frac{2(2^n-1)}{2^n} = \frac 1 {2^{n-1}}$

PS

It is simpler to eliminate all exponents and GP etc, by instead solving for a $K$ player tournament, knowing that in a normal tournament $K$ is in the form $2^n$, then the indicated probability is simply $\frac{K-1}{\binom{K}2}$ as indicated in the first part of the answer to get $\frac2 K$, eg for a $16$ player tournament, it will be $\frac 2{16} = \frac1 8$


答:
Here is the picture for the situation:

One possibility to "see the height" is as follows.

Of course, we have an isosceles trapezium $ABCD$.
In the triangle $\Delta ABC$ the sides are known, $9, 10, 17$, the semi-perimeter is $(9+10+17)/2=18$, so Heron gives its area $[ABC]=A_1+A_3$ as
$$
[ABC]=A_1+A_3=\sqrt{18(18-9)(18-10)(18-17)}=36\ .
$$
Since $AB=9$, the heights $DD'$ and $CC'$ of the isosceles trapezium have both length $8$. This gives then $AC'=BD'=15$, since there is the pythagorean triple $(8, 15, 17)$. We get $BC'=AD'=6$ finally. (And one can take this as a start).

Let us now denote by $A_1$ and $A_2$ the areas of the similar triangles $\Delta XAB$ and $\Delta XCD$. It is easy to compute all areas now. We may also want to compute the position of $X$, the intersection of the diagonals, on the diagonals:
$$
\begin{aligned}
A_1 + A_3 &= [ABC] = 36 \ ,\\
\frac {A_1}{A_3} &= \frac{XA}{XC} =\frac{AB}{CD}=\frac {9}{21}=\frac37\ ,\\
\frac {A_1}{A_1+A_3} &= \frac3{3+7}=\frac3{10}\ ,\\
A_1 &= (A_1+A_3)\cdot \frac3{10}=36\cdot  \frac3{10}=\frac{54}5\ .
\end{aligned}
$$

$\square$

@Note: Checks, and other possibilities to get $A_1$:

The area $[ABCD]$ is $\frac 12\cdot 8(9+21)=120$, and $A_1=54/5$, $A_3=126/5$, $A_2=294/5$. We have

$A_1+A_3=36$, as it should be $\frac 12\cdot 9\cdot 8$,

$A_2+A_3=84$, as it should be $\frac 12\cdot 21\cdot 8$,

$A_1:A_2=9/49=(3/7)^2=(9/21)^2=(AB/CD)^2$, as it should be from the similarity.

$A_1+A_2+2A_3=[ABCD]=120$.

Title: Find Area of shaded region.

Body:
In the given figure $AB$ = $9cm$, $AD$ = $BC$ = $10cm$, $DB$ = $AC$ = $17cm$. If area of the shaded
region is ${m\over n}$, where ($m, n$) = $1$, then evaluate m + n .

My Work :-

Well initially I had ideas of joining $D$ and $C$ and then maybe using something related to similar triangles. Well then I had another idea of maybe finding the height of the trapezium $ABCD$ and then maybe find height of the shaded triangle and similar ones. However whenever I am approaching either ways, I am just getting stuck in variables.

So I need an advice as to how to approach this problem.


问:
Image of Union is Equal Union of Images
I would hope someone can help me verify my solution for this seemingly basic set theory problem. We are given that $f : X\longrightarrow Y$ such that $A\subseteq X$ and $B\subseteq X$ and are asked to show:
$$
f(A\cup B)=f(A)\cup f(B)
$$
I have proved it this way :
\begin{align*}
y\in f(A\cup B)&\iff\exists x\in A\cup B\;;\;y=f(x) \\
&\iff (x\in A)\lor(x\in B) \\
&\iff[f(x)\in f(A)]\lor[f(x)\in f(B)] \\
&\iff f(x)\in(f(A)\cup f(B))
\end{align*}
Is this proof correct? Also I wish to ask what is the difference between in these two notations for a function of a set :
$$
f(X)\qquad\text{and}\qquad f[X]
$$
and if they have same meaning which one is the better to use?

答:
The proof is much too sloppy to be correct.

y in f(A $\cup$ B)
iff

exists x in A $\cup$ B with y = f(x)

iff

exists x with ((x in A or x in B)

and y = f(x))

iff

exists x in A with y = f(x)

or exists x in B with y = f(x)

iff

y in f(A) or y in f(B)

iff

y in f(A) $\cup$ f(B)

Title: Image of Union is Equal Union of Images


问:
Concept not clear: Permutations and Combinations
Q: There are $5$ shirts all of different colors, $4$ pairs of pants all of different colors, and $2$ pairs of shoes with different colors. In how many ways can Amy and Bunny be dressed up with a shirt, a pair of pants, and a pair of shoes each ? ( Question from Brilliant App ).

Solution: $(5C2 \times 2!) \times (4C2 \times 2!) \times (2C2 \times 2!) = 480$

My approach:
No of permutations $= 5 \times 4 \times 2 = 40$ ways of ordering

Choosing 2 from these $40$ orderings $= 40C2 \times 2! = 1560$

Please let me know what is wrong with my approach.

答:
Your answer is completely wrong , because when you multiply $5,4,3$ , you find the possible choice for only one individual.The rest is also confusing , as far as i could understand , you try to have random $2$ combinations from among these $40$ different combination.The problem is that you try to pick up $2$ things from tne number of permutation.

Be careful!Permutation does not give you subsets, it only gives the number of ordering,so you cannot choose somethings from permutation you can make choosing only from sets.This was the key sentence for you.


问:
$\int_{e}^{\infty} \frac{x+1}{x^3\ln ^2(x)} dx$
I am trying to solve the given integral (on my own, not a homework problem.)

$$\int_{e}^{\infty} \frac{x+1}{x^3\ln ^2(x)} dx$$

I am having problem with a part of using Feynman's technique which I need resolved.

I can rewrite this as:

$$\int_{e}^{\infty} \frac{x^{-2}+x^{-3}}{\ln ^2(x)} dx$$

And can create the function:

$$I(s,t) = \int_{e}^{\infty} \frac{x^{-s}+x^{-t}}{\ln ^2(x)} dx$$

Note that our original integral is $I(2,3)$. Here's where my confusion first sets in. I am attempting to get rid of the $\ln^2(x)$, so I can either compute $I_{tt} (s,t), I_{ss}(s,t),$ or $I_{st} (s,t)$, and these will ALL get rid of the $\ln^2(x)$ in the denominator. The first confusion is why this is. Here's my working after computing $\frac{\partial ^2I}{\partial s \partial t}$.

$$\frac{\partial ^2I}{\partial s \partial t}= \int_{e}^{\infty} (x^{-s} + x^{-t})dx = \frac{e^{1-t}}{1-t} + \frac{e^{1-s}}{1-s}$$.

$\int \frac{e^x}{x} dx$ doesn't have an elementary antiderivative, so call this integral $\text{Ei} (x)$, as it is often named.

Integrate with respect to $t$ yields:

$$\frac{\partial I}{\partial s} = -\text{Ei} (1-t) + \frac{te^{1-s}}{1-s} + f_1 (s)$$

Integrate with respect to $s$.

$$I(s,t) = -s\text{Ei} (1-t) - t\text{Ei} (1-s) + \int f_1(s) ds + f_2(t).$$

Here is my other question, which relates to Feynman's technique with a multivariable function in general, not necessarily with this question. When we differentiate under the integral sign to evaluate an integral, we use one parameter, and use a clever value for the parameter to evaluate the constant $C$ after integrating. When we integrate and there's partials involved, we have to introduce a whole FUNCTION with the other variable. How can we find this, and how in general can we use $I(s,t)$ to evaluate the integral? You can't simply put a nice value of $s$ or $t$ for a function.

答:
Concerning the first integral, let $x=e^t$ to end with
$$I=\int \frac{x+1}{x^3\log ^2(x)} dx=\int \frac{e^{-2 t}}{t^2} \,dt+\int \frac{e^{-t}}{t^2} \,dt$$ Using one integration by parts
$$\int \frac{e^{-k t}}{t^2} \,dt=-k\, \text{Ei}(-k t)-\frac{e^{-k t}}{t}$$ This makes
$$I=-\frac{2 t \,\text{Ei}(-2 t)+t \,\text{Ei}(-t)+e^{-2 t}+e^{-t}}{t}$$ Integrating from $1$ to $\infty$, this gives
$$\int_{e}^{\infty} \frac{x+1}{x^3\log ^2(x)} dx=2\,\text{Ei}(-2)+\text{Ei}(-1)+\frac{1}{e^2}+\frac{1}{e}$$

You could also make the problem more general using, if $k>0$,
$$\int t^{-n} \,e^{-kt}\,dt=-k^{n-1} \Gamma (1-n,k t)$$
$$\int_1^\infty t^{-n} \,e^{-kt}\,dt=k^{n-1} \Gamma (1-n,0)$$


问:
Do The Eigenvectors of a Positive Semi-Definite matrix span the column space.
If we have a matrix B=A'A , where ' denote transpose, is a Symmetric Positive Semi definite matrix.
Does the Eigenvectors of B spans the column space of B?

Thank you.

答: Let $$A=\pmatrix{0&0\cr1&2\cr}$$ Then $$B=A^tA=\pmatrix{0&1\cr0&2\cr}\pmatrix{0&0\cr1&2\cr}=\pmatrix{1&2\cr2&4\cr}$$ $B$ has eigenvalues $0$ and $5$ with eigenvectors $(2,-1)$ and $(1,2)$, respectively, so it is positive semi-definite. Those eigenvectors span all of ${\bf R}^2$, but the column space of $B$ is one-dimensional, spanned by $(1,2)$.


答:
I'd suggest to consider 2 approaches. The first one will help you to understand why your approach is not correct. Then second will help to understand how is it related to combinations and permutations.

The numbers are small. That's why first you can apply a straight forward approach, without combinations and permutations. How may choices of short do you have for Amy? It is 5. If you have chosen one color for Amy, Bunny can choose one of remaining colors. How many remain? 4. Thus, you have 5x4 pairs of shirt colors. Same about pants: Amy has 4 choices, for Bunny remain 3 choices. Thus there are 4x3 pants choices. For shoes Amy has 2 choices, but Bunny has no choice (i.e. only 1 choice) after Amy has chosen. That's why we have 2x1 choices of shoes. Thus the total number of ways is (5x4) x (4x3) x (2x1) = 480.

Now we can consider it from the point of view of combinations and permutations. We have 2 people. From 5 shirt colors we have to chose 2. How many ways are there to chose 2 out of 5? There are 5C2 ways. When we have chosen 2 shirt colors, in how many ways we can assign these colors to 2 persons? In 2! ways. If we had more colors and more persons, it would be easier to see, why we are talking about permutations here. E.g. if we had chosen 7 colors out of 20, the number of way to assign 7 colors to 7 persons would be the number of permutations, i.e. 7x6x5x4x3x2x1 = 7!. For 2 persons we have 2!.
So, for shirts we have 5C2x2! For pants we have 4C2 ways to chose 2 colors out of 4. And there are 2! ways to assign them to Amy and Bunny. Thus there are 4C2x2! ways for pants.
The same for shoes: 2C2x2!.
The total number of ways is (5C2x2!) x (4C2x2!) x (2C2x2!) = 480.

