问:
$f:\mathbb R\to \mathbb R$ s.t. $f(x) = 1$ for $x\in\Bbb Q$ and $f(x) = 0$ for $x\notin \Bbb Q$ is discontinuous

$f:\mathbb R\to \mathbb R$ s.t. $f(x) = 1$ for $x\in\Bbb Q$ and $f(x) = 0$ for $x\notin \Bbb Q$ is discontinuous.

So the function above is obviously discontinuous, and I am looking to prove the same. One way that comes to mind is constructing a sequence $x_n\to x$ s.t. $f(x_n)\not\to f(x)$.

How about $$x_n = 2  - \frac{\sqrt 2}{n}$$
$x_n$ is irrational for every $n$, but $x_n\to 2\in \Bbb Q$. In other words, $f(x_n) = 0$ for all $n$, but $f(x) = 1$. This suffices, right?

Also, I have trouble understanding a proof of the same result in Carothers' Real Analysis:

(Example $5.2$) $f^{-1}(B(1,1/3)) = \mathbb Q$ and $f^{-1}(B(0,1/3)) = \mathbb R\setminus\mathbb Q$. Thus, $f$ cannot be continuous at any point of $\mathbb R$ since $\Bbb Q$ and $\Bbb R\setminus\Bbb Q$ do not contain intervals.

$B(x,r)$ is the open ball of radius $r$ centered at $x$. I don't see how $f^{-1}(B(1,1/3)) = \mathbb Q$ and $f^{-1}(B(0,1/3)) = \mathbb R\setminus\mathbb Q$ come about, and from there the rest of the argument.

Thank you!

答:
For the last question:

The inverse of continuous function transforms open sets into open sets. $B(1,1/3)$ is an open set but $f^{-1}(B(1,1/3))=\Bbb Q$ cannot be open since it does not contain any interval (so, in particular, it can not contain a neighborhood of any point).

The same occurs for $f^{-1}(B(0,1/3))=\Bbb R\setminus\Bbb Q$.

For the first one the argument is correct.

You can also do it in the opposite way: take your favourite irrational number say $\pi$. Antd take $(x_n)$ to be the next sequence: $$\{3,\, 3.1,\, 3.14,\, 3.141,\, 3.1415,\, 3.14159,\,\ldots\}$$
It is clear that $x_n\in\Bbb Q$ so $f(x_n)=1$ but $x_n\to\pi$ and $f(\pi)=0$-


问:
Intuition/example on the smallest $\sigma$-algebra
I have recently begun to learn about $\sigma$-algebra. Here, the set $\mathcal{F}\in P(\Omega)$ (where $P(\Omega)$ denotes the power set of $\Omega$) is a $\sigma$-algebra if $\mathcal{F}$ has the following properties:
\begin{align}
(1):& \ \ \emptyset\in\mathcal{F} \\
(2):& \ \ A\in\mathcal{F}\implies A^c\in\mathcal{F} \\
(3):& \ \ A_1,A_2,...\in\mathcal{F}\implies \bigcup_{i=1}^\infty A_i\in\mathcal{F}
\end{align}
I am struggling to understand the idea of the smallest $\sigma$-algebra. I have read that given a family $U$ of subsets of $\Omega$, the smallest $\sigma$-algebra is the intersection of all $\sigma$-algebras containing $U$. To help my understanding, I considered an example.

Let $\Omega=[0,1]$ and $U=\{[0,0.2],[0.8,1]\}$. Then the smallest $\sigma$-algebra, denoted by $\sigma_U$, is
$$\sigma_U=\{\emptyset\ ,[0,0.2],[0.8,1]\}.$$
Is this correct? I am quite lost.

答:
I'll try to explain this from probability theory perspective.

Maybe someone will correct me here

In probability of discrete sample spaces we denote by $\Omega$ the sample space. For example by speaking of one dice rolling the sample space can be the possible results: $\Omega = {1,2,3,4,5,6}$ and then we can simply take $2^\Omega = \{ \emptyset,\{1\},\{2\},..,\{1,2\},\{1,3\},...,\{1,2,3,4,5,6\} \}$ to be every possible event.

But, when there is a need to speak of not countable sample spaces (which is the need for example when you investigate probability of events that occur on the interval $(0,1)$) we cant take $2^\Omega$ (There are going to be sets like Vitally sets that will cause problems)

Hence, we need a subset of $2^\Omega$ that maintains the important properties: empty event, complement of an event and union of events.

Now, the concept of minimal $\sigma$-algebra is important because intersection of $\sigma$-algebras is also $\sigma$-algebra and we can prevent ourselves from including problematic sets as Vitally sets and take for example the minimal sigma algebra that includes all the open subsets of $(0,1)$ and that is good enough for many applications.

(Of course we still have to prove that each set of this sigma algebra can have probability, but this can be done by Lebesgue measure).

ContentLicense: CC BY-SA 4.0


问:
Guess the production rule for this sequence
In this question, I’ll temporarily bypass MSE rules and wait a little bit
before revealing the context, because I believe people looking at it with
a fresh eye will be able to see several things that I'm unable to see now.

I’ve computed the six first terms of a sequence that looks fairly regular :

$$
\begin{eqnarray}
u_1 &=& (3,3,2) \\
u_2 &=& (15,24,41) \\
u_3 &=& (57,45,122) \\
u_4 &=& (213,336,1247) \\
u_5 &=& (795,627,2954) \\
u_6 &=& (2967,4680,26729)
\end{eqnarray}
$$

The coefficients seem to grow sort of exponentially. Can anybody guess
a closed form, or an induction rule for $u_n$ ?

 UPDATE 06/01/2013  : Abel’s enlightening comments below allowed me
to improve my algorithm and compute some more values, up to the 16th value.

Those values confirm the guesses about the linear recurrence relations guessed by
Abel for the first and second coordinate.

For the third coordinate, one obtains the following sequence :

$$
2, 41, 122, 1247, 2954, 26729, 58610, 502655, 1059602, 8816873, 18146666, 148093919, 299943578, 2414936489, 4834990562, 38541995519
$$

One can check that this $16$-element sequence does not satisfy any linear recurrence
of order $8$. Does anybody see a pattern here ?

答:
For the first component at least, the recursion $n_{i+2} = 4n_{i+1}-n_i$ by Abel leads to the differential equation $f''(x) = 4f'(x) - f(x)$, which gives the exponential generating function:

$c_1e^{-(\sqrt{3} - 2)x} + c_2e^{(2 +\sqrt{3})x}$

The initial conditions $n_0$ = 3 and $n_1$ = 15 give the values

$c_1 = \frac{3}{2} - \frac{3\sqrt{3}}{2}$ and $c_2 =\frac{3}{2} + \frac{3\sqrt{3}}{2}$.

So the terms in the sequence are simply the coefficients of $\frac{x^n}{n!}$ in the series expansion of $c_1e^{-(\sqrt{3} - 2)x} + c_2e^{(2 +\sqrt{3})x}$, which should just be:

$c_1(2 -\sqrt{3})^n + c_2(2 + \sqrt{3})^n$.

ContentLicense: CC BY-SA 3.0


问:
Continuity of $x^2y^2/(x^2+xy+y^2)$ at $(0,0)$
I want to check whether the function $f(x,y)=\frac{x^2y^2}{x^2+xy+y^2}$ with $f(0,0)=0$ is continuous at $(0,0)$.

But due to term $xy$ in addition to squares in the denominator, I am unable to proceed.

Can one give some hint for it?

I tried to find partial derivatives.

Both partial derivatives are $0$ on $x$ as well as $y$ axis.

But at other points, we get as: $f_x=\frac{x^2y^3+2xy^4}{(x^2+xy+y^2)^2}$; I couldn't get any direction to see whether this partial derivative w.r.t. $x$ is bounded near $(0,0)$.

答:
Take polar coordinates
$$\lim_{r\to 0} \frac{r^4\cos^2\theta \sin^2\theta}{r^2(\cos^2\theta + \cos\theta\sin\theta + \sin^2\theta)} = \lim_{r\to 0} r^2 \frac{\cos^2\theta \sin^2\theta}{\cos^2\theta + \cos\theta\sin\theta + \sin^2\theta}$$

For this limit to exist, we have to able to approach the same value for every direction. Does it happens in this case?


问:
Problem regarding non-abelian groups and their $Z(G)$
If $G$ is a non-abelian group and $a,b\in G$ such that for every $c\in G\setminus Z(G)$ we have $acb=bca$, then $a,b$ commute, i.e. $ab=ba$.

So far, I have noticed that if we suppose that $ab\ne ba$, then $b\notin Z(G)$, therefore $ab^2=b^2a$. Analogously, we get $a^2b=ba^2$. But I did not yet manage to obtain a contradiction.

Later edit: I had to correct the initial statement; it is $acb=bca$. Sorry!

答:
If $a^{-1}b \in Z(G)$, then $a^{-1}ba=aa^{-1}b=b$, so $ab=ba$.

If $a^{-1}b \notin Z(G)$, then apply the hypothesis with $c=a^{-1}b$ : it gives you $aa^{-1}bb=ba^{-1}ba$, so $b=a^{-1}ba$, so $ab=ba$.


Title: Continuity of $x^2y^2/(x^2+xy+y^2)$ at $(0,0)$

Body:
I want to check whether the function $f(x,y)=\frac{x^2y^2}{x^2+xy+y^2}$ with $f(0,0)=0$ is continuous at $(0,0)$.

But due to term $xy$ in addition to squares in the denominator, I am unable to proceed.

Can one give some hint for it?

I tried to find partial derivatives.

Both partial derivatives are $0$ on $x$ as well as $y$ axis.

But at other points, we get as: $f_x=\frac{x^2y^3+2xy^4}{(x^2+xy+y^2)^2}$; I couldn't get any direction to see whether this partial derivative w.r.t. $x$ is bounded near $(0,0)$.

Body:
Since $2|xy|\le x^2+y^2$
$$
\begin{align}
\left|\frac{x^2y^2}{x^2+xy+y^2}\right|
&=|xy|\,\frac{2|xy|}{(x+y)^2+x^2+y^2}\\
&\le|xy|
\end{align}
$$


问:
Minimum and maximum values
I am not sure if my method for this question is correct:
Given the function $$f(x,y)=x^2+y^2+2x+y$$, find it's minimum and maximum values about a closed disc of radius 2 centred at the origin.
I took the partial derivatives, $$f_x=2x+2; f_y=2y+1$$ and solved for $x$ and $y$. The minimum value is $f(0,0)$ and the maximum value is $f(-1,-1/2)$. Am I doing the correct thing? The answer I got is wrong according to the key.  </p>

答:
By "completing the squares" for the terms in each coordinate variable, we obtain

$$ f(x,y) \ = \ x^2 \ + \ y^2 \ + \ 2x \ + \ y \ \ = \ \ (x \ + \ 1)^2 \ - \ 1 \ + \ \left(y \ + \ \frac{1}{2} \right)^2 \ - \ \frac{1}{4} $$ $$= \ \ (x \ + \ 1)^2  \ + \ \left(y \ + \ \frac{1}{2} \right)^2 \ - \ \frac{5}{4} \ \ , $$

so there is indeed a critical point at $ \ \left( -1 \ , \ -\frac{1}{2} \right) \ $ where the value of the function is $ \ -\frac{5}{4} \ \ . $ This is the only critical point in the interior of the disk; since we see immediately that $ \ f(0 \ , \ 0) \ = \ 0 \ \ , $ this is a local minimum for the function.  [We can confirm this through the usual multivariate test:  $ \ f_{xx}f_{yy} - [f_{xy}]^2 \ = \ 2 · 2 \ - \ 0^2 \ > \ 0 \ $ and $ \ f_{xx} \ > \ 0 \ . $ ]   There is nothing special about the function at the origin.

Since the region of interest is the closed disk $ \ x^2 \ + \ y^2 \ \le \ 4 \ ,  \ $ we will also want to explore the boundary.  There are a number of ways to do so:  here, we'll look at a couple not already discussed in the other postings.  As we are working on the boundary $ \ x^2 \ + \ y^2 \ = \ 4 \ ,  \ $ we could write the function as $ \  f(x,y) \ = \ 4 \ + \ 2x \ + \ y \  ,  \ $ and focus only on the last two terms.

If we use Cartesian coordinates, we can form the single-variable function $ \ \phi(x) \ = \ 2x \ + \ \sqrt{4 \ - \ x^2} \ $ , for which the extremization gives us

$$ \ \phi'(x) \ = \ 2  \ - \ \frac{x}{\sqrt{4 \ - \ x^2}} \ = \ 0 \ \ \ \Rightarrow \ \ \ 16 \ - \ 4x^2 \ = \ x^2 \ \ \ \Rightarrow \ \ x \ = \ \pm \frac{4}{\sqrt{5}} $$ $$ \Rightarrow \ \ \ y \ = \ \pm \frac{2}{\sqrt{5}} \ \ . $$  The absolute maximum on the disk is then $$ \  f\left(\frac{4}{\sqrt{5}} \ , \ \frac{2}{\sqrt{5}} \right) \ = \ 4 \ + \ 2·\frac{4}{\sqrt{5}}  \ + \ \frac{2}{\sqrt{5}} \ = \ 4  \ + \ \frac{10}{\sqrt{5}} \  = \ 4  \ + \ 2\sqrt{5}  \ \approx \ 8.472 $$

and the local minimum on the boundary is $ \  f\left(-\frac{4}{\sqrt{5}} \ , \ -\frac{2}{\sqrt{5}} \right) \ = \ 4  \ - \ 2\sqrt{5}  \ \approx \ -0.472 \ \ . $  So in fact the critical point at $ \ \left( -1 \ , \ -\frac{1}{2} \right) \ $ is the location of the absolute minimum value  $ \ -\frac{5}{4} \   $ for the region.

