问:
How to solve equation like this?
Is there any good method to solve equations like this?
$$
\begin{cases} x^2=y^2\\-6xy+10y^2=16\end{cases}
$$
From the first equation, I see:
$$
1) x=1, y=1
$$
$$
2)x=1, y=-1
$$
$$
3)x=-1, y=1
$$
$$
4)x=-1, y=-1
$$
When I plug 2) and 3) to the second equation, I see it works.

So now I have 2 solutions, but wolfram found more...
Is there any good method for equations like this?

答:
From the first equation we can just conclude $x=\pm y$, nothing more.

If $x=y,$ from the second equation
 $-6(y)y+10y^2=16\implies 4y^2=16\implies y^2=4\implies  y=\pm 2$

If $x=-y, $ from the second equation

$-6(-y)y+10y^2=16\implies y^2=1\implies y=\pm 1$

Title: How to solve equation like this?


问:
Induction for binomial coefficients
I would like some help to prove the following equality :
$$\sum_{i=0}^n \binom{n}i^2=\binom{2n}n$$
I wanted to do a proof by induction :
$$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=1}^{n+1} \binom{n+1}i^2=1 + \sum_{i=0}^{n} \binom{n+1}{i+1}^2=1+\sum_{i=0}^{n} \bigg(\binom{n+1}i+\binom{n}{i+1}\bigg)^2$$
$$\sum_{i=0}^{n+1} \binom{n+1}i^2=1+\sum_{i=0}^{n} \bigg(\binom{n}i^2+2\binom{n}i\binom{n}{i+1}+\binom{n}{i+1}^2\bigg)=1+\binom{2n}n+ \sum_{i=0}^{n} 2\binom{n}i\binom{n}{i+1} +(\sum_{i=0}^n \binom{n}i^2-1) $$
$$\sum_{i=0}^{n+1} \binom{n+1}i^2=2\binom{2n}n+\sum_{i=0}^{n} 2\binom{n}i\binom{n}{i+1}=2\bigg(\sum_{i=0}^n \binom{n}i^2(1+\frac{n-i}{i+1})\bigg)=2(n+1)\bigg(\sum_{i=0}^n \binom{n}i^2\frac{1}{i+1}\bigg)$$

But now I'm stuck.

答:
One way to proceed is to prove a more general identity by induction, and then deduce the identity in the problem statement as a corollary.

The more general identity is called "Vandermonde's Identity":
$$\sum_{i=0}^k \binom{m}{i} \binom{n}{k-i} = \binom{m+n}{k} \tag{1}$$
for non-negative integers $k$, $m$ and $n$.  We give a proof by induction over $n$. The case $n=0$ reduces to $\binom{m}{k} = \binom{m}{k}$.  Now suppose $(1)$ holds for some $n \ge 0$.  Then the $n+1$ case is
$$\begin{align}
\sum_{i=0}^k \binom{m}{i} \binom{n+1}{k-i} &= \sum_{i=0}^k \binom{m}{i} \left( \binom{n}{k-i-1} + \binom{n}{k-i} \right)  \tag{2}\\
&=\sum_{i=0}^k \binom{m}{i}  \binom{n}{k-i-1} + \sum_{i=0}^k \binom{m}{i}  \binom{n}{k-i} \\
&=\binom{m+n}{k-1} + \binom{m+n}{k} \tag{3} \\
&= \binom{m+n+1}{k} \tag{4}
\end{align}$$
This completes the proof by induction.  At steps $(2)$ and $(4)$ we used the identity
$$\binom{n}{m} = \binom{n-1}{m-1} + \binom{n-1}{m}$$ and at step $(3)$ we used the inductive hypothesis.

To prove $$\sum_{i=0}^n \binom{n}{i}^2 = \binom{2n}{n}$$
from $(1)$, take the special case $k=n$, $m=n$, and apply the identity
$$\binom{n}{i} = \binom{n}{n-i}$$

ContentLicense: CC BY-SA 4.0

Title: Induction for binomial coefficients


问:
Flux of $v=(y_1e^{y_2}-e^{y_1},y_2e^{y_3}-e^{y_2},y_3e^{y_1}-e^{y_3})$ through a sphere

$v=(y_1e^{y_2}-e^{y_1},y_2e^{y_3}-e^{y_2},y_3e^{y_1}-e^{y_3})$

$Y = (y_1^2+y_2^2+y_3^2=r^2, y_2>=0, y_3>=0)$

Calculate flux of v through Y.

My first intuition was to use spherical coordinates so I set $\varphi(x)=(r\sin x_2\cos x_1, r\sin x_2 \sin x_1, r\cos x_2)$, with $x_1$ between $0$ and $\pi$ and $x_2$ between $0$ and $\pi/2$.

Then I wanted to use the Stokes-Ampère theorem, but I don't really know how to use it. I saw that div v = $0$ and found $\omega_v^2$. Then I found $\omega_u^1$ with $u=(0,y_3e^{y_1}-y_1e^{y_3},y_1y_2e^{y_3}-y_1e^{y_2})$.

But what now? How do I set the integral? In class we saw an example with polar coordinates and a simple u and it seemed easy with just one variable. But here I have spherical coordinates and the vector u is very confusing.

Could someone please explain how to use this theorem correctly and how do I integrate from here?

答:
Since the divergence is zero, we can close the surface and say

$$\iint\limits_S\cdots + \iint\limits_{y_2=0}\cdots +\iint\limits_{y_3=0}\cdots = 0 \implies \iint\limits_S \cdots= -\iint\limits_{y_2=0}\cdots-\iint\limits_{y_3=0}\cdots$$

Those two integrals are easy to compute, they are constants being integrated over half circles

$$-\iint\limits_{y_1^2+y_3^2=r^2\cap y_3\geq 0}(\cdots, 0 - e^0,\cdots)\cdot (0,-1,0)dA - \iint\limits_{y_1^2+y_2^2=r^2\cap y_2\geq 0}(\cdots, \cdots, 0 - e^0)\cdot (0,0,-1)dA $$

$$=\iint\limits_{y_1^2+y_3^2=r^2\cap y_3\geq 0}1\:dA + \iint\limits_{y_1^2+y_2^2=r^2\cap y_2\geq 0}1\:dA = \frac{\pi r^2}{2}+\frac{\pi r^2}{2} = \pi r^2$$


问:
Search for simpler (other) idea to show $(1.001)^{1000}>2$
The below question asked me to solve: (without using logarithm) which one is bigger $$2 \ , \ (1.001)^{1000}$$ I use two Idea to show $(1.001)^{1000}$ is bigger.

First: $$\lim_{n \to \infty}\left(1+\frac 1n\right)^n=e=2.71...\\$$and $1000$ is big enough to say $$(1.001)^{1000} \sim e>2$$

Second: Using binomial expansion $$(1.001)^{1000}=(1+0.001)^{1000}=\\1+1000\times 1^{999} \times (0.001) +\frac {1000\times 999}{2}1^{998}(.001)^2+...=\\1+1+\frac{999}{2000}+...>2$$   Now ,my question is there any simpler or other Idea to show $(1.001)^{1000}>2$

Thanks in advance for any hint.

Body:
You can us bernoulli inequality

$(1,001)^{1000}=(\frac{1001}{1000})^{1000}=$

$(1+\frac{1}{1000})^{1000}\geq  1+\frac{1} {1000}×1000$


答:
Add the equations.

You get $x^2-6xy+9y^2=16 \implies(x-3y)^2=16 \implies (x-3y)= \pm4$.And you have $x= \pm y$ .

ContentLicense: CC BY-SA 3.0

Title: How to solve equation like this?

Body:
Is there any good method to solve equations like this?
$$
\begin{cases} x^2=y^2\\-6xy+10y^2=16\end{cases}
$$
From the first equation, I see:
$$
1) x=1, y=1
$$
$$
2)x=1, y=-1
$$
$$
3)x=-1, y=1
$$
$$
4)x=-1, y=-1
$$
When I plug 2) and 3) to the second equation, I see it works.

So now I have 2 solutions, but wolfram found more...
Is there any good method for equations like this?


问:
Example of torsion modules over integral domain that has zero annihilator
I was able to prove that if $R$ is an integral domain and $M$ is a finitely generated $R$-module is torsion if and only if $Ann(M) \neq 0$, but so far, I have failed to come up with an example that this fact does not hold for arbitrary modules over $R$, more specifically, I am looking for a torsion (non-finitely generated) module $M$ with $Ann(M) = 0$.

Here is a couple of definitions:

An element $m \in M$ is a torsion element if $\exists r \in R$ such that $rm = 0$

$M$ is a torsion module if every elements of M is torsion.

$Ann(M) = \{r \in R \mid \forall m \in M, rm = 0 \}$

Any help would be greatly appreciated.

答:
Hint:

One might observe that

$ann(\mathbb Z/(2))=(2)$

$ann(\mathbb Z/(2)\oplus \mathbb Z/(3))=(6)$

$ann(\mathbb Z/(2)\oplus \mathbb Z/(3)\oplus \mathbb Z/(5))=(30)$

What might you conjecture about the annihilator of $\bigoplus_{p\in P} \mathbb Z/(p)$ where $P$ is the set of positive prime integers?  Don't forget to consider the question of whether it is torsion or not...


问:
Determinant as a combination of ALL minors of dimension $N-1$.
Let $A = (a_{i,j})_{1\leq i,j\leq N}$ be a matrix of dimension $n$. If we denote by $A_{i,j}$ the minor of the matrix, it is well known that $$ |A| = \sum_{i=1}^N (-1)^{i+j}A_{i,j}a_{i,j}$$ for every $1\leq j \leq N.$ I remember to read in a book that, in fact,
$$ |A| = \sum_{i=1}^N \left( \sum_{m=1}^N (-1)^{i+m}A_{i,m} \right) a_{i,j}$$ for every $1\leq j \leq N$, but I cannot get a proof of this result. Really, I do not know if this result is true at all, I have checked it for $N=2$ and $N=3$ but I have some problems to get the result for an arbitrary dimension $N$.

Anyone can help me? I just need some references to check the proof!

Thank you very much!

EDIT: After some research, I found a result in 'Dale & Vein, Determinants and their Applications in Mathematical Physics' called the Alien Cofactor Theorem that states:
$$
\sum_{i=1}^N a_{i,j}A_{i,k} = 0$$ for every $1\leq j \leq N$, $1\leq j \leq N$, $k\neq j.$ A direct application of this result yields the result, but the book does not include any references about the proof. Does anyone know any references for the proof?

答:
The Alien Cofactor Theorem states:

$$ \sum_{i=1}^N (-1)^{i+m}A_{im}\,a_{ij}=0 \qquad\text{ if }\quad m\neq j, \tag{1}$$

and symmetrically swapping rows with columns. Combining this with Laplace's theorem, we can write

$$ \sum_{i=1}^N (-1)^{i+m}A_{im}\,a_{ij}=\det(A)\,\delta_{mj}, $$

where $\,\delta_{mj}\,$ is the Kronecker delta.

Proof of $(1)$:$\quad$ Fixed $\,m\,$ and $\,j,\,$ let $\,A'=(a'_{hk})\,$ be the matrix defined by:

$$ a'_{hm}=a_{hj} \qquad\text{and}\qquad a'_{hk}=a_{hk} \quad\text{for}\quad k\neq m, $$

i.e. $A'$ is obtained from $A$ by replacing the $m$-th column with the $j$-th column, leaving the other columns unchanged (including the $j$-th). Note that the $(i,m)-$minor $A'_{im}$ of $A'$ is equal to the minor $A_{im}$ of $A$, and $\,(1)\,$ gives then

$$ \sum_{i=1}^N(-1)^{i+m}A_{im}\,a_{ij} = \sum_{i=1}^N(-1)^{i+m}A'_{im}\,a'_{im} = \det(A') = 0 $$

by the Laplace's Theorem, and because $\,\det(A')=0,\,$ having $\,A'\,$ two equal columns, $\quad$ q.e.d.

Now your thesis follows:

\begin{align}
\sum_{i=1}^N\left(\sum_{m=1}^N(-1)^{i+m}A_{im}\right)a_{ij} &= \sum_{m=1}^N\left(\sum_{i=1}^N(-1)^{i+m}A_{im}a_{ij}\right) = \\[1.5ex]
&=\sum_{m=1}^N\det(A)\delta_{mj} = \det(A).
\end{align}


问:
Hint about proving $\lim_{n \to \infty}( \sqrt{\log{(x_n + a)}} - \sqrt{\log{x_n}} ) = 0$
I need to prove that for any $a \in \mathbb{R}$ and while $\lim x_n = \infty$, $\lim_{n \to \infty}( \sqrt{\log{(x_n + a)}} - \log{\sqrt{x_n}} ) = 0$. While a full proof would be nice (it may avoid future questions), I am only requesting a simple hint about how to simplify $\sqrt{\log{(x_n + a)}} - \log{\sqrt{x_n}}$. This is an exercise from "Analise Real. Funções de Uma Variável - Volume 1" by Elon Lima. Thanks in advance.

Edit: Yes, this is typo in the translation I was reading. My apologies to everyone. $\sqrt{\log{(x_n + a)}} - \sqrt{\log{x_n}}$ is the correct expression. I modified the title accordingly.

答:
Assuming a typo in the question,

$$\sqrt{\log{(x_n + a)}} -\sqrt{\log{{x_n}}}=\frac{\log{(x_n + a)} -\log{{x_n}}}{\sqrt{\log{(x_n + a)}}+\sqrt{\log{{x_n}}}}.$$

The numerator tends to $0$ and the denominator to $\infty$.


问:
Calculating the Method of Moments estimator

Consider a random sample $X_1,\ldots , X_n$ that are independent and identically distributed according to some $f_X(\cdot|\alpha)$ where:
$$f_X(x|\alpha)=\frac{\text{exp}\{\alpha x-e^{\alpha}\}}{x!}$$
We also know $\alpha \in \mathbb{R}$, $\alpha$ is unknown,  and E$[X]=e^{\alpha}$.

My (possibly flawed) understanding of method of moments is that we let the sample mean equal the first moment, i.e.: $\frac{1}{n}\sum_{i=1}^nX_i=\bar{X}=e^{\alpha}$, so our estimator $\hat{\alpha}_{MM}=\ln(\bar{X})$.

I'm doubting myself because when I then examine the bias which I define to be E$[\hat{\alpha}_{MM}]-\alpha$ I end up with $\ln(\bar{X})-\alpha$ which I can't seem to infer anything from. Where is my understanding of this topic falling short?