Title: Concept not clear: Permutations and Combinations

Body:
Q: There are $5$ shirts all of different colors, $4$ pairs of pants all of different colors, and $2$ pairs of shoes with different colors. In how many ways can Amy and Bunny be dressed up with a shirt, a pair of pants, and a pair of shoes each ? ( Question from Brilliant App ).

Solution: $(5C2 \times 2!) \times (4C2 \times 2!) \times (2C2 \times 2!) = 480$

My approach:
No of permutations $= 5 \times 4 \times 2 = 40$ ways of ordering

Choosing 2 from these $40$ orderings $= 40C2 \times 2! = 1560$

Please let me know what is wrong with my approach.


问:
Problem in proof that $ \lim_{x\to0}\frac{\sin(x)}{x}$=$1$
I'm having some confusion on understanding the proof that

$$\lim_{x\to 0}\frac{\sin x}{x}=1.$$

In the proof on khanacademy and on other various sources we always end up with the inequality
$$\cos(x)\le\frac{\sin(x)}{x}\le1\
$$
And then we apply the squeeze theorem.

However in the proof we always say that it is valid for any non zero x however if this is true how can we ever say that the inequality is less than or equal to each other my question is why isn't the inequality this instead

$$\cos(x)<\frac{\sin(x)}{x}<1
$$
No one seems to explain how they make the less than or equal to statement true so which one is correct?

答:
The answers thus far are missing an important point. It's not merely allowed to switch to the weaker, $\leq$ inequality; it is necessary to do so.

Suppose $f(x) < g(x)$ for all $x$, and that $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ both exist.

In this case it is true that $\lim_{x\to a}f(x) \leq \lim_{x\to a}g(x)$.

However, It is not necessarily true that $\lim_{x\to a}f(x) < \lim_{x\to a}g(x)$.

As an example, consider

$$f(x) = 0$$

$$g(x) = |x| \text{ (if $x\neq 0$)}$$
$$g(x) = 1 \text{ (if $x = 0$)}$$

Here, $f(x) < g(x)$ for all $x$ but $\lim_{x\to 0}f(x) = 0 = \lim_{x\to 0}g(x)$

Frequently, when taking limits on both sides of a strict inequality, we're forced to weaken the inequality from $<$ to $\leq$.

In the case of $\cos(x)<\frac{\sin(x)}{x}<1$, if we wrongly insist that strict inequality holds for the limits of the functions, we end up with $1 < \lim_{x\to 0}\frac{\sin(x)}{x} < 1$, which is clearly nonsense.

There's an exercise in Calculus by Spivak that explores some of this in more detail: chapter 5, exercise 12 (3rd ed.). I mention it here because I know @The homeschooler has access to Spivak :). Here are the relevant parts:

5-12.(a) Suppose that $f(x) \leq g(x)$ for all $x$. Prove that $\lim_{x\to a} f(x) \leq \lim_{x\to a} g(x)$, provided that these limits exist.

(c) If $f(x) < g(x)$ for all $x$, does it necessarily follow that $\lim_{x \to a} f(x) < \lim_{x \to a} g(x)$?

For part (a), you can begin by assuming $\lim_{x\to a} f(x) > \lim_{x\to a} g(x)$ and then showing how this leads to a contradiction.

Part (c) can be dealt with by giving any counter-example (the $f$ and $g$ mentioned above will work, and are from the Answer Book). Part (c) emphasizes that $\leq$ is the best we can do; we can show that $\lim_{x\to a} f(x) \not> \lim_{x\to a} g(x)$, and that is all. We cannot show strict inequality holds for the limits, even though it holds for the functions for all $x$.

Though this exercise is slightly different than the case homeschooler is asking about, the arguments are largely applicable.

Edit: I've neglected an important point myself! There is a moment when we do actively choose to weaken the inequality.

Given $\cos(x)<\frac{\sin(x)}{x}<1$ in some neighborhood around $0$, the above arguments can get us to $1\leq \lim_{x \to 0}\frac{\sin(x)}{x} \leq 1$, but only if we assume that this middle limit exists, which of course we shouldn't do because that's what we're trying to prove!

Generally speaking, if $f(x) \leq g(x) \leq h(x)$ and $\lim_{x \to a} f(x) = \lim_{x \to a} h(x)$, then $\lim_{x \to a} g(x)$ exists and is equal to the other two.

If we weaken the inequalities to read $\cos(x)\leq \frac{\sin(x)}{x} \leq 1$, we can use the above theorem to get $\lim_{x \to 0}\frac{\sin(x)}{x} = 1$, all while avoiding any assumptions about existence.

Proving this general "limits+squeeze theorem" theorem is Spivak's next exercise, 5-13.

Suppose $f(x) \leq g(x) \leq h(x)$ and $\lim_{x \to a} f(x) = L = \lim_{x \to a} h(x)$.

For any $\varepsilon > 0$ we can find a $\delta > 0$ such that for all $x$, if $0 < |x - a| < \delta$, then $L - \varepsilon < f(x) < L + \varepsilon$, and also $L - \varepsilon < h(x) < L + \varepsilon$.

Therefore we have

$$L - \varepsilon < f(x) \leq g(x) \leq h(x) < L + \varepsilon$$

which shows $|g(x) - L| < \varepsilon$.

Anyway, sorry for the confusion and I'm guessing I've managed somehow to provide simultaneously too much and not enough in this answer.

Edit 2: reading it over again, I don't think we actually need to choose to weaken the inequalities for $\cos(x)<\frac{\sin(x)}{x}<1$. The above "limits+squeeze theorem" proof works even if we replace all the $\leq$'s with strict inequalities.

I'm relieved. Choosing to weaken it is too clever for me.

In summary,

(1) You can take limits of both sides of an inequality if you know the limits exist, but doing so turns strict $<$ into $\leq$. This is because...

(2) Two functions that are unequal everywhere can still have equal limits.

(3) If you have some function sandwiched between two other functions and those functions have the same limit, then so does the middle one.

(4) The requirement "...if $a \leq b$..." is certainly met by $a <b$.


答:
Others have specifically addressed where you went wrong.

The standard way of attacking the problem is as follows:

Let

$$\binom{n}{k} = \frac{n!}{[k!][(n-k)!]} ~\text{where}~
n \in \mathbb{Z^+}, ~k \in \{0,1,2,\cdots, n\} ~\text{and}~ 0! = 1.$$

Here, $\binom{n}{k}$ represents the number of different ways of selecting $k$ items from a group of $n$ items.

Let $A, B, C$ denote the # of different ways that Amy and Bunny can select shirts, pants, and shoes, respectively.

Then the final answer will be

$$A \times B \times C.$$

If Amy and Bunny are selecting $2$ items, from a group of $n$
items, superficially, you would suppose that the number of ways of doing this is $\binom{n}{2}$.  However, this overlooks that once the two items (say one green, one red) are selected, then you have to distinguish between [Amy selecting the green item, Bunny selecting the red item] and vice-versa.

Therefore, when computing $A,B,C$, the general formula for each variable will be

$$\binom{n}{2} \times 2.$$

Thus, since the number of shirts, pants and shoes are $5,4,2$ respectively, you have that

$$A = \binom{5}{2} \times 2 = 20,$$

$$B = \binom{4}{2} \times 2 = 12,$$

$$C = \binom{2}{2} \times 2 = 2.$$


问:
The relationship between covariance and inner product
It looks like covariance has some relationship between inner product. I have checked that covariance is bilinear and symmetric, but not positive definite (it is positive semi-definite). Although in WIKI, there are some information about the relationship between covariance and inner product, I still cannot really get the point. Hope someone will give me a more detailed answer about the relationship.</p>

答: If $X$ and $Y$ are random variables with finite second moments on a probability space $(\Omega, \mathcal F ,P)$  and $L^{2}$ is the space of square integrable random variables on $(\Omega, \mathcal F ,P)$ with the inner product $ \langle X, Y \rangle=EXY$ then covariance of $X$ and $Y$ is the inner product of $X-EX$ and $Y-EY$.


问:
Existence of a linear transformation $T:\mathbb{R}^3\to \mathbb{R}^3$
I am trying to prove / disprove the existence of a linear map $T:\mathbb{R}^3\to \mathbb{R}^3$ such that $T\neq 0$ and:
$$T^3=-T$$
This should be pretty easy but for some reason I just can’t figure it out.

Hints will be much appreciated!

答:
We can use the concept of a companion matrix, which given a monic polynomial $f$, provides a square matrix $T$ of dimension equal to the degree of $f$, whose charcteristic polynomial is equal to $f$.

Since $T^3 + T = 0$, a matrix with characteristic polynomial $x^3+x$ satisfies the requirements of the question, by the Cayley-Hamilton theorem , so long as that matrix is non-zero.

Therefore, by the companion matrix construction, we write our polynomial as $0+1x+0x^2+x^3$, and the resulting companion matrix is :
$$
T = \begin{pmatrix}
0 & 0 & 0 \\
1 & 0 & -1 \\
0 & 1 & 0
\end{pmatrix}
$$

which satisfies the conditions.


问:
Transfer of separability and Lindelöf properties to a nonempty open subset.

Let $O$ be a nonempty open subset of a separable (Lindelöf) topological space $( X,\tau)$. Prove that $O$ as a subspace of $X$, is also separable (Lindelöf).

(1-separable) Given a countable dense subset of $X$, if I intersect with the open subset of $X$, we get a countable dense subset of the open subset.

(2-Lindelöf)  Let $U$ be an open covering of the subspace $O$. Since all the elements of $U$ are open are open $O$, they equal the intersection of some family of open sets with $X$, call it $U'$.

I am kind of in the clouds about this proof. Please help me.
Thank you!
Klara

答:
There exists a Lindelöf space $X$ with an open subspace $O$ which is not Lindelöf.

Let $X=[0,\omega_1]$. It is compact, and hence it is Lindelöf. Let $O=\{x\in X: x \text{ is isolated in } X \}$. It is not difficult to see $O$ is an uncountable open subspace of $X$. The cover $\mathscr U=\{\{x\}: x\in O\}$ of $O$ has no countable subcover, which witnesses that $O$ is not Lindelöf.


答: Given the triangle’s base $9$ and its rational area $\frac mn$, its height is rational; so are likely the height and the base of the triangle DAB per similar triangles. Note that $(8,15,17)$ and $(6,8,10)$ are Pythagorean’s triplets. It is determined that $8$ and $15$ fit its height and base, respectively, which leads to $\frac{\frac 92}{15}\cdot 8=\frac{12}{5}$ as the height of the shaded triangle. Thus, its area is $\frac12\cdot\frac{12}5\cdot 9=\frac{54}5 $.

Title: Find Area of shaded region.


答:
Following my comment above, any singular matrix of the form
$$\begin{pmatrix} a&b&0\\c&-a&0\\0&0&0\end{pmatrix}$$ with $bc+a^2=-1$ satisfies your hypothesis.