We could also find the extrema on the boundary without calculus, by employing a method we learn for working with trigonometric functions.  Using polar coordinates as suggested by Cameron Buie, we can write our one-variable function as  $ \ \phi(\theta) \ = \ 2 · 2 \cos \theta \ + \ 2 \sin \theta \ \ . $  Extracting a factor of $ \ \sqrt{4^2 \ + \ 2^2} \   $ produces

$$ \ \phi(\theta) \ = \ \sqrt{20} \ \left[ \ \underbrace{\frac{4}{\sqrt{20}}}_{\cos \alpha} \ \cos \theta \ + \ \underbrace{\frac{2}{\sqrt{20}}}_{\sin \alpha} \ \sin \theta \ \right] \ \ = \ \ 2\sqrt{5} \   \cos (\theta  \ -  \  \alpha)   \ \ , $$

making use of an "angle-addition" formula, where $ \ \alpha \ $ has the indicated trigonometric values.  We see directly that $ \ \phi(\theta) \ $ has the extremal values $ \ \pm \ 2\sqrt{5} \ $ , so the corresponding values of our function $ \ f(x) \ $ on the boundary are $ \ 4 \ \pm \ 2\sqrt{5} \ \ . $

(While it is not required for the given problem, should we wish to know the locations of the extremal points, they are at a radius of $ \ 2 \ $ from the origin at the angles corresponding to

$$ \tan \alpha \ \ = \ \ \frac{\frac{2}{\sqrt{20}}}{\frac{4}{\sqrt{20}}} \ \ = \ \ \frac{1}{2} \ \ , $$

with the maximum at $ \ \theta \ = \ \alpha \ $ and the minimum at $ \ \theta \ = \ \alpha \ + \ \pi \ \ . $  This agrees with the rectangular coordinates we found earlier.)

Title: Minimum and maximum values


问:
Open subsets of $[0,1]$
Denote $I=[0,1]$ and endow $I$ with the subspace topology as a subset of $\mathbb R$. Then the only open subsets of $I$ are of the form:

$(a,b)$ with $0\le a\le 1$ and $0\le b \le 1$ and $a<b$.

$(a,1]$ with $0\le a<1$

$[0,b)$ with $0<b\le 1$.

Is this correct?

Edit : I meant the basic open sets.

答:
Those are the basic open sets. All open sets are unions of these, so also $(\frac12,\frac34) \cup (\frac78,1]$ is open etc.
But in many proofs it suffices to only consider basic open sets.

Title: Open subsets of $[0,1]$


问:
Сardinality of the set of increasing functions from Z to Z?
Could you show me how to "calculate" the cardinality of the set of increasing (not necessarily strictly) functions  $\ f: \mathbb{Z} \rightarrow \mathbb{Z}$ ?</p>

答:
Hint: For every $A\subseteq\Bbb Z$ you can construct a function which increases exactly at the points of $A$ (i.e. $g(n)<g(n+1)$ if and only if $n\in A$).

(Bonus points: show that if $A\subseteq\Bbb N$ is infinite, then we can use $\{x\in\Bbb Z\mid x\in A\lor -x\in A\}$ to construct a strictly increasing function, conclude that the strictly increasing functions also make a family of size $2^{\aleph_0}$.)


问:
Prove $\frac{2ab}{a+b}\leq\sqrt {ab}$
$a$ and $b$ are both positive real numbers. I'm supposed to work backwards (i.e. start with what I'm trying to prove and change it until something is absolutely true, then start from what is absolutely true in my proof).

Here's my attempt:

$\frac{2ab}{a+b}\leq\sqrt {ab}$

$\frac{2a^2b^2}{(a+b)^2}\leq{ab}$

$\frac{2a^2b^2}{a^2+2ab+b^2} - ab \leq 0$

$\frac{2a^2b^2-ab(a^2+2ab+b^2)}{a^2+2ab+b^2}\leq 0$

$\frac{2a^2b^2-a^3b-2a^2b^2-ab^3}{a^2+2ab+b^2}\leq 0$

$\frac{-a^3b-ab^3}{a^2+2ab+b^2}\leq0$

Because $a$ and $b$ are positive, this guarantees the left side is negative, making the inequality true.

答:
Let $a,b\in\mathbb{R}$ with $a,b>0$.

$$\begin{align*}\frac{2ab}{a+b}\leq \sqrt{ab}&\Longleftrightarrow \frac{2}{a+b}\leq\frac{\sqrt{ab}}{ab}\\&\Longleftrightarrow \frac{2}{a+b}\leq\frac{1}{\sqrt{ab}}\\&\Longleftrightarrow\sqrt{ab}\leq\frac{a+b}{2}\\&\Longleftrightarrow 2\sqrt{ab}\leq a+b\\&\Longleftrightarrow 0\leq a-2\sqrt{ab}+b\\&\Longleftrightarrow 0\leq (\sqrt{a}-\sqrt{b})^2\end{align*}$$ where the last inequality is always true.

Title: Prove $\frac{2ab}{a+b}\leq\sqrt {ab}$


答:
Consider $a^{-1}b$. If $a^{-1}b \in Z(G)$ then $aa^{-1}b=a^{-1}ba$. Therefore $ab=ba$.

If $a^{-1}b \notin Z(G)$ then apply the hypothesis to give $a(a^{-1}b)b=b(a^{-1}b)a$. So $b^{2}=ba^{-1}ba$ and $ab=ba$.

Title: Problem regarding non-abelian groups and their $Z(G)$

Body:
If $G$ is a non-abelian group and $a,b\in G$ such that for every $c\in G\setminus Z(G)$ we have $acb=bca$, then $a,b$ commute, i.e. $ab=ba$.

So far, I have noticed that if we suppose that $ab\ne ba$, then $b\notin Z(G)$, therefore $ab^2=b^2a$. Analogously, we get $a^2b=ba^2$. But I did not yet manage to obtain a contradiction.

Later edit: I had to correct the initial statement; it is $acb=bca$. Sorry!


问:
$2$-universal family of hash functions and $\varepsilon$-good
Let $V=\{0,1\}^m$ and $H\subseteq V\to V$ a $2$-universal family of hash functions. Fix two sets $A,B\subseteq V$. Call a hash function $h\in H$ $\varepsilon$-good for $A,B$ if:
$$
|\Pr_{x\in V} [x\in A \cap h(x)\in B]-\rho(A)\rho(B)|\leq \varepsilon
$$
where $\rho(C) = \frac{|C|}{|V|}$.
Prove that for any $A,B\subseteq V, \varepsilon>0$,
$$
\Pr_{h\in H} [h \space is \space not \space \varepsilon-good \space for \space A,B] \leq \frac{\rho(A)\rho(B)(1-\rho(B))}{\varepsilon^2 |V|}
\leq \frac{1}{\varepsilon^2 |V|}$$
Any suggestions?

答:
Note that the problem requires strong $2$-universality of $H$, i.e. pairwise independence; this implies uniformity (a key is equally likely to map to any hash).

Instead of $V$, suppose $H$ maps from an arbitrary set $M$ to an arbitrary set $N$. Define an indicator random variable $I_x$ for each key $x\in M$ that is $1$ when $x\in A$ and $h(x)\in B$.

Because of uniformity, $\Pr(I_x)=\frac{|A|}{|M|}\cdot\frac{|B|}{|N|}$; call this probability $p$. Thus $\mathbb E\left[\sum_xI_x\right]=p|M|$.

Because of strong $2$-universality, the $I_x$ are pairwise independent and $\mathsf{Var}\left[\sum_xI_x\right]=\sum_x\mathsf{Var}[I_x]=p(1-p)|M|$.

Apply Chebyshev's inequality:
$$\Pr(h\text{ is not }\varepsilon\text{-good})=\Pr\left(\left|\sum_xI_x-p|M|\right|\ge\varepsilon|M|\right)\le\frac{p(1-p)|M|}{\varepsilon^2|M|^2}=\frac{p(1-p)}{\varepsilon^2|M|}<\frac1{\varepsilon^2|M|}$$
In the case where $M=N=V$, then:
$$\Pr(h\text{ is not }\varepsilon\text{-good})\le\frac{\rho(A)\rho(B)(1-\rho(A)\rho(B))}{\varepsilon^2|V|}<\frac1{\varepsilon^2|V|}$$


答:
To tell that a open set is a basic open set makes no sense unless you previously specify a basis.

The question you are asking should say "Is this family of open sets a basis for the topology?"

The answer is "yes".

Notice that is a basis.

Title: Open subsets of $[0,1]$

Body:
Denote $I=[0,1]$ and endow $I$ with the subspace topology as a subset of $\mathbb R$. Then the only open subsets of $I$ are of the form:

$(a,b)$ with $0\le a\le 1$ and $0\le b \le 1$ and $a<b$.

$(a,1]$ with $0\le a<1$

$[0,b)$ with $0<b\le 1$.

Is this correct?

Edit : I meant the basic open sets.


问:
Inverse Fourier transform of $\exp(A \operatorname{sinc} (B t) )$
I need to evaluate the integral
$$ I = \int_{-\infty}^\infty \exp\Big({-i x t} + A \operatorname{sinc}(Bt)\Big)dt $$
either exactly or for any level of approximation beyond first order for $B\rightarrow 0$. Expanding the $\operatorname{sinc}$ function, the first order approximation is
$$ I =  \int_{-\infty}^\infty \exp\Big({-i x t} + A \Big[1-\frac{(Bt)^2}{3!}+ \cdots \Big]\Big)\,dt \approx e^A \sqrt{\frac{3! \pi}{A B^2}}\exp\Big(-\frac{3! x^2}{4 A B^2}\Big) . $$
The next order approximation would involve a convolution between this Gaussian solution and the inverse transform of a function like $e^{kt^4}$. I think this diverges.

Therefore I'm asking for advice: is there a closed form for the initial integral $I$ in terms of special functions? Or if not, is there a means to evaluate
$$ J = \int_{-\infty}^\infty e^{-itx + k t^4}dt$$
in terms of special functions? Does this integral even converge? Any thoughts are appreciated !

A thought: is this related to a stable distribution? https://encyclopediaofmath.org/wiki/Stable_distribution#:~:text=The%20characteristic%20function%20of%20a%20strictly%2Dstable%20distribution%20with%20exponent,only%20be%20a%20Cauchy%20distribution.

答:
For $$J = \int_{-\infty}^{+\infty} e^{-itx + k t^4}\,dt$$ provided that $k<0$,a CAS gives in terms of a generalized hypergeometric function
$$J=\frac{2\, \Gamma \left(\frac{5}{4}\right) }{(-k)^{1/4}}\,\,
   _0F_2\left(;\frac{1}{2},\frac{3}{4};-\frac{x^4}{4^4 k}\right)+\frac{ \Gamma \left(-\frac{1}{4}\right)}{16 (-k)^{3/4}}x^2\,\, _0F_2\left(;\frac{5}{4},\frac{3}{2};-\frac{x^4}{4^4 k}\right)$$


问:
How many solutions are there for $3x_1+3x_2+x_3+x_4=30$?

Find how many solutions there are for $$ 3x_1+3x_2+x_3+x_4=30$$

I know how to solve this: $x_1+x_2+x_3+x_4=30$
and I read this link, but I am still not sure about my answer.
I wrote:
$x_4+x_3=30-3(x_1+x_2) $
then I get this: $0 \le x_1+x_2 \le 10$ (that I know to solve: $66$ combination)
and now I'm not sure what to do with $x_3+x_4$.

答:
Using stars and bars method, it is easy to establish this lemma:

Lemma. The number of solutions of equation $$x_1+x_2+...+x_k =n$$ where $x_1,x_2,...x_k$ are nonnegative integers, is ${n +k-1\choose n}$.

For each $k\in \{0,1,...,10\}$ the number of solution to $x_1+x_2= k$ is ${k+1\choose 1}=k+1$ and the number of solution to $x_3+x_4 = 30-3k$ is ${31-3k\choose 1} = 31-3k$.

So the total number of solutions is \begin{align}\sum _{k=0}^{10}(31-3k)(k+1)&=\sum _{k=0}^{10}(31 +28k-3k^2)\\
&=341 + 28\sum _{k=0}^{10}k-3\sum _{k=0}^{10}k^2\\
&=341 +28 {10\cdot 11\over 2} -3{10\cdot 11\cdot 21\over 6}\\
&= 726
\end{align}


问:
How was this really handy bijection thought up of?
Here's a proof from my combinatorics textbook.

Theorem 1.6.5. For any positive integer $k$,

$$(1-x)^{-k}=\sum_{n\geq 0}\binom{n+k-1}{k-1}x^n$$

Proof: Let $S_n$ be the set of all $k$-tuples $(a_1,\dots ,a_k)\in N^k_{\geq 0}$ with $a_1+\cdots +a_k=n$. We previously showed that

$$|S_n|=[x^n](1-x)^{-k}$$Let $T_n$ denote the set of all $(k-1)$-element subsets of ${1,2,\dots ,n+k-1}$, thus

$$|T_n|={n+k-1 \choose k-1}$$
We will establish a bijection between $T_n$ and $S_n$.

For $A=\{a_1,a_2,\dots ,a_{k-1}\}\in T_n$, with $a_1<a_2<\cdots<a_{k-1}$, we define $f(A)=\sigma$ where

$$\sigma=(a_1-1,a_2-a_1-1,\dots ,a_k-a_{k-1}-1,(n+k)-a_k-1)$$

(proof that it is bijection follows)

My question is, how in earth would somebody derive such a useful bijection. Just by staring at $\sigma$, where this came from is very opaque - leading to this theorem being really hard to proof without giving you the bijection to use.

答:
Think of writing out $a_1+\cdots+a_k=n$ where the numbers in the left hand side are written in unary notation, that is $a_i$ is represented by a sequence of $a_i$ vertical strokes (the sequence is empty if $a_i=0$). Then you've always got $n+k-1$ symbols in the left hand side ($n$ strokes and $k-1$ plus signs). So you could find all such expressions by each time writing down $n+k-1$ vertical strokes in a row, and choosing any subset of $k-1$ of them to which you add a horizontal stroke to form a "$+$".

(A variation of this argument is called "stars and bars". However I find that argument presented in its Wikipedia article rather roundabout when applied to the current situation, as it amounts to taking $n+k$ stars, choosing $k-1$ of the $n+k-1$ spaces between the stars to place a bar in, and then removing one star form each of the $k$ non-empty groups of stars formed. Also the article has the names $n$ and $k$ interchanged.)

This shows that $|S_n|=\binom{n+k-1}{k-1}$, but using unary notation seems terribly primitive, so we can deduce what the numbers $a_i$ would be if we crossed the strokes at positions $p_1<p_2<\cdots<p_{k-1}$ (starting with position$~1$). Clearly $a_1=p_1-1$ and $a_n=(n+k-1)-p_{k-1}$, while all intermediate numbers are given by $a_i=p_i-p_{i-1}-1$. This gives the bijection you described (although you called the $p_i$ by the same name $a_i$).