EDIT $1$: When analysing the bias I am assuming E$[\hat{\alpha}]=\text{E}[\ln(\bar{X})]=\ln(\bar{X})$ since once we have our random samples, then $\bar{X}$ is just a number and E$[c]=c$

答:
you are right.

If you know that

$$\mathbb{E}[X]=\mu=e^\alpha$$

this means that

$$\alpha=\log \mu$$

thus using the Method of Moments you get

$$\hat{\alpha}_{\text{MM}}=\log \overline{X}_n$$


问:
Question regarding a "Without loss of generality" statement in proof of optimal mean-square estimator.
The following question is from a statement in a proof (p. 287 in Shiryaev I Third edition) that the conditional expectation of $\eta \mid \xi=x $ is the optimal estimator of $\eta$ in terms of $\xi$.

Given a pair of random variables $( \eta, \xi )$, we may try to find a Borel function $\phi$ such that $\phi(\xi)$ estimates $\eta$. The optimal such estimator is the one that minimizes the mean-squared error. There is a theorem that states that if $\mathrm{E} \eta^2 < \infty$ then
$$
\phi(x) = \mathrm E (\eta \mid \xi = x )
$$
can be taken as the optimal estimator.

In the first line of the proof it is stated that: "Without loss of generality we may consider only estimators $\phi(\xi)$ for which $\mathrm \phi^2(\xi) < \infty$." I have a hard time figuring out what is meant with this?

I can see that, if $\phi ^2 (\xi ) = \infty $, then
$$
\mathrm E \left[ \eta - \phi (\xi) \right] ^2 = \mathrm E \eta^2 - 2 \mathrm E \eta \phi(\xi) + \mathrm E \phi^2(\xi) = \infty
$$
(assuming $\mathrm E \phi(\xi) < \infty$ and that we have defined addition with extended real numbers $r < \infty$ such that $r + \infty = \infty$), and if there is no better estimator then perhaps the term "optimal" does not give much information?

I have included a screenshot of the whole section below.

答:
For any estimator with $\mathbb{E}\left[\varphi^{2}\left(\xi\right)\right]=\infty$, you have shown that $\mathbb{E}\left[\left(\eta-\varphi\left(\xi\right)\right)^{2}\right]=\infty$.

Now consider a naive estimator $\varphi_{0}\left(\xi\right)\equiv0$. We have $\mathbb{E}\left[\left(\eta-\varphi_{0}\left(\xi\right)\right)^{2}\right]=\mathbb{E}\left[\eta^{2}\right]<\infty$ which is guaranteed by assumption of Theorem 1. Therefore, $\varphi_{0}\left(\xi\right)$ is always better than $\varphi\left(\xi\right)$ in terms of MSE.

In other words, for any estimator $\varphi\left(\xi\right)\not\in L_{2}$, at least the naive estimator $\varphi_{0}\left(\xi\right)\in L_{2}$ dominates it, hence it cannot be optimal. That's why we only need to consider estimators in $L_{2}$.


问:
Second Linear ODE
Confused on how I can approach this ODE, and I just need some help on figuring out what to do

$y''-4y'-21=5e ^{-t}$ with initial conditions $y(0)=0$ and  $y'(0)=4$

After Laplace transforming, I got:
$(\frac{(5)}{(s+1)}{+ 4+\frac{(21)}{(s+1)}}) /( s^2-4s) $

And now I am stuck on what to do here

Title: Second Linear ODE

Body:
Because you didn't write down your process i'll solve the whole thing here.

Let $y''-4y'-21=5e^{-t}:=f(y(t))=f(y)$

Take the Laplace Transform of each term.

Plug in initial value(s) of $y^{(i)}(t_0)$ into the equation.

Solve for $Y(s)$.

Take the Inverse Laplace Transform.

$$\mathscr{L}\{f(y)\}=\mathscr{L}\{y''-4y'-21=5e^{-t}\}=[s^2Y(s)-sy(0)-y'(0)]-4[sY(s)-y(0)]-\dfrac{21}{s}=5\dfrac{1}{s+1}$$

You can only arrive to $\frac{1}{s-a}$ if you have a $e^{at}$ function.

$$\mathscr{L}\{f(y,y(0),y'(0))\}=[s^2Y(s)-0-4]-4[sY(s)-0]-\dfrac{21}{s}=\dfrac{5}{s+1}$$
$$s^2Y(s)-4-4sY(s)-\dfrac{21}{s}=\dfrac{5}{s+1}$$

$$s^2Y(s)-4sY(s)=\dfrac{5}{s+1}+\dfrac{21}{s}+4$$
$$Y(s)(s^2-4s)=\dfrac{5}{s+1}+\dfrac{21}{s}+4$$
$$Y(s)=\dfrac{5}{(s+1)(s^2-4s)}+\dfrac{21}{s(s^2-4s)}+\dfrac{4}{(s^2-4s)}$$

$$Y(s)=\dfrac{5}{s(s+1)(s-4)}+\dfrac{21}{s^2(s-4)}+\dfrac{4}{s(s-4)}$$

$$\mathscr{L}^{-1}\{Y(s)\}=y(t)=-\dfrac{5}{4}e^{0t} + e^{-t} + \dfrac{1}{4}e^{4t} - e^{0t} + e^{4t} -\dfrac{21}{16}-\dfrac{21}{4}t+\dfrac{21}{16}e^{4t}$$

$$y(t)=-\dfrac{5}{4} + e^{-t} + \dfrac{1}{4}e^{4t} - 1 + e^{4t} -\dfrac{21}{16}-\dfrac{21}{4}t+\dfrac{21}{16}e^{4t}$$

$$y(t)=-\dfrac{20}{16} + e^{-t} + \dfrac{4}{16}e^{4t} - \dfrac{16}{16} + \dfrac{16}{16}e^{4t} -\dfrac{21}{16}-\dfrac{21}{4}t+\dfrac{21}{16}e^{4t}$$

$$\boxed{y(t)=\dfrac{41}{16}e^{4t}-\dfrac{57}{16}+e^{-t}-\dfrac{21}{4}t}$$

Which checks out by @Lukas' answer.


问:
Finding $a+b$ for real $a$ and $b$ such that $f(x)=ax+b$ and $f(bx+a)=x$
How do I solve this question on functions?

There are real numbers $a$ and $b$ for which the function $f$ has the properties that
$f(x) = ax + b$ for all real numbers $x$, and $f(bx + a) = x$ for all real numbers $x$. What
is the value of $a + b$?

I was working on this math problem and I couldn't figure it out,
I though that I could say $f(x) = bx + a$, but I quickly realized that was wrong

Any help would be appreciated

答: $bx+a$ is a real number in its own right so $f(bx+a) = a(bx+a) + b = x$, where the first equals sign is the condition $f(x) = ax+b$ for all $x \in \mathbb{R}$


问:
Prove that $ta+(1-t)b\in R$ for $a,b\in R$ and $t\in [0,1]$
I have that $R$ is the rectangle $R=[\alpha,\beta]\times[\gamma,\delta]\ and\ a=(a_1,a_2), b=(b_1,b_2)$
$$\alpha\leq a_1\leq \beta, \gamma \leq a_2 \leq \delta$$ because $a\in R$
and
$$\alpha\leq b_1\leq \beta, \gamma \leq b_2 \leq \delta$$
because $b\in R$.

I need to show that $ta+(1-t)b\in R$.

答:
Hint: let $c=(c_1,c_2)=ta+(1-t)b$, then you need to show $c_1\in[\alpha,\beta]$ and $c_2\in[\gamma,\delta]$, so 4 inequalities in total. For example:
$$
\beta=t\beta+(1-t)\beta\geq\underbrace{ta_1+(1-t)b_1}_{c_1}\geq t\alpha+(1-t)\alpha=\alpha.
$$


问:
Show that if $G$ is 8-edge-connected ; if ${E}'$ is a sub set of $E$ edges of $G$ and $|{E}'|=8$ then $w(G-{E}')\leq 2$
This is an example similar to Bondy and Murty graph theory book.
in that exercise $G$ is k-edge-connected and here it's 8 edge connected.

Show that if $G$ is 8-edge-connected, and if ${E}'$ is a sub set of $E$ edges of $G$, and $|{E}'|=8$ then $w(G-{E}')\leq 2$

$W$ is number of $G$ components.

答:
Let $G=(V, E)$ and let $S$ be an arbitrary subset of $E$ containing exactly $7$ edges. As $G$ is $8$-edge-connected, the graph $G' = (V, E \setminus S)$ is connected. That means that $w(G') = 1$. Now choose any $e \in E \setminus S$ and remove it from $G'$. As $G'$ was connected, the graph $G'' = (V, E \setminus (S \cup \{e\}))$ is either connected or has exactly two components, because deleting a single edge can only increase the number of components by one. That means that $w(G'') \leq 2$.

As we chose both $S$ and $e$ arbitrarily, we have shown that removing any subset of edges of $E$ containing exactly $8$ edges can generate a graph with at most two components.


问:
Properties of a set of all isomorphisms $ f: G \to G $
I'm kinda stuck with this task.

Let $G$ be a group and $ S $ the set of all isomorphisms $ f: G \to G$. I first want to show that $ (S, \circ) $ is also a group.

I believe I've shown that all the properties of a group is fulfilled with $ (S, \circ) $:

i) Assume that $ x \in $ and $ f_1,f_2 \in S$. Then $f_2(x) = y \in G$, since $ f_1 $ is an ismorphism. $ f_2(y) = z \in G $ since $ f_2 $ is an ismorphism. Then $ f_1(f_2(x)) = f_1(y) = z = f_1 \circ f_2(x),$ hence $f_1, f_2 \in S \longrightarrow f_1 \circ f_2 \in S.$

ii) $id$ is an isomorphism $ \longrightarrow id \in S$.

iii) $ f $ is an isomorphism $ \longrightarrow \exists f^{-1}$, one can show that $ f^{-1} $ is also an isomorphism, $ \longrightarrow f^{-1} \in S$.

Now, assume that $ | S | = 1$, then I want to show that $ G $ is abelian and each element has an order of 1 or 2.

Im kinda lost with that $ | S | = 1 $. If $ f \in S $ then $f^{-1} \in S $ due to previous result, should not $ | S | = 1 $ imply that $ f = f^{-1} = id$? And how do I move forward from this? Any hints is muy appreciated!

答: Hint: conjugation by any element of the group is an isomorphism in any group and taking inverse is an isomorphism in any abelian group.</p>


答:
Note that because $f(x)=ax+b$ for all real $x$, we also have
$$
f(bx+a) = a(bx+a) + b = abx + a^2 + b.
$$
It follows that $x = abx + a^2 + b$ for all real $x$.

For $x=0$, we find that $a^2 + b = 0$.

For $x=1$, we find that $1 = ab + a^2 + b = ab$, so $ab = 1$.

From $a^2+b=0$ we have $b=-a^2$, and we obtain $-a^3 = 1$, so $a=-1$. It then follows that $b=-a^2=-1$, so $a+b=-2$.

Title: Finding $a+b$ for real $a$ and $b$ such that $f(x)=ax+b$ and $f(bx+a)=x$

Body:
How do I solve this question on functions?

There are real numbers $a$ and $b$ for which the function $f$ has the properties that
$f(x) = ax + b$ for all real numbers $x$, and $f(bx + a) = x$ for all real numbers $x$. What
is the value of $a + b$?

I was working on this math problem and I couldn't figure it out,
I though that I could say $f(x) = bx + a$, but I quickly realized that was wrong

Any help would be appreciated


答:
If $f(x)=\color{blue}ax+\color{green}b$ for all real number $x$ means that it continues to hold for any real argument of the function $f$ and so $f(\color{red}{bx+a})=\color{blue}a(\color{red}{bx+a})+\color{green}b=abx+a^2+b$ but this last  is required by the problem to be $=x$, so:
$$abx+a^2+b=x\iff \begin{cases}
ab=1\\
a^2+b=0
\end{cases} \implies \begin{cases}
-a^3=1\\
b=-a^2
\end{cases}\implies a=-1; b=-1
$$


问:
Show that a certain net converges in a von Neumann algebra.
Consider the abstract von Neumann algebra $M = \ell^\infty\text{-}\bigoplus_{i \in I} B(H_i)$. Moreover, we assume $\dim H_i< \infty$ for all $i \in I$. Let $x_i$ be the identity on $B(H_i)$ and let $z_F = \sum_{i\in F} x_i$ where $F$ is a finite subset of $I$. If $\omega:M \to \mathbb{C}$ is a $\sigma$-weakly functional, then is it true that
$$\lim_F \sup_{\|m\| \leq 1}|\omega(m)-\omega(z_Fm)|= 0?$$

I tried to estimate
$$\sup_m |\omega(m)-\omega(z_Fm)| \leq \|\omega\|\|z_F-1\|$$
so it suffices to show that
$$\lim_F \|z_F-1\| = 0.$$

Let $\epsilon > 0$. I want to prove that there is a finite subset $F_0$ of $I$ such that
$$F \supseteq F_0 \implies \|z_F-1\| < \epsilon$$
but I have no idea how to construct $F_0$. This approach does not use the $\sigma$-weak continuity of $\omega$ though so it might be flawed.

答:
Notice that $1-z_F$ is a nonzero projection in $M$, so  $\|z_F-1\| = 1$, and hence it is impossible to prove that
$\lim_F \|z_F-1\| = 0.$  Nevertheless it is true that
$$
  \lim_F \sup_{\|m\| \leq 1}|\omega(m)-\omega(z_Fm)|= 0,
  $$
and the reason is as follows: the $\sigma $-weakly continuous linear functionals of $M$ are of the form
$$
  \omega (m) = \sum_{i\in I} \text{tr}(m_ih_i), \quad\forall m = (m_i)_i\in M,
  $$
where
$$
  h = (h_i)_i\in  \prod_{i\in I}B(H_i)
  $$
is such that
$$
  \infty >  \sum_{i\in I}\text{tr}(|h_i|) = \Vert \omega \Vert .
  \tag {1}
  $$
Given any such $\omega $, one has for every finite $F\subseteq I$, and every   $m$ in $M$,  that
$$
  |\omega(m)-\omega(z_Fm)| =
  \Big|\sum_{i\in I} \text{tr}(m_ih_i) - \sum_{i\in F} \text{tr}(m_ih_i)\Big| = $$$$ =
  \Big|\sum_{i\in I\setminus F} \text{tr}(m_ih_i) \Big| \leq
  \sum_{i\in I\setminus F} |\text{tr}(m_ih_i)| \leq  $$$$\leq
  \sup_i\Vert m_i\Vert   \sum_{i\in I\setminus F} \text{tr}(|h_i|) =
  \Vert m\Vert   \sum_{i\in I\setminus F} \text{tr}(|h_i|).
  $$
Therefore
$$
  \lim_F \sup_{\|m\| \leq 1}|\omega(m)-\omega(z_Fm)|\leq
  \lim_F   \sum_{i\in I\setminus F} \text{tr}(|h_i|) = 0,
  $$
by (1).


