问:
Collatz Conjecture: does an integer $4x+1$ resolve in $n$ steps if and only if an odd $x$ resolves in $n$ steps
When analyzing the collatz conjecture in reverse. I came across this argument that attempts to show that an integer $4x+1$ resolves in $n$ steps if $x$ is odd and resolves in $n$ steps and likewise if $x$ is odd an resolves in $n$ steps, then $4x+1$ likewise resolves in $n$ steps.

Is my reasoning correct?

Let:

$\nu_2(x)$ be the 2-adic valuation of $x$

$x_1, x_2, \dots, x_n$ be $n$ distinct odd integers such that:

$x_{i+1} = \dfrac{3x_i + 1}{2^{\nu_2(3x_i+1)}}$

$x_1$ resolves in $n$ steps if $\dfrac{3x_n+1}{2^{\nu_2(3x_n + 1)}} = 1$

Observation 1:  if an odd integer $x$ resolves to $1$ in $n$ steps, then there exist $n$ positive integers $a_1, a_2, \dots, a_n$ where $x = f(a_1, a_2, \dots, a_n)$

(1)  If $n=1$, then $\dfrac{3x+1}{2^{a_1}} = 1$ and $x = f(a_1) = \dfrac{2^{a_1}-1}{3}$

(2)  Assume that for any $x$ that resolves in $n-1$ steps, there exist $n-1$ integers $a_1, \dots, a_{n-1}$ with $f_{n-1}(a_1, a_2, \dots, a_{n-1})= x$

(3)  If $x$ resolves in $n$ steps, there exists a positive integer $a_n$ such that $\dfrac{3x+1}{2^{a_n}}$ is an integer that resolves in $n-1$ steps and $x = f(a_1, a_2, \dots, a_n) = \dfrac{3f_{n-1}(a_1, a_2, \dots, a_{n-1}) + 1}{2^{a_n}}$

Observation 2: if an odd integer $x$ resolves in $n$ steps, then it follows that $4x+1$ also resolves in $n$ steps

(1) From Observation 1, there exist $a_1, a_2, \dots, a_n$ such that $\dfrac{