问:
Universal property of universal bundles.
A classifying space for a group $G$ is a topological space $BG$ with a principle
$G$-bundle $p : EG \to BG$ where $EG$ is contractile, so that $BG = EG/G$. A classifying space is universal in the sense that if $q : E \to B$ is a principle $G$-bundle, then there is a
continuous map $B \rightarrow BG$, unique up to homotopy, such that $E$ is the fiber product $f^*EG$. Why this definition implies tha there is a one-to-one corrispondence between the isomorphism classes of $G$-bundle on a space $X$ (that I denote with $\mathcal{P}_G(X)$) and the homotopy classes of maps $X \to BG$ ([X,BG])? In practice, how can I show that
$$ [\mathcal{P}_G(X)] \simeq [X,BG] $$</p>

答:
The property you described gives you a map
$$
[X, BG] \rightarrow \mathcal{P}_G(X)
$$
that is surjective.

To check that it is injective, note that if $f: X \rightarrow BG$ and $g: X \rightarrow BG$ correspond to isomorphic bundles over $X$, then we can form a bundle over $X \times I$ by gluing together the bundle on $X \times [0,3/4)$ classified by $f$ and the bundle on $X \times (1/4, 1]$ classified by $g$. This is classified by a map $X \times I \rightarrow BG$ and the restriction to $X \times 0$ and $X \times 1$ is $f$ and $g$, resp., thus giving a homotopy between $f$ and $g$, which is what we wanted.


问:
What is the chance to get two pairs from a hand of five cards if I have extra card (♠ king, ♠ queen and ♦ queen) added to the standard cards
What is the chance to get two pairs if I have extra card (♠ king, ♠ queen and ♦ queen) added to the standard cards, thus I have 55 cards? Can anybody help me !!

Is my solution right?

$$\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1} + \binom{13}{2}\binom{5}{2}\binom{5}{2}\binom{11}{1}\binom{5}{1} + \binom{13}{2}\binom{6}{2}\binom{6}{2}\binom{11}{1}\frac{\binom{6}{1}}{\binom{55}{5}}$$

First, I chose $2$ ranks of $13$ and then $2$ suits of $4$ (this is for hearts and clovers). Then I chose $2$ ranks of $13$ and then $2$ suits of $5$ suits (this is for diamonds, but I am not sure). Then I chose $2$ ranks of $13$ and then $2$ suits of $6$ suits (this is for spades, but I am not sure).

Lastly, I divided all on $\binom{55}{5}$

答:
It seems wise to count hands with pairs of kings or queens separately:

Hands with a pair of kings and not a pair of queens:

$$\binom{5}{2}\binom{11}{1}\binom{4}{2}\binom{46}{1}$$

because we have to pick $2$ kings out of the $5$, then one other type of pair not king or queen (so $11$ ranks to choose from), then $2$ cards out of the $4$ available in that rank, and finally, $1$ card out of the remaining $55  - 5- 4 = 46$ cards (subtracting the ranks (kings, and the other one) from the total number of cards we now have). The latter single remaining card I do not see occur in your count at all, which seems fishy to me.

Hands with a pair of queens but not a pair of kings:

$$\binom{6}{2}\binom{11}{2}\binom{4}{2}\binom{45}{1}$$

Which has been simularly computed as the previous type, but now we have $6$ queens to choose $2$ from and now $55-6 - 4 = 45$ final cards to choose from.

Hands with both a pair of queens and a pair of kings:

$$\binom{6}{2}\binom{5}{2}\binom{55-11}{1}$$

Simpler because there are no extra ranks for pairs to choose and more cards left at the end (just kings and queens, $11$ now, are forbidden).

Finally hands with neither pairs of kings nor queens:

$$\binom{11}{2}\binom{4}{2}^2\binom{55- 8}{1}$$

Where we first choose the two ranks to form pairs from and then twice the $2$ pairs themselves, and finally a term for the final card.

Now add these $4$ numbers to get all hands with two pairs and indeed divide by $\binom{55}{5}$ to get the probability (all possible choices of $5$ out of the total expanded deck).


问:
Is the category of abelian presheaves on a topos closed?
Take the category of presehaves of abelian groups on a topos $\mathcal{C}$. That is, an object of our category is a functor $F: \mathcal{C}^{\operatorname{op}} \to \operatorname{Ab}$.

We have a clear definition of a symmetric tensor product by $(F \otimes G)(S) = F(S) \otimes G(S)$.

However, defining an appropriate internal hom seems trickier. Taking inspiration from the internal sheaf hom, it seems like it would make sense to define

$${\mathscr{H}\kern -.5pt om} (F,G)(S) = \operatorname{Hom}(F_{| \mathcal{C}/S}, G_{|\mathcal{C}/S})$$

where $\mathcal{C}/S$ is the slice category. However, I can't seem to prove the adjunction (if it is even true). The main problem is that the slice category is more difficult to work with than in a general topos. Could there be some stronger requirements on the topos to make sure the adjunction is satisfied?

答:
The category of abelian presheaves is locally finitely presentable, and for each abelian presheaf $F$, the functor $F \otimes {-}$ preserves colimits, so we may apply the accessible adjoint functor theorem to obtain a right adjoint $\mathscr{H}om (F, -)$.
Representability of the functor $H \mapsto H (S)$ can be used to determine the values of $\mathscr{H}om (F, G)$:
$$\mathscr{H}om (F, G) (S) \cong \textrm{Hom} (F \otimes \mathbb{Z} h_S, G)$$
Here $h_S = \mathcal{C} (-, S)$ and $\mathbb{Z} h_S$ is the free abelian presheaf generated by $h_S$.

Actually, your formula is also correct.
The point is that you need to verify that
$$\textrm{Hom} (F \otimes \mathbb{Z} h_S, G) \cong \textrm{Hom} (F |_{\mathcal{C}_{/ S}},  G |_{\mathcal{C}_{/ S}})$$
which can be done essentially the same way as the Yoneda lemma.


问:
Why decimal form of $\frac{A}{2^m\times5^n\times3^p\times7^q\cdots}$ has both repeating and non repeating numbers?
I read in my book that, decimal form of the fraction $\cfrac{A}{\underbrace{2^m\times5^n}_{\text{at least one}}\times\underbrace{3^p\times7^q}_{\text{at least one}}\times\cdots}\quad(A\in\mathbb{N})$, has both repeating numbers and non repeating numbers. I don't understand the reason of that. I know the decimal form of $\cfrac{A}{2^m\times5^n}$ is determinable because we can rewrite it as $A\times(0.5)^m\times(0.2)^n$. hence with other prime factors in denominator like $3^p$ and $7^q$ decimal is not determinable so it has repeating numbers. but for $\cfrac{A}{{2^m\times5^n}\times{3^p\times7^q}\times\cdots}$ which is a combination of two forms I don't know how to show it mathematically that contain both repeating and non repeating numbers in decimal form. I can rewrite it as :
$$A\times(0.5)^m\times(0.2)^n\times\frac{1}{3^p\times 7^q\cdots}$$
It is product of determinable and repeating decimals but it doesn't help me.

答:
I just came up with this answer:

Consider fraction $\cfrac{A}{2^m\times5^n\times3^p\times7^q}$ , Without Loss Of Generality assume $2^m\times5^n<3^p\times7^q$. Let's decompose it:

$$A\times\cfrac{1}{2^m.5^n.3^p.7^q}=A\times\cfrac{1}{3^p.7^q-2^m.5^n}\left(\cfrac{3^p.7^q-2^m.5^n}{2^m.5^n.3^p.7^q}\right)$$

$$=\cfrac{A}{3^p.7^q-2^m.5^n}\times\left(\cfrac{1}{2^m.5^n}-\cfrac{1}{3^p.7^q}\right)$$

If we write $\cfrac{1}{2^m.5^n}-\cfrac{1}{3^p.7^q}$ in decimals, we have subtraction of  determinable decimal and repeating decimal, it is not hard to see the result is a decimal number which has initial segments followed by repeating segments. multiplying it by $\cfrac{A}{3^p.7^q-2^m.5^n}$ doesn't influence on the conclusion.


问:
$(0,1]$ is not homeomorphic to $(a,b)$ - Carothers Example 5.6(g)
I'm reading Carothers' Real Analysis, and I'm currently looking at homeomorphisms. The author says "two intervals that look different, are different" - i.e. they are not homeomorphic. The proof is done for the case $(0,1]$ and $(a,b)$, where the first interval is semi-open so we expect that the two are not homeomorphic.

I have some difficulty following the book's argument which I paraphrase here (for convenience) with inline questions.

Proof by contradiction. Suppose $(0,1]$ and $(a,b)$ are homeomorphic. Then by removing $1$ from $(0,1]$ we get $(0,1)$, and by removing its image $f(1) = c$ from $(a,b)$ we have that $(0,1)$ is homeomorphic to $(a,c)\cup (c,b)$.

Wait, why is that? I understand that we are deleting an element each from the domain and codomain, so the bijection remains a bijection. What about the continuity of $f$ and $f^{-1}$ though? We literally threw away an element, how do we know that continuity still holds?

But $(0,1)$ is homeomorphic to $\Bbb R$, so $\Bbb R$ would have to be homeomorphic to $(a,c)\cup (c,b)$ too. So $\Bbb R$ can be written as the disjoint union of two non-trivial open sets, which is impossible.

Two questions:

Why is $(0,1)$ homeomorphic to $\Bbb R$? We really need to produce a homeomorphism, that's all I'm asking.

How to prove that $\Bbb R$ cannot be written as the disjoint union of two non-trivial open sets? The idea that comes to my mind is - we already know that any open subset of $\Bbb R$ can be written as the disjoint union of unique open maximal intervals. Since $\Bbb R$ is open in $\Bbb R$, we can obviously do the same for it as well, and clearly, our choice of intervals would not be maximal in the event that we could write $\Bbb R$ as the union of two disjoint open intervals. Does this make sense?

Thank you!

答:
Continuity still holds when you remove a point, because the restriction of a continuous map is still continuous.

A homeomorphism between $(0,1)$ and $\mathbb{R}$ is given by
$$x \mapsto \tan \left(-\frac{\pi}{2}+t\pi \right)$$

$\mathbb{R}$ is connected, so cannot be written as the disjoint union of two non-trivial open sets.


问:
A question on logic - where intuition can fail
Suppose I have two predicates $P(x)$ and $Q(x)$, such that $\overline{P(x)\wedge Q(x)}$ holds for all $x$.

Now, if $\displaystyle \bigwedge_{x\in A}P(x)$ for a set $A$, it must be certainly true, that $\displaystyle \overline{\bigwedge_{x\in A}Q(x)}$ by intuition. But suppose, that $A = \{\;\;\}$, then obviously both $\displaystyle \bigwedge_{x\in A}P(x)$ and $\displaystyle\bigwedge_{x\in A}Q(x)$.

Where is my mistake? I guess it is in the assumption that
$$\bigwedge_x\overline{P(x)\wedge Q(x)} \rightarrow \left(\bigwedge_{x\in A}P(x) \rightarrow \overline{\bigwedge_{x\in A}Q(x)}\right)?$$

*Note: $\displaystyle \overline{P(x)\wedge Q(x)} \equiv \lnot(P(x)\land Q(x))$

$\displaystyle \quad\quad\quad\quad \overline{\bigwedge_{x\in A}Q(x)} \equiv \lnot \left(\bigwedge_{x\in A}Q(x)\right)$

答: You are given $\forall x \lnot(P(x)\land Q(x))$, which is the same as $\forall x \  \lnot P(x) \lor \lnot Q(x)$. Then you are given $\forall x \in A\  P(x)$.  You are right that you can then derive $\forall x \in A \lnot Q(x)$.  If $A$ is empty, it is also true (without any of the preceding) that $\forall x \in A \ Q(x)$.  But for empty $A$, this is not a contradiction, as it expands to $\forall x (x \in A) \implies Q(x)$  As the antecedent is always false, the implication is always true.</p>


问:
Definition of diffeomorphism functions
I see the definition of Diffeomorphism in Wikipedia homepage, but I don't understand whether "the differential of f (Dfx : Rn → Rn) should be bijective at each point x in U" or "f itself" should be bijective?</p>

答:
To answer the question regarding the what the Wikipedia article say; you need that $Df_{x}$ is bijective for all $x\in U$. This is due to a theorem by Hadamard, and is in essence saying that if $f$ is locally diffeomorphic and $f$, $U$, and $V$ are "nice" then $f$ will in fact be globally diffeomorphic. The answer given by t.b to this question gives a nice exposition and references on this.

That said this is not the definition of a diffeomorphism, but is instead a more advanced characterization of when a local diffeomorphism might in fact be a global diffeomorphism.  A map $f:U\rightarrow V$ is a differomorphism if $f$ is bijective and smooth and $f^{-1}$ is also smooth.

Title: Definition of diffeomorphism functions


答:
Your approach seems fine but there are a few mistakes.

Number of ways to choose two pairs -

i) A pair of queen and a pair of king

$\displaystyle \binom{6}{2}\binom{5}{2}\binom{44}{1}$

ii) A pair of queen and a pair from other $11$ ranks (not king)

$\displaystyle \binom{6}{2} \binom{11}{1} \binom{4}{2} \binom{45}{1}$

iii) A pair of king and a pair from other $11$ ranks (not queen)

$\displaystyle \binom{5}{2} \binom{11}{1} \binom{4}{2} \binom{46}{1}$

iv) Two pairs from other $11$ ranks (other than queen and king)

$\displaystyle \binom{11}{2} \binom{4}{2} \binom{4}{2} \binom{47}{1}$

Now the probability is sum of $(i), (ii), (iii), (iv)$ divided by $\displaystyle {55 \choose 5}$.

Title: What is the chance to get two pairs from a hand of five cards if I have extra card (♠ king, ♠ queen and ♦ queen) added to the standard cards

Body:
What is the chance to get two pairs if I have extra card (♠ king, ♠ queen and ♦ queen) added to the standard cards, thus I have 55 cards? Can anybody help me !!