答:
Given that: $ta+(1-t)b=t(a_1,a_2)+(1-t)(b_1,b_2)=(ta_1+(1-t)b_1,ta_2+ (1-t)b_2) \tag 1$

Note that $\alpha\leq a_1\leq \beta, \gamma \leq a_2 \leq \delta\implies t\alpha\leq ta_1\leq t\beta, t\gamma \leq ta_2 \leq t\delta$

Similarly,  $\alpha\leq b_1\leq \beta, \gamma \leq b_2 \leq \delta\implies (1-t)\alpha\leq (1-t)b_1\leq (1-t) \beta, (1-t)\gamma \leq (1-t)b_2 \leq (1-t)\delta$

Add the above inequalities to get: $(1-t)\alpha+t\alpha\leq (1-t)b_1+ta_1\leq (1-t) \beta+t\beta \implies \alpha\le (1-t)b_1+ta_1\le \beta$

Similarly, from the rest two inequalities, you get $\gamma\le ta_2+ (1-t)b_2 \le \delta$

Hence proved. (By $(1)$)

Title: Prove that $ta+(1-t)b\in R$ for $a,b\in R$ and $t\in [0,1]$

Body:
I have that $R$ is the rectangle $R=[\alpha,\beta]\times[\gamma,\delta]\ and\ a=(a_1,a_2), b=(b_1,b_2)$
$$\alpha\leq a_1\leq \beta, \gamma \leq a_2 \leq \delta$$ because $a\in R$
and
$$\alpha\leq b_1\leq \beta, \gamma \leq b_2 \leq \delta$$
because $b\in R$.

I need to show that $ta+(1-t)b\in R$.


问:
If $a_n\to+\infty$, $b_n\to0$, how about $\prod_{k=1}^n(1+b_ne^{-a_k})$?
If $a_n$ and $b_n$ are real series, and $a_n\to+\infty$, $b_n\to0$, how is the limit
$$\underset{n\to\infty}{\lim}\prod_{k=1}^n(1+b_ne^{-a_k})?$$

There could be some specific requirements on $a_n$ and $b_n$, but I wish a more general result. Thanks~

Thanks for all the answers below, but I think I need a restriction like this:
$$a_n\sim C_1n^{\alpha},\qquad b_n\sim C_2n^{-\beta},$$
where $C_1,C_2,\alpha,\beta$ are constants, and $\alpha$, $\beta$ are positive.

I really need this problem to be solved, hope someone can help, thanks~

答:
Take $b_n = 0$, then the limit is equal to $1$.

Take $a_n = \log(\sqrt{n})$ and $b_n = \log^{-1}(n)$, then the limit is now $\infty$.

Thus, no general result exists.


问:
$\mathbb{Q}[x]$ is dense in $\mathbb{R}^\mathbb{R}$?
I know that $\mathbb{R}^\mathbb{R}$ is separable by the general theorem on the products of separable spaces, but my professor said that a countable dense subset of this space is $\mathbb{Q}[x]$, i.e. the set of polynomial functions with rational coefficient, which is easy to verify to be countable, but I searched for a proof that is also dense and I was unable to find one, can someone provide such a proof to me?

答:
Yes, $\Bbb Q[x]$ is indeed dense in this power in the product topology (here we regard these polynomials as defining functions only, not as elements of the ring).
Sketch of the argument:

Let $\prod_{r \in \Bbb R} U_r$ be a basic non-empty open set of $\Bbb R^{\Bbb R}$, so that we have a finite subset $F$ of $\Bbb R$ so that $U_r = \Bbb R$ when $r \notin F$ and $U_r$ is a non-empty open subset (usual topology) in $\Bbb R$ for $r \in F$.

It's easy to find a polynomial over $\Bbb R$, say $p(x)=\sum_{i=0}^n a_i x^i$ (of degree $|F|$ at most) such that $p(r) \in U_r$ for all $r \in F$ (we only have $|F|$ many constraints, see a.o. Lagrange interpolation).

Then approximate the coefficients $a_i$ by rational ones $q_i$ so that $q(x)=\sum_{i=0}^n q_i x^i$ also maps all $r \in F$ into rational values in $U_r$; this is a bit technical but some continuity/density arguments and maybe a version of Stone-Weierstraß can be used to achieve this.

Then $q(x)$ is a function from $\Bbb Q[x]$ in this basic subset and that shows $\Bbb Q[x]$ is indeed dense.


问:
Prove that $f(X \cap f^{-1} (Y)) = f(X) \cap Y$
Prove that $f(X \cap f^{-1} (Y)) = f(X) \cap Y$

Given that $A$ and $B$ are non-empty sets and that $f:X \to Y$ is a function. Let $X \subseteq A$ a non-empty subset of $A$, and $Y \subseteq B$ a non-empty subset of $B$.

答:
Take $f(a) \in f(X \cap f^{-1} (Y))$ where $a \in X \cap f^{-1}(Y)$. Then, $a \in X$ and $f(a) \in Y$. Thus, $f(a)$ is in the intersection and this shows the first inclusion.

Take $f(a) \in f(X) \cap Y$ where $a \in X$. Then, $f(a) \in Y$. So, $a \in f^{-1} (Y)$, and, $a \in X$. therefore, $a \in X \cap f^{-1} (Y)$ and so $f(a) \in f(X \cap f^{-1} (Y))$. This shows the other inclusion and the proof is done.


问:
Wikipedia's definition of a computable numbers
According to Wikipedia,

computable numbers are the real numbers that can be computed to within any desired precision by a finite, terminating algorithm.

I'm somewhat confused here. I always thought that given some fixed number of decimal places $n$, one can compute any number up to at least that precision.

Do they mean that for computability of a given number ('described' in some way, just not necessarily by an algorithm, e.g. Chaitin's constant), there needs to be a fixed finite, terminating algorithm which works for any $n$?

But once we have chosen some $n$, we can still compute up to that precision, just not necessarily by the initially picked algorithm anymore?

答:
Yes, your understanding is correct. For a real number $x$ to be computable you need to have a fixed algorithm that, given $n$, outputs the $n$-th decimal digit of $x$. It is true that for every real $x$ and every $n$, there is an algorithm that outputs exactly the first $n$ decimal digits of $x$, but there are real numbers s.t. no single algorithm works for every $n$.

As an additional note, the description of the real number $x$ plays an important role here. The idea of "being able to produce the $n$-th decimal digit" is essentially formalized by saying that, for every $n$, you can produce a rational number $q$ s.t. $|x-q| \le 2^{-n}$. This is the so-called Cauchy representation of $x$. However, this is not the only way to represent a real. For example, you can identify $x$ with a monotonically increasing sequence of rational numbers $(q_n)_{n\in\mathbb{N}}$ s.t. $\sup_n q_n = x$. This is the so-called left-cut representation of $x$. The set of Cauchy-computable real numbers is strictly smaller than the set of left-computable real numbers.


问:
Solve: $\left(1-\sin2x\right)\left(\cos x-\sin x\right)=1-2\sin^{2}x$
My attempt:

$$1-2\sin^{2}x=\cos2x$$
Let $$\cos x-\sin x=t$$
Thus, on squaring: $$\sin2x=1-t^{2}$$

I tried simplifying further, but that $cos2x$ is giving me trouble.

Please help!

答: After a rewrite, we get $$(\cos x-\sin x)^3 = (\cos x-\sin x)(\cos x+\sin x)$$Can you solve from here?


答:
Assuming that $b_n$ is sufficiently small, more precisely $1+b_ie^{-a_J}>0,$ we can take the logarithm and use the asymptotic $\ln (1+x)\asymp x.$ The convergence of the product is equivalent to the convergence of $c_n=b_n\sum_{k=1}^ne^{-a_k}.$ Obviously, $c_n$ does not usually converge. One may take $b_n=\frac{1}{\ln \ln{n}}$ and $a_k=\ln k.$ However, for some particular sequences one may get the results about convergence. For instance, if $b_n\asymp \frac{1}{n}$ then $c_n=\frac{\sum_{k=1}^ne^{-a_k}}{n}\to 0$ by Cesàro mean theorem.

Edit (continuation for an updated question).

Using the conditions you mentioned above $c_n$ can be rewritten as
$$c_n\asymp n^{-\beta}\left(\sum_{k=1}^n a^{n^{\alpha}}\right),$$
where $a=e^{-C_1}.$ Now we apply Stoltz- Cesaro theorem to compute the limit:
$$\lim_{n\to\infty}c_n=\lim_{n\to\infty}\frac{\left(\sum_{k=1}^n a^{n^{\alpha}}\right)}{n^{\beta}}=\lim_{n\to\infty}\frac{a^{(n+1)^{\alpha}}}{(n+1)^{\beta}-n^{\beta}}=\lim_{n\to\infty}\frac{a^{(n+1)^{\alpha}}}{\beta n^{\beta-1}}.$$
You can now easily analyze convergence of the last limit depending on the values of $\alpha,\beta$ and $a.$

Title: If $a_n\to+\infty$, $b_n\to0$, how about $\prod_{k=1}^n(1+b_ne^{-a_k})$?

Body:
If $a_n$ and $b_n$ are real series, and $a_n\to+\infty$, $b_n\to0$, how is the limit
$$\underset{n\to\infty}{\lim}\prod_{k=1}^n(1+b_ne^{-a_k})?$$

There could be some specific requirements on $a_n$ and $b_n$, but I wish a more general result. Thanks~

Thanks for all the answers below, but I think I need a restriction like this:
$$a_n\sim C_1n^{\alpha},\qquad b_n\sim C_2n^{-\beta},$$
where $C_1,C_2,\alpha,\beta$ are constants, and $\alpha$, $\beta$ are positive.

I really need this problem to be solved, hope someone can help, thanks~


答:
$\sin2x=2\cos x\sin x \Rightarrow 1-\sin2x=cos^2x+sin^2x-2\sin x \cos x=(cosx-sinx)^2$

$1-\sin2x=(cosx-sinx)^2 \Rightarrow (1-\sin2x)(cosx-sinx)=(cosx-sinx)^3=1-2\sin^2x$

$(cosx-sinx)^3=1-2\sin^2x=\sin2x$

$cosx-sinx=t \Rightarrow t^3=1-t^2$

Title: Solve: $\left(1-\sin2x\right)\left(\cos x-\sin x\right)=1-2\sin^{2}x$

Body:
My attempt:

$$1-2\sin^{2}x=\cos2x$$
Let $$\cos x-\sin x=t$$
Thus, on squaring: $$\sin2x=1-t^{2}$$

I tried simplifying further, but that $cos2x$ is giving me trouble.

Please help!


问:
Homotopy fixed points and ordinary fixed points
Currently I know nothing about homotopy fixed points except for its definition: given a $G$-space $X$, the set of homotopy fixed points is defined as the space of equivariant maps from $EG$ to $X$.

I was wondering under which (maybe simple) conditions, homotopy fixed points could be “identified” with (ordinary) fixed point sets, in a suitable sense.

I ask this question since I know that if the action of $G$ on $X$ is free, then the Borel construction $(EG\times X)/G$ is homotopy equivalent to $X/G$. This is how I understand equivariant (co)homology as a good replacement of the (co)homology of $X/G$ in case the action is not free.

It seems the condition dual to “free” (no kernel) should be “transitive” (no cokernel). Is there a good correspondence between homotopy fixed points and ordinary fixed points when $G$ acts transitively on $X$? Or any other conditions and results?

If the above naive attempt is not in the correct way, how should I understand homotopy fixed points as a natural generalization/replacement of ordinary fixed points?

答:
If $G$ is discrete and $T$ is a transitive $G$-orbit with $\left\lvert T \right\rvert > 1$, then $T^{hG} \cong T^G$, but in this case both fixed points are empty.  It's easy to see that the latter is empty, and for the former note that the only maps from the connected space $EG$ to the discrete space $T$ are the constant maps, and they aren't $G$-equivariant if $\left\lvert T \right\rvert > 1$.  So this case is not very interesting.

The correct dual to "free" should be "cofree": a $G$-space $C$ is cofree if there is some ordinary space $\bar{C}$ such that $$\operatorname{Map}_G(X,C) \simeq \operatorname{Map}(X, \bar{C})$$ for all $G$-spaces $X$ in a natural way.  For example, $$C = \operatorname{Map}(G, \bar{C})$$ is cofree for any $\bar{C}$.

From this definition, we can show that if $C$ is cofree, then $$C^{hG} = \operatorname{Map}_G(EG, C) \simeq \operatorname{Map}(EG, \bar{C})  \simeq \bar{C}$$ since $EG \simeq *$ non-equivariantly.

Here's one way to think about homotopy fixed points intuitively.  Homotopy fixed points can be regarded as a relaxed version of ordinary fixed points.  Instead of requiring a fixed point $x$ to satisfy equalities $x = gx$ for all $g \in G$, we simply require $x$ and $gx$ to be connected by paths in a coherent manner (these paths are part of the data) for it to be a homotopy fixed point


问:
Express $(\theta, \phi)$ of a torus using $(x,y,z)$
I know how to write the $(x,y,z)$ coordinates of a torus using $(\theta, \phi)$:
$$
\begin{align}
x(\theta, \phi) &= (R+r\cos(\theta))\cos(\phi) \\
y(\theta, \phi) &= (R+r\cos(\theta))\sin(\phi) \\
z(\theta, \phi) &= r\sin(\theta)
\end{align}
$$

For $\theta, \phi\in [0, 2\pi]$. How can I find the inverse relation? I would like to find expressions
$$
\theta(x,y,z) = \ldots \\
\phi(x,y,z)=\ldots
$$

Given that I have a point on the torus in cartesian coordinates $(x, y,z)$. How can I find its expression in terms of angles $(\theta, \phi)$?