Title: Existence of a linear transformation $T:\mathbb{R}^3\to \mathbb{R}^3$

Body:
I am trying to prove / disprove the existence of a linear map $T:\mathbb{R}^3\to \mathbb{R}^3$ such that $T\neq 0$ and:
$$T^3=-T$$
This should be pretty easy but for some reason I just can’t figure it out.

Hints will be much appreciated!


问:
Need Help! Evaluating a complicated integral
For this problem, I can't seem to figure how to start. I tried to do u-substitution but it doesn't seem to work...

$$\int \frac{x^5 + 1}{x^3 -3x^2 - 10x} \mathrm{d}x$$

答:
Hint:

As the comments suggest, you might try the following technique:

First factor the denominator as $x^3 - 3x^2 - 10x = (x-5)(x+2)x$.

Then you should rewrite $\frac{x^5 +1}{(x-5)(x+2)x}$ using
partial fractions:

$$x^2 + 3x + 19 + \frac{3126/5}{x-5} - \frac{31/14}{x-2} - \frac{1/10}{x}.$$

Do you see how you might evaluate the integral of the above function?

I hope this helps ^_^


问:
Skewed Trigonometric Function
What would be an expression for a periodic function (period $2\pi$) that essentially behaves just like a negative sine function, but it has the following quirk:

It's $0$s lie on the usual places (even integer multiples of $\frac \pi 2$), but it's maximum and minimum values (of $\pm 1$), instead of lying on odd integer multiples of $\frac\pi 2$, lie deviated by an angle $\alpha$ from the even integer multiples of $\frac \pi 2$. These are it's only maximums and minimums.

It's maximums and minimums can be graphically represented as follows:

In Orange we can see the function $-\sin (x)$, and the Red points represent the maximums and minimums of the function (the Gray lines just represent connections between the points, not the actual function).

Much appreciated.

答:
Considering the variables x, y, skew, t, α and k, and setting:

(x, y, skew, t, α) ∊ ℝ

k ∊ ℤ

f = (x, y) = (t + skew × sin t, - sin t), a parametric equation

α = skew + π/2

We get for skew = (0, -1, -π/2, 2) ⇒ α = (π/2, -1 + π/2, 0, 2 + π/2)
and 0 ≤ t ≤ 2π :

Note that ∫f = ∫(x, y) = (t + skew × sin t, (1/2) × (cos t) × (2 + skew × cos t)). For 0 ≤ t ≤ π the area above the curve is always 2, independent of the value of skew.

Note also that the function minima (-1) occurs at t = π/2 + 2kπ (as y = - sin t), for any value of skew.
As an example, for skew = 5π/32 and t = π/2 we get:

Title: Skewed Trigonometric Function


问:
Identity for categorical product is terminal?
It is well-known that for a category $\mathbf{C}$ with binary products and terminal object $\mathbf{1} \in \mathbf{C}$, we have $\mathbf{1} \times A \cong A$ naturally in $A$. I want to know if the converse holds, i.e. if $B \times A \cong A$ for all objects $A$, does this mean $B$ is terminal?

答:
The answer is no. Consider for instance $\mathbf{C}$ the full subcategory of $\mathbf{Ab}$ (the category of abelian groups) consisting of only the single object $\mathbb{Z}^\mathbb{N}$. Now, $\mathbb{Z}^\mathbb{N}$ is isomorphic to $\mathbb{Z}^\mathbb{N}\times\mathbb{Z}^\mathbb{N}$ in $\mathbf{Ab}$, and we can exploit this fact to show that $\mathbf{C}$ is closed under binary products. To see this, define projection maps $\pi_1,\pi_2:\mathbb{Z}^\mathbb{N}\to\mathbb{Z}^\mathbb{N}$ by
\begin{align}
\pi_1(x_0,x_1,\dots)&= (x_0,x_2,x_4,\dots)\\
\pi_2(x_0,x_1,\dots)&= (x_1,x_3,x_5,\dots).
\end{align}
I'll leave it to you to show that this indeed induces a product structure on $\mathbb{Z}^\mathbb{N}$. In particular, your desired condition holds for $B=\mathbb{Z}^\mathbb{N}$ in $\mathbf{C}$, but $\mathbb{Z}^\mathbb{N}$ is certainly not a terminal object in $\mathbf{C}$; it has many non-identity endomorphisms.


问:
How does $\lim\limits_{n\to\infty} \sum_{i=1}^{n^2} 1$ result into 1+3+5...
How can $\lim\limits_{n\to\infty} \sum_{i=1}^{n^2} 1$ result into 1+3+5...? Why odd numbers?</p>

答:
Note that the definition of summation
$
\sum_{k=1}^m a_k
$
$$
\sum_{k=1}^m a_k=a_1 +a_2+ \ldots +a_m.
$$
So, for eachpositive integer $n$, we have
$$
\sum_{k=1}^{n^2} 1=\sum_{k=1}^{n^2} (1+0k)=1+ 1+\ldots +1(n^2 {\rm times})=n^2.
$$
On the other hand, for the same integer $n$, we have
$$
{\rm sum\,\, of \,\,n-positive\,\, odd \,\,numbers}(=1+3+5+ \ldots+ (2n-1))=n^2.
$$
Thus, we can write
$$
\sum_{k=1}^{n^2} 1=1=3+5+\ldots+(2n-1) \tag 1
$$
Letting $n\to \infty$ in (1) we get the asked expression.</p>


问:
Equating two operators in different bases
Suppose

$$[A]_B = \begin{bmatrix}
1 & x\\
x & 7
\end{bmatrix}$$
where $A$ is some linear operator and $B$ is a basis for $\mathbb{R}^2$. Also, suppose
$$[C]_D = \begin{bmatrix}
-2 & 0\\
0 & 5
\end{bmatrix}$$
is another linear operator and $D$ is another basis for $\mathbb{R}^2$. For what values of $x$ is $A = C$?

How would I approach this problem?

答: You require $M[A]_BM^{-1}=[C]_D$ i.e. $[A]_B,[C]_D$ should be similar, where $M$ is the transformation matrix from basis $B\to D$. Since similar matrices have the same eigenvalues, you would need $-2,5$ to be the eigenvalues of $[A]_B$. Is there an $x$ which makes this possible?


问:
$a_{ii}=0, a_{ij}+a_{ji}=1, 1\leq i<j\leq n$. This matrix has rank $\geq n-1$?

Let $A$ be an $n\times n$ matrix with entries $a_{ij}$. Assume that $a_{ii}=0, a_{ij}+a_{ji}=1, 1\leq i<j\leq n$. This matrix has rank $\geq n-1$?

How to prove then? What is known is that $A+A'$ is invertible.

答:
supposing this is in reals:

compute the same thing 2 different ways -- if rank is $\leq n-2$, apply rank-nullity-- that means $\dim\ker\big(A\big)\geq 2$, so you can find a unit length vector $\mathbf v$ in the kernel of $A$ that is orthogonal to the ones vector, $\mathbf 1$.  Then

$0= \mathbf v^T \big(A+A^T\big)\mathbf v =\mathbf v^T \big(\mathbf {11}^T-I_n \big)\mathbf v =\mathbf v^T \big(-I_n \big)\mathbf v=-1$

which is a contradiction.

the technique is nicely illustrated in miniature 8

https://kam.mff.cuni.cz/~matousek/stml-53-matousek-1.pdf


问:
Existence of a non constant analytic function
Assume $f:\mathbb{D} \to \mathbb{C}$ is non-constant analytic function. Which of the following are possible:

i. $|f(z)|=1\ \forall\ z\ s.t. 0 < |z|<\frac{1}{2}$

ii. $f(z)=0$ for infinitely many z such that $0 < |z|<\frac{1}{2}$

iii. $f(z)\in \mathbb{R}\  \forall\ z\ s.t. 0 < |z|<\frac{1}{2}$

iv. $|f(z)| < |z|\ \forall z\ s.t.0 < |z|<\frac{1}{2}$

ii. is false as $f^{-1}\{0\}$ is a set with limit point and hence by identity theorem $f$ is constant.

iii. is false as $f$ is open map and $f(0<|z|<\frac{1}{2})$ is not open.

What about i. and iv?
Thank you.

答:
i. $|f|$ attains its maximum value of the disk of radius $\frac 1  2$ around $0$. By MMP $f$ is a constant $c$ on this disk. This implies that it is a  constant on $\mathbb D$ also. [Apply Identity Theorem to $f-c$].

1v. $f(z)=\frac z 2$ satisfies this inequality.


问:
does $\int_0^{\infty} e^{(-t^2 -1/t^{2})} $ exists as improper riemann integral or lebesgue integral

$\int_0^{\infty} e^{(-t^2 -1/t^{2})} $ dt

This question is from Apostol's mathematical analysis and the author has given many questions of this type in exercises of Ch-10 . I have done many of them but this is 1 of the question which I was not able to solve.

I used the defination of 10.32 for improper riemann integral and theorem 10.33 for lebesgue integral in most of the questions along with comparison tests for reimann integral but in this question I was not able to use any of the above bcoz of the type of function in the integrand.

So, I need help in how to approach this question.

Thanks!

答:
I will prove it is Lebesgue integrable. Clearly your issue is near $0$, since by comparison with $e^{-t^2}$ you can integrate it on $(\epsilon, \infty)$ for any $\epsilon > 0$ so we will focus on integrating on $(0,1)$.

Using the substitution $s = 1/t$ you get

$$ \int_{0}^1 e^{-t^2 -1/t^2}dt = \int_{1}^{\infty} e^{-s^2 - 1/s^2} s^{-2} ds $$

And now just use comparison test with $e^{-s^2}$.


问:
How to prove case 1 of Theorem 4.3 in "Calculus of Several Variables" by Serge Lang?
I am reading "Calculus of Several Variables 3rd Edition" by Serge Lang.

How to prove case 1 of Theorem 4.3 in "Calculus of Several Variables" by Serge Lang?

In Theorem 4.1, for any two points $P, Q$ in $U$, $\int_{P,C}^{Q} F$ was independent of the path $C$ in $U$ joining $P$ and $Q$.

But in Theorem 4.3, Lang specifies the path from $(1,0)$ to $X$ for $\int_{(1,0)}^{X} F$.

So I cannot mimic the proof of Theorem 4.1.

Theorem 4.3

Theorem 4.1

答:
Even though the path is specified, you can still mimic the proof of Theorem 4.1.

Focus on case 1. We need to show that the gradient of $\phi$ is indeed $F=(f,g)$, i.e., $D_1\phi=f$ and $D_2\phi=g$. I’ll show how to mimic the proof that $D_1\phi=f$. (Showing $D_2\phi=g$ is very similar; I leave that to you.)

By definition, $D_1\phi(x)$ is

$$\lim_{h\to0}\frac{\phi(x+he_1)-\phi(x)}{h}$$

Denote by $c_1$ the specified path from $(1,0)$ to $x$, and denote by $c_3$ the specified path from $(1,0)$ to $x+he_1$. Then, by definition of $\phi$, the numerator in the difference quotient above can be written as a difference of line integrals:

$$\int_{c_3}F-\int_{c_1}F$$

In theorem 4.1 we simplify this expression by picking $c_3$ cleverly, but we don’t have that freedom here: $c_3$ is fixed for us. The trick is to notice that $c_3$ and $c_1$ can be combined with a third curve $c_2$ to give a closed path from $(1,0)$ back to itself, where $c_2$ is the line segment running from $x$ to $x+he_1$. In particular, the path $c_1+c_2-c_3$ is closed.