Is my solution right?

$$\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{11}{1}\binom{4}{1} + \binom{13}{2}\binom{5}{2}\binom{5}{2}\binom{11}{1}\binom{5}{1} + \binom{13}{2}\binom{6}{2}\binom{6}{2}\binom{11}{1}\frac{\binom{6}{1}}{\binom{55}{5}}$$

First, I chose $2$ ranks of $13$ and then $2$ suits of $4$ (this is for hearts and clovers). Then I chose $2$ ranks of $13$ and then $2$ suits of $5$ suits (this is for diamonds, but I am not sure). Then I chose $2$ ranks of $13$ and then $2$ suits of $6$ suits (this is for spades, but I am not sure).

Lastly, I divided all on $\binom{55}{5}$


问:
Is there also an other way to show the equality: $\left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$?
I want to show that:

$$ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$$

That's what I have tried:

$ \left\lfloor \frac{n}{2}\right\rfloor=\max \{ m \in \mathbb{Z}: m \leq \frac{n}{2}\}$

$\left\lceil \frac{n}{2} \right\rceil= \min \{ m \in \mathbb{Z}: m \geq \frac{n}{2}\}$

If $n=2k,k \in \mathbb{Z}$,then: $\frac{n}{2} \mathbb{Z}$,so
$$\left\lfloor \frac{n}{2}\right\rfloor=\max \left\{ \frac{n}{2}, \frac{n-2}{2}, \dots \right\}=\frac{n}{2} \\ \left\lceil \frac{n}{2} \right\rceil=\min \left\{ \frac{n}{2}, \frac{n+2}{2}, \dots \right\}=\frac{n}{2}$$

Therefore, $ \left\lfloor \frac{n}{2}\right\rfloor + \left\lceil \frac{n}{2} \right\rceil=n$.

If $n=2k+1, k \in \mathbb{Z}$,then $\frac{n}{2} \notin \mathbb{Z}$.So:

$$\left\lfloor \frac{n}{2}\right\rfloor=\max \left\{ \frac{n-1}{2}, \frac{n-3}{2}, \dots \right\}=\frac{n-1}{2}\\ \left\lceil \frac{n}{2} \right\rceil=\min \left\{ \frac{n+1}{2}, \frac{n+3}{2}, \dots \right\}=\frac{n+1}{2}$$

Therefore, $ \lfloor \frac{n}{2}\rfloor + \lceil \frac{n}{2} \rceil=\frac{n-1}{2}+\frac{n+1}{2}=n$

Is there also an other way to show the equality or is it the only one?

答: Let $x,y$ be any two real numbers. Then $\lfloor x\rfloor+\lceil y\rceil>x+y-1$, since $\lfloor x\rfloor>x-1$ and $\lceil y\rceil\geq y$. Similarly $\lfloor x\rfloor+\lceil y\rceil<x+y+1$. So $\lfloor x\rfloor+\lceil y\rceil$ differs from $x+y$ by strictly less than $1$. Since $\lfloor x\rfloor+\lceil y\rceil$ is always an integer, if $x+y$ is also an integer the two must be equal. This holds in particular for $x=y=n/2$.


问:
"Playable" games on an uncountable set
Almost all games in real life are based on a finite set of integers (for example, we can index each possible movement in chess by an integer). But it is still pretty interesting to think about games that goes beyond the integers, to the uncountable, for example the Banach–Mazur game, which depends on the topological structure of the underlying set. We can play this game on $\mathbb R.$

However, to judge the winner, we have to move countably many steps.

My question is: does there exists some "uncountable" or "topological" games which can actually be played in real life? After all, many infinite things have finite analogs - so maybe it can somehow made to work in real life?

EDIT: Here is what I mean by the phrase "finite analogs" above. Of course with our countable language we cannot play games on an uncountable space. But if a nice finite or countable analog makes sense, then it would be great.

答:
How about this:

Take an irregular shaped (bounded) region in the plane. Players take turns placing disks (considerably smaller than the region) into the region. Once a disk is placed, it stays in its position. Disks must lie completely within the region, and disks may not overlap.  The winner is the last player to successfully place a disk according to the rules.

Because of the free placement, there are uncountably many game trees/game positions. However the game will end after finitely many moves, assuming that all the disks are the same size--or at least that the disk size doesn't go to $0$ as the game progresses.

You could play this game with other shapes than disks.

[See also Sam Loyd's 'egg game': http://www.jwstelly.org/CyclopediaOfPuzzles/PuzzlePage.php?puzzleid=Pz169.1 ]

Title: "Playable" games on an uncountable set


问:
The quest for a more natural distance function $d(x,A)$
In Carothers' Real Analysis, the distance of a point $x$ from a set $A$ is defined as
$$d(x,A) := \inf\{d(x,a): a\in A\}$$
While this distance enjoys some nice properties, I think it is a little unnatural in the following sense:

It is not necessarily true that $d(x,A) > 0$ when $x\notin A$. For example, $d(x,\Bbb Q) = 0$ for any $x\in \Bbb R$.

Can we find a distance function that satisfies:

$d(x,A) > 0 \Longleftrightarrow x\notin A$

$d(x,A) = 0 \Longleftrightarrow x\in A$

That seems like a more natural choice! The nice properties that the current definition of $d$ has are:

A. $d(x,A) = 0 \Longleftrightarrow x\in \bar A$.

B. $|d(x,A) - d(y,A)| \le d(x,y)$.

Thanks! Looking forward to insightful discussions on this.

答:
Let $(X, d)$ be the metric space and $A \subset X$. If you require that
$$ \tag{*}
 \forall x, y \in X: |d(x,A) - d(y,A)| \le d(x,y)
$$
then $d(\cdot, A): X \to \Bbb R$ is a continuous function. If, in addition, $d(x ,A) = 0$ for $x \in A$ then it follows that $d(x, A) = 0$ also for $x \in \overline A$. This shows that
$$
d(x,A) = 0 \iff x\in A
$$
cannot be satisfied together with $(*)$ (unless $A = \overline A$, of course).


问:
Convert a non-standardized test data into a standard report for every student taking test of similar difficulty more than once.
Context :

Suppose that there are 5 sets of exams being conducted, each for the
same purpose. Since they are being conducted over a period of 5 days,
the papers are different but a student can appear at most once. The
number of students appearing on each day is different and so is the
difficulty of the paper depending on the student.

Questions :

Given the test reports of every student, how would you scale the scores (ie, generate the test report of each student) as if every student is writes the same question paper?

Now suppose a subset (possibly improper) of these students take the test (which is conducted over a period of 5 days and has the "same" difficulty but a student can appear at most once) again. Given their test reports, how would you find the score of each student if we were to consider his/her best score in the two tests (upon proper scaling) and generate a single final test report?

In [2] above, the best score possibly also requires a proper scaling. I am essentially trying to see how we can solve [2]. But that requires us to solve [1] and this is precisely why it looks like a 2-part question (which is against the rules of Stack, if I am not mistaken). I am trying to use the non-standardized test data for a very large (hypothetically, so the misc. dependence factors are reduced) pool of students from the same place into a standardized report for every student.

Please feel free to suggest or edit the topic and add more suitable tags. I do not have a very strong grasp of the subject.

答:
First, determine if there is a significant difference between exams.  Fit a Generalized Linear Model (GLM) with an appropriate link function $g$ for your exam.  Perhaps this is a Beta regression problem if the scores are on the 0-100% scale.

$$g(Y_i) = \beta_0 + \beta_1 x_i + \epsilon_i$$

where $x_i$ is a categorical variable for the exam (so it is really four variables)

Second, if the overall Likliehood Ratio Test for the model is significant, then there are at least two exams that are different.

Third, if there is a significant exam effect, then scale the exams by the mean exam score for that group so that the mean translates to a particular target score (80%).

Something like this in R:

```<code>set.seed(123454)
# number of students
N <- 100
# randomly select a test
x <- sample(c("A","B","C","D","E"), size = N, replace = TRUE)
# randomly create a score for each student
y <- ifelse(x == "A", rbeta(N, 3, 2), ifelse(
  x == "B", rbeta(N, 4, 2), ifelse(
    x == "C", rbeta(N, 5, 2), ifelse(
      x == "D", rbeta(N, 6, 2), ifelse(
        x == "E", rbeta(N, 7, 2), NA)
      )
    )
  )
)

hist(y, breaks = 10)

# fit a simple quasibinomial model with a logit link
#   better to do a beta regression here, but quasibinomial is illustrative
lm1 <- glm(y ~ x, family = quasibinomial(link = "logit"))
lm2 <- glm(y ~ 1, family = quasibinomial(link = "logit"))

anova(lm1, lm2, test = "LRT")

# therefore there are at least two exams that are different

# now scale
mean_values <- by(y, x, mean)
target_mean <- 0.8

# one method of normalizing so that no one goes over 100%
new_scores <- 1 - c((1-y) / (1-mean_values[x]) * (1-target_mean))
mean(new_scores)
by(new_scores, x, mean)
hist(new_scores, breaks = 10)

<span class="math-container">```
```

Body:
First, determine if there is a significant difference between exams.  Fit a Generalized Linear Model (GLM) with an appropriate link function $g$ for your exam.  Perhaps this is a Beta regression problem if the scores are on the 0-100% scale.

$$g(Y_i) = \beta_0 + \beta_1 x_i + \epsilon_i$$

where $x_i$ is a categorical variable for the exam (so it is really four variables)

Second, if the overall Likliehood Ratio Test for the model is significant, then there are at least two exams that are different.

Third, if there is a significant exam effect, then scale the exams by the mean exam score for that group so that the mean translates to a particular target score (80%).

Something like this in R:

set.seed(123454)
# number of students
N <- 100
# randomly select a test
x <- sample(c("A","B","C","D","E"), size = N, replace = TRUE)
# randomly create a score for each student
y <- ifelse(x == "A", rbeta(N, 3, 2), ifelse(
  x == "B", rbeta(N, 4, 2), ifelse(
    x == "C", rbeta(N, 5, 2), ifelse(
      x == "D", rbeta(N, 6, 2), ifelse(
        x == "E", rbeta(N, 7, 2), NA)
      )
    )
  )
)

hist(y, breaks = 10)

# fit a simple quasibinomial model with a logit link
#   better to do a beta regression here, but quasibinomial is illustrative
lm1 <- glm(y ~ x, family = quasibinomial(link = "logit"))
lm2 <- glm(y ~ 1, family = quasibinomial(link = "logit"))

anova(lm1, lm2, test = "LRT")