From the last expression I can get
$$
\theta = \arcsin\left(\frac{z}{r}\right)
$$
and then I guess from the first equation one could find
$$
\phi = \arccos\left(\frac{x}{R + r\cos(\theta)}\right)
$$
which works when the denominator is not zero. To avoid having the denominator being zero we need
$$
R+r\cos(\theta) \neq 0 \implies \theta \neq \arccos\left(-\frac{R}{r}\right)
$$

答:
I'm going to answer the question you meant to ask (and indeed, did ask in a comment to Michał Miśkiewicz's answer), namely:

And basically all I want to do now is, given my point (0,0,0) I want to find (0,0) so that I can plug them in the two expressions above to find the basis vectors of the tangent plane in terms of (,,)

I'm going to first rephrase that:

Given a point $P = (x, y, z)$ on a torus with main radius $R$ and tube radius $r$ (both known), find "the" basis for the tangent plane at $P$, where I mean the basis where one arrow points "around the tube" and the other points "around the big circle".

I confess that I'm only conjecturing that this is the basis you want, but it is the one given by the vectors $\frac{\partial{x,y,z}}{\partial \phi}$ and $\frac{\partial{x,y,z}}{\partial \theta}$, so that seems likely.

The second one is easiest: a vector pointing "around the big circle" at location $(x, y, z)$ is given by
$$
{\mathbf v}_\theta = \pmatrix{-y\\ x\\ 0}.
$$
That's easy to see if you look from above (i.e., along the $z$-axis) and consider the problem as being just in the $xy$-plane.

The first is slightly messier, but not much. If we let $\mathbf u$ be the unit vector pointing from the origin to $(x, y, 0)$, then the "tube slice" at $P$ lies in the $uz$ plane (with positive $u$ coordinate!). In the $uz$-coordinate system, your point $P$ lies at $(u, z) = (R + r\cos \phi, r \sin \phi)$ and by taking the derivative of this with respect to $\phi$, we get the vector (again in $uz$ coordinates)
$$
\pmatrix{-r\sin \phi \\ r \cos \phi}.
$$
The first coordinate here is just the negative $z$ coordinate of our point $P$; the second is the $u$-coordinate of $P$ (namely $\sqrt{x^2 + y^2}$), minus $R$, divided by $r$. So because
$$
\mathbf u = \frac{1}{\sqrt{x^2 + y^2}}\pmatrix{x\\y\\0},
$$
we get that
\begin{align}
\mathbf v_\phi
&= -z\mathbf u + \left(\frac{\sqrt{x^2 + y^2} - R}{r}\right)\hat{\mathbf z}\\
&= \frac{-z}{\sqrt{x^2 + y^2}}\pmatrix{x\\y\\0} + \left(\frac{\sqrt{x^2 + y^2} - R}{r}\right)\pmatrix{0\\0\\1}\\
\end{align}
Because you didn't ask for a unit vector, I guess I can multiply through by the constant $\sqrt{x^2 + y^2}$ and say instead that
\begin{align}
\mathbf v_\phi
&= {-z}\pmatrix{x\\y\\0} + \left(\frac{(x^2 + y^2) - R\sqrt{x^2 + y^2}}{r}\right)\pmatrix{0\\0\\1}\\
&= \pmatrix{-zx\\-zy\\ \frac{(x^2 + y^2) - R\sqrt{x^2 + y^2}}{r}}
\end{align}
instead.

Notice that the numbers $\theta$ and $\phi$ don't enter into this anywhere, so you don't need to worry about the domain of arctan, arccos, etc.


问:
Difficulty with Definition of Disconnectedness w.r.t. closed sets
Here's what I know:

When is a (metric) space disconnected?

Consider a metric space $(M,d)$. $M$ is disconnected if there exist non-empty disjoint open subsets $A,B \subset M$ such that $A\cup B = M$. $\{A,B\}$ is called a disconnection of $M$.

When is a subset of $M$ disconnected?$\color{red}{^1}$

Consider $E\subset M$. $E$ is disconnected if there exists non-empty disjoint open subsets $A,B \subset M$ such that $E\subset A\cup B$, $A\cap E\neq \varnothing$ and $B\cap E\neq \varnothing$.

I have seen that using the first definition, people often say that $M$ is disconnected if there exist non-empty disjoint closed sets such that $A\cup B = M$. This makes sense here, since $A^c = B$ and $B^c = A$, so if $A,B$ are open, they are also closed.

What about the second definition though, i.e. when is $E$ disconnected? Can we come up with something in terms of closed sets here?

My problem is that in the second definition, $A^c$ is not necessarily $B$ - so nothing useful comes out of that apparently. Could someone help me out here, and better clarify notions of connectedness/disconnectedness for me? Thank you very much.

Footnote:

$\color{red}{1.}$ In Carothers' Real Analysis, this characterization of disconnectedness of subsets of $M$ has been motivated by the relative metric, and notions of open sets in $E$.

答:
$E \backslash A$ is closed IN $E$ under the subspace topology since $E \backslash A = A^c \cap E$. Similarly with $B$. In particular, the same idea works for $E \subseteq M$, you just have to be careful with the subspace topology.

Another very useful definition for disconnectedness goes as follows: $X$ is disconnected if there exists a nonempty, proper subset of $X$ that is both open and closed.


答:
$E\subset M$ is disconnected if there exist two non empty sets $A,B$ such that $A\cap \bar B$ and $\bar A\cap B$ are empty such that $E=A\cup B$. Here, $\bar X =X\cup X'$, where $X'$ is the set of all limit points of $X$.

If $A$ and $B$ are closed, then $A'\subseteq A, B'\subseteq B $, so your definition follows from here.

Title: Difficulty with Definition of Disconnectedness w.r.t. closed sets

Body:
Here's what I know:

When is a (metric) space disconnected?

Consider a metric space $(M,d)$. $M$ is disconnected if there exist non-empty disjoint open subsets $A,B \subset M$ such that $A\cup B = M$. $\{A,B\}$ is called a disconnection of $M$.

When is a subset of $M$ disconnected?$\color{red}{^1}$

Consider $E\subset M$. $E$ is disconnected if there exists non-empty disjoint open subsets $A,B \subset M$ such that $E\subset A\cup B$, $A\cap E\neq \varnothing$ and $B\cap E\neq \varnothing$.

I have seen that using the first definition, people often say that $M$ is disconnected if there exist non-empty disjoint closed sets such that $A\cup B = M$. This makes sense here, since $A^c = B$ and $B^c = A$, so if $A,B$ are open, they are also closed.

What about the second definition though, i.e. when is $E$ disconnected? Can we come up with something in terms of closed sets here?

My problem is that in the second definition, $A^c$ is not necessarily $B$ - so nothing useful comes out of that apparently. Could someone help me out here, and better clarify notions of connectedness/disconnectedness for me? Thank you very much.

Footnote:

$\color{red}{1.}$ In Carothers' Real Analysis, this characterization of disconnectedness of subsets of $M$ has been motivated by the relative metric, and notions of open sets in $E$.


问:
For which rational fraction of pi the tangent function is rational?
We know all the fractions of $\pi$ for which $sin(\frac{k}{n} \pi)$ is rational thanks to Nievens theorem, same for cosine.

But I wonder if there is a similar theorem for the tangent $tan(\frac{k}{n} \pi)$?

Apart $tan(\frac{1}{4} \pi) = 1$ and $tan(0 \pi) = 0$, I fail to find any other case

The beautiful identities at https://mathworld.wolfram.com/Tangent.html all lead to algebraic irrational for small fractions of $\pi$. Is it a known result?

答:
Forget it, it's a case of blindness.

It's covered by Nievens theorem, as reported on wikipedia page https://en.wikipedia.org/wiki/Niven%27s_theorem


问:
Show the surface area of revolution of $e^{-x}$ is finite
I need to show that the surface area of revolution of $e^{-x}$ is finite when the region from $x=0$ to $x=\infty$ is rotated about the x-axis.

I tried using the surface area formula, but got stuck on the integration, so I thought that maybe there was a trick to show that it was infinite without actually finding the definite integral.

This is where I got with my integration:

$$
\begin{align}
SA &= 2\pi\, \int^b_a\, f(x)\sqrt{1+f'(x)^2}\, \mathrm{d}x \\
&= 2\pi\, \int^\infty_0\, e^{-x}\sqrt{1+e^{-2x}}\, \mathrm{d}x \\
&= \lim_{t \to \infty} 2\pi\, \int^t_0\, e^{-x}\sqrt{1+e^{-2x}}\, \mathrm{d}x \\
\end{align}
$$

$$
\begin{align}
\mathrm{let\,\,} u &= e^{-x}.\\
\mathrm{d}u &= -e^{-x}\, \mathrm{d}x\\
\mathrm{d}x &= \frac{-1}{e^{-x}} \,\mathrm{d}u\\
\mathbf{But:}\, u &= e^{-x}, \,\mathrm{so}\\
\mathrm{d}x &= \frac{-1}{u} \mathrm{d}u.
\end{align}
$$
When $x=0$, $u=1$ and when $x=t$, $u=e^{-t}$. However, as $t \to \infty$, $u \to 0$. (Is this a bad assumption to make here?) So,

$$
\begin{align}
SA &= -2\pi\int_1^0 \sqrt{1+u^2}\, \mathrm{d}u.
\end{align}
$$

This looks like it could be a nice enough integral, but I can't immediately see how to do it, and Wolfram|Alpha seems to use magic (and $\sinh^{-1}$, which is almost the same thing... =P)

Remember that I have only been asked to show the surface area is finite. I don't actually have to find the surface area. (But if you know a method that doesn't involve too many advanced tricks, I'm happy to hear it.)

Can we just say that the area under that curve is obviously finite, because it is the area bounded by a continuous curve between $x=0 $ and $x=1$? Should we perhaps say that this area is less than the area under $-2\pi \int_1^0 1+x^2\,\mathrm{d}x$, which is obviously finite, and thus this area itself is finite? What is the best way to answer this?

答:
We want to calculate the surface area $S(M)$ from $0$ to $M$, and see what happens to $S(M)$ as $M$ gets large.  You had done all the necessary work. We have
$$S(M)=\int_0^M 2\pi e^{-x}\sqrt{1+e^{-2x}}\,dx.$$
Note that $\sqrt{1+e^{-2x}}\le \sqrt{2}$ in our interval. So the function we are integrating is between $0$ and $2\sqrt{2}\pi e^{-x}$.

The integral
$$\int_0^\infty 2\sqrt{2}\pi e^{-x}\,dx$$
is an upper bound for $S(M)$, so
$$\int_0^\infty 2\pi e^{-x}\sqrt{1+e^{-2x}}\,dx$$
converges.

Remark: There are various ways to find explicit expressions for the integral. The most pleasant is through a hyperbolic function substitution. Quite a bit uglier, but doable, is your substitution, followed by $u=\tan t$.

Your substitution process that leads to $\int_0^1 \sqrt{1+u^2}\,du$ can be justitified. The part that made you justifiably nervous can be dealt with by noting that for $S(M)$  we are integrating from $u=e^{-M}$ to $1$.

And there is no issue of convergence for $\int_0^1 2\pi \sqrt{1+u^2}\,du$. We are integrating a continuous function on a bounded interval.


答:
A subset $E \subseteq M$ is disconnected exactly when $(E,d)$ in the relative metric (or as a topologist would say, the subspace topology) is disconnected.

We can also use the same reformulation as you have given but with closed sets instead.
So

$E$ is disconnected if there exists non-empty disjoint closed subsets $A,B \subset M$ such that $E\subset A\cup B$, $A\cap E\neq \varnothing$ and $B\cap E\neq \varnothing$.

The reason that this works is that $A \cap E$ and $B \cap E$ are then open and closed in $(E,d)$ just as in the open $A$,$B$ case.

Sidenote: for general (non-metric) spaces we don't demand that $A \cap B = \emptyset$ but $A \cap B \cap E = \emptyset$, to avoid weird examples). So the sets need only be disjoint on $E$, not in the whole space.


答:
@LordCommander did it the way that was asked, but it is actually a really simple variable substitution problem.  Original problem:

$$ y'' - 4y' = 5e^{-t} + 21 $$

Substitute $w = y'$:

$$ w' - 4w = 5e^{-t} + 21 \\
dw - 4w\, dt = 5e^{-t}\,dt + 21\,dt $$

Multiply by an integrating factor of $e^{-4t}$:

$$ e^{-4t}\, dw - 4 e^{-4t}\, dt = 5e^{-5t}\, dt + 21e^{-4t}\, dt \\
\int e^{-4t}\, dw - 4 e^{-4t}\, dt = \int 5e^{-5t}\, dt + 21e^{-4t}\, dt
 \\
w\,e^{-4t} = e^{-5t} - (1/4) 21 e^{-4t} + C \\
w = e^{-t} - (21/4) + Ce^4t $$

Now we can substitute $y' = w$ back in and solve again.

$$
y' = e^{-t} - (21/4) + Ce^4t \\
dy = e^{-t}\,dt - (21/4)\,dt + Ce^4t\,dt \\
\int dy = \int e^{-t}\,dt - (21/4)\,dt + Ce^4t\,dt \\
y = e^{-t} - (21/4)t + Ce^4t + D
$$

Title: Second Linear ODE

Body:
Confused on how I can approach this ODE, and I just need some help on figuring out what to do

$y''-4y'-21=5e ^{-t}$ with initial conditions $y(0)=0$ and  $y'(0)=4$

After Laplace transforming, I got:
$(\frac{(5)}{(s+1)}{+ 4+\frac{(21)}{(s+1)}}) /( s^2-4s) $

And now I am stuck on what to do here


问:
Double integrate the following function
$$ \int_{0}^{a} \int_{0}^{\sqrt{a^2-y^2}} xy(x^2+y^2)^\frac{3}{2}dxdy
  $$
any way to evaluate the integrand tried this with multiple substitution unable to evaluate

答:
First, the order of $dy \, dx$ is wrong.

Now regarding the solution, draw a picture of the region you are integrating. It is the first quadrant of the circle $\{(x,y) : x^2+y^2=a^2\}$. So, in polar coordinates $x=r \cos t$ , $y=r \sin t$ , $dx dy = r dr dt$ your integral is $$\int_0^a \int_0^{\pi/2} r \cos (t) r \sin (t) r^3 r dt dr = \left( \int_0^ar^6dr\right)\left(\int_0^{\pi/2}\sin(t)\cos(t) dt \right)=\frac{a^7}{7}\cdot \frac{1}{2} = \frac{a^7}{14}$$

Title: Double integrate the following function


问:
Help me visualize a Lindelöf space and a corresponding theorem
I am self studying Metric Spaces from the book by Satish Shirali and H.L. Vasudeva. I am having trouble in visualizing how a Lindelöf space looks like. The book defines it as:

A metric space is said to be Lindelöf if each open covering of $X$
contains a countable subcovering.

I am actually having difficulty in wrapping my head around the concept of a countable subcovering inside a covering. A relevant diagram regarding the same would be highly useful. Also, if possible , please give an example of a Lindelöf space and one of a non-Lindelöf space. It would be extremely appreciated.