Now, for sufficiently small $h$, we can enclose the path $c_1+c_2-c_3$ in an open rectangle that does not contain the origin. This means that we can apply Theorem 2.1, which says that if $D_2f=D_1g$ throughput a rectangle, then $F$ has a potential in that rectangle. (Ignore how that potential may or may not relate to the $\phi$ we are studying.) You can see now why the assumption that $D_2f=D_1g$ is crucial: we need it to invoke Theorem 2.1.

Now that we know $F$ has a potential in the rectangle, we can apply Theorem 3.1 to conclude the integral of $F$ around the closed curve $c_1+c_2-c_3$ is zero. That means

$$\int_{c_2}F=-\int_{c_1-c_3}F=\int_{c_3-c_1}F$$

But the right-hand side is precisely the numerator of our difference quotient above!

The upshot is that the numerator works out to be exactly the same numerator as in Theorem 4.1, the integral of $F$ along the line segment from $x$ to $x+he_1$. The rest of the proof proceeds the same.

I’ll remark that we could also prove that $F$ vanishes over the closed curve by using Green’s theorem on the region enclosed by the closed curve, but Lang doesn’t prove that until later in the book.


问:
How to frame this set of linear equations?
I have the following set of equations, as an example

$2x + 1y + 2z = A$

$0x + 2y + 2z = A$

$1x + 2y + 1z = A$

I assume this can be rewritten as a matrix? How can I check if a solution exists such that x, y, and z are nonnegative? In this case I don't believe a solution exists but how can I verify it without manually testing values?

答:
Use an augmented coefficient matrix, and obtain row-echelon form (using elementary row operations), to see if a solution exists, and/or if the system is inconsistent. If inconsistent, then no solution exists.

$2x + 1y + 2z = A$

$0x + 2y + 2z = A$

$1x + 2y + 1z = A$

$$
M = \begin{pmatrix}
2 & 1 & 2 & A \\
0 & 2 & 2 & A \\
1 & 2 & 1 & A
\end{pmatrix}
$$

Subtract 1/2 × (row 1) from row 3

Multiply row 3 by 2

Swap row 2 with row 3

Subtract 2/3 × (row 2) from row 3

Multiply row 3 by 3

Subtract 1/3 × (row 3) from row 1

Subtract 1/3 × (row 2) from row 1

Divide row 1 by 2

Divide row 2 by 3

Divide row 3 by 6

$$\text{Result}:\quad
\begin{pmatrix}
1 & 0 & 0 & A/6\\
0 & 1 & 0 & A/3\\
0 & 0 & 1 & A/6
\end{pmatrix}
$$

If you row reduce carefully, (and you should attempt this so you can gauge your success in being able to do so), you should obtain the following:

$$x = A/6,\; y = A/3, \; z = A/6$$

So for any given value of A, you will have a unique solution for $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} A/6 \\ A/3 \\ A/6 \end{pmatrix}$.

So long as $A\geq 0$, the solution will be non-negative.

For review: see row echelon form and row operations.


问:
Is every metric space with countable base a separable metric space?
Theorem: Every compact metric space $K$ has a countable base and therefore $K$ is separable.

A: Since every infinite subset of $K$ has a limit point in $K$ as $K$ is a compact metric space, it follows that $K$ is separable and since every separable metric space has a countable base so does $K$. This proves the theorem.

Now, let's say that this A is not known to us so we proceed as below:

Choose $\delta \gt 0$,
Clearly $K\subseteq \cup_{x\in K} N_\delta (x)$. Since, $K$ is compact, there must exist a finite subset $K_f\subset K$ such that $K\subseteq \cup_{x\in K_f} N_\delta (x)=B$

Let $G\subseteq K$ be any open subset of $K$. For some $t\in G$, $\exists \epsilon_t\gt 0$ such that $N_{\epsilon_{t}}(t)\subset G$.

Clearly, $t\in B$ and hence $d(t,x)\lt \delta$ for some $x\in K_f$

If $a\in N_\delta (x)$ then $d(a,x)\lt \delta$

Hence, $d(a,t)\le d(a,x)+d(x,t)\lt 2\delta$

Since $\delta\gt 0$  is arbitrary, we choose $\delta =\epsilon_t/2\implies d(a,t)\lt \epsilon_t\implies a\in G$. Thus we have proved that for any $x\in G, x \in N_\delta (x) \text{(for some $\delta \gt 0$)}\subset G$. Hence, $B$ is a countable base of $K$.

I am stuck at proving (if this is true at all):

$K$ has a countable base $\implies  K$ is separable. If this statement is not true, then $A$ can be employed to prove $K$ is separable.

Please help. Thanks.

答:
There are multiple problems with what you’ve written. I’ll leave the ones in A for the end.

It is true that $\{N_\delta(x):x\in K\}$ is an open cover of $K$ and that there is a finite $K_f\subseteq K$ such that $K\subseteq\bigcup_{x\in K_f}N_\delta(x)$. However, all of the sets $N_\delta(x)$ are subsets of $K$, so in fact $\bigcup_{x\in K_f}N_\delta(x)=K$: the set that you called $B$ is simply $K$. Thus, it makes no sense later on to say that $B$ is a countable base for $K$: $B$ is just the set $K$ and is not a base for $K$ at all.

What you have actually shown here is that for each $\delta>0$ there is a finite $K_\delta\subseteq K$ such that for each $x\in K$ there is at least one $y\in K_\delta$ such that $d(y,x)<\delta$. You could now use this directly to get a countable dense subset of $K$, as follows. For $n\in\Bbb Z^+$ let $D_n=K_{1/n}$, and let $D=\bigcup_{n\in\Bbb Z^+}D_n$. $D$ is the union of countably many finite sets, so $D$ is countable.

To show that $D$ is dense in $K$, let $x\in K$ and $\epsilon>0$. There is an $n\in\Bbb Z^+$ such that $\frac1n<\epsilon$, and there is a $y\in D_n$ such that $d(x,y)<\frac1n$, so $y\in D\cap N_\epsilon(x)$. Thus, $x\in\operatorname{cl}D$, and $D$ is dense in $K$.

Thus, you don’t need to show that every space with a countable base is separable, though this is actually very easy. Suppose that $\mathscr{B}$ is a countable base for a space $X$. For each $B\in\mathscr{B}$ let $x_B\in B$, and let $D=\{x_B:B\in\mathscr{B}\}$. Clearly $D$ is countable. If $x\in X$ and $U$ is any open nbhd of $x$, there is a $B\in\mathscr{B}$ such that $x\in B\subseteq U$, so $x_B\in D\cap U$, $x\in\operatorname{cl}D$, and $D$ is dense in $X$.

Showing that every separable metric space has a countable base is actually harder. It looks like you may have been trying to do something like that in the two paragraphs after you took the finite subcover $K_f$, but what you actually have there simply doesn’t make much sense. (E.g., for any $x\in K$ and $\delta>0$ it’s always true that $x\in N_\delta(x)$; this is not something true just for points $x$ in some open set $G$ and does not require any argument at all, since it’s true by definition.)

One way to prove that result is to show that $\left\{B_{1/n}(x):x\in D\text{ and }n\in\Bbb Z^+\right\}$ is a countable base for the topology. You can use a similar idea to prove directly from compactness that $K$ has a countable base, as the author of this question did.

Finally we come to A. It is true that the compactness of $K$ implies that every infinite subset of $K$ has a limit point, but the fact that every infinite subset of the space has a limit point does not by itself imply that the space is separable: there are non-separable spaces in which every infinite subset has a limit point. (The space of countable ordinals is one.) You really do need to use the fact that $K$ is metric to show that it is separable.


问:
Probability of catching subway that shows up randomly in a uniform distribution given any hour of the day at a certain stop.
A blue tram shows up randomly in a uniform distribution given any hour of the day at a certain stop. A person shows up independently within this same hour. If they are only willing to wait 10 minutes maximum, what is the probability the tram will show up?

Attempt: If the tram shows up at say 5:01, the probability is $1/60$. Similarly, 5:02 would have probability $2/60$. This trend continues until the 11 minute mark where it would remain at a probability of $10/60$. If the tram stops at any time between 5:10 and 6:00 the probability will remain $10/60$.

My answer: $(5*9+10*51)/60*60$ or $\Pr = .1542$

Questions: First off, is this intuition right. Secondly, is there some kind of theory behind this problem to simplify it?

答:
I don't know Asymptote or any other graphics rendering language, so you're out of luck with that. I'll do my best to describe what the situation looks like.

Let the $x$-axis vary from $ 0 $ to $ 60 $ to denote the time of the hour at which the man arrives. Note that if the man arrives at time $ x $, then the tram must arrive at time $ t $ that satisfies $ x \le t \le t + 10 $, for the most part.

Now, plot let the $y$-axis denote $ t $. Your area of interest will measure $ 60 \times 60 $. For $ 0 \le x \le 50 $, the interval of $ t $ satisfying inequality has width $ 10 $. Hence, the area of interest for $ x \le 50 $ is $ 50 \cdot 10 $; to be precise, it is a parallelogram bounded by the equations $ x = 0 $, $ x = t, x = t + 10, x = 50 $. For the interval $ 50 \le x \le 60 $, the area of interest is a triangle; it is bounded by $ x = 50, t = 60, x = t $. It has area $ 50 $. Hence, the probability is $ \frac{500 + 50}{60 \cdot 60} = \frac{11}{72} $.

This assumes that if the man arrives at, for example, 11:50, then the bus tram cannot arrive at 12:00 to pick him up because that is in a different hourly interval. If the tram comes once per hour at random constantly, then the probability is simply $ \frac{1}{6} $.


问:
What should I do to get how much juice is to be added?
A certain brand of beverage contains $60\%$ orange and $40\%$ mango juice. How much mango juice must be added to $2.4$ liters of the beverage in order to get a mixture which is $30\%$ orange and $70\%$ mango juice ?

答:
If you have $V_t = 2.4 $ litres of your beverage at first then it means that there is $V_o=0.6\times 2.4 = 1.44$ litre of orange juice and $V_m =0.4 \times 2.4 = 0.96$ litre of mango juice .

You wonder what is the volume $x$ of mango juice to add in order you to have $30\%$ of orange juice and $70 \%$ of mango juice in your beverage.

So you can solve:

$$V_o=30 \% (V_t+x)=0.3(V_t+x)$$

Therefore:

$$ x = \dfrac{V_0}{0.3}-V_t = 2.4 \text{ litres}$$

But this could have done more quickly, because if you only add mango juice then if you want the proportion of orange juice in the beverage to be half what it was, it means you just need to add mango juice for the total volume to be the double as it was originally.


答:
For first one:

If you consider $f = u + iv$ then first condition implies that $u^2 + v^2=1$ then by partial differentiation we have
$$
u_xu+v_xv = 0
$$
and
$$
u_yu+v_yv=0
$$

Now use Cauchy Riemann equations $u_x = v_y $ and $u_y = -v_x$ and get a contradiction.