# therefore there are at least two exams that are different

# now scale
mean_values <- by(y, x, mean)
target_mean <- 0.8

# one method of normalizing so that no one goes over 100%
new_scores <- 1 - c((1-y) / (1-mean_values[x]) * (1-target_mean))
mean(new_scores)
by(new_scores, x, mean)
hist(new_scores, breaks = 10)

```


问:
Volume of open box and of closed box
In the book Analysis 2, Terence Tao defined the notion of an open box and its volume:

Definition 7.2.1 (Open box)

An open box (or box for short) $B$ in $\mathbb{R^n}$ is any set of the form:
$B = \prod_{i: 1 \to n} (a_i,b_i) := $ {$(x_1,..,x_n)\in \mathbb{R^n}: x_i \in (a_i,b_i) \, for \, all \, 1 <= i < = n$} ($a_i, b_i$ are real numbers).

We define the volume $vol(B)$ of this box to be the number:

$vol(B) := \prod_{i = 1 \to n} (b_i - a_i) =  \, (b_1 - a_1)(b_2 - a_2)...(b_n - a_n)$

By this definition, I understood that the notion of $volume$ is only defined for an open box. The author does not say anything about the volume of a closed box, or a half-closed box.

How can I rigorously extend the notion of a volume of an open box to the notion of a volume of a:

closed box

i.e. $B = \prod_{i: 1 \to n} [a_i,b_i] := $ {$(x_1,..,x_n)\in \mathbb{R^n}: x_i \in [a_i,b_i] \, for \, all \, 1 <= i < = n$}

half-closed box

$i.e.$ $B = \prod_{i: 1 \to n} (a_i,b_i] := $ {$(x_1,..,x_n)\in \mathbb{R^n}: x_i \in (a_i,b_i] \, for \, all \, 1 <= i < = n$}

EDIT:
What is the point of defining the notion of volume of an open box and not other kind of box (closed, half-closed) ? Is it personal preference of the author or there is some motivations behind ?

答:
These formulae are exactly the same for the closed and half-closed boxes.

In my old measure theory course we used $[a_i, b_i)$-sided-boxes (defined on a semi-algebra and then extended via standard theorems).. It all comes down to the same here.

Usually (I don't know Tao's approach) we define a first approximation of the measure on a small subcollection of all the sets that we eventually want to define it on (here at least the Borel sets) and then apply theory to extend it to that larger collection. Probably Tao can prove the formulae for the other kinds of boxes later, after proving the desired extension exists and is unique, and then he doesn't need to define it here. But I know that these formulae will be the correct ones in the end (and you do too, as you've learnt about volume in primary school, right?)

The starting type of boxes is a matter of preference as to what way of extending Tao will choose. He will have his reasons, that hopefully become clear later in the text.

Title: Volume of open box and of closed box


问:
Are inverse unique in unital algebra?
Let $A$ be a unital algebra over a field $F$ with unity $1$. If $a,b,c\in A$ such that $ab=ba=1=ac=ca$, does this imply that $b=c$?

We have $c(ab)=c$. However, algebras are not necessarily associative so we can't conclude $c=c(ab)=(ca)b=b$.

Are there examples of algebras over a field with elements having multiple inverses?

答:
Are there examples of algebras over a field with elements having multiple inverses?

Yes. Consider $S=\{e,a,b\}$ with the following multiplication table

$$
\begin{array}{c|c|c|c}
& e
& a
& b \\\hline
e
& e
& a
& b \\\hline
a
& a
& e
& e \\\hline
b
& b
& e
& e \\
\end{array}$$

which is nonassociative with unity and with multiple inverses.

Now let $F$ be any field and consider $F[S]$ which is a vector space over $F$ with $S$ as a basis. The multiplication on $F[S]$ is induced linearly from the multipication on $S$, i.e.

$$\bigg(\sum_{i=1}^n \lambda_i s_i\bigg)\cdot \bigg(\sum_{i=1}^n\tau_i s_i\bigg):=\sum_{i,j=1}^n\lambda_i\tau_j s_is_j$$

This makes it a unital, nonassociative algebra over $F$, but $a$ has at least two inverses: itself and $b$.

Title: Are inverse unique in unital algebra?


问:
Two real numbers $a$ and $b$ are equal iff $| a - b |\leq \epsilon$ for all $\epsilon\gt0$

Two real numbers $a$ and $b$ are equal iff $|a - b | \leq \epsilon$ for all $\epsilon >0$

I know that it's true if it says "iff $| a - b | < \epsilon$". But if I use $\leq$, isn't that statement false? Because intuitively, since $\epsilon$  can't be $0$, $| a - b |$ might not be $0$ for all $\epsilon$

答: If you know that it is true for $<$, then you can either adapt the proof for $\le$ or simply argue that if $\mid a - b \mid \leq \epsilon $  $\forall \epsilon >0$, then also $\mid a - b \mid < \epsilon $  $\forall \epsilon >0$. In fact, given $\epsilon >0$, then we know that $\mid a - b \mid \leq \epsilon/2 < \epsilon $.


问:
Conjugacy classes of non-normal subgroups
Let $G=P \ltimes Q=QM$ be a finite non-nilpotent group which is solvable. ($P$ and $Q$ are Sylow subgroups of $G$ such that $P$ is cylic and $|Q|=q^4$). Let $M=C_G(P)=N_G(P)=P \times (Q \cap M)$ is a non-cyclic maximal subgroups of $G$ which is abelian( $Q \cap M \trianglelefteq G$). Also we have $Q \cap M$ contains a non-cyclic subgroup $T \cong C_q \times C_q$ ($q$ is a prime) which is maximal in $Q \cap M$ and $Q \cap M$ is maximal in $Q$. Now i want to find at least three non-conjugate subgroups of $G$ which are non-normal and non-cyclic in $G$.

$\bf{My try:}$ By assumption, $M$ is non-normal non-cyclic subgroups of $G$. Also since $PT < N_G(P)$, we have $PT$ is non-normal non-cylic subgroups of $G$. Also we have $M \le C_G(Q \cap M)\le G$. If $M =C_G(Q \cap M) \trianglelefteq N_G(Q \cap M)=G$, then $M \trianglelefteq  G$, a contradiction. So $Q \cap M \le Z(G)$. Since $G/Q$ is abelian, $G^{\prime} \le Q$.Thus we deduce from maximality of $M$ that $G=G^{\prime}M$, and so $Q=G^{\prime}(Q \cap M)$. Please help me to find another non-normal non-cyclic subgroup of $G$ which is not conjugate to $M$ and $PT$.

答:
Since $Q \cap M \le Z(G)$, $Q$ must be abelian, so $Q = [P,Q] \times (Q \cap M)$, with $|[P,Q]| = q$.

Let $[P,Q] = \langle x \rangle$ and $T = \langle y,z \rangle$. Then we can take the subgroup $\langle xy,z \rangle$.


问:
Linear Differential Equation Using Integrating Factor
Here is problem I am trying to solve for my differential equations class. I have spent several hours trying to solve it and have not been successful. I have tried breaking up the fraction various ways and using different techniques of integration but can't seem to get a solution that is valid near $t=\frac{1}{2}$. Here is the problem;

Solve the IVP $$\frac{dy}{dt}= t + \frac{t}{t^2−1}y ; \quad y\left(\frac{1}{2}\right) = 6.$$
Hint: Be careful with your integrating factor as you need to describe a solution that is valid near $t = 1/2.$

I am trying to solve the problem in the form: $$\frac{dy}{dt} - \frac{t}{t^2−1}y = t $$ using an integrating factor as the hint suggests.

If anyone could suggest some further hints as to how to solve this problem I would appreciate it. I would like to figure it out without having the solution and steps given to me.

答:
$$\frac{dy}{dt}= t + \frac{t}{t^2−1}y$$
In the present case the method of integrating factor isn't the simplest. Nevertheless we will use it as requested.

You wrote : I am trying to solve the problem in the form  $\frac{dy}{dt}- \frac{t}{t^2−1}y=t$ using an integrating factor.

This is a bad start because the correct form to start is
$$N(t,y)dt+M(t,y)dy=0$$
You should start with this form :
$$\left(t + \frac{t}{t^2−1}y\right)dt-dy=0$$
Then one have to find an integrating factor $\mu(t,y)$ so that the equation
$$\mu(t,y)\left(t + \frac{t}{t^2−1}y\right)dt-\mu(t,y)dy=0\quad\text{be exact}.$$

HINT:

Try the simplest cases of function $\mu(t,y)$. For example try $\mu$ function of $t$ only. You will find
$$\mu=\frac{1}{\sqrt{t^2-1}}$$


问:
Let $C[a,b]$ be the set of continuous functions. Show that $\psi : C[a,b] \to \Bbb{R}, \psi(f) = \int_{a}^{b}f(x) \ dx$ is continuous.

Let $C[a,b]$ be the set of continuous functions with the sup norm. Show that $\psi : C[a,b] \to \Bbb{R}, \psi(f) = \int_{a}^{b}f(x) \ dx$ is continuous.

I have that $\forall \varepsilon >0 $ there exists $\delta >0$ s.t $\|f-g\|<\delta \implies |\psi(f)-\psi(g)| < \varepsilon.$ Now $$|\psi(f)-\psi(g)|=|\int_{a}^{b}f(x) \ dx - \int_{a}^{b}g(x) \ dx| \leqslant |\int_{a}^{b}f(x) -g(x) \ dx| \leqslant \int_{a}^{b}|f(x)-g(x)|\ dx$$

but I'm not sure how I can continue from here. What should I do to find $\delta$ that would work here?

答:
With sequences: Let $(f_n)$ be a sequence in $C[a,b]$ and $f \in C[a,b]$ such that $||f_n-f||_{\infty} \to 0.$ That means that $(f_n)$ converges uniformly to$f$.

Thus

$$\int_{a}^{b}f_n(x) \ dx \to \int_{a}^{b}f(x) \ dx .$$

This means

$$ \psi(f_n) \to  \psi(f).$$


问:
Removing a point from a set of i.i.d. points.
Let $n$ datapoints are sampled iid from a distribution $\mathcal{D}$. If we remove a single point (point having maximum leverage score) after looking at all the points, can we say the remaining $n-1$ points are iid?

If we sample $n$ points from a Gaussian $\mathcal{N}(0,1)$ and remove all the points less than zero, then the remaining points can not be said to be sampled from a Gaussian, but are they independent? What is the mathematical characterization of the remaining set of points?

Also, if possible, please mention any reference to read about such problems.

答: Consider reading about censoring and truncation.</p>


问:
Evaluate $\lim\limits_{x \to { \infty } } (\frac{x}{x+2})^x$. Need for explanation.
My question is about the procedure for this limit problem:
$$\lim\limits_{x \to  { \infty } } (\frac{x}{x+2})^x$$

My solution was like that:

$$(\frac{x}{x+2})^x=e^{x\ln\frac{x}{x+2}}
= e^u$$ with $\ u = x \ln(\frac{x}{x+2})$.

Then $$\lim\limits_{x \to  { \infty } } x\ln(\frac{x}{x+2})
=\lim\limits_{x \to  { \infty } }{\ln{x\over x+2}\over {1\over x}}$$

Applying L'Hôpital's rule:

$$\lim\limits_{x \to  { \infty } } -{{2\over x(x+2)}\over
 {1\over x^2}} = \lim\limits_{x \to  { \infty } } -{2x\over x+2} = -2 $$

$$\lim\limits_{x \to  { \infty } } u = -2 $$

∴ $\lim\limits_{u \to  { \ -2 } } e^{u} = e^{-2} = {1\over e^{2}} $

However, according to my answer sheet, the correct answer is $e^{2}$. So, Please I need to know where's my mistake here.

Thank you.

答:
Your answer is correct! Possibly easier proof
\begin{align}
\lim_{x\to\infty}\left(\frac{x}{x+2}\right)^x
= \lim_{x\to\infty}\frac{1}{\left(\frac{x+2}{x}\right)^x}
= \lim_{x\to\infty}\frac{1}{\left(1+\frac{2}{x}\right)^x}
= \frac{1}{e^2} = e^{-2}
\end{align}


问:
Trouble understanding differential equation using Python
I want to use python to solve a set of equations describing a thermistor, a electronic component where the electrical resistance changes with temperature. The upper equation G(T) describes the conductance through the thermistor, dependent on the temperature T and the bottom equation how T varies with time, where $T_{ON}$ is an initial condition and v is the applied voltage, which I want to be a sinusoidal voltage.
\begin{equation}
    G(T) = \bigg( R_{ON} \cdot e^{\beta_N \big(\frac{1}{T}-\frac{1}{T_{ON}} \big)} \bigg)^{-1}
\end{equation}
\begin{equation}
    \frac{dT}{dt} = \frac{\delta_N}{H_{CN}}(T_{ON}-T) + \frac{G(T)}{H_{CN}}v^2
\end{equation}

Using for example "scipy.odeint" I pass in dT/dt, an initial value, $T_{ON}$, and a time array. But I struggle to understand how to do it in this case..

If I wanted to plot G(T) as a function of applied voltage, do I pass in the arguments mentioned above and update G(T) for each iteration and store these G(t) values?

I try to pass in a vector of sinusoidal values v, for the time array, but get an error saying the time array must be monotonically increasing. Am I mixing up things here? The differential equation is not dependent on the time step, just the voltage v?

Any pointers or tips on where to look would be of great help!

答:
This should be pretty straightforward. You define the helper functions as described,

```<code>def G(T): return 1/(R_ON*exp(1/T-1/T_ON))
def v(t): return V_0*sin(W*t)
</code>```

Add the necessary import statements for sin and exp, from math or numpy, both should work. (The ODE solver only uses single-point evaluations in default mode, numpy has overhead to detect and handle arrays.)

Then define the ODE function as given

```<code>def odefun(t,T): return dta_N/H_CN*(T_ON-T) + G(T)/H_CN*v(t)**2
</code>```

and pass it to the ODE solver as usual. To plot $G(T(t))$ against $v(t)$, solve for a time array and then compute these functions for the solution (now one needs the numpy variants)

```<code>t = linspace(t0,tf,N+1)
T = odeint(odefun, T0, t, tfirst=True)
plot(v(t), G(T[:,0]))
</code>```

Body:
This should be pretty straightforward. You define the helper functions as described,

def G(T): return 1/(R_ON*exp(1/T-1/T_ON))
def v(t): return V_0*sin(W*t)

Add the necessary import statements for sin and exp, from math or numpy, both should work. (The ODE solver only uses single-point evaluations in default mode, numpy has overhead to detect and handle arrays.)

Then define the ODE function as given

def odefun(t,T): return dta_N/H_CN*(T_ON-T) + G(T)/H_CN*v(t)**2

and pass it to the ODE solver as usual. To plot $G(T(t))$ against $v(t)$, solve for a time array and then compute these functions for the solution (now one needs the numpy variants)

t = linspace(t0,tf,N+1)
T = odeint(odefun, T0, t, tfirst=True)
plot(v(t), G(T[:,0]))


问:
Equation with high exponents
I would appreciate any help with this problem:
$ x^8+2x^7+2x^6+5x^5+3x^4+5x^3+2x^2+2x^1+1x^0=0 $

I know that when $x$ isn't zero $x^0=1$ so the equation could be re-written as $ x^8+2x^7+2x^6+5x^5+3x^4+5x^3+2x^2+2x+1=0 $. I am not sure what to do from here. I have tried using wolframalpha to get the solutions so I know real ones exist (2 of them, actually), but I have no idea how to get them.

答:
As the equation is Reciprocal Equation of the First type,

Divide either sides by $x^4,$ to get $$x^4+\frac1{x^4}+2\left(x^3+\frac1{x^3}\right)+2\left(x^2+\frac1{x^2}\right)+5\left(x+\frac1x\right)+3=0$$

Put $x+\frac1x=y$ to reduce the equation the degree $\frac82=4$

Can you take it form here?

Title: Equation with high exponents


问:
How to find vector that is parallel to parametric vector?
Hello so I have a question on the following:

The line r = i + (  1 + 3t )j - ( 3 - 4t )k passes through $p_1=(1,1,-3)$ and is parallel to v = 3j + 4k.

But why can we say that this is parallel? And how can I apply this for different problems.

If someon could help me, it would be very much appreciated

###thanks in advance###

答:
$\mathbf r=\hat i+\hat j-3\hat k+t(3\hat j+4\hat k),t\in\Bbb R$ represents the equation of a straight line that passes through the point with position vector $\hat i+\hat j-3\hat k$ (take $t=0$). Take any two distinct points on the line, say for $t_1\ne t_2\in\Bbb R$,$$\begin{align*}\mathbf{r_1}=\hat i+\hat j-3\hat k+t_1(3\hat j+4\hat k)\\\mathbf{r_2}=\hat i+\hat j-3\hat k+t_2(3\hat j+4\hat k)\end{align*}$$Then $\bf r_1-r_2$ is a vector that is parallel to the line, i.e. the line is parallel to $(t_1-t_2)(3\hat j+4\hat k)$ or put simply, the line is parallel to $3\hat j+4\hat k$.

The information about a vector that is parallel to a line is as important as the information about the normal to a plane. It comes in handy in almost all applications involving lines, such as finding the distance of a point from the line, finding the image of a point with respect to a line, finding the angle between a line and a plane, etc..


答:
For the final question: you just pick your favorite type of box as your starting point. For this type of box we first define the "usual naive" notion of volume. Then, the idea is that once you properly define the Lebesgue measure (for example by first defining the Lebesgue outer measure using these boxes and then using Caratheodory's procedure to restrict to an appropriate $\sigma$-algebra), you can then prove that for any type of box, the Lebesgue measure of the box equals the product of its side lengths. So ultimately it doesn't matter which way you start.

Although here are just some small details to take note of:

Boxes with one side open one side closed, like $[a_1,b_1)\times \cdots \times [a_n,b_n)$ are nice because if you take two "neighbouring" boxes, then they will be disjoint (for example, $[0,1)$ and $[1,2)$ are disjoint) so maybe in the beginning stages of developing the theory this may come in handy. Another nice thing about such boxes is that any open set in $\Bbb{R}^n$ can be written as a countable disjoint union of such boxes.