Besides, there is a proposition regarding Lindelöf spaces, which seems to be  all the more abstract.

Proposition: Let $(X, d)$ be a metric space. If $X$ satisfies the second axiom of countability, then every open covering $\{U_a\}_{a \in \Lambda}$ of $X$ contains a countable subcovering. In other words, a second countable metric space is Lindelöf.

Proof: Let $\{G_i : i = 1, 2, \cdots\}$ be a countable base of open sets for $X$. Since each $U_a$ is a union of sets $G_i$ , it follows that a subfamily $\{G_{i_j} : j = 1, 2, \cdots \}$ of the base
$\{G_i : i = 1, 2, \cdots \}$ is a covering of $X$. Choose $U_{i_j}
  \supseteq G_{i_j}$ for each j. Then
$\{U_{i_j} : j = 1, 2, \cdots \}$ is the required countable subcovering.

Please help me understand this diagrammatically! I am completely lost.

Thank You !

答:
The objects studied  in Math often  come in many sizes,  some of which might be too big for
us to deal with.  So we end up introducing notions of smallness in order to rule out the scary monsters we're afraid of.

In Linear Algebra,  this is done e.g. by introducing finite dimensionality.  In Group Theory, finite groups are
often easier to study, and perhaps also the finitely generated ones.

In measure theory, people sometimes prefer to deal with measure spaces whose total measure is finite and, if that is
to restrictive, $\sigma $-finite measure spaces are considered instead.

In topology, the most restrictive size limitation is perhaps to stick to finite spaces, but that should probably be
thought of as size XXS (extra-extra-small),  and hence does not fit too many people!

The next size up the chart is perhaps occupied by  compact spaces, which indeed
present many of the features of finite sets.

Moving up a notch, one may look at the $\sigma $-compact spaces, namely those
which can be expressed as a countable  union of compact subsets, aka the $\sigma $-compact spaces.

If $X$ is a $\sigma $-compact space,  say
$$
  X=\bigcup_{n\in {\bf N}}K_n,
  $$
where the $K_n$ are compact subsets,  then every open
covering $\{U_i\}_{i\in I}$ of $X$ is also an open covering for each $K_n$,  so we may find finite subsets $I_n\subseteq I$, such that
the finite collection $\{U_i\}_{i\in I_n}$ covers $K_n$.  Clearly $J=\bigcup_{n\in N}I_n$ is countable and
$\{U_i\}_{i\in J}$ is a countable subcovering for $X$.

In other words, every $\sigma $-compact space is Lindelöf, so the latter may be thought of as the next size up the size
chart for topological spaces.


问:
Separation of Variable PDE
Consider the PDE :

$$u_{xxxx} + u_{xxyy} + u_{yyyy} = 0.$$

Let $u(x,y) = X(x)Y(y).$

After the working shown in this wikipedia article under the heading https://en.wikipedia.org/wiki/Separation_of_variables#Example:_mixed_derivatives , we have

$X''(x) = -\lambda_1X(x),\,\, X^{(4)}(x) = \mu_1X(x).$ My question is how did we get  $\mu_1 = \lambda_1^{2}?$

答:
You end up with
$$0 =\frac{X^{(4)}}{X} + \frac{ X^{(2)}}{X}\frac{Y^{(2)}}{Y} + \frac{Y^{(4)}}{Y},$$
where, as you say, $X$ is a function of $x$ and $Y$ is a function of $y$. Start with the middle term: by the usual argument you need
\begin{align}
X^{(2)} / X &= a, \\
Y^{(2)} / Y &= b
\end{align}
where $a$ and $b$ are constants. That means $X^{(2)} = aX$. Differentiating twice you get $X^{(4)} = aX^{(2)} = a(aX) = a^2 X$, and it follows the term on the left is $a^2$. Do the same thing for the term on the right and you end up with
$$ 0 = a^2 + ab + b^2.$$

Your problem probably has some conditions bounding the solution as $x$ or $y$ or both tend to $\infty$ and that'll limit your choices of $a$ and $b$.

Title: Separation of Variable PDE


问:
Let $f_n \in L_p(X, \mathbb{X}, \mu)$, $1 \leq p < \infty$, and let $\beta _n$ defined for $E \in \mathbb{X}$ by
Let $f_n \in L_p(X, \mathbb{X}, \mu)$, $1 \leq p < \infty$, and let  $\beta _n$ defined for $E \in \mathbb{X}$ by

$\beta _n = (\int_E |f_n|^pd \mu)^{\frac{1}{p}}$

and suppose that $f_n$ is a Cauchy sequence. If $\epsilon >0$, then there exists a $\delta(\epsilon) >0$ such that if $E \in \mathbb{X}$ and $\mu(E) < \delta(\epsilon)$, then $\beta_n (E) < \epsilon$ for all $n \in \mathbb{N}$.

Hint: Use that if $f \in M^+$, and $\lambda$ is defined on $\mathbb{X}$ by  $\lambda (E) = \int_E fd \mu$, then  $\lambda$ is a measure and is absolutely continuous with respect to $\mu$.

This is the 6.S exercise of Bartle the elements of integration and Lebesgue Measure. I thought it would only be an application from the previous exercise (6.R), but my teacher said that this exercise is very difficult, so I have no idea how to solve it. Someone can help

Note: I was able to solve the exercises below. Can these be applied in solving this?

答:
I came across this exercise in a measure theory and integration course and I actually solved it using only theorems, lemmas and corolaries from Bartle itself.

First, notice that the measures

$$\lambda_{n}(E) = \left\{ \int_{E} |f_{n}|^{p} d\mu \right\}^{1/p}, \ n \in \mathbb{N}$$

are absolutely continuous with respect to the measure $\mu$(this is a direct aplication of corolaries 4.9, 4.10 and 4.11).

Now, you have an equivalence of being absolutely continuous with respect to another measure $\mu$ given by:

Let $\lambda$ and $\mu$ be finite measures in a $\sigma$- algebra $\mathcal{A}$. Then $\lambda$ is absolutely continuous with respect to $\mu$ if and only if $\forall \varepsilon > 0$ there exists $\delta(\varepsilon) > 0$ such that $\forall X \in \mathcal{A}$ with $\mu(X) < \delta(\varepsilon)$ we have $\lambda(X) < \varepsilon$.

If you apply this equivalence you get that $\forall \varepsilon > 0$, $\exists X \in \mathcal{A}$ such that if $\mu(X) < \delta(\varepsilon)$ we have $\lambda_{n} < \varepsilon$, which translates as

$$\lambda_{n}(X) = \left\{ \int_{X}|f_{n}|^{p} \right\}^{1/p} < \varepsilon, \ \forall n \in \mathbb{N}$$

or,

$$\int_{X}|f_{n}|^{p} < \varepsilon^{p}, \ \forall n \in \mathbb{N}$$

as we wanted.

I strongly think that this equivalence is in Bartle but as I couldn't find it there so there you go!

To prove the equivalence mentioned, one of the sides is easy, let $\lambda$ satisy the condition, then if $\mu(X) = 0$ we have that $\lambda(X) < \varepsilon$ for every $\varepsilon > 0$, proving that $\lambda(X) = 0$ and so $\lambda$ is absolutely continuous with respct to $\mu$.

Now suppose that there exists $\varepsilon > 0$ and sets $X_{n} \in \mathcal{A}$ such that $\mu(X_{n}) < 2^{-n}$ and $\lambda(X_{n}) \geqslant \varepsilon$. Let $Y_{n} = \underset{k = n}{\overset{\infty}{\bigcup}}X_{k}$. That being said, notice that

$$\mu(Y_{n}) = \sum_{j = n}^{\infty} \mu(X_{j}) < \sum_{j = n}^{\infty}2^{-j} = 2^{-n} \sum_{l = 0}^{\infty}2^{-l} = 2^{-(n - 1)}$$

and $\lambda(Y_{n}) \geqslant \varepsilon$. As $(Y_{n})$ is a decreasing sequence of measurable sets we have that

$$\mu(\bigcap_{n = 1}^{\infty} Y_{n}) = \lim \mu(Y_{n}) = 0$$

and

$$\lambda(\bigcap_{n = 1}^{\infty} Y_{n}) = \lim \lambda(Y_{n}) \geqslant \varepsilon$$

that way, $\lambda$ is not absolutely continuous with respect to the measure $\mu$.


问:
Show that $2r^2-3$ is never a square, $r=2,3,...$
Show that $2r^2-3$ is never a square, $r=2,3,...$

I know that no perfect square can have $2, 3, 7$, or $8$ as its last digit. I'm not sure how to do this with congruence/mod notation. Any hints or solutions are greatly appreciated. I also tried to assume it was equal to a perfect square and through some algebraic manipulations arrive at a contradiction, but that proved to be difficult. I'm curious to know if it can be done algebraically.

答: Each perfect square is congruent to $0$, $1$ or $4$ modulo $8$. But each number of the form $2n^2-3$ is congruent to $5$ or $7$ modulo $8$.


问:
If $(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24]=(z+x-7)[y(z+x)+24]$, find $x^2+y^2+z^2$

Let x, y, z be pairwise distinct real numbers, if $$(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24] $$ $$=(z+x-7)[y(z+x)+24]$$, find $x^2+y^2+z^2$

I've tried many ways but couldn't find a working way to solve it. I tried letting $(x+y-7)[z(x+y)+24] = k$ and $t = x + y + z$ but none of these gives useful transformation as far as I can see. Could somebody shed some lights on this? Thanks in advance.

答: let $f_z=(x+y-7)(z(x+y)+24)$,$f_y,f_x$ defined similarly $$f_x=f_y=f_z=k$$ $$f_z-f_y=0\to -(y - z) (x^2 - 7 x - y z - 24)=0$$  so $$x^2-7x-yz=24$$and similarly for others yield  $$p_x=p_y=p_z=24$$ where $p_x=x^2-7x-yz$ and others defined similarly $$p_x-p_y=0\to (x-y)(x+y+z-7)=0$$ so $$x+y+z=7\to z=7-x-y$$ Plug this into orginal condition to get $$z^3-7z^2-24z-k=x^3-7x^2-24x-k=y^3-7y^2-24y-k=0$$ thus $x,y,z$ are the roots of cubic $$p(t)=t^3-7t^2-24t-k$$ Now by vieta you can easily find $x^2+y^2+z^2$


答:
Let $S=x+y+z$. Given is
$$P(x)=(S-x-7)\{x(S-x)+24\}$$
$$=x^3-x^2(2S-7)+x(S^2-7S-24)+24S-168$$
has same value as $P(y), P(z)$.

This means $x,y,z$ are roots of the polynomial, $P(x)+\text{constant}$.

By Vieta's, sum of roots is $S=2S-7 \Rightarrow S=7$. Also
$$xy+yz+zx=-(7^2-7\cdot 7 -24)=24$$

Hence $x^2+y^2+z^2=7^2-2\cdot 24=1$

Title: If $(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24]=(z+x-7)[y(z+x)+24]$, find $x^2+y^2+z^2$

Body:
Let x, y, z be pairwise distinct real numbers, if $$(x+y-7)[z(x+y)+24]=(y+z-7)[x(y+z)+24] $$ $$=(z+x-7)[y(z+x)+24]$$, find $x^2+y^2+z^2$

I've tried many ways but couldn't find a working way to solve it. I tried letting $(x+y-7)[z(x+y)+24] = k$ and $t = x + y + z$ but none of these gives useful transformation as far as I can see. Could somebody shed some lights on this? Thanks in advance.


答:
I write $u = y + z$, $v = z + x$, $w = x + y$ and $s = (u + v + w) / 2 (= x + y + z)$.

The equation becomes $$(u - 7)(us - u^2 + 24) = (v - 7)(vs - v^2 + 24) = (w - 7)(ws - w^2 + 24).$$

Taking the difference of the first and the second term, we get $$(u - v)((u + v - 7)s - (u^2 + uv + v^2 - 7(u + v) - 24)) = 0.$$

Since $x, y, z$ are all different, we know that $u, v, w$ are all different. This leads to $$(u + v - 7)s = u^2 + uv + v^2 - 7(u + v) - 24 \tag{1}$$ and symmetrically $$(v + w - 7)s = v^2 + vw + w^2 - 7(v + w) - 24.$$ Taking the difference of the above two identities, we get $$(u - w)s = (u - w)(u + w + v - 7)$$ which leads to $$s = 2s - 7$$ and thus $s = 7$.

Now (1) becomes $$u^2 + uv + v^2 - 14(u + v) + 25 = 0. \tag{2}$$ Taking the symmetric sum of (2), we get $$2\sum u^2 + \sum uv - 56s + 75 = 0. \tag{3}$$

If we write $t = \sum x^2$, then we have $\sum xy = (s^2 - t)/2 = (49 - t)/2$. It follows that $$\sum u^2 = \sum (y^2 + 2yz + z^2) = 2\sum x^2 + 2\sum xy = 49 + t$$ and $$\sum uv = ((2s)^2 - \sum u^2) / 2 = (147 - t) / 2.$$

Thus the equation (3) reads $$2(49 + t) + (147 - t) / 2 = 317$$ and we deduce that $t = 97$.

Finally, it is possible to give examples of $x, y, z$ satisfying these equalities.

E.g. $x = 3, y = 2 + 2\sqrt{10}, z = 2 - 2\sqrt{10}$.


答: $2r^2-3$ is odd, so let $(2y+1)^2=4y^2+4y+1=2r^2-3$, then $4y^2+4y=2r^2-4$ so $2\mid r$ giving $r=2s\to y^2+y=2s^2-1$.  But $y^2+y$ is always even and $2s^2-1$ is always odd, so there is no solution in integers $y,r$.

Body:
Show that $2r^2-3$ is never a square, $r=2,3,...$

I know that no perfect square can have $2, 3, 7$, or $8$ as its last digit. I'm not sure how to do this with congruence/mod notation. Any hints or solutions are greatly appreciated. I also tried to assume it was equal to a perfect square and through some algebraic manipulations arrive at a contradiction, but that proved to be difficult. I'm curious to know if it can be done algebraically.


问:
Disproving uniform convergence
Given a sequence of functions $f_n$ which is known to be pointwise convergent how would you go about showing that is not uniform convergent?

The particular example I'm working with is $f_n (x)=\frac {nx} {x^2+n^2}$ and I've tried using Theorem 7.9 from Rudin (uniform convergence $\iff M_n=(\sup |f_n(x)-f(x)|) \rightarrow 0$ as $n\rightarrow \infty$) but using that I found $M_n$ to be $\frac {\sqrt n} {1+n}$ by maximising $f_n$ but this gives the wrong result.

edit:
$f:\mathbb R \rightarrow \mathbb R $, sorry about that

答:
It looks like you've mis-computed $M_n$.