Third one also very easy to do with the help of C-R equations. Observe that third condition implies that $v=0$ and consequently $v_x = v_y = 0$. Further by C-R equations we have $u_x = u_y =0$ which would imply that $u=0$ and ofcourse this is a contradiction.

Title: Existence of a non constant analytic function

Body:
Assume $f:\mathbb{D} \to \mathbb{C}$ is non-constant analytic function. Which of the following are possible:

i. $|f(z)|=1\ \forall\ z\ s.t. 0 < |z|<\frac{1}{2}$

ii. $f(z)=0$ for infinitely many z such that $0 < |z|<\frac{1}{2}$

iii. $f(z)\in \mathbb{R}\  \forall\ z\ s.t. 0 < |z|<\frac{1}{2}$

iv. $|f(z)| < |z|\ \forall z\ s.t.0 < |z|<\frac{1}{2}$

ii. is false as $f^{-1}\{0\}$ is a set with limit point and hence by identity theorem $f$ is constant.

iii. is false as $f$ is open map and $f(0<|z|<\frac{1}{2})$ is not open.

What about i. and iv?
Thank you.


答:
Hoo boy. You're in for a treat; this theorem was known to Pythagoras and his disciples.

Theorem: The sum of the first $n$ odd numbers is equal to $n^2$.

Proof: By picture.

The original image is part of a page containing two proofs of this identity: http://www.9math.com/book/sum-first-n-odd-natural-numbers. It is not mine.

Title: How does $\lim\limits_{n\to\infty} \sum_{i=1}^{n^2} 1$ result into 1+3+5...

Body: How can $\lim\limits_{n\to\infty} \sum_{i=1}^{n^2} 1$ result into 1+3+5...? Why odd numbers?</p>


问:
How to find free generators of a subgroup generated by conjugates of certain elements
Reviewing old qualifying exams in algebraic topology, I see a lot of the following type of question: let $F$ be the free group on generators $a$ and $b$. Describe a set of free generators of the subgroup generated by all conjugates of $S$, where $S$ is some small set of elements in $F$.

EDIT: For a concrete example, I've seen a solution when $S = \{aba^{-1}b^{-1}\}$. I have seen the Cayley graph that is drawn, but am not sure why they chose that specific graph.

I know that I have to draw a certain Cayley (Schreier?) graph, but as it is phrased I don't know which one to draw. For instance, some questions like this are phrased slightly differently: let $h : F \to G$ be a surjective homomorphism sending $a \mapsto x, b \mapsto y$ where $x, y$ are generators of $G$. Then I know that I should be drawing the Cayley graph of the group $G$. However, when the question is phrased like in the above paragraph, I am at a loss. Perhaps to be more clear, the involvement of the phrase "all conjugates of" throws me off.

Also, is there any way to know a priori if the given subgroup is normal? Or would I have to check that the relevant covering is normal/regular?

答:
I am afraid that this answer might not be what you are looking for, because I am thinking algebraically rather than topologically.

In your example, the subgroup is the commutator subgroup $[F,F]$. The vertices of the Schreier graph can be labelled by a transversal to the subgroup, and the obvious choice is $\{a^ib^j: i,j \in {\mathbb Z}\}$. You have edges labelled $a$ and $b$ from $a^ib^j$ to $a^{i+1}b^j$ and $a^ib^{j+1}$, respectively.

The associated free subgroup generators (often called the Schreier generators) are $\{ a^ib^jab^{-j}a^{-i-1} : i,j \in {\mathbb Z}, j \ne 0 \}$.

In a general problem of this type, the vertices of the Schreier graph correspond to the elements of the quotient group $F/\langle S^F \rangle$ of $F$, which is the group defined by the presentation $\langle a,b \mid S \rangle$.


问:
Leinster's Category theory model answers
Is there a source of model answers for the examples in Leinster's Basic Category theory?

The book has an excellent set of questions but sadly no apparent answers are provided, despite the book being aimed at those with no previous experience learning outside of lectures. As a graduate moving into mathematics from physics the ability to see the rigorous formulation of the answers would be a godsend (Never mind those that leave me stumped).

答:
If anyone is interested, I also wrote down solutions to all exercises in the book (I was not aware of the other set of solutions above).

You can find them here: https://positron0802.wordpress.com/basic-category-theory-leinster/


问:
Solutions manual for Analysis On Manifolds
A few months ago,I wanted to learn something fundmental about manifolds. From highly recommend , I decided to choice Analysis on Manifolds by James R.Munkres as my self-learning textbook.Until now ,I have finished the first two chapter's solutions. But I am not sure my answer to this exerices abusolutly right.Is there   some solutions manual for this book? Can anyone  provides free downloads ?  I appreciate it indeed!

答:
If anyone is interested, I wrote down solutions to all exercises in Chapters 1 and 2 (The Algebra and Topology of $\mathbb{R}^n$ and Differentiation).

You can find them here: https://positron0802.wordpress.com/analysis-on-manifolds-munkres/


问:
Solutions to questions in "Categories for the Working Mathematician"
Does anyone have solutions for the exercises in "Categories for the Working Mathematician"? I'm working my way through them, and want to check my answers.</p>

答:
If anyone is interested, I also wrote down solutions to exercises in Chapters 1, 2 and 3 (so far).

You can find them here: https://positron0802.wordpress.com/categories-for-the-working-mathematician-mac-lane/


问:
Why ${x}_{1}={a}_{1} {e}^{i \omega t}$ and ${x}_{2}={a}_{2} {e}^{i \omega t}$ ansatzs work in solutions to coupled oscillator DE?
in two coupled harmonic oscillator differential equations, we use ansatzs of the form ${x}_{1}={a}_{1} {e}^{i \omega t}$ and ${x}_{2}={a}_{2} {e}^{i \omega t}$ . And it works by giving us the frequencies as the roots of zeroes of polynomials. This trick works. Can I know exactly why this trick works?
Coupled DE:
$m \ddot{x}_{1}+k x_{1}+k^{\prime}\left(x_{1}-x_{2}\right)=0$
$m \ddot{x}_{2}+k x_{2}-k^{\prime}\left(x_{1}-x_{2}\right)=0$

答: If you understand why it works for a differential equation of the form $$y''+ay'+by=0,$$ then you should be able to see why it works for systems of such equations (of course, $a$ and $b$ have to be constants). So why does it work for the above equation? First rewrite it in the form $$y''+ay'=-by,$$ and you're asking yourself what function of $x$ which when differentiated once, twice gives you back a constant multiple of itself? Well, from elementary calculus we know that this is the property of the exponential functions, which leads us to trying $y=e^{mx}$ for some constant $m$ to be determined.


问:
Given two relations on a set, I want to see if one is reflexive/transitive/symmetric, and find some relation compositions.
As there is limit for title to put the full question, I will put the full question below.

Question: Given $R$ and $S$ are relations on $A = \{1,2,3,4\}$ such that $R = \{(2,2),(2,3),(2,4),(3,2),(3,3),(3,4),(4,4)\}$ and $S = \{(1,1),(1,3),(3,4)\}$...

Is $R$ reflexive? Transitive? Symmetric?

Find $S \circ R$ and $S \circ S$.

Well..Since $1$ from set $A$ is not included in set $R$, can I straightly say that since $1$ is not included in $R$ so $R$ is none of the symmetric, transitive and reflexive?

And also are $S \circ R = \{(2,4),(3,4)\}$ and $S\circ S = \{(1,1),(1,4)\}$?

答:
$
\newcommand{\CC}{\mathcal{C}}
$
Observe that, if we represent your initial relations $R,S$ as a directed graph (where $x$ points to $y$ iff $(x,y) \in R$ or $S$, depending on the relation), then we have these graphs respectively:

In graphs like these, certain properties can be intuitively and visually observed:

Reflexivity appears if every node points to itself.

Symmetry happens if, whenever $x$ points to $y$ (and thus a path exists), then $y$ points to $x$ (and thus a path to "backtrack" exists). Basically, there are no "stray," one-way edges - only closed loops.

Transitivity is visible when, if there exists paths $x$ to $y$ and $y$ to $z$, then there is a path $x$ to $z$. For distinct nodes, then this means that, whenever two sides of a triangle are filled in, the last one is too, though the direction depends on the direction of the other sides in question.

Your equivalence classes are all of the nodes in each of the disjoint graphs you have. (Two graphs are disjoint if, basically, one is entirely separate from another, totally isolated.)

Is $R$ reflexive? Transitive? Symmetric?

$R$ is not reflexive, since $1$ is not pointing to itself, i.e. $(1,1) \not \in R$.

$R$ is not symmetric, since while $2$ and $3$ point to $4$, they do not point back. That is, $(2,4) \in R$ but $(4,2) \not \in R$; similarly, $(3,4) \in R$ but $(4,3) \not \in R$.

$R$ is transitive, since every necessary "triangle" is closed, with the third side always pointing in the right direction.

Note that the three properties above are independent of each other, to some degree. A relation can be any one of these while not the others. For instance

Also note that "$R$ is not reflexive" is not necessarily the same as "$R$ is irreflexive", and similar for symmetry and antisymmetry. A relation can be both, or neither, or one, or the other.

Note in particular: while $(1,1) \not \in R$ shows that $R$ is not reflexive, it does not show symmetry or transitivity (or, rather, the lack thereof), because $(x,1),(1,x) \not \in R$ for any $x \in A$ in the first place, so it's not really a concern.

Find $S \circ R$ and $S \circ S$.

Personally, when possible, I like to approach this sort of problem visually. Write out the elements of the relevant set, in columns -- three for any given composition of relations, but since we're finding multiple compositions we can do this more compactly with four columns to find both compositions.

For convenience, I will also label the columns $\mathcal C_i$, for $i=1,2,3,4$. Then, in the gaps between $\CC_1$ and $\CC_2$, draw arrows between the two, where $x$ in $\CC_1$ points to $y$ in $\CC_2$ if and only if $(x,y) \in R$. Next, draw arrows similarly in the gap for $\CC_2$ and $\CC_3$ representing the relation $S$: $x$ in $\CC_2$ points to $y$ in $\CC_3$ if and only if $(x,y) \in S$. Do this exact same thing for the next column.

In short: the arrows in the first gap represent the first "mapping" for the relation under $R$, and those in the subsequent gaps represent that for $S$. As you might imagine, following from $\CC_1$ to $\CC_3$ gives you $S \circ R$; followings from $\CC_2$ to $\CC_4$ gives you $S \circ S$. (This in particular might be more intuitive if you read $S \circ R$ as "$S$ after $R$," for instance, and think about function composition - in a definition sense, functions are just special relations.)

Anyhow, the diagram you get for your case is this:

You might be guessing what comes next. Essentially, if $x$ in $\CC_1$ points to $y$ in $\CC_2$, and then to $z$ in $\CC_3$, then $(x,z) \in S \circ R$. Or put differently: if there is a path from $x$ in $\CC_1$ to $z$ in $\CC_3$ you can walk along, then $(x,z) \in S \circ R$.

Similarly, the existence of such a path for $x$ in $\CC_2$ to $z$ in $\CC_4$ gives $(x,z) \in S \circ S$.