Sometimes, closed boxes may be nicer to work with because they're compact. THis is a crucial fact which is used in the proof I know for the fact that the Lebesgue (outer) measure of any type of box (closed/open/mixed etc) is simply the product of side-lengths. More precisely, $B$ is an open box and $\overline{B}$ is the corresponding closed box, then one can show that for any subset $B\subset S\subset \overline{B}$, we have the outer measure satisfying $m^*(S)=\text{"usual volume" of $B$}$.

Title: Volume of open box and of closed box

Body:
In the book Analysis 2, Terence Tao defined the notion of an open box and its volume:

Definition 7.2.1 (Open box)

An open box (or box for short) $B$ in $\mathbb{R^n}$ is any set of the form:
$B = \prod_{i: 1 \to n} (a_i,b_i) := $ {$(x_1,..,x_n)\in \mathbb{R^n}: x_i \in (a_i,b_i) \, for \, all \, 1 <= i < = n$} ($a_i, b_i$ are real numbers).

We define the volume $vol(B)$ of this box to be the number:

$vol(B) := \prod_{i = 1 \to n} (b_i - a_i) =  \, (b_1 - a_1)(b_2 - a_2)...(b_n - a_n)$

By this definition, I understood that the notion of $volume$ is only defined for an open box. The author does not say anything about the volume of a closed box, or a half-closed box.

How can I rigorously extend the notion of a volume of an open box to the notion of a volume of a:

closed box

i.e. $B = \prod_{i: 1 \to n} [a_i,b_i] := $ {$(x_1,..,x_n)\in \mathbb{R^n}: x_i \in [a_i,b_i] \, for \, all \, 1 <= i < = n$}

half-closed box

$i.e.$ $B = \prod_{i: 1 \to n} (a_i,b_i] := $ {$(x_1,..,x_n)\in \mathbb{R^n}: x_i \in (a_i,b_i] \, for \, all \, 1 <= i < = n$}

EDIT:
What is the point of defining the notion of volume of an open box and not other kind of box (closed, half-closed) ? Is it personal preference of the author or there is some motivations behind ?


问:
How to fully factorise $2x^4+7x^3+4x^2-4x$?

How to fully factorise $2x^4+7x^3+4x^2-4x$?

I'm struggling to factorise polynomials like this one. I'm not sure how to best approach this problem. I've tried using the remainder and factor theorems but they seem to work only when at least one factor is known. Since I only know the polynomial itself, I'm struggling to efficiently solve these types of questions. Any help would be greatly appreciated

答:
Hint:

First factor out $x$:
$$2x^4+7x^3+4x^2-4x=x(2x^3+7x^2+4x-4)$$
Next, you can factor the cubic polynomial using the Rational roots theorem. After you've found a root $\alpha$, divide  by $x-\alpha$. The quotient will be a standard quadratic polynomial, of which the roots are easy to determine.


答:
Taking out a factor of $x$, we get
$$
x(2x^3+7x^2+4x-4) \, .
$$
To factorise this further, we need to find a root of the equation
$$
2x^3+7x^2+4x-4 = 0 \, .
$$
The rational root theorem states that if $p/q$ is a root of a polynomial, then $p$ must be an integer factor of the constant term, and $q$ must be an integer factor of the leading coefficient. Here, this means that $p$ must go into $-4$, and $q$ must go into $2$. Once you have found a root, you can factorise the cubic further, and you are left with a quadratic, which can be broken down further into two linear factors.

Body:
How to fully factorise $2x^4+7x^3+4x^2-4x$?

I'm struggling to factorise polynomials like this one. I'm not sure how to best approach this problem. I've tried using the remainder and factor theorems but they seem to work only when at least one factor is known. Since I only know the polynomial itself, I'm struggling to efficiently solve these types of questions. Any help would be greatly appreciated


问:
Prove that $\sum_{n=1}^{\infty}n!x^n$ only converges if $x=0$
The question title basically sums up the goal of this question, that is, how to show that the power series $$\sum_{n=0}^{\infty}n!x^n $$ converges only if $x=0$.

I will show my closest attempt at proving this, using the root test as all other methods, like divergence test or ratio test, seemed to be inconclusive or out of reach for me to prove.

Attempt

Although I cannot prove this (and it would be amazing if someone could offer a proof outline or show a proof of this), it is known that $$n!\geq \left(\frac{n}{3}\right)^n\tag{1}$$ for all $n\in\mathbb{N}$ which gives $$\sqrt[n]{n!}\geq \frac{n}{3}\tag{2}$$ for all $n\in\mathbb{N}$.

Assuming this, let $x\neq 0$. Then $$\lim_{n\to\infty}\sqrt[n]{n!|x|^n }=|x|\cdot\lim_{n\to\infty}\sqrt[n]{n!}=\infty $$ since $$\lim_{n\to\infty}\frac{n}{3}=\infty$$ thus the power series diverges by the root test.

Questions

I have two questions.

First: Is this valid assuming $(1)$ and $(2)$?

Second: How would one prove $(1)$ and $(2)$. I cannot do it inductively. Also, if there is a different or simpler method, please let me know or guide me in the right direction. Thanks

答:
In this case, you don't really need to use anything other than the fact that a series can only converge if its general term goes to zero. When $|x| \ge 1$, it is immediate to conclude that $\displaystyle \lim_{n \to \infty} n! x^n \ne 0$ and, therefore the series is divergent. If we show that when $x \in (-1,1)\setminus \{0\}$ the limit is also not zero, we are done. Denoting $y = \frac 1x$, we want to analyse
$$ \lim \dfrac{n!}{y^n}, \quad |y| > 1. $$

For $y>1$, using Stirling's formula, we get
$$
\lim \dfrac{n!}{y^n} = \lim \dfrac{\sqrt{2 \pi n} (n/e)^n}{y^n} = \lim \sqrt{2 \pi n} \left(\frac{n \cdot e}{y} \right)^n = +\infty.
$$


问:
Image of the idele class group and its subgroup of idelic norm 1
[Sorry if the title isn't specific, it was too long.]

My question is: Why does $J_{K}/J_{K}^{1}\cong
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ imply that $J_{K}/K^{\ast }$ and $J_{K}^{1}/K^{\ast }$ have the same image
under any continuous homomorphism from $J_{K}$ to a discrete group $G$?

Here $J_{K}$ denotes the group of ideles on a number field $K$ and $%
J_{K}^{1}$ the subgroup of elements with adelic norm $1$. I know it has
something to do with the fact that $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is connected, but don't know how to proceed from there.

Thank you.

答:
If $\phi: H \to G$ is a map from a topological group to a discrete topological group, then $\phi$ kills the connected component of the identity in $H$ (proof: the inverse image of the identity of $G$ is closed and open and contains the identity, hence contains the connected component of the identity).

But anything in $J_K$ can be moved to something in $J_K^1$ by multiplying by something in the connected component of the identity. (e.g. by modifying just one archimedean place.)


答:
The common $x$ is a dead give-away so we start with writing it as $x(2x^3 + 7x^2 + 4x -4)$.

Then you basically hope and pray that the remaining third degree term has an integer or easy rational root which you can then pull out as a linear term. Any third degree polynomial over $\Bbb R$ has a real root after all. But applying classic methods like Cardano can be nasty.. So hope for a nice root.

Candidates follow from the rational root theorem: divisors of $-4$ so $-1,1,-2,2$ (maybe halved because of the leading $2$).
It turns out that $x=-2$ is a root. So $x+2 = (x - (-2))$ divides your polynomial

and we get $$x(x+2)(2x^2 + 3x -2)$$

Now the final quadratic also has a root $x=-2$, found the same way

and we finally get $$x(x+2)^2(2x-1)$$

as the factorisation.

Title: How to fully factorise $2x^4+7x^3+4x^2-4x$?


答:
Assume there is an $\epsilon$ such that $|a-b|=\epsilon$. Then with any smaller $\epsilon'$, you would have $|a-b|>\epsilon'$, which contradicts the hypothesis "for all".

In fact, equality never occurs because $\epsilon$ is nonzero.

Title: Two real numbers $a$ and $b$ are equal iff $| a - b |\leq \epsilon$ for all $\epsilon\gt0$

Body:
Two real numbers $a$ and $b$ are equal iff $|a - b | \leq \epsilon$ for all $\epsilon >0$

I know that it's true if it says "iff $| a - b | < \epsilon$". But if I use $\leq$, isn't that statement false? Because intuitively, since $\epsilon$  can't be $0$, $| a - b |$ might not be $0$ for all $\epsilon$


问:
finding UMVUE for $\theta_x$/$\theta_y$
Let Xi ~ Exp($\theta_x$), Yj ~ Exp($\theta_y$), i = 1; ... ; n1, j = 1;...;n2. Find UMVUE of $\theta_x$/$\theta_y$.

Since $\bar{X}$ and $\bar{Y}$ are compelete sufficient statistic, by using Lehmann-scheffe theorem, UMVUE should be found by solving E($\bar{X}$/$\bar{Y}$), but due to inverse of exponential family, is it possible to find it?

答:
Assuming independence between $X_i $ and $Y_j$ the calculation of the following expectation

$$\mathbb{E}\left[\frac{\overline{X}_n}{\overline{Y}_m}  \right]$$

(I used $n,m$ instead of $n_1,n_2$ to simplify the notation...)

is very easy being

$$\mathbb{E}\left[\frac{\overline{X}_n}{\overline{Y}_m}  \right]=\frac{m}{n}\mathbb{E}\left[\frac{\Sigma_iX_i}{\Sigma_iY_i}  \right]=\frac{m}{n}\mathbb{E}\left[\Sigma_iX_i\cdot\frac{1}{\Sigma_iY_i}  \right]=\frac{m}{n}\mathbb{E}\left[\Sigma_iX_i\right]\cdot\mathbb{E}\left[\frac{1}{\Sigma_iY_i} \right]$$

and as known

$$\frac{1}{\Sigma_iY_i}\sim \text{Inverse Gamma}$$

Title: finding UMVUE for $\theta_x$/$\theta_y$


问:
Showing that $\frac{aA+bB+cC}{a+b+c}\geq\frac\pi3$ for $A$, $B$, $C$ the angles, and $a$, $b$, $c$ the sides, of a triangle

Let $A,B,C$ be the magnitude (in radians) of the three angles of a triangle, and let $a,b,c$ be the lengths of the sides opposite to these angles respectively. Prove that $$\frac{aA+bB+cC}{a+b+c}\geq\frac\pi3$$

Attempt:

WLOG, $A\geq B\geq C$.

As $A+B+C=\pi$, this implies $\frac{A+B+C}3=\frac\pi3$. We also note that by sine rule: $$\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}=d$$

for some $d$. This implies $a=d\sin A$, $b=d\sin B$, $c=\sin C$. Using these substitutions:

$$\frac{Ad\sin A+Bd\sin B+Cd\sin C}{d\sin A+d\sin B+d\sin C}\geq\frac{A+B+C}3$$

$$A\sin A+B\sin B+C\sin C\geq\frac{A+B+C}3(\sin A+\sin B+\sin C)$$

$$\frac{A\sin A+B\sin B+C\sin C}3\geq\frac{A+B+C}3\frac{\sin A+\sin B+\sin C}3$$

If we assume $A,B,C\leq\frac\pi2$, then we note that in the interval $\left[0,\frac\pi2\right]$, $\sin x$ is monotonically increasing, hence $\sin A\geq\sin B\geq\sin C$. Then by Chebyshev's inequality, the above inequality is true.

But if $A>\frac\pi2$, it is not necessarily true that $\sin A\geq\sin B\geq\sin C$, and I cannot apply Chebyshev's inequality! What should I do in such a case?

I did wonder if I could apply Jensen's inequality, but the function $x\sin x$ has an inflection point in the interval $[0,\pi]$ so I could not use it.

答:
Suppose $a \geqslant b \geqslant c,$ then $A \geqslant B \geqslant C.$ By the Rearrangement inequality, we have
$$3(aA+bB+cC) \geqslant (A+B+C)(a+b+c) = \pi(a+b+c),$$
or
$$\dfrac{aA+bB+cC}{a+b+c} \geqslant\frac{\pi}{3}.$$