In general, let $g$ be a function continuous at $0$, and define $g_n(x)= g(x/n)$.

Then $g_n(x) \rightarrow g(0)$ as $n\rightarrow\infty$ for all $x$.

But $M_n=\sup_x |g_n(x)-g(0)| = \sup_x |g(x)-g(0)|=M_1$ for all $n$, so $M_n\rightarrow 0$ if and only if $M_1=0$ - that is, $g$ is constant.  So $g_n$ does not converge uniformly to $g(0)$ unless $g$ is constant.

In your case, $f_n(x) = \frac {x/n}{1+(x/n)^2}$

Title: Disproving uniform convergence


问:
Search for simpler (other) idea to show $(1.001)^{1000}>2$
The below question asked me to solve: (without using logarithm) which one is bigger $$2 \ , \ (1.001)^{1000}$$ I use two Idea to show $(1.001)^{1000}$ is bigger.

First: $$\lim_{n \to \infty}\left(1+\frac 1n\right)^n=e=2.71...\\$$and $1000$ is big enough to say $$(1.001)^{1000} \sim e>2$$

Second: Using binomial expansion $$(1.001)^{1000}=(1+0.001)^{1000}=\\1+1000\times 1^{999} \times (0.001) +\frac {1000\times 999}{2}1^{998}(.001)^2+...=\\1+1+\frac{999}{2000}+...>2$$   Now ,my question is there any simpler or other Idea to show $(1.001)^{1000}>2$

Thanks in advance for any hint.

Body:
The inequality is equivalent to

$$2^{\frac 1{1000}} <1+\frac 1{1000} $$

But this is true according to GM-AM:

$$2^{\frac 1{1000}} = \sqrt[1000]{1^{999}\cdot 2}\stackrel{GM-AM}{<}\frac{999+2}{1000} = 1+\frac 1{1000}$$


问:
What is the probability of a random COVID antigen test detecting someone with symptoms
I am trying to figure out the probability of an individual (let's call him Bob) returing a positive covid antigen result if he is a member of a group of 100 people being tested in groups of about 12 over 8 weeks.

Facts:

The Antigen test is most reliable for 5 days from when symptoms are
first displayed.

Symptoms are first displayed after 4 or 5 days of
being infected.

100 people are being tested in groups over 8 weeks
on working days only.

Each person only gets tested once in that period

The question is: What is the probablility of Bob returning a positive result for his test?

EDIT: I'll try to add more information (assumptions) here...

If somebody gets the virus, they would normally recover after 14 days.
The antigen test is only likely to return a positive result for this person when the viral load is greatest which from day 5 until day 10 of the infection.

So the probability of the antigen test giving a positive result for somebody infected is 5/14 (the 14 days is probably irrelevant)

Let's make the assumption that Bob does actually get infected at some point during the 8 week testing period.

In the 8 weeks, Bob would only get tested on working days. That's 40/56.

Working days on which he could be tested = 40/56
Days on which he could test positive if he had the virus 5/56

So do I now multiply these two numbers 40x5 / 56x56?
That would be approx 1/16

Where I think this may be wrong:
Weekends are obviously 2 says. Does it make any difference on which day Bob  gets infected. Let's say he gets it on a Saturday. The Antigen test would detect the virus from Thursday. He could be tested positive on Thursday, Friday or Monday. Does that make any difference to the 40/56 days? Should I be taking the 14 days of infection into account?

What about the days that he would test negative - does that have any influence on the maths?

答:
As stated, your question does not give enough information to answer the question. I will try to give an answer that encompasses a few different interpretations, and hopefully makes it clearer which information is needed to give an exact answer to the problem.

As I understand your post, the testing setup is as follows:

If a person is infected on day $n$, they will return a positive antigen test on days $n+5, n+6, \ldots, n+10$ if they are tested on any of those days, and will otherwise get a negative result. (Perhaps you meant "day $5$" to be four days after "day $1$", the date of infection? It doesn't really affect the results, though.)

Bob will be tested exactly once over an $8$-week period, and the date of the test will be selected uniformly at random from one of the $40$ weekdays in that period.

The above setup is enough for us to assess the probability that Bob will receive a positive antigen test, if we know the probability of infection on a given day. Without this information, it is impossible to give an answer: if Bob does not plan to interact with another human being during the 8-week period, his probability of a positive test is effectively zero, and if he deliberately times his infection to overlap with an entire work week, his probability could be as high as $5/40=1/8$. (It cannot be higher than this, if he only  gets infected once.) Again, to re-emphasize, you need to specify the probability Bob gets infected on a given day if you want this problem to have a well-defined answer.

Suppose that we are given such a series of probabilities for each day: $p_i$ is the chance of infection on day $i$.

If we start our numbering of the eight weeks with Monday as day $0$, and the final Sunday as day $55$, then the chance of being tested on day $n$ is

$$f(n)=\begin{cases}
\frac1{40} & n\equiv 0,1,2,3,4 \pmod{7}\\
0 & n\equiv 5,6 \pmod{7}
\end{cases}$$

The chance of being at a stage of infection to return a positive result if tested on day $n$ is

$$p_{n-5}+p_{n-6}+\ldots+p_{n-10}$$

Then, we can compute

$$P(\text{positive test on day $n$}) = P(\text{gets tested on day $n$}) \cdot P(\text{positive test on day $n$}\ |\ \text{gets tested on day $n$}) $$
$$= f(n)\cdot (p_{n-5}+\ldots+p_{n-10})$$

Now, because we know that there is at most one positive result, the total probability of a positive result is just the expected number of positive results, or the sum of the probabilities of a positive test on each day (by linearity of expectation). If there were more than one test given over the 8-week period, the calculations would be somewhat more involved.

So our total probability is

$$\sum_{n=0}^{55}\left(f(n)\sum_{i=n-10}^{n-5}p_i\right)$$

You state in your post

Let's make the assumption that Bob does actually get infected at some point during the 8 week testing period.

If we interpret this as meaning that $p_i=\frac1{56}$ for all $i$ from $0$ to $55$, and $p_i=0$ otherwise, then the the probability turns out to be $\frac3{28}\approx 0.107143$.

If we interpret this as meaning that $p_i=\frac1{61}$ for all $i$ from $-10$ to $50$, and $0$ otherwise (i.e. Bob is infected at a random day from among the days where he might be detected by an antigen test at some point during the testing period), then the resulting probability is $\frac{261}{2440}\approx 0.10697$. (These numbers are not very different, which should be expected since the setups are quite similar.)

However, note that our model of choosing the probability of infection on each day separately is a little misleading. If we ignore the relatively small chance of reinfections over the course of the $8$-week period, then it probably makes more sense to think about the probability $q_i$ that Bob is exposed to a source that would have infected him if he were susceptible, but will do nothing if he has already been infected. Furthermore, let's suppose that Bob has some initial chance $q'$ of having been exposed already on day $0$.

Then, we can obtain our series of $p_i$s from this: the chance of getting infected on day $i$ is the probability of exposure $q_i$, multiplied by the combined probabilities of not being exposed on day $0$ and of avoiding exposure each previous day. So we have

$$p_i = q_i\cdot (1-q')\cdot\prod_{j=0}^{i-1}(1-q_j)$$

(We can use a similar formula if we want to model this from an earlier starting point before day $0$.)

I have written some simple Python code here, which you can modify to customize some of the assumptions in the problem. I am still not sure exactly what the question you want to answer is, but I hope that this will help you in pursuit of it.


问:
What is a closed form of $\int_0^1\ln(-\ln x)\ \text{li}\ x\ dx$
Let $\operatorname{li} x$ denote the logarithmic integral:
$$\operatorname{li} x=\int_0^x\frac{dt}{\ln t}.$$
Is it possible to find a closed form of the following integral?
$$\int_0^1\ln(-\ln x) \operatorname{li} x\ dx$$
I tried integration by parts, but only ended up with more complicated expressions.</p>

答:
This integral can be expressed in a closed form:
$$\int_0^1\ln(-\ln x)\operatorname{li}x\ dx=\frac{\pi^2}{12}+\frac{(\ln2)^2}2+\gamma\,\ln2,\tag1$$
where $\gamma$ is the Euler-Mascheroni constant.

Proof:

Let
$$\mathcal{I}(a)=\int_0^1(-\ln x)^a\operatorname{li}x\ dx,\tag2$$
then, taking a derivative with respect to $a$, we get
$$\mathcal{I}'(0)=\int_0^1\ln(-\ln x)\operatorname{li}x\ dx.\tag3$$
Mathematica can express the integral $(2)$ in a closed form for explicitly given non-negative integer values of $a$, but not in a general form with a parameter. We could use those values to get a conjectured general formula, and indeed one could get the result $(1)$ this way. Numerical integration shows that
$$\Bigg|\frac{\pi^2}{12}+\frac{(\ln2)^2}2+\gamma\,\ln2-\int_0^1\ln(-\ln x)\operatorname{li}x\ dx\Bigg|<10^{-1000}\tag4,$$
so the result looks plausible.

But here we will follow a different approach to get a rigorous proof (modulo possible bugs in Mathematica tuned exactly to hide that tiny difference). Take the formula $(12)$ from the MathWorld article:
$$\operatorname{li}x=\operatorname{Ei}(\ln x),\tag5$$
where $\operatorname{Ei}z$ is the exponential integral, and plug it into $(2)$. In this form the integral can be evaluated in Mathematica, that gives us
$$\mathcal{I}(a)=-\frac{_2F_1(a+1,a+1;a+2;-1)\ \Gamma(a+1)}{a+1},\tag6$$
and, taking a derivative with respect to $a$,
$$\mathcal{I}'(0)=(1+\gamma)\ln2-\left[\frac{d}{da}{_2F_1}(a+1,a+1;a+2;-1)\right]_{a=0}.\tag7$$
To find the derivative of the hypergeometric function we need to replace it with its definition (formula $(8)$ in this MathWorld article):
$$_2F_1(a+1,a+1;a+2;-1)=\sum_{n=0}^\infty\frac{(a+1)\ \Gamma(a+n+1)}{(a+n+1)\ \Gamma(a+1)\ n!}(-1)^n.\tag8$$
So,
$$\left[\frac{d}{da}{_2F_1}(a+1,a+1;a+2;-1)\right]_{a=0}=\sum_{n=0}^\infty\frac{(n+1)(\gamma+\psi(n+1))+n}{(n+1)^2}(-1)^n,\tag9$$
where $\psi(n+1)=\frac{\Gamma'(n+1)}{\Gamma(n+1)}=-\gamma+\sum_{m=1}^n\frac1m\,$ is the digamma function. Evaluating this sum with Mathematica we get
$$\left[\frac{d}{da}{_2F_1}(a+1,a+1;a+2;-1)\right]_{a=0}=\ln2-\frac{(\ln2)^2}2-\frac{\pi^2}{12}\tag{10}.$$
Plugging this back in $(7)$ and $(3)$ we get the final result $(1)$.


问:
How do integrals add up all the areas within their bounds?
Before I proceed,
I am a high school student and have just started learning calculus.

OK

Now, I understand that an integral is basically the opposite of a derivative. I also understand that it finds the area under the graph of a function.

Based on what I understood, while integrating you take a very small change in the variable $x$ (or maybe $\mathrm{d}x$) such that the change in $f\left(x\right)$ is constant and therefore calculate the area of the rectangle and add the areas...

However, my question here is how does the integration tool include all the areas
bound under its limit, it makes less sense to me.
For example:
$$\int_2^4 x\,\mathrm{d}x$$

Solving the integral:

You get
$$\frac{x^2}{2}$$

with limits $4$ and $2$.
So, now you substitute the limits and subtract them .....
but this is what feels weird, it looks like I'm only subtracting the area at $4$ and $2$,
How does this tool work

(Any response would be helpful)
EDIT:
(i want to make my question a bit more clear)

[My question is basically, how does integration include all the areas between the interval, i have seen a couple of proofs on the fundamental theorem of calculus
however, that just seemed to explain why we subtract while integrating (to remove the constant term (i think).The problem i guess i am facing is that there is no hint of a summation of areas happening,and this is exactly what my question is -how do we know the areas are included??]

答:
The way to understand this is to step back from thinking about integrals as areas.  Fundamentally, integrals are not areas, they are infinite sums of infinitely small values.  To understand how it works, let's look at the opposite direction - differentials.  Take the equation:

$$ y = x^2 $$

Now, let's differentiate it:

$$ dy = 2x\,dx $$

On the left, you have an infinitely small change in $y$.  On the right, you have infinitely small values, which are also equivalent to the same infinitesimal change in $y$.

Now, if you add up all of the infinitesimal changes in $y$ between two $x$ values, you will get the TOTAL change in $y$, correct?  Therefore, if you had the original function, you could just subtract the $y$ values at those points to get what the total change is.

A way of stating that mathematically is:

$$ \int_{x_0}^{x^1} dy = y(x_1) - y(x_0) $$

So that's what an integral is.  It is the sum of all of the infinitely small changes of an expression.  So how does it relate to areas?  Let's go back to your problem, where you want the area under the curve $y = x$.

With areas, an area of a rectangle is width times height.  So, each area will be $dx$ wide and $y$ high.  Therefore, each rectangle will be:

$$ y\,dx $$

Now, we can imagine a function $z$ such that its differential, $dz$, is the same value as this area.  So we set:

$$ dz = y\,dx $$

So, using the reasoning we used before, we know that the TOTAL of all of $dz$ is going to be equal to the difference between $z$ evaluated at $x_0$ and $x_1$.  In other words:

$$ \int_{x_0}^{x_1} dz = \int_{x_0}^{x_1} y\,dx = \int y\,dx\, \biggr|_{x_0}^{x_1} $$

So, since $z(x_1) - z(x_0)$ is equal to the sum of the areas, so is $ \int y\,dx\, \biggr|_{x_0}^{x_1}$.

Now, we can't integrate $y\,dx$ directly, but, thankfully, we have a formula for $y$, which is $y = x$.  Therefore, this becomes:

$$ \int_{x_0}^{x_1} dz = \int_{x_0}^{x_1} y\,dx = \int y\,dx\, \biggr|_{x_0}^{x_1} = \int x\,dx\, \biggr|_{x_0}^{x_1} = \frac{x^2}{2} \biggr|_{x_0}^{x_1} $$

So, what we did was:

Recognize that the integral is an infinite sum of infinitely small pieces

Recognize that the difference between two points on a function is equal to the infinite sum of the differentials of that function.