It will usually be easiest to just start in $\CC_3$, and see if you can trace backwards to $\CC_1$ (and similar for $\CC_4$ and $\CC_2$), but that's up to you. Anyhow, with this in mind, it becomes clear that

$$S \circ R = \big\{ (2,4),(3,4) \big\} \qquad
S \circ S = \big\{ (1,1),(1,3),(1,4) \big\}$$

Thus, your answers in this respect are correct (once we account for your comments in Bram28's answer).

...granted, this question is quite old, so I imagine you don't need help now. But hopefully this helps someone in the future, and, if nothing else, gets this question out of the unanswered queue.


问:
Why does the following formula of multiplying and adding 1 always work?
I was coming back from somewhere on the bus. I was playing with my calculator and managed to realized this formula:

9 * 9 = 81

10 * 10 = 100

100 - (9 + 9 + 1) = 81

So the formula is that you multiply x by x, then multiply x + 1 by x + 1, then (x + 1) * (x + 1) - (x + x + 1) will always give the result of x * x. I wonder why.

So I use the programming language python to check whether this always works or not, it does always work!

```<code>x = 4 # assign x as 4
a = x * x # assign a as x times x
b = (x + 1) * (x + 1) # assign b as (x + 1) times (x + 1)
c = b - (x + x + 1) # assigns c as b minus (x + x + 1)
print(a == c) # checks whether a equals to c
print(c) # outputs c
print(a) # outputs a
</code>```

And the above code outputs:

```<code>True
16
16
</code>```

I wonder why it's like this, I tested with all numbers and it always works.

Could anyone explain to me the logic?

答:
This is just the development of

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

When you sum two numbers and multiply the result by a third one, you obtain the same result when you make the sum of the first one multiplied by the third one plus the second one multiplied by the third one.


答:
You have that $x \times x=x^2$

and $(x+1)\times (x+1)-(x+x+1)=x^2+2x+1-(2x+1)=x^2$

These two are equivalent when you simplify them which is why they  are equal for all numbers/inputs.

Title: Why does the following formula of multiplying and adding 1 always work?

Body:
I was coming back from somewhere on the bus. I was playing with my calculator and managed to realized this formula:

9 * 9 = 81

10 * 10 = 100

100 - (9 + 9 + 1) = 81

So the formula is that you multiply x by x, then multiply x + 1 by x + 1, then (x + 1) * (x + 1) - (x + x + 1) will always give the result of x * x. I wonder why.

So I use the programming language python to check whether this always works or not, it does always work!

```<code>x = 4 # assign x as 4
a = x * x # assign a as x times x
b = (x + 1) * (x + 1) # assign b as (x + 1) times (x + 1)
c = b - (x + x + 1) # assigns c as b minus (x + x + 1)
print(a == c) # checks whether a equals to c
print(c) # outputs c
print(a) # outputs a
</code>```

And the above code outputs:

```<code>True
16
16
</code>```

I wonder why it's like this, I tested with all numbers and it always works.

Could anyone explain to me the logic?


Body:
Rewrite $(x+1)(x+1)=x^2+2x+1$
as $(x+1)(x+1)-2x-1= x^2$.


问:
Ito-Doeblin Formula (a step involved in the derivation)
This is an abbreviated version of Exercise 4.14 from Stocastic Calculus for Finance II by Steven Shreve but the context of the question should be sufficient to answer the question.

Define $Z_j = f''(W(t_j))[(W(t_{j+1}) - W(t_j))^2 - (t_{j+1} - t_{j})]$

where $W(t)$ is a Brownian motion , $t \geq 0.$ Show that $\mathbb{E}\sum_{j=0}^{n-1}Z_j = 0.$

My attempt:

\begin{align}
\mathbb{E}\sum_{j=0}^{n-1}Z_j
&= \mathbb{E}\sum_{j=0}^{n-1}f''(W(t_j))[W(t_{j+1}) - W(t_j)]^2 - \mathbb{E}\sum_{j=0}^{n-1}f''(W(t_j))(t_{j+1}-t_j)\\
&= \sum_{j=0}^{n-1}\mathbb{E}f''(W(t_j))\mathbb{E}[W(t_{j+1}) - W(t_j)]^2 - \mathbb{E}\sum_{j=0}^{n-1}f''(W(t_j))(t_{j+1}-t_j)\\
&= \sum_{j=0}^{n-1}\mathbb{E}f''(W(t_j))(t_{j+1}-t_j)  - \mathbb{E}\sum_{j=0}^{n-1}f''(W(t_j))(t_{j+1}-t_j)\\
&= 0.
\end{align}
where in the second equality, I have used the fact that $f''(W(t_j))$ is $\mathcal{F}(t_j)$- measurable while $[W(t_{j+1}) - W(t_j)]^2$ is independent of $\mathcal{F}(t_j)$ and in order to get the third equality, I have simply plugged in the value of $\mathbb{E}[W(t_{j+1}) - W(t_j)]^2$ which is $(t_{j+1}-t_{j}).$ Please comment if I'm right.

The official solution goes as follows:

$Z_j$ is $\mathcal{F}(t_{j+1})$- measurable and $\mathbb{E}[Z_j|\mathcal{F}(t_j)] = 0.$ I'm okay with both these steps and have easily proved it. But then they do the following working which I'm not clear with:

$\mathbb{E}\big[\sum_{j=0}^{n-1}Z_j\big] = \mathbb{E}\big[\sum_{j=0}^{n-1}\mathbb{E}[Z_j|\mathcal{F}(t_j)]\big] = 0.$

My question regarding their solution is the following:

$Z_j$ is $\mathcal{F}(t_{j+1})$- measurable so $\mathbb{E}[Z_j|\mathcal{F}(t_{j+1})] = Z_j$ so how can they replace $Z_j$ with $\mathbb{E}[Z_j|\mathcal{F}(t_{j})]$ ?

I know that $\mathbb{E}[Z_j|\mathcal{F}(t_{j})]$ is $\mathcal{F}(t_{j})$- measurable but can't understand the logic used in their solution. Could be a fundamental misunderstanding so any help highly appreciated.

答: They are using the Law of Iterated Expectations.</p>


问:
A question regarding uniform boundness on norm spaces

Let $X,Y,Z$ be normed-spaces above same field.
Assume that $Y$ is a banach space.
Let $F:X×Y\to Z$ a function that satisfy:

$\forall x\in X$ $F(x,-)$ linear and continuous above $Y$.
$\forall y\in Y$ $F(-,y)$ continuous.
Assume that $(x_n)_n \subset X$ converges to 0 and $(y_n)_n\subset Y$ converges to 0.
$\forall n\in N$ define:
$T_n:Y\to Z$ , $T_n=F(x_n,-)$.

A.Show that $\forall n\in N$ , $T_n\in L(X,Y)$ and that there is $0 \leq M<\infty$ such that $||T_n||\leq M$ $\forall n\in N$.

B.Prove that: $lim_{n\to \infty} F(x_n,y_n)=0$.

My work:

Part A.
$T_n$ is linear:
Let $y_1,y_2\in Y$ and $a,b\in Z$ then:
$T_n(ay_1+by_2)=F(x_n,ay_1+by_2)=${ F(-,y) is linear} = $aF(x_n,y_1)+bF(x_n,y_2)=aT_n(y_1)+bT_n(y_2)$
But

$T_n$ is continuos:
This is concluded from the fact that $F(x,-)$ is continuous for all $x\in X$.So $T_n\in L(X,Y)$.
$T_n$ is continuous thus bounded so by definition we get that there's a constant $C$ such that $||T_n(y)||\leq C*||y||$ then $||T_n||\leq C$.
But, how to use the uniform bounded theory to finish it ?

Part B.
$|F(x_n,y_n)|=||T_n(y_n)||\leq ||T_n||*||y_n||\leq C*||y_n||$.
Here we used part A.
Now, since $(y_n)_n$ converges to 0, then $(y_n)$ is bounded. So $|y_n)|\leq \epsilon/C$ .
We get that $||F(x_n,y_n)||\leq \epsilon$, thus $F(x_n,y_n)\to 0$.
Or we can say directly that, $||y_n||\ to 0$ since $(y_n)$ converges to zero and the norm is continuous. So $F(x_n,y_n)\to 0$ as $n\to \infty$.

Thanks in advance for any help.

答: Hints: The linear map $T_n$ is continuous because $F(x_n, .)$ is continuous. For fixed $y$ use the  facts that $(x_n)$ is a bounbded sequence and $F(.,y)$ is continuous to conclude that $\sup_n \|T_ny\| <\infty$. By UBP we get $\sup \|T_n|| <\infty$. Let $M= \sup_n \|T_n\| $. Conclude that $|F(x,y) | \leq M\|x\| \|y||$ to prove part B.


问:
James R. Munkres' TOPOLOGY, 2nd edition: How to check my work?
I'm trying to learn, or revise, some topology from James R. Munkres' TOPOLOGY, 2nd edition. I'm working alone; that is, I'm self-learning. It is quite fun. But the problem is how do I check if I've managed to arrive at a correct solution to an exercise problem? Can I get hold of a solution manual? Or, can I find someone over the Internet with whom I can discuss my solutions? Of course, putting up every other problem at Mathematics Stack Exchange, it seems to me, is not so practical! </p>

<p>What would be the best possible for me? </p>

答:
If anyone is interested, I wrote down solutions to exercises in Chapters 2 and 3.

You can find them here: https://positron0802.wordpress.com/topology-munkres/


问:
Let $f :\mathbb{R}→ \mathbb{R}$ be a function such that $f^2$ and $f^3$ are differentiable. Is $f$ differentiable?
Let $f :\mathbb{R}→ \mathbb{R}$ be a function such that $f^2$ and $f^3$ are differentiable. Is $f$ differentiable?

Similarly, let $f :\mathbb{C}→ \mathbb{C}$ be a function such that $f^2$ and $f^3$ are analytic. Is $f$ analytic?

答: If you do want $f^{p}$ to represent the $p^{\text{th}}$ iterate, then you can let $f$ denote the characteristic function of the irrational numbers.  Then $f^2$ and $f^3$ are both identically zero, yet $f$ is nowhere continuous, let alone differentiable.</p>


问:
Which person will leave in last?
Can anyone give me the quickest idea to solve the question below?

Read the information below and answer the question that follows.

In a mathematical game, one hundred people are standing in a line and they are required to count off in fives as "one, two, three, four five, one, two, three, four, five," and so on from the first person in the line. The person who says "five" is taken out of the line. Those remaining repeat this procedure until only four people remain in the line.

What was the original position in the line of the last person to leave?

A. 93
  B. 96
  C. 97
  D. 98

If any one can elaborate with more similar kind of example then it will be very nice.

答:
After the first cycle, 80 people will remain.

As a result, 99th will be last.

Similarly,  after the second cycle, 64 People will remain and 98th person will be the last.

Now, even after the subsequent cycles, the 98th person will not leave the queue as the number of remaining people will never be a multiple of 5.

Hence 98 is the required answer.

NOTE:- The last person leaves the queue iff, the remaining people is a multiple of 5.

Hope you understood and enjoyed the solution.
Thanks for your patience to read.

Title: Which person will leave in last?


问:
Finding the homology of space comprised of $S^2$ and an intersecting torus.
I am reviewing old algebraic topology qualifying exams, and I need to compute the fundamental group and the homology of the following space.