问:
How do you solve this simultaneous equation?
Is there some kind of algorithm that I can use to find the exact values for v, u, e and f so that I can solve for a and b in the equation below?

```<code>av + bu = 5

ae + bf = 9
</code>```

If I were to apply random values to v, u, e and f, Most of the time, I am able to solve for a and b. But there are some times that I am not able to solve for a and b because the random values applied to v, u, e and f made the equation unsolvable (Haven't you guys encountered problems like this before where you can't solve for the unkown values of an equation because of its coefficients v,u,e and f?). Is there an exact method that I can use to initialize v, u, e and f so a and b is always solvable?

答:
These are two linear Diophantine equations. It is well known that $av+bu=5$ is solvable in $\Bbb Z$ if and only if $\gcd(u,v) \mid 5$, and similarly $ae+bf=9$ is solvable if and only if $\gcd(e,f)\mid 9$.

Reference:

How to find solutions of linear Diophantine ax + by = c?


问:
Topology of the intersection of two topological spaces.
If we are given two topological spaces $(X,\tau_X)$ and $(Y,\tau_Y)$. Then obviously we can form the set
$X\cap Y$ as the set of elements which belong to both $X$ and $Y$. My question is are we allowed to talk about the topological space $X\cap Y$ when $X$ and $Y$ are not subsets of a third space $Z$? and if yes what is the topology we usually put on $X\cap Y$ when $X$ and $Y$ are not subsets of a third space $Z$?

答:
There is no usually as this is not commonly done.

Yes, if both $X$ and $Y$ are subspaces of a common space $Z$ and their topology is the subspace topology wrt that $Z$. In that case, the topology on $X \cap Y$ is just the subspace topology of that $X \cap Y$ wrt that same $Z$, and $X \cap Y$ will also have the subspace topology wrt $X$ and $Y$, as everything is coherent and initial topologies are transitive.

If there is no relation like that such things are not done unless in the case of $Y=X$, so two topologies on the same set. (There even is such a thing as a bitopological space, a space with two topologies on it).


问:
Is a non-differentiable function convex near a strict minimum?
Let $f:(-1,1) \to \mathbb R$ be a continuous function, which is not differentiable at zero. Suppose that $f$ is strictly  decreasing on $(-1,0]$, and is strictly increasing on $[0,1)$.

Is $f$ convex in some neighbourhood of zero?

I know that if $f$ is smooth at zero, then it does not need to convex near zero. (One can construct examples using bump functions, see here and here.)

答:
Does this functions fit to your conditions?
$$
f(x) = \sqrt{|x|}
$$


问:
Good General Strategy For Solving For The Core In A Coalitional Game?
Does anyone have a good general approach to solving for the core of a coalitional game (or finding that there is no core)?</p>

答:
As it was correctly mentioned in the previous answer, the task to check whether the core of a game is non-empty is NP-complete, which is much easier to accomplish than the computation of the core itself. However, to directly check that the core is non-empty is still NP-hard, it requires in general to solve a LP.

It was also emphasized in the previous answer to focus on indirect strategies to infer that the core is empty or not. This is certainly the better way, but I cannot recommend the proposed strategies in that answer. To focus on other solution concepts may be much more harder than solving the non-emptiness issue directly. For instance, finding the nucleolus is in general much harder than to determine that the core is empty or not, since one has to solve iteratively a sequence of LPs. Thereafter, one has to check that the array of excesses is on each coordinate negative to conclude that the nucleolus is inside of the core or not. In negation of the latter case, this allows us to conclude that the core does not exist. Contrasted to the Shapley value, which may not belong to the core even if the core exists. Hence, the Shapley value is not suitable as an indirect device to infer on the non-emptiness of the core. The bargaining set is the hardest problem to solve of the listed problems, and gives also no indication about the existence of the core, since it may exist even if the core is empty. To compute an element of the Kernel is from the listed problems the easiest solution to determine. However, similar to the nucleolus, one has to determine the excess array to infer that it is a core element.

I recommend to referring to some game properties like convexity, average-convexity, or on the veto rich player property as an indirect device to conclude on the existence of the core. This is much easier to accomplish and more save. Though this approach does not cover all possible cases. Thus, sometimes one is forced to use the direct method.


问:
A question about symmetry of real random variables.
Let $X$ be a real random variable and let $x\in\mathbb{R}$ a real number such that $X$ is centrally symmetric with respect to $x$, i.e., $X-x$ is equally distributed tan $x - X$. My question is whether or not $|X - x|$ is equally distributed than $|x - X|$.

The inverse is false, but this implication i doubt.

Thanks in advance.

答:
I think it is true for the next reason. Let $X$ and $Y$ two real random variables equally distributed.

By definition of the absolute value, we have that
$$\mathbb{P}(|X|\leq t) = \mathbb{P}(-t\leq X\leq t)$$
for any $t\geq 0$. Then, by hypothesis
$$\mathbb{P}(-t\leq X\leq t) = \mathbb{P}(-t\leq Y\leq t) = \mathbb{P}(|Y|\leq t).$$

The case $t < 0$ it is clear that $\mathbb{P}(|X|\leq t) = 0 = \mathbb{P}(|X|\leq t)$.


问:
Showing that a function has a fixed point in a given interval
I want to be able to show that the function $g(x) = 2^{-x}$ has a unique fixed point contained within the interval $[1/3,1]$, and that the fixed-point iteration $$x_{n+1} = g(x_n), \qquad n \in \mathbb{N}$$

converges to this unique fixed point for all starting points $x_0 \in [1/3, 1]$.

How can one go about demonstrating something like this? Any help/ guidance would be appreciated.

答:
@Semmah: I made it by desmos.com , an ilustration with $$g(x)=2^{-x}\\\frac13=a\leq x_1\leq1\\
x_2=g(x_1)=g(a)\\
x_3=g(x_2)=g(g(a))\\
x_4=g(x_3)=g(g(g(a)))\\...$$ by scrolling $a$ you can see te fix point of $x=2^{-x}$

https://www.desmos.com/calculator/thgnnppzq0


问:
Matrix norm induced from vector explanation
Can someone explain as thoroughly as possible what a matrix induced norm is? A concrete example would for this norm would help a lot...

$A \in C^{n \times n}, v \in C^n$

$|| A || = \displaystyle\sup_{v \neq 0}\frac{||Av||}{||v||}$

Thank you.

答:
One can view the induced norm as the largest factor by which $A$ can stretch any vector (while also changing its direction).

In other words, if we denote $\varphi := \|A\|$, then $\|Av\| \le \varphi \|v\|$, for all $v$, and $\varphi$ is the smallest nonnegative real number with such a property.

Note that the "stretch" above means "make longer by a factor (shorter if that factor is less than 1)" and "length" is dependent of the choice of the vector norm.

This, of course, is not a formal definition, but  more of a layman explanation if you have a hard time of grasping the concept.


问:
Prove that in any metric space $(X,\rho)$ closed disks are closed in the metric topology.
Here is my attempt:

Let $(X,\rho)$ be a metric space. By definition, a subset of a metric space is said to be closed in the metric topology if its compliment is open in $(X, \rho)$. Suppose $D_r(a)=\{x\in X : \rho(a,x)\leq r\}$ is the set of closed disks with center $a$ and radius $r$. Let $y\in X\setminus D_r(a)$. Then $\epsilon=\rho(x,y)-r>0$. Next, if $x\in D_r(y,\epsilon)$ then by the triangle inequality $\rho(x,z)\geq \rho(x,y)-\rho(y,z)>\rho(x,y)-\epsilon=r$. Which shows $(D_r(a))^\complement$ is open.

I'm usually wrong and haven't been doing amazing in this class so if someone could help me out with this I would really appreciate it. We are using the text "Elementary Topology Problem Textbook" by O.Ya. Viro, O. A. Ivanov, N. Yu. Netsvetaev, and V. M. Kharlamov and are in chapter 1 section 4 titled metric spaces.

答:
First mistake: $y \in X \setminus D(a,r)$ implies $\rho (y,a )> r$.  Your definition of $\epsilon$ does not make sense because you did not say what $x$ is.

Second mistake: In the last part you did not say what $z$ is.

Correct proof: $\rho (y,a )> r$. Let $\epsilon =\rho (y,a )- r$. Let us show that $B(y,\epsilon) \subset  X \setminus D(a,r)$. Pick $z \in B(y,\epsilon)$. Then $\rho (z,y) <\epsilon= \rho (y,a )- r$. Hence,  $\rho(z,a) \geq \rho (y,a)-\rho (y,z) >\rho (y,a)-\epsilon=r$. This finishes the proof.


问:
Let $C[a,b]$ be the set of continuous functions. Show that $\psi : C[a,b] \to \Bbb{R}, \psi(f) = \int_{a}^{b}f(x) \ dx$ is continuous.

Let $C[a,b]$ be the set of continuous functions with the sup norm. Show that $\psi : C[a,b] \to \Bbb{R}, \psi(f) = \int_{a}^{b}f(x) \ dx$ is continuous.

I have that $\forall \varepsilon >0 $ there exists $\delta >0$ s.t $\|f-g\|<\delta \implies |\psi(f)-\psi(g)| < \varepsilon.$ Now $$|\psi(f)-\psi(g)|=|\int_{a}^{b}f(x) \ dx - \int_{a}^{b}g(x) \ dx| \leqslant |\int_{a}^{b}f(x) -g(x) \ dx| \leqslant \int_{a}^{b}|f(x)-g(x)|\ dx$$

but I'm not sure how I can continue from here. What should I do to find $\delta$ that would work here?

答:
Assuming $a \neq b$, then pick $\delta = \frac{\varepsilon}{(b-a)}$. We want to show $\|f-g\|_{\infty}<\delta \implies |\psi(f)-\psi(g)| < \varepsilon.$ Indeed notice that:
$$|\psi(f)-\psi(g)|=|\int_{a}^{b}f(x) \ dx - \int_{a}^{b}g(x) \ dx| =|\int_{a}^{b}f(x) -g(x) \ dx| \leqslant \int_{a}^{b}|f(x)-g(x)|\ dx \leq (b-a) ||f-g||_{\infty} \leq (b-a) \frac{\varepsilon}{(b-a)} = \varepsilon $$


答:
Relying on Banach-Caccioppoli fixed-point theorem, the following result can be proven:

Theroem: Let $f$ be a function which is continuous on a closed interval $[a,b]$, and such that $f(x) \in [a,b]$ $\forall x \in [a,b]$. Then $f$ has at least one fixed-point $\alpha \in [a,b]$.

If $f \in C^{1}(a,b)$ and there exist a constant $\beta <1$ such that $|f'(x)| \leq \beta <1 \forall x \in (a,b)$ then we can conclude that there is a single fixed-point in $[a,b]$. In this case, for every $x_{0} \in [a,b]$ the iteration $x_{n+1}=f(x_{n})$ converges to the unique fixed-point $\alpha$.

So, in your case: $g(x)=2^{-x}$ is continuous in the interval $[\frac{1}{3},1]$. We know that: $g(\frac{1}{3})=2^{-\frac{1}{3}}<1$ and $g(1)=\frac{1}{2}>\frac{1}{3}$. The function $g$ is streactly decreasing, so $g(x)\in [\frac{1}{3},1]$ $\forall x \in [\frac{1}{3},1]$. Furthermore: $|g'(x)|=|-\frac{log(2)}{2^x}|\leq \frac{log(2)}{2^{\frac{1}{3}}}<1$ $\forall x \in [\frac{1}{3},1]$. You can conclude applying the theorem we've seen before.

Title: Showing that a function has a fixed point in a given interval

Body:
I want to be able to show that the function $g(x) = 2^{-x}$ has a unique fixed point contained within the interval $[1/3,1]$, and that the fixed-point iteration $$x_{n+1} = g(x_n), \qquad n \in \mathbb{N}$$

converges to this unique fixed point for all starting points $x_0 \in [1/3, 1]$.

How can one go about demonstrating something like this? Any help/ guidance would be appreciated.


问:
Necessary and sufficient condition for two polynomials to generate coprime ideals
This question concerns the ring of polynomials with integer coefficients, $R=\mathbb Z[X]$. $R$ is not a principal ideal domain, and two elements $p,q\in R$ that are coprime does not necessarily generate coprime ideals, i.e. it is possible to have $\langle p \rangle+\langle q \rangle\neq R$.

I am investigating whether there is a simple way to decide whether two coprime elements $p,q$ generate coprime ideals. For example, the $\langle 2x+3 \rangle+\langle 4 \rangle\neq R$.

I have tried a few examples, but it turns out that every time I am directly expanding the coefficients:
$$
a(x)p(x)+b(x)q(x)=\sum_{r=0}^\infty \left(\sum_{i+j=r} (a_ip_j+b_iq_j) \right) x^r=1.
$$
So we may deduce that $a_0p_0+b_0 q_0=1$, so $p_0, q_0$ are coprime. And then the four integers $p_1, q_1, p_0, q_0 $ are $\mathbb Z$-linearly dependent, and so on. This is essentially a system of linear equations of infinitely many unknowns, and I don't know how to tackle the inverse of an infinite matrix - after all, it is NOT a complex matrix, but an integer matrix.

I am not quite getting the essence of the issue here. Are there any simple ways to decide whether of not $\langle p \rangle+\langle q \rangle= R$? Is there an algorithm for it?

In general, how to judge if two elements generate coprime ideals in a ring?

答:
If the principal ideals generated by $p,q$ are coprime, then their highest degree coefficients are coprime, since if $g$ is the greatest common divisor of the highest degree coefficients of $p,q$ then $g$ also divides the highest degree coefficient of every element of $(p)+(q)$. Suppose $\deg p \le \deg q=n$.

Lemma.
The highest degree coefficients of $p,q$ are coprime if and only if there is a monic polynomial of the form $ap+bq$ with degree at most $n$.

Proof.
Multiplying $p$ by $x^k$ for some $k$, we can make the degree of $x^kp$ equal to the degree of $q$. Then the degree $n$ coefficients of $x^kp,q$ are coprime iff there is an integer linear combination making them 1, iff there is an integer linear combination of $x^kp,q$ that is monic.
$\blacksquare$

Here is an algorithm:

Check that the highest degree coefficients of $p,q$ are coprime. If not, $(p)+(q)\ne (1)$, so stop.

Otherwise, there is a monic polynomial of degree $n$ of the form $ap+bq$, divide $p$ by $ap+bq$ and replace $p$ by the remainder.

Repeat.

This produces a sequence of generators of the ideal $(p)+(q)$ of decreasing degree. Either it terminates in some round of step 1 showing that the polynomials do not generate coprime ideals, or we get a monic polynomial of degree 0 in $(p)+(q)$, i.e. $(p)+(q)=(1)$.

In your example, $2x+3,4$ do not have highest degree coefficients coprime, so $(2x+3)+(4)\ne (1)$ as you said.


问:
Natural number solutions to $\frac{xy}{x+y}=n$ (equivalent to $\frac 1x+\frac 1y=\frac 1n$)
I have a question about the following problem from a Putnam review:

Let $n\in \mathbb{N}$. Find how many pairs of natural numbers $(x, y)\in \mathbb{N}\times \mathbb{N}$ solve
$$
\frac{xy}{x+y}=n.
$$

I have found some solutions for all $n$, such as $(2n, 2n)$ and $(n+1, n(n+1))$, but I feel as though this is the wrong approach, as the question is only asking for the number of solutions to the equation.

I don't really want a complete solution, but any hints would be greatly appreciated.

答:
HINT:

Utilize  $$(x-n)(y-n)=n^2$$

Body:
HINT:

Utilize  $$(x-n)(y-n)=n^2$$


答:
The AC method works with this polynomial. First factor an $x$ out to get $x(2x^3+7x^2+4x-4)$, and multiply the cubic by $4$ to get $8x^3+28x^2+16x-16 = (2x)^3 + 7(2x)^2 + 8(2x) - 16$. Then you can substitute $x = 2X$ to get $X^3 + 7X^2+8X-16$, where it is apparent that $X = 1$ is a root by the factor theorem.

Once you have one divisor, use a method such as polynomial long division to get the other factor, which must be a quadratic. Using synthetic division:

$1$
$1$
$7$
$8$
$-16$

$+$

$1$
$8$
$16$

$1$
$8$
$16$
$0$

thus $X^3 + 7X + 8X - 16 = (X - 1)(X^2 + 8X + 16)$. Find the roots of this polynomial, and finally use $x = 2X$ to get the original roots back. $x = 0$ also from earlier.


问:
On the monotone class theorem
In Real analysis for graduate students by R F Bass,

he proves that the intersection of two sets, either both chosen from $M$ or one is chosen from $M$ and the other $A_0$, belongs to $M$ again (by introducing sets $N_2$ and $N_3$). How proving for two sets implies for a countable intersection to make $M$ a $\sigma$-algebra? I also believe not all elements of $M$ are in a single chain of sets.

答:
I'm not completely sure if this is what you are asking for. But every monotone class $\mathcal{M}$ that is closed under complements and finite intersections must be a $\sigma$-algebra.

You can see it like this: Let $A_1, A_2, \ldots \in \mathcal{M}$. Then we also have
$$B_n := \bigcap_{k=1}^n A_k \in \mathcal{M}$$
for all $n\geq 1$, because $\mathcal{M}$ is closed under finite intersections (*). Furthermore we clearly have $B_1 \supseteq B_2 \supseteq B_3\supseteq \ldots$ so it follows that
$$\bigcap_{n=1}^\infty A_n = \bigcap_{n=1}^\infty B_n \in \mathcal{M}$$
(using that $\mathcal{M}$ is a monotone class).

Hence $\mathcal{M}$ is also closed under countable intersections (and complements, and thus also countable unions), and is therefore a $\sigma$-algebra.

(*) Or, if you only want to use intersections of two sets, do it inductively:
$B_1 = A_1 \in \mathcal{M}$ and $B_n \in \mathcal{M} \implies B_{n+1} = B_n\cap A_{n+1} \in \mathcal{M}$.

Title: On the monotone class theorem


答:
Just pointing out that the example in the first linked post is easily modified to produce a function that is non-differentiable at $0$, not convex at $0$, and $f(0)=0$ is a strict minimum (i.e. the smoothness was supposed to make it "harder" to prove):

$$f(x)=\begin{cases}x^2 + \max(0,x)^2\sin^2 \frac1x & x\neq 0\\ 0 & x=0\end{cases}$$
The $x^2$ makes it so that the unique zero of $f$ is $x=0$, and $f>0$ otherwise. Its not convex to the right of $0$ because $f$ "jumps" between the $x^2$ curve and $2x^2$ curve infinitely often. That it is not differentiable at $0$ is more or less standard.

Title: Is a non-differentiable function convex near a strict minimum?

Body:
Let $f:(-1,1) \to \mathbb R$ be a continuous function, which is not differentiable at zero. Suppose that $f$ is strictly  decreasing on $(-1,0]$, and is strictly increasing on $[0,1)$.

Is $f$ convex in some neighbourhood of zero?

I know that if $f$ is smooth at zero, then it does not need to convex near zero. (One can construct examples using bump functions, see here and here.)


问:
Kernel of bipartite linear map
Let V be a vector space. A linear map T : V →V is said to be bipartite if there are subspaces W1, W2 ⊂ V such that V = W1 ⊕W2 (internal direct sum). and such that T(W1) ⊂ W2 and T(W2) ⊂ W1. Let T be
bipartite with respect to the decomposition V = W1 ⊕W2. Show that dimKerT ≥ |dimW1 -dimW2|.

I have proved that the bases for W1 and W2 should be the bases for V. But I do not know how to convert this conclusion to the objective.

答: We assume WLOG that $\dim (W_1)\geq \dim (W_2)$ then we can look at the restriction of the map $T$ to $W_1$. Now by the rank nullity theorem https://en.m.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem used on the restriction we have that $$\dim(W_1)=\dim(\ker(T|_{W_1}))+\dim(Im(T|_{W_1}))$$ But as $V$ is bipartite we get by the condition on $T$ that $Im(T|_{W_1})\subset W_2$ hence $\dim(W_1)=\dim(\ker(T|_{W_1}))+\dim(Im(T|_{W_1}))\leq \dim(\ker(T|_{W_1}))+\dim (W_2)\leq \dim(\ker(T))+\dim(W_2) \Leftrightarrow \dim(\ker(T))\geq \dim(W_1)-\dim(W_2)$

Title: Kernel of bipartite linear map


问:
If $(a_n)$ is the sequence of non zero real numbers such that $\frac{a_{n+1}}{a_n}$ is a constant, then $\sum_{n=1}^{\infty}a_n$ is a geometric series
Let $(a_n)$ the sequence of non zero real numbers such that $\frac{a_{n+1}}{a_n}$ is a constant. Show that $\sum_{n=1}^{\infty}a_n$ is a geometric series.

I would like to know if my proof holds and have a feedback, please.

Let $k\in \mathbf{R}^*$ a constant of $\dfrac{a_{n+1}}{a_n}$. Then, $\frac{a_{n+1}}{a_n}=k$. Then, we can define the sequence $(a_n)$ recursively: $a_{n+1}=a_n\cdot k$.

First, note that $a_n=(a_n-a_{n-1})+...+(a_1-a_0)+a_0$ and that $\forall p\ge1$:

$a_p-a_{p-1}=k(a_{p-1}-a_{p-2})=...=k^{p-1}(a_1-a_0)$.

Thus,

$a_n=k^{n-1}(a_1-a_0)+k^{n-2}(a_1-a_0)+...+a_1$ and $\sum_{n=1}^{\infty}a_n=(a_1-a_0)\sum_{n=1}^{\infty}k^n+a_0$. We conclude that $\sum_{n=1}^{\infty}a_n$ is a geometric series.

答: What to have done is correct but unnecessarily complicated. (No point in using the differences $a_p-a_{p-1}$). Use induction to show that $a_n= a_1 k^{n-1}$. So $\sum k^{n}=\sum a_1k^{n-1}$ which is a geometric series.


Title: If $(a_n)$ is the sequence of non zero real numbers such that $\frac{a_{n+1}}{a_n}$ is a constant, then $\sum_{n=1}^{\infty}a_n$ is a geometric series

Body:
Let $(a_n)$ the sequence of non zero real numbers such that $\frac{a_{n+1}}{a_n}$ is a constant. Show that $\sum_{n=1}^{\infty}a_n$ is a geometric series.

I would like to know if my proof holds and have a feedback, please.

Let $k\in \mathbf{R}^*$ a constant of $\dfrac{a_{n+1}}{a_n}$. Then, $\frac{a_{n+1}}{a_n}=k$. Then, we can define the sequence $(a_n)$ recursively: $a_{n+1}=a_n\cdot k$.

First, note that $a_n=(a_n-a_{n-1})+...+(a_1-a_0)+a_0$ and that $\forall p\ge1$:

$a_p-a_{p-1}=k(a_{p-1}-a_{p-2})=...=k^{p-1}(a_1-a_0)$.

Thus,

$a_n=k^{n-1}(a_1-a_0)+k^{n-2}(a_1-a_0)+...+a_1$ and $\sum_{n=1}^{\infty}a_n=(a_1-a_0)\sum_{n=1}^{\infty}k^n+a_0$. We conclude that $\sum_{n=1}^{\infty}a_n$ is a geometric series.

Body:
$$A_{n+1}=k A_n$$
$$A_1=k A_0$$
$$A_2= k A_1$$
$$A_3=k A_2$$
...............
$$A_{n-1}=k A_{n-2}$$
$$A_{n}=k A_{n-1}$$
Multiplying them we get
$$A_n=k^n A_0.$$
Which$nth$ term of a GP. GP


Title: Evaluate $\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$

Body:
Evaluate

$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$$

My attempt : I put  $t= \sqrt{1-r^2}$ now  $dt/dr= \frac{-r}{2\sqrt {1-r^2}}$ $$\implies dr=\frac{2\sqrt {1-r^2}}{r}dt$$

$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}\frac{2\sqrt {1-r^2}}{r}dt$$

$$= 2\int_{0}^{1} r^2 dt$$

$t= \sqrt{1-r^2}\implies r^2= t^2-1$

$$2\int_{0}^{1} t^2-1 dt= 2[\frac{t^3}{3} -t]_{0}^{1}==-4/3$$

Body:
First of all, since $t= \sqrt{1-r^2}$, we should have $dt/dr = \frac {-r}{\sqrt{1-r^2}}$ and $t^2 + r^2 = 1$.

Now

$$\int_0^1 \frac {r^3}{\sqrt{1-r^2}}dr = \int_{\color{red}1}^{\color{red}0} \frac {r^3}{\sqrt{1-r^2}}\frac {\sqrt{1-r^2}}{-r}dt = \int_0^1r^2dt = \int_0^1(\color{red}{1-t^2})dt = [t-\frac{t^3}3]_0^1 = \frac23$$


答:
As an alternative: often it is shown that $d: X \times X \to \Bbb R$ is in fact a continuous function when $X$ is given the metric topology.
It follows that $d_a: X \to \Bbb R, x \to d(x,a)$ is also continuous for any fixed $a \in X$.

Then $D_r(a) = \{x\mid d(x,a) \le r\}= (d_a)^{-1}[(-\infty, a]]$ is closed as the inverse image of a closed set under $d_a$, using the order topology on $\Bbb R$.

Title: Prove that in any metric space $(X,\rho)$ closed disks are closed in the metric topology.

Body:
Here is my attempt:

Let $(X,\rho)$ be a metric space. By definition, a subset of a metric space is said to be closed in the metric topology if its compliment is open in $(X, \rho)$. Suppose $D_r(a)=\{x\in X : \rho(a,x)\leq r\}$ is the set of closed disks with center $a$ and radius $r$. Let $y\in X\setminus D_r(a)$. Then $\epsilon=\rho(x,y)-r>0$. Next, if $x\in D_r(y,\epsilon)$ then by the triangle inequality $\rho(x,z)\geq \rho(x,y)-\rho(y,z)>\rho(x,y)-\epsilon=r$. Which shows $(D_r(a))^\complement$ is open.

I'm usually wrong and haven't been doing amazing in this class so if someone could help me out with this I would really appreciate it. We are using the text "Elementary Topology Problem Textbook" by O.Ya. Viro, O. A. Ivanov, N. Yu. Netsvetaev, and V. M. Kharlamov and are in chapter 1 section 4 titled metric spaces.


答:
An example similar to daw's is:

$$f(x)=\begin{cases}\arcsin(-2x-1) + \frac{\pi}{2} & -1\leq x< 0\\   \arcsin(2x-1) + \frac{\pi}{2} & 0\leq x\leq 1  \end{cases}$$


Body: With $r=\sin t$ it's$$\int_0^{\pi/2}\sin^3tdt=\int_0^{\pi/2}(\sin t-\sin t\cos^2t)dt=[\tfrac13\cos^3t-\cos t]_0^{\pi/2}=\tfrac23.$$


问:
"Tooth" function and its Lipschitz property
Suppose that $\varphi:[0,1]\to \mathbb{R}$ is given by $$\varphi(x) =
\begin{cases}
x, & \text{if } x\in[0,1/2],\\
1-x, & \text{if } x\in [1/2,1].
\end{cases}$$ Let's continue this function over $\mathbb{R}$ with period $1$ and label the new function as $f:\mathbb{R}\to\mathbb{R}$. This function is used when we want to construct an example of continuous, nowhere differentiable function (van der Waerden function). I am going to prove that $|f(x)-f(y)|\leq C|x-y|$ and probably $C=1$, i.e. $f(x)$ is Lipshitz  function.

I have some difficulties to prove it. It is enough to consider the case when $|x-y|<\frac{1}{2}$. Suppose that $x\in [n,n+1)$. Then $x\in[n,n+\frac{1}{2})$ or $x\in [n+\frac{1}{2},n+1)$. So suppose that $x\in[n,n+\frac{1}{2})$ Since $|y-x|<\frac{1}{2}$, then $y\in (n-\frac{1}{2},n+1)$. Then we can consider the following 3 cases:

I.  $y\in (n-\frac{1}{2},n)$; II. $y\in [n,n+\frac{1}{2})$; III. $y\in [n+\frac{1}{2},n+1)$.

For example, in the first case we know that: $f(x)=x-n$ and $f(y)=n-y$. Then $|f(x)-f(y)|=|x-n-(n-y)|=|x+y-2n|$.

How to show that $|f(x)-f(y)|\leq |x-y|$ in that case?

I'd be thankful for help!

答: Here is an interesting proof: Let $g(x)=1$ for $0 \leq x \leq \frac 1 2$ and $g(x)=-1$ for $\frac 1 2 \leq x \leq  1 $. Extend $g$ to a peridic function with period $1$. Then you can check that $f(x)=\int_0^{x} g(t)dt$. Hence, $|f(x)-f(y)| \leq |\int_x^{y} g(t)dt| \leq |x-y|$.


问:
Projection of matrix onto subspace
I am confused about why the orthogonal projection of matrices onto subspaces is given by a change-of-basis-like formula. For example, in the below image from these notes, why is the orthogonal projection of matrix A onto the subspace $V_m$ given by $V_m^* A V_m$?

When I search online, I come across a formula like  (for example, from the Wikipedia article), which does not look at all like $V_m^* A V_m$.

答:
I have the same question, but don't have the reputation to comment.

It's worth noting that you have two different $A$ matrices in your question - the $A$ in the standard projection formula corresponds to your $V_m$. Because the column-vectors of the subspace are orthonormal, $V_m^T V_m = I$, and so the projection matrix (in this notation) is $P \equiv V_m V_m^T$. Here is where I get stuck.

Title: Projection of matrix onto subspace


问:
A simple question about nonnegativity of square roots
Say I have a sequence $(x_n)$ and also that $x_n \geq 0 $. Let $x = \lim_{n\to\infty}x_n$. I also know that $x \geq 0$. My question is,

Is $\sqrt{x_n} + \sqrt{x}$ a nonnegative quantity? I believe this is so because the square root operator returns the principal (nonnegative root)? Is this reasoning correct?

答: If $a,b \ge 0$, then $\sqrt{a},\sqrt{b} \ge 0$ hence $\sqrt{a}+\sqrt{b} \ge 0.$


答: Yes your reasoning is correct. The square root operator returns the principal (non-negative) root.</p>

Title: A simple question about nonnegativity of square roots

Body:
Say I have a sequence $(x_n)$ and also that $x_n \geq 0 $. Let $x = \lim_{n\to\infty}x_n$. I also know that $x \geq 0$. My question is,

Is $\sqrt{x_n} + \sqrt{x}$ a nonnegative quantity? I believe this is so because the square root operator returns the principal (nonnegative root)? Is this reasoning correct?