Because of (2), integrating differentials is the same as performing an infinite sum of them.

If we can write the formula for an infinitely small area as a differential, we can use (1)-(3) to generate a formula for a function such that the function's change is equal to the sum of the infinitely small values between those points.


问:
strict order property
I wanted to prove that the theory of ordered abelian group has a strict order property.
I know by the theory of Kikyo and Shellah we have:

a theory is unstable iff it has SOP or NIP

and by the :
https://www.jstor.org/stable/1999281?seq=1

the ordered abelian group does not have IP but is there any way to prove SOP directly?

答:
It is immediate from the definition that the formula $x\leq y$ witnesses SOP in the theory of ordered abelian groups. For example, in $\mathbb{Z}$, look at the family of sets defined by $x\leq n$ for various $n$. These sets form an infinite chain of proper containments.

In  fact, the definition of SOP is a direct generalization of the property of having a definable linear order (which is infinite in some model). More precisely, a theory has SOP iff it has a definable preorder with infinite strict chains (in some model).

Also, a correction. You wrote "I know by the theory of Kikyo and Shellah we have:"

A theory is unstable iff it has SOP or NIP

This is a theorem of Shelah. It appeared in Classification  Theory in 1978. There is also a well-known theorem of Kikyo and Shelah (which appeared in their only joint paper, in 2002), but that theorem says that if $T$ is an SOP theory, then the theory of models of  $T$ expanded by an automorphism has no model companion.

Title: strict order property


问:
Find an invertible matrix $Q \in M_3(\mathbb{R})$ such that $Q^{-1}AQ$ is a upper triangular matrix

Let be  \begin{align} A= \begin{bmatrix} 1 & 1 & 0\\  0 & 1 & 0\\  1 &
 1 & 1 \end{bmatrix} \in M_3(\mathbb{R}) \end{align}

a) Find the minimal polynomial of $A$

b) Determine if $A$ is diagonizable over $\mathbb{R}$

c) Find an invertible matrix $Q \in M_3(\mathbb{R})$ such that $Q^{-1}AQ$ is a upper triangular matrix

My attempt:

a) First we find the characteristic polynomial
\begin{align}
det (A-tI)=(1-t)^{3}
\end{align}
Then, the minimal polynomial of $A$ is $(1-t)$

b) To verify if $A$ is diagonalizable, we have to verify if the dimension of the eigenspace associated to the eigenvalue 1, is 3. That is, to check if the algebraic multiplicity is equal to the geometric multiplicity of the eigenvalue 1.

\begin{align} \begin{pmatrix} 1-1 & 1 & 0\\  0 & 1-1 & 0\\  1 &
 1 & 1-1 \end{pmatrix}\begin{pmatrix}x\\ y\\z\end{pmatrix}&= \begin{pmatrix} 0 & 1 & 0\\  0 & 0 & 0\\  1 &
 1 & 0 \end{pmatrix}\begin{pmatrix}x\\ y\\z\end{pmatrix}=\begin{pmatrix}0\\ 0\\0\end{pmatrix} \\ \end{align}

\begin{align}
\Longrightarrow \begin{pmatrix}x\\ y\\z\end{pmatrix}=\begin{pmatrix}0\\ 0\\0\end{pmatrix}
\end{align}

$\Longrightarrow$ The dimension of the eigenspace associated to the eigenvalue 1, is zero. Then, $A$ is not diagonalizable.

My doubts:

Am I correct in a) and b)?

How can I find the matrix of c)? I would really appreciate your help

答:
For $(a)$, it's easy to check by inspection that your minimal polynomial is not correct.

For $(b)$, an eigenvector associated to the eigenvalue $1$ must satisfy $Av = v$, or $(A-I)v = 0$ as you put it.
If $v = (x,y,z)$ then $(A-I)v = (y, 0, x+y)$, so $(A-I)v = 0$ implies $y = 0$ which then implies $x+y = x = 0$.
Notice that this leaves $z$ free, and indeed it's easy to see $(0,0,1)$ is an eigenvector.

Of course, that still means $A$ is not diagonalizable.

For $(c)$, if $A$ were diagonalizable, say $A = QDQ^{-1}$ with $D$ diagonal, then you could write $D = Q^{-1}AQ$ and since $D$ is diagonal it is also upper triangular.

Now, you've shown in $(b)$ that $A$ is not diagonalizable, but every matrix has a certain normal form that one could say is the next best thing.
Do you think you can take it from here?


问:
Math Olympiad question: Quadratic Equations.

Math Olympiad question:

If the equation in $x$ has real roots, then find the value of $a$ and $b$.
$x^2 + 2(1 + a)x + (3a^2 + 4ab + 4b^2 + 2) = 0$

Approach:

For at least one real root:

$$b^2 -4ac \ge 0$$

So:

$$[2(1+a)]^2 - 4(3a^2 +4ab+4b^2+2)(1)\ge 0$$

$$1+a^2+2a - 3a^2-4ab-4b^2-2\ge 0$$

$$a^2 +2ab+2b^2 -a+1\le 0$$

$$(a+b)^2 +b^2-a+1\le 0$$

What to do further? I don't think a graphical approach would work, and I'm also not able to find the value which might satisfy the above. Is there a better method?

Also I'm not sure if the last step helps.

答: You have an error in the line $a^2 +2ab+2b^2 -a+1\le 0$. It should be $$a^2 +2ab+2b^2 -a+\frac 1 2 \leq 0$$ and it can be rewritten as $$\frac12 (a + 2b)^2 + \frac12 (a - 1)^2 \leq 0.$$ Now you should know what to do.


问:
Book recommendation for formal, proof-based, theoretical books on time series and panel data
I'd like to ask you if there are any formal, proof-based and mainly theoretical books on time series and panel data. When searching for books on time series and panel data I got really good recommendations but all the books were more leaning towards the applied aspect. Either the books focused more on the computer science aspect (how to plot time series, how to execute Dickey-Fuller test, how to run regressions on panel data) accompanied with intuition but there are no proofs. Sometimes basic stuff is proven (like AR(1) is stationary if and only if the coefficient next to the laggued variable is between -1 and 1).

But, I've already seen a similar question asked here and the only recommended book that meets my needs is this one : Time Series Theory and Methods by Peter J. Brockwell, Richard A. Davis. The one that suggested this book said, I quote, "This used to be (still is?) the main reference for Time Series back in the day for those theoretically inclined." And so I'd like to know if there are any other books of the sort that are more recent or, even if they're old but were heavily used back in the days.

Thank you in advance.

答:
I am not really in a position to answer this question, since I was about to ask something similar when I stumbled upon it...

Also, I am not entirely clear on what you are looking for exactly...

And not knowing the subject very well, I do not know what I am looking for exactly either...

Having said that, let's say we are interested in mathematical treatments of stochastic processes indexed by $\mathbb{Z}$ which are stationary in the $L^2$ sense (i.e. covariance stationary or wide-sense stationary).

Then I have selected these three books

Shiryaev - Probability (Chapter 6)

Koralov & Sinai - Theory of Probability and Random Processes (Chapter 15)

Gikhman & Skorokhod - The Theory of Stochastic Processes I (Chapter 4)

There are really a lot of books trating these topics from the point of view of a statistician, engineer, economist and so on. But I find them so (mathematically) sloppy
that I basically cannot read them.

I will try to build-up some knowledge on the subject and hopefully add suggestions in the future.

In the meantime I hope this helps. Let me know.


问:
Integration notation - integrating over $k'$ instead of $k$
My physics professor has defined a Fourier transform as:
$$\tag{1}
D_{i}(\boldsymbol{r}, t)=\frac{1}{(2 \pi)^{2}} \iiiint D_{i}\left(\boldsymbol{k}^{\prime}, \omega^{\prime}\right) e^{i\left(\boldsymbol{k}^{\prime} \cdot \boldsymbol{r}-\omega^{\prime} t\right)}\, d \omega^{\prime}\, d^{3} k^{\prime}
$$
with the inverse transform of eq. $(1)$ being:
$$\tag{2}
D_{i}(\boldsymbol{k}, \omega)=\frac{1}{(2 \pi)^{2}} \iiiint D_{i}(\boldsymbol{r}, t) e^{-i(\boldsymbol{k} \cdot \boldsymbol{r}-\omega t)}\, d t\, d^{3} r
$$
What is the purpose of using the notation $\omega'$ and $k'$ in eq. $(1)$ while simply integrating over $t$ and $r$ in eq. $(2)$?

答: It's all about not conflating a free variable with a bound dummy integration variable. If (1) holds, (2)'s RHS is$$\begin{align}&\int_{(\Bbb R^4)^2}D_i(k^\prime,\,\omega^\prime)\frac{e^{i((k^\prime-k)\cdot r-(\omega^\prime-\omega )t)}}{(2\pi)^4}dtd^3rd^3k^\prime d\omega^\prime\\&=\int_{\Bbb R^4}D_i(k^\prime,\,\omega^\prime)\delta^{(3)}(k^\prime-k)\delta(t^\prime-t)d^3k^\prime d\omega^\prime\\&=D_i(k,\,\omega).\end{align}$$


问:
Two closed subsets $A$ and $B$ in $\mathbb{R}$ with $d(A,B)=0$
I am looking for two closed subsets A and B (with $A\cap B = \emptyset$) of $\mathbb{R}$ with $d(A,B)=0$. I found a solution in $\mathbb{R}^2$, namely $A=\{(x,\frac{1}{x})\mid x>0\}$ and $B=\{(x,0)\mid x>0\}$. I know that those subsets have to be unbounded because there is a theorem that says: the distance between two closed subsets of which one is bounded, is greater than 0.</p>

答: Set $A=\mathbb{N}\setminus \{1\}$ and  $B=\{x+\frac{1}{x}:x\in A\}$</p>


答: You got the right characteristic polynomial. But the dimension of the eigenspace associated with an eigenvalue is always greater than $0$. In this case, it is $1$: that eigenspace is spanned by $(0,0,1)$. So, let $e_1=(0,0,1)$. Now, you can check that a solution of $A.(x,y,z)=(x,y,z)+e_1$ is $(1,0,0)$. So, let $e_2=(1,0,0)$. And you can check that a solution of $A.(x,y,z)=(x,y,z)+e_2$ is $(-1,1,0)$. So, let$$Q=\begin{bmatrix}0&1&-1\\0&0&1\\1&0&0\end{bmatrix}$$(the columns of $Q$ are $e_1$, $e_2$, and $e_3$), and$$Q^{-1}.A.Q=\begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}.$$

Body:
Let be  \begin{align} A= \begin{bmatrix} 1 & 1 & 0\\  0 & 1 & 0\\  1 &
 1 & 1 \end{bmatrix} \in M_3(\mathbb{R}) \end{align}

a) Find the minimal polynomial of $A$

b) Determine if $A$ is diagonizable over $\mathbb{R}$

c) Find an invertible matrix $Q \in M_3(\mathbb{R})$ such that $Q^{-1}AQ$ is a upper triangular matrix

My attempt:

a) First we find the characteristic polynomial
\begin{align}
det (A-tI)=(1-t)^{3}
\end{align}
Then, the minimal polynomial of $A$ is $(1-t)$

b) To verify if $A$ is diagonalizable, we have to verify if the dimension of the eigenspace associated to the eigenvalue 1, is 3. That is, to check if the algebraic multiplicity is equal to the geometric multiplicity of the eigenvalue 1.

\begin{align} \begin{pmatrix} 1-1 & 1 & 0\\  0 & 1-1 & 0\\  1 &
 1 & 1-1 \end{pmatrix}\begin{pmatrix}x\\ y\\z\end{pmatrix}&= \begin{pmatrix} 0 & 1 & 0\\  0 & 0 & 0\\  1 &
 1 & 0 \end{pmatrix}\begin{pmatrix}x\\ y\\z\end{pmatrix}=\begin{pmatrix}0\\ 0\\0\end{pmatrix} \\ \end{align}

\begin{align}
\Longrightarrow \begin{pmatrix}x\\ y\\z\end{pmatrix}=\begin{pmatrix}0\\ 0\\0\end{pmatrix}
\end{align}

$\Longrightarrow$ The dimension of the eigenspace associated to the eigenvalue 1, is zero. Then, $A$ is not diagonalizable.

My doubts:

Am I correct in a) and b)?

How can I find the matrix of c)? I would really appreciate your help


问:
Proving Young's inequality for n numbers whitout using AM-GM inequality

Let $\alpha_1, ..., \alpha_n \geq 0$, and $p_1,...,p_n \geq 0$ such
that $\sum_{i=1}^{n} p_i^{-1} =1$. Show that $\alpha_1...\alpha_n \leq p_{1}^{-1}\alpha_1^{p_1}+...+p_{n}^{-1}\alpha_n^{p_n}$.

It is Young's inequality for n numbers. I know the proof that uses AM-GM, but in this case, I can't use it.

My attempt: I tried to solve it by induction, I suppose that it was true for $n-1$ numbers and summed all the possibilities to obtain $\sum_{k=1}^{n} \prod_{i \in ([k]-\{k\})} \alpha_i \leq (n-1)(p_{1}^{-1}\alpha_1^{p_1}+...+p_{n}^{-1}\alpha_n^{p_n})$, but don't seem to be that the left right hand is greater or equal to $\alpha_1...\alpha_n$.

答:
The usual way is by Jensen's, but you can do this by induction without Jensen's or AM-GM if you want as follows:

Two variables is obvious from non AM-GM means, e.g., consider areas with the curve $x_2=x_1^{p_1-1}$ (which is the same as $x_1=x_2^{p_2-1}$) on nonnegative reals.

Then inductively, suppose we have prove the $n$-variable case for all $n\leq N$: all $p=(p_1,\dots,p_n)$ and all $\alpha=(\alpha_1,\dots,\alpha_n)$ satisfying the condition we have Young's inequality holds.
To prove the case for the $N+1$ variable case $p=(p_1,\dots,p_N,p_{N+1})$ and $\alpha=(\alpha_1,\dots,\alpha_{N+1})$ (so $N\geq 2$), first use the two-variable case on $\tilde{p}=(p_N/C, p_{N+1}/C)$ and $\tilde{\alpha}=(\alpha_N^C,\alpha_{N+1}^C)$ (where the constant $C=1/(p_N^{-1}+p_{N+1}^{-1})$ is chosen to make sure the hypothesis is satisfied):
$$
\alpha_N^C\alpha_{N+1}^C\leq Cp_N^{-1}\alpha_N^{p_N}+Cp_{N+1}^{-1}\alpha_{N+1}^{p_{N+1}}
$$
and so with the $N$-variable case $\alpha'=(\alpha_1,\dots,\alpha_{N-1},\alpha_N\alpha_{N+1})$ and $p'=(p_1,\dots,p_{N-1},C)$:
\begin{align*}\require{color}
\alpha_1\dots\alpha_{N-1}\alpha_N\alpha_{N+1}
&\leq p_1^{-1}\alpha_1^{p_1}+\dots+p_{N-1}^{-1}\alpha_{N-1}^{p_{N-1}}+C^{-1}(\alpha_N\alpha_{N+1})^C\\
&\leq p_1^{-1}\alpha_1^{p_1}+\dots+p_{N-1}^{-1}\alpha_{N-1}^{p_{N-1}}+C^{-1}\color{red}{(Cp_N^{-1}\alpha_N^{p_N}+Cp_{N+1}^{-1}\alpha_{N+1}^{p_{N+1}})}\\
&=p_1^{-1}\alpha_1^{p_1}+\dots+p_{N-1}^{-1}\alpha_{N-1}^{p_{N-1}}+p_N^{-1}\alpha_N^{p_N}+p_{N+1}^{-1}\alpha_{N+1}^{p_{N+1}}
\end{align*}
as desired.


问:
Help me visualize a Lindelöf space and a corresponding theorem
I am self studying Metric Spaces from the book by Satish Shirali and H.L. Vasudeva. I am having trouble in visualizing how a Lindelöf space looks like. The book defines it as:

A metric space is said to be Lindelöf if each open covering of $X$
contains a countable subcovering.

I am actually having difficulty in wrapping my head around the concept of a countable subcovering inside a covering. A relevant diagram regarding the same would be highly useful. Also, if possible , please give an example of a Lindelöf space and one of a non-Lindelöf space. It would be extremely appreciated.

Besides, there is a proposition regarding Lindelöf spaces, which seems to be  all the more abstract.

Proposition: Let $(X, d)$ be a metric space. If $X$ satisfies the second axiom of countability, then every open covering $\{U_a\}_{a \in \Lambda}$ of $X$ contains a countable subcovering. In other words, a second countable metric space is Lindelöf.

Proof: Let $\{G_i : i = 1, 2, \cdots\}$ be a countable base of open sets for $X$. Since each $U_a$ is a union of sets $G_i$ , it follows that a subfamily $\{G_{i_j} : j = 1, 2, \cdots \}$ of the base
$\{G_i : i = 1, 2, \cdots \}$ is a covering of $X$. Choose $U_{i_j}
  \supseteq G_{i_j}$ for each j. Then
$\{U_{i_j} : j = 1, 2, \cdots \}$ is the required countable subcovering.

Please help me understand this diagrammatically! I am completely lost.

Thank You !

答:
Ruy already said a lot about motivation that I won't repeat here, but anyway, having a countable subcover (instead of a finite one, as in the definition of compact) for every open cover (which could be huge!) is the "next best thing", it allows us to work with countable covers for which we can do recursion arguments, make sequences etc. And analysis prefers countable things (sequences, series, sigma-finite measures) because most tools are built for that.

A typical non-Lindelöf metric space is $\Bbb R$ in the discrete metric $d(x,y)=1$ when $x \neq y$, then $B(x,1)= \{x\}$ for all $x$ and the open cover $\{x\}, x \in \Bbb R$ of the space has no proper subcover at all: if we omit some $\{p\}$ then $p$ won't be covered anymore, and as $\Bbb R$ is uncountable, this open cover has no countable subcover.

For metric spaces we have that being Lindelöf is equivalent to being separable (having a countable dense subset) and to being second countable. So most of the favourite spaces in analysis are indeed Lindelöf.

The theorem's proof is somewhat chaotic and badly written.
Suppose $X$ is a second countable space, with countable base $\mathcal{B}= \{B_n\mid n \in \Bbb N\}$.
Let $\mathcal{U}$ be any open cover of $X$.
For each $x \in X$ there is some $U_x \in \mathcal{U}$ with $x \in U_x$. This uses that we have a cover. As $U_x$ is open and $x \in U_x$ there is some $n(x) \in \Bbb N$ such that $x \in B_{n(x)} \subseteq U_x$. This uses that $\mathcal{B}$ is a base.

Now let $M=\{n(x)\mid x \in X\}$ which is a countable set as a subset of $\Bbb N$.
For each $m \in M$ we pick $x_m$ so that $n(x_m)= m$ and then we note that

$\{U_{x_m}\mid m \in M\}$ is a countable subcover of $\mathcal{U}$. That it's a subset of $\mathcal{U}$ is clear by definition, so let $p \in X$ be arbitrary. Then we have $p \in B_{n(p)} \subseteq U_p$ "by construction", and $n(p) \in M$ by definition. So $n(p) =: m_0 = n(x_{m_0})$ also by construction but then $p \in B_{n(p)} = B_{n(x_{m_0})} \subseteq U_{x_{m_0}}$, so that $p$ is indeed covered.

Bit technical index juggling with some have use of AC (axiom of choice) but the idea is clear: we just use countably many base elements in a refinement of the cover and we use some expansion of that refinement as subcover. Your proof had sort of the same idea, but in other notation.


问:
Linear Transformation, if $T(x) \neq 0$ for some $x \in V_1$, prove that $Tx \neq 0 \iff x \neq 0$
I don't know if it's true, but it seems to be, altought I cannot prove it yet.

I'm trying to find a way of write any other vector as a sum of $\textbf{x}$ with another vector (I don't know if it will work because I don't know if $T\textbf{y} = 0, \textbf{y} \neq \textbf{x}$) and multiplication by scallar, but don't know how.

I think it's pretty simple but I can't figure out.

答: This is false.  All you need for a counterexample is a nonzero linear transformation which is not injective.  For instance, the one whose matrix rel the standard basis for $\Bbb R^3$ is $\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}$.


问:
Natural projection of tangent bundle is submersion
I am wondering how to see that the natural projection from $\pi:TX \to X$ is a submersion. To see that $\pi$ is surjective I can identify $TX$ as product space $X \times F$ where $F$ is a vector space with the same dimension as $X$ and so for all $(x,v) \in TX \simeq X \times F$, $\pi(x,v) = x$ is surjective. Correct?
But I don't now how to proof that $d\pi$ is surjective. Do I need some concept of the double tangent space Or can I avoid  the map $d\pi: TTX \to TX$?   </p>

答:
I want to provide a easier alternative proof here.

Suppose dim$M=n.$

Let $(U,(x^i))$ be a chart on $M$ and $p\in U$. Note that $(x^i)$ provides a smooth local coordinates on the open subset $U\subseteq M$. And then we get the corresponding natural coordinates $(x^i,v^i)$ on $\pi ^{-1}(U)\subseteq TM$. The coordinate representation of $\pi$ is $\hat{\pi}(x,v)=x.$ Note that the top-left $n\times n$ submatrix in the Jacobian Matrix of $\hat{\pi}$ is $I_n$, the $n\times n$ identity matrix. So the rank of the Jacobian Matrix of $\hat{\pi}$ is $n$ because its rank is bounded by the minimum of $\{ \text{dim}M, \text{dim}TM\}$.

The corresponding matrix of the linear map $d\pi$ w.r.t these two bases is exactly the Jacobian Matrix of $\hat{\pi}$. By the discussion above and the fact that $\dim T_p M=\dim M=n$, it is easy to see that for every $\alpha \in\pi^{-1}(U)$ with $\pi(\alpha)=p$, we have that $d\pi[T_\alpha(TM)]=T_p M$. And since $p$ is arbitrary, we conclude that the map $d\pi$ is surjective. Hence $\pi$ is a smooth submersion.

$\tag*{$\blacksquare$}$


问:
Fourier series of a measurable function
Let $f: [0,2\pi] \to \mathbb{R}$ be a bounded Lebesgue measurable function that satisfies $\int_{0}^{2\pi}f(x)\sin( kx) = 0$ and $\int_{0}^{2\pi}f(x)\cos( kx) = 0$ for every $k\in \mathbb{N}$. It is my guess that this implies $f$ is constant almost everywhere.

However, I have much trouble proving it or finding an counter-example.  The only relevant result which I can think of is that the Fourier series of a continuous function $f$ on closed interval converges pointwise almost everywhere to $f$. So if we make the stronger assumption of $f$ being continuous (instead of just measurable), the statement should be true.

I would appreciate any thoughts whether the original guess is indeed true, and if so, how to go about proving it.

答:
Hint

Let $g(x)=f(x)-a_0$. Show that the Parseval Identity for $g$ gives
$$
\int_{0}^{2\pi} |g(x)|^2 d x=0
$$
Deduce from here that $g=0$ almost everywhere.

Title: Fourier series of a measurable function


问:
Show that $\int_{(a,b]}G(x)dF(x)+\int_{(a,b]}F(x)dG(x)=F(b)G(b)−F(a)G(a)$ where F and G are distribution functions
$\mathbf {The \ Problem \ is}:$ If $F$ and $G$ are two bounded distribution functions on the interval $[a,b]$ with no common points of discontinuities,  then show that
$\int_{(a,b]} G(x)dF(x) + \int_{(a,b]} F(x)dG(x) =F(b)G(b) - F(a)G(a)$

$\mathbf {My \ approach}:$ Actually,I was trying to apply Integration By Parts formula in the Riemann-Stiltjes Integral as $F$ is non-decreasing on
$[a,b]$ .

But, I am confused about where to apply the fact that their set of discontinuities are disjoint .

A hint is very much required at this time, thanks in advance .

答: Hints: Let $\mu$ and $\nu$ be the measures associated by $F$ and $G$. Apply Fubini-Tonelli Theorem to compute $\mu\times\nu( A )=\int\int 1_A(x,y) d\mu(x) d\nu(y)$, where $ A = (a,b]\times (a,b]$. Write $A = A_1 \cup A_2$, where $A_1 = \{(x,y)\in A\mid x\leq y\}$ and $A_2 =\{(x,y)\in A \mid x>y\}$.


问:
Prove basis for product topology is collection of all sets with finitely many factors restricted to open sets
Prove a basis for the product topology on $\prod_{a \in \lambda}X_{a}$ is the collection of all sets of the form $\prod_{a \in \lambda}U_a$ where $U_a$ is open in $X_a$ for each $a$ and $U_a=X_a$ for all but finitely many $a$. Given that the product topology is defined as the topology generated by the subbasis $S=\bigcup\limits_{B \in \lambda}S_B$. Where $S_B=\{\pi_B^{-1}(U_B)|U_B \ \ \text{open in}\ \ X_B \}$

Attempt:Let $S=\bigcup\limits_{B \in Y}S_B$ be a subbasis for the product topology, where $S_B=
\{ \pi^{-1}_{B}(U_B)|U_{B} \ \ \text{open in}\ \ X_B \}$. Then each of the factors of elements of $S$ are of the form $U_a=X_a$ for all $a \neq B$ and $U_B$ for $B \in Y$.So only one factor is not necessarily equal to $X_a$. Then a basis element is a finite intersection of subbasis elements $\pi^{-1}_i(U_i)$ where $i=i_1,...,i_k$. So $U_a=X_a$ for all $i\neq i_1,...,i_k$.

答:
Yes, that's essentially the idea: more succinctly:

(standard fact about bases and subbases): if $\mathcal{S}$ is a subbase, then the set of finite intersection of elements from $\mathcal{S}$ form a base.

A finite intersection from $\mathcal{S}$ looks like $\pi_{a_1}^{-1}[U_1] \cap \pi_{a_2}^{-1}[U_2] \ldots \cap \pi_{a_n}^{-1}[U_n]$, where $a_1, \ldots a_n$ form a finite subset of $\lambda$ (all distinct, because we can use $p_a^{-1}[U] \cap \pi_a^{-1}[V]= \pi_a^{-1}[U \cap V]$ to take sets with the same projection together, a minor point that you omitted) and this is just the set $$\prod_{a \in \lambda} O_a$$ where $O_a = U_i$ if $a = a_i$ for some $i \in \{1,\ldots n\}$ (the finite exception set) and $O_a = X_a$ otherwise. So these form exactly the base you had to prove to be a base: it's just the standard base from the subbase.


问:
Two continuous functions on closed interval and smallest element
Assume $f,g\colon [0,1]\to [0,1]$ are continuous functions and $$ g(0)<f(0) \ \ \text{and}\ \ g(1)>f(1)$$Is it true that there exists smallest $t\in(0,1)$ such that $$ g(t)\geq  f(t)$$
I know continuous will have max and min since it defined on closed interval. But I know If we could ensure such smallest element.

答: The set $\{t: g(t) \ge f(t)\}$ is closed and bounded so it has a minimum.


问:
Multivariable Taylor and partial derivatives
Any idea how to prove this? I know it's Taylor related, and the Taylor expansion of $(u^3-v)$ it's $(u^3-v)$ because it's a polynomial. But I don't how what to do with the sin...

Let $f(u,v)=(u^3-v)\cdot\sin(u+v) \Longrightarrow \dfrac{\partial^n f}{\partial u^j \partial v^{n-j}}(0,0)=0 \;\; \; \forall \; n\in \mathbb{N} \; \;\text{odd}.$

答:
Start expanding
$$\left(u^3-v\right) \sin (u+v)=$$ $$u^3 \sin (u) \cos (v)+u^3 \cos (u) \sin (v)-v \sin (u) \cos (v)-v \cos (u) \sin(v)$$Now use the series around $u=0$ and $v=0$.

This is Taylor expansion for multiple variables.


问:
Is the following inequality in the form of integer linear programming?
Since I'm using an element of an array as an index of another array, is the following inequality still considered as ILP representation?

$1 - M[s[i], s[k]] + M[s[i], s[j]] >= 1$

M and S are respectively 2d and 1d arrays and S's elements are in the range of M's indexes.

答:
It is not valid because $s$ is not constant. If $M$ is fixed, you can use $\sum_{i,j} M_{ij} x_{ij}$ with $x_{ij}$ binary to select elements of $M$. For example, you can model $M_{s_i,s_j}$ as $\sum_{i,j} M_{ij} x_{ij}$ by using the following constraints on $x_{ij}$ that ensure that only $x_{s_i,s_j}$ is $1$ and the other elements of $x_{ij}$ are $0$:
$$\sum_{i,j} i x_{ij} =s_i$$
$$\sum_{i,j} j x_{ij} =s_j$$
$$\sum_{i,j} x_{ij} =1$$
$$x_{i,j} \in \{0,1\}$$