First, a brief calculation of $\pi_1(X)$. Let $C$ be the sphere and let $D$ be the torus unioned with some small path connected open neighborhood of the intersection of $C$ and $D$. Note that $C\cap D$ (path-connected) will look like an oval on the sphere with two holes in it. It will deform rectract to $S^1\vee S^1$. Thus we have that $\pi_1(C)$ is trivial, $\pi_1(D)=\langle a,b| a b a^{-1} b^{-1}\rangle$ and $\pi_1(C\cap D)=\langle c,d| \hspace{3mm}\rangle$. We then have $(i_C)_*(c)=(i_C)_*(d)=0$ where $i_C:C\cap D\hookrightarrow C\cong S^2$ is the inclusion. $(i_D)_*(d)=(i_D)_*(d)=a\in \pi_1(T^2)$ where $i_D:C\cap D\hookrightarrow D\cong T^2$ is the inclusion. So SVK's theorem gives
$$ \pi_1(X)= \langle a, b | a b a^{-1} b^{-1}, a\rangle \cong \mathbb{Z}$$
Next, I need to calculate the homology of $X$. My first idea is to use the Mayer-Vietoris long exact sequence. I know that $H_0(X)$ is just $\mathbb{Z}$ (one path component) and that $H_1(X)=\pi_1^{ab}(X)\cong \mathbb{Z}$, but I would like to calculate $H_1(X)$ and $H_2(X)$ by hand for practice. Using $A\cong S^2$ and $B\cong T^2$ (perphaps with some small retractible neighborhoods so that $X=\textrm{int}(A)\cup \textrm{int} (B))$.

$(\Psi_1)_*$ will map $[a]=[(1,0)]\in \pi_1(T^2)$ to $[0]\in H_1(X)$ because that loop on the torus is homotopic to a constant map when it is seen as a loop in $S^2$. $(\Psi_1)_*$ will map $[b]=[(0,1)]\in \pi_1(T^2)$ to $[1]\in H_1(X)$. However, this analysis is based on knowledge of $\pi_1(X)$, which is isomorphic to $H_1(X)$, so this kind of reasoning seems contingent on the fact that I already know that $H_1(X)=\langle b| \hspace{3mm}\rangle$.  I think that that $(\Phi_1)_*$
will be neither injective nor surjective. It does seem to split and map one copy of $\mathbb{Z}$ to $0$ and the other copy of $\mathbb{Z}$ to the second summoned of $H_1(T^2)=\mathbb{Z}\oplus \mathbb{Z}$. $\textrm{Im}(\Phi_1)_*\cong\mathbb{Z}\cong\textrm{Ker} (\Psi_1)_*$ by exactness. So I'm unsure how to compute $H_1(X)$ without relying on the fundamental group.

for $n=2$ we have

It is immediate that $(\Phi_2)_*$ is injective. I know that given $[z]\in H_2(X)$ ($[z]$ is a cycle) , I can write $z=x+y$ with $x,y\in C_2(A)\oplus C_2(B)$, then $0=\partial([z])=\partial(z)+\partial(y)$ and $\delta_2([z])=[\partial(x)]=-[\partial(y)]$. I haven't decided on a cell decomposition for $X$, so I feel like I cannot proceed any further with the analysis of these maps. Is there a straight-forward way of proceeding with my long exact sequence, or would I be better off looking for a cell decomposition and computing the homology that way?

答:
As mentioned by JHF, $X$ is actually $S^1 \vee S^1 \vee S^2 \vee S^2 \vee S^2$. To see that, take a look at #1 here: https://www.math3ma.com/blog/clever-homotopy-equivalences, and contract what you called the oval, so that both halves of the torus become spheres with two identified points.

Your approach does lead to a result as well, but you have to be more careful. First of all, I don't think that $\pi_1(D)=\langle a,b| a b a^{-1} b^{-1}\rangle$. (Your definition of $D$ is recursive but I think I got it.) $D$ is more than a torus, and, accordingly, you seem to have lost one generator: "from the side" a torus looks like a big circle whereas $D$ looks like a wedge of two circles due to the component you've added.

Next, I don't think that to say what you've said about $(\Psi_1)_*$ is to use $\pi_1(X)$. One generator is mapped to zero and the image of the other generates a free subgroup - that's all you need to say here, no $\pi_1$ references. Then you can follow through with this calculation: first, by looking at what happens in $H_0,$ you can find the image of $\delta_1$ (as the kernel of $(\Phi_0)_*$). Then you can also find its kernel as the image of $(\Psi_1)_*$. Both will be $\mathbb{Z}$, which yields the right answer.

$H_2$ is even simpler. $(\Phi_2)_*$ is injective, of course, but you can say more: it's obviously zero. So $(\Psi_2)_*$ is injective. Thus you have a $\mathbb{Z}\oplus \mathbb{Z}$ subgroup in $H_2(X)$ and the quotient over it is the image of $\delta_2$, which is the kernel of $(\Phi_1)_*$, which is $\mathbb{Z}$. So, $H_2(X)=\mathbb{Z}^3.$ (Both times I'm using that extensions with the quotient a free group are trivial.)


问:
Why is a manifold called a manifold?
Manifold in English means many and various. (https://www.merriam-webster.com/dictionary/manifold)

How did this come to mean a topological manifold with a differentiable structure?

答: According to Aguirre, Foster, and Merali's Questioning the Foundations of Physics, the term "manifold" comes from the German term "Mannigfaltigkeit" used by Riemann, which he used to describe the set of possible values of a variable under several constraints, because the variable can have many values. English mathematician and philosopher William Kingdon Clifford then translated this to "manifoldness."

Title: Why is a manifold called a manifold?


问:
K-theory - dependence of algebraic structure: What is the K-theory of a direct product?
I want to figure out the dependence of K-class of an finitely generate projective ring and its algebraic struture.

For example consider $K_0(\mathbb{C})\cong \mathbb{Z}$ and $K_0(\mathbb{R}\times\mathbb{R})$,
where the algebraic on $\mathbb{C}$ is the standard multiplication and on $\mathbb{R}\times\mathbb{R}$ the multipation is given by: $(r_1,r_2)\times (r_1',r_2')\mapsto (r_1*r_1',r_2*r_2')$.
Are  $K_0(\mathbb{C})$ and $K_0(\mathbb{R}\times\mathbb{R})$ isomorphic?
What can we say in general?

答:
The grothendieck group of a field $k$ is always equal to $\mathbb{Z}$ - the ring of integers. The isomorphism is as follows:

$i: K_0(k) \cong \mathbb{Z}$ with $i([V]):=dim(V)$.

In general there is for arbitrary comutative unital rings $A_j, j=1,..,l$ an isomorphism

D1. $K_i(A_1\oplus \cdots \oplus A_l)\cong K_i(A_1)\oplus \cdots \oplus K_i(A_l)$

for any integer $i\geq 0$, hence $K_0(\mathbb{R}\times \mathbb{R})\cong \mathbb{Z}\times \mathbb{Z}$.

Question: "Are K0(C) and K0(R×R) isomorphic? What can we say in general?"

Answer: The following holds:

$K_0(\mathbb{C})\cong \mathbb{Z} \neq K_0(\mathbb{R}\times \mathbb{R})$.


问:
How to find a condition which makes box topology have a better properties?
Lately I've been working on box and product topology. I want to find some better conditions.Here is one of my thoughts.

There is an easy theorem about product topology.

$$ f:A{\rightarrow}{\Pi_{{\alpha\in J}}X_\alpha}\ \ \ \ \ \ f(a)=(f_{\alpha}(a))_{{\alpha}\in J} \\ \text{For each} \ {\alpha}, f_{\alpha}:A{\rightarrow}X_{\alpha}$$
If $\Pi_{{\alpha\in J}}X_\alpha$ has a product topology, then $f$ is continuous iff each $f_\alpha$ is continuous.

Now I am considering the box topology circumstance:
It is obvious that: If $f$ is continuous, then each $f_\alpha$ is continuous.
Now I want to find a better condition, which makes $f$ continuous when each $f_\alpha$ is continuous.

Could anyone give me some suggestions or examples?

答:
There is a unique topology on $X=\prod_{j\in J} X_j $ such that

(1) For every $A$, $f:A\to X$ is continuous iff $f_j:A\to X_j$ are continuous,

which is the product topology.

Let $A$ be any directed set with a maximal element $1$. Consider the topology on $A$ generated by sets like $(\lambda,1]$ and sets like $ \{\lambda\}$ with $\lambda\neq 1$.

Now suppose (1) holds. (1) implies that,  $(\forall \{\lambda_\eta\in A\}_{\eta\in H})(\lambda_\eta \to \lambda\Rightarrow f(\lambda_\eta)\to f(\lambda))\Leftrightarrow(\forall \{\lambda_\eta\in A\}_{\eta\in H})(\forall j\in J)(\lambda_\eta\to \lambda\Rightarrow f_j(\lambda_\eta)\to f_j(\lambda)))$

Fix a directed set $\Lambda$ (which has no maximal elements) and a net $\{f_\lambda\}_{\lambda\in \Lambda}$. Equip $A=\Lambda\cup\{1\}$ with the above topology and define $f(\lambda)=f_\lambda$, $f(1)=f$.

First assume $f_\lambda \to f$. For every $\{\lambda_\eta\in A\}_{\eta\in H}$, if $\lambda_\eta\to \lambda$, then either $\lambda=1$ and $\lambda_\eta\to 1$ or $\lambda\in \Lambda$ and $\lambda_\eta=\lambda$ for sufficiently large $\eta$. Whichever it is, we have, $(\forall \{\lambda_\eta\in A\}_{\eta\in H})(\lambda_\eta \to \lambda\Rightarrow f(\lambda_\eta)\to f(\lambda))$. This shows $f_\lambda(j)=f_j(\lambda)\to f_j(1)=f(j)$ for every $j$ . The converse is similar.


问:
What is meant by a "copy" of one field in another?
My first language isn't English so I need an explanation for this problem:

Every field of characteristic zero contains a copy of the rationals.

In the proof they defined a homomorphism between rationals and a field and proved it was injective!
What does copy mean?

My guess is that they mean the field has a subset isomorphic to the rationals. But then how does proving there is a homomorphism between rationals and the field help?

答: If you have an injective homomorphism $\varphi\colon\Bbb Q\longrightarrow k$, where $k$ is your field, then $\varphi(\Bbb Q)$ and $\Bbb Q$ are isomorphic fields and $\varphi(\Bbb Q)$ is a subfield of $k$. But it is also isomorphic to $\Bbb Q$. Therefore, it behaves as a copy of $\Bbb Q$ inside $k$.


答: Saying that an object $X$ is a copy of an object $Y$ means that $X$ is isomorphic to $Y$. In your example it can be shown that each field $K$ of characteristic zero contains a subfield $ Q \subset K$ which is isomorphic to the field $\mathbb Q$ of rationals. The proof you mentioned constructs a homomorphism $f : \mathbb Q \to K$ and shows that it is injective. Then $Q = f(\mathbb Q)$ is the desired subfield.

Title: What is meant by a "copy" of one field in another?

Body:
My first language isn't English so I need an explanation for this problem:

Every field of characteristic zero contains a copy of the rationals.

In the proof they defined a homomorphism between rationals and a field and proved it was injective!
What does copy mean?

My guess is that they mean the field has a subset isomorphic to the rationals. But then how does proving there is a homomorphism between rationals and the field help?


问:
Double integral: domain with function
and happy new year!

I'm working on a double integra where there is a function at the domain: cos(x)

$ \int_D $ at the domaine $ \{(x,y) | 0 \leq x \leq \frac{\pi}{2} \text{ and } 0 \leq y \leq cos(x) \}  $

Which I have written as

$\int_0^{cos(x)}\int_0^{\pi/2} x \; dx dy $

But: Does this mean I have to integrate w.r.t the dy part first? (normally I would do dx first)

Hope it makes sense!

答:
No, you are supposed to integrate dy first, and then over dx, as you have set it your integral as a type-1 region, where y is bounded between two different functions of x. ( y=cos(x) and y=0 in your case).

Had you rewritten the function as a type-2 regions, where x is bounded by two functions of y, the correct approach would have been to integrate it over dx first and then dy.

Title: Double integral: domain with function


问:
Can we have a closed form or approximation of following combination summation?Or when it is positive or not?
Can we have any closed form or approximation of the summation? And can we know when it is positive?

$S(n,t)=\sum_{i=0}^{t}(-1)^{i}\tbinom{n}{i}$

We can assume that $0<t<n/2$.

This is the sum of the $D(n,t)=\tbinom{n}{t}-\tbinom{n}{t+1}$.  A simpler form of $D(n,t)$ may help the problem.

Till now, I can not find any help from Stirling's approximation.

答:
Welcome to MSE!

It is well known that your sum $S(n,t) = (-1)^t \binom{n-1}{t}$. There are a variety of proofs here for instance. Among other things, this means that the sum is nonnegative if and only if $t$ is even.

As for asymptotics, there are lots of known approximations for $\binom{n-1}{t}$ which depend on how quickly $t$ and $n$ are growing. One useful set of bounds is

$$
\frac{n^k}{k^k} \leq \binom{n}{k} \leq \frac{n^k}{k!}
$$

Which one can easily use to approximate $S(n,t) = (-1)^t \binom{n-1}{t}$.

Asymptotically, one can use stirling's approximation to obtain the sharpest result which is always true as follows:

First expand the definition of the closed form of $S(n,t)$:

$$S(n,t) = (-1)^t \binom{n-1}{t} = (-1)^t \frac{(n-1)!}{t!(n-1-t)!}$$

Then apply the stirling approximation:

$$
(-1)^t \frac{(n-1)!}{t!(n-1-t)!} \sim
(-1)^t \frac{\sqrt{2 \pi (n-1)}(n-1)^{n-1}/e^{n-1}}{\left ( \sqrt{2 \pi t} t^t / e^t \right ) \left ( \sqrt{2 \pi (n-1-t)} (n-1-t)^{n-1-t}/e^{n-1-t}\right )}
$$

Now you can cancel and simplify this to

$$
(-1)^t \sqrt{\frac{n-1}{2 \pi t (n-1-t)}} \frac{(n-1)^{n-1}}{t^t (n-1-t)^{n-1-t}}
$$

This expression, while correct, is slightly unwieldy. If you know more information about how $t$ grows (for instance, is $t \approx \sqrt{n}$? $kn$? $n^{0.1}$?) you can simplify the above expression (sometimes dramatically). Google has lots of results in this vein.

I hope this helps ^_^


问:
Understanding weighted inner product and weighted norms
I am reading  this book where at page 27 following definitions about weighted inner product and weighted  norms are given.

Let $M$ and $N$ be Hermitian positive definite matrices of order $m$ and $n$ respectively. The weighted inner products in $\mathbb{C}^{m}$ and $\mathbb{C}^{n}$ are

$(x, y)_{M} = y^{*}Mx$ , $x, y \in \mathbb{C}^{m}$ and $(x, y)_{N} = y^{*}Nx$ , $x, y \in \mathbb{C}^{n}$ ....$(1)$

The definitions of weighted vector norms are

$\|x\|_{M} = (x, x)^\frac{1}{2}_{M} = (x^{*}Mx)^\frac{1}{2} = \|M^\frac{1}{2} x\|_{2}$, $x\in\mathbb{C}^{m}$ ....$(2)$

$\|x\|_{N} = (x, x)^\frac{1}{2}_{N} = (x^{*}Nx)^\frac{1}{2} = \|N^\frac{1}{2} x\|_{2}$, $x\in\mathbb{C}^{n}$ ....$(3)$

The definitions of weighted matrix norm are

$\|A\|_{MN} = \max_{\|x\|_{N} = 1}{\|Ax\|_{M}},\;  x \in\mathbb{C}^n and ~~A\in \mathbb{C}^{m\times n}$

$\|B\|_{NM} = \max_{\|x\|_{M} = 1}{\|Bx\|_{N}},\;  x \in\mathbb{C}^n and ~~B\in \mathbb{C}^{n\times m}$

Such a norm is sometimes called an operator norm subordinate to vector norm. It is easy to verify that

$\|A\|_{MN} =  \|M^\frac{1}{2} A  N^\frac{-1}{2} \|_{2}$ ....$(4)$

$\|B\|_{NM} =  \|N^\frac{1}{2} B  M^\frac{-1}{2} \|_{2}$ ....$(5)$

Could anybody explain me about the significance of weighted norms? Why we need weighted norm? In $(2)$ how we got $\|M^\frac{1}{2} x\|_{2}$? How could we find square root of matrix $M$?  How did we got equation $(4)$ and $(5)$.

I would be very much thankful for the help and suggestions.

答:
The significance of "weighted norms", why we need them:

note that norms you consider are induced by Hermitian bilinear forms, therefore we should discuss the Hermitian bilinear forms.

From the point of view of linear algebra there is nothing special about the "standard" Hermitian bilinear form $\langle x,y\rangle=x^{*}Iy$. All Hermitian bilinear forms are bilinear forms with additional Hermitian property. It can be shown that for every Hermitian bilinear form $\langle\cdot,\cdot\rangle$ (i.e. an inner product) there exist a Hermitian positive definite matrix $M$ such that $\langle x,y\rangle=x^{*}My$. And that is all. None of those inner products is in any way "better" than the other.

The only reason that you distinguish the "standard" inner product $\langle x,y\rangle=x^{*}Iy$ is the choice of a particular basis of your space (i.e. coordinate system). You perceive it as a "best" inner product because its matrix "$M$" is the identitiy matrix $I$, the simplest of all Hermitian positive definite matrices.

So in fact the distinction between the "standard" norm and weighted norms is artificial and has no real meaning.


问:
Stochastic Dominance Inequality
Let $(X_i)_{1\le i\le n}$ and $(Y_i)_{1\le i\le n}$ be two sequences of random variables that only take non-negative integer values. Let $X=\sum_{1\le i\le n}X_i$ and $Y=\sum_{1\le i\le n}Y_i$.
If for all real numbers $t$ and $1\le i\le n$, we have
$$\mathbb{P}(X_i\le t)\le \mathbb{P}(Y_i\le t),$$
is it true that
$$\mathbb{P}(X\le t)\le \mathbb{P}(Y\le t)?$$

答:
No. Consider $X_1=1-X_2$ with distribution
$$\mathbb P(X_1=0) = \mathbb P(X_1=1) = \frac12=\mathbb P(X_2=0) = \mathbb P(X_2=1)$$
and independent r.v.'s $Y_1, Y_2$  with distribution
$$\mathbb P(Y_1=0) = \mathbb P(Y_2=0) = \frac23, \quad \mathbb P(Y_1=1) = \mathbb P(Y_2=1)=\frac13.$$
Then for all real $t$
$$
\mathbb P(X_1 \leq t) \leq \mathbb P(Y_1 \leq t),
$$
$$
\mathbb P(X_2 \leq t) \leq \mathbb P(Y_2 \leq t)
$$
but $X_1+X_2=1$ and
$$
\mathbb P(Y_1+Y_2 = 2) =\frac19,
$$
so
$$
\mathbb P(X_1+X_2 \leq 1.5) =1 > \mathbb P(Y_1+Y_2 \leq 1.5) = \frac89.
$$

Title: Stochastic Dominance Inequality


问:
Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x > 0, x \in \mathbb{R}: f(x) \geq \frac{1 + 2\ln(2)}{4}$
Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x\in(0,+\infty): f(x) \geq \frac{1 + 2\ln(2)}4$.

That would be same as proving that $x^2 + \ln (\frac{x + 1}{x}) - \frac{1 + \ln(4)}{4} \geq 0$, isn't?

I have also found the derivative $f'(x) = \frac{2x^3 + 2x^2 - 1}{x^2 + x}$ and that there is a solution of $f(x)$ in the interval of $\left[\frac12, \frac1{\sqrt 2}\right]$.

But from there I am stuck, I am not able to see how to link all the concepts.

Could someone give me a hint?

Thank you and Happy New Year 2021 :)

答:
Hint: Note that over $(0,\infty)$, $x^2$ is increasing and $\ln(1+\tfrac{1}{x})$ is decreasing.

Hence, for $x \ge 1$, we have $x^2 \ge 1^2 = 1$ and $\ln(1+\tfrac{1}{x}) \ge \displaystyle\lim_{x \to \infty}\ln(1+\tfrac{1}{x}) = \ln(1) = 0$.

Also, for $0 < x < 1$, we have $x^2 \ge 0^2 = 0$ and $\ln(1+\tfrac{1}{x}) \ge \ln(1+\tfrac{1}{1}) = \ln(2)$.

See if you can put all these pieces together to prove a stronger bound than the one you need for the problem.


问:
Existence of a non constant analytic function
Assume $f:\mathbb{D} \to \mathbb{C}$ is non-constant analytic function. Which of the following are possible:

i. $|f(z)|=1\ \forall\ z\ s.t. 0 < |z|<\frac{1}{2}$

ii. $f(z)=0$ for infinitely many z such that $0 < |z|<\frac{1}{2}$

iii. $f(z)\in \mathbb{R}\  \forall\ z\ s.t. 0 < |z|<\frac{1}{2}$

iv. $|f(z)| < |z|\ \forall z\ s.t.0 < |z|<\frac{1}{2}$

ii. is false as $f^{-1}\{0\}$ is a set with limit point and hence by identity theorem $f$ is constant.

iii. is false as $f$ is open map and $f(0<|z|<\frac{1}{2})$ is not open.

What about i. and iv?
Thank you.

答:
Your A to (iii) is right but your reason is wrong. If $f$ is not constant on $E=\{z:0<|z|<1/2\}$ then its image is non-empty and open in $\Bbb C$ (as you said), but no non-empty subset of $\Bbb R$ is open in $\Bbb C.$

Another way, with Re$(f)=u$ and Im$(f)=v.$ If $v$ is $0$ on $E$ then  $u_x=u_x+iv_x=f'=-iu_y+v_y=-iu_y,$ so $f'$ is both real and imaginary so $f'=0$ on $E$. And $E$ is a connected open set. So $f$ is constant on $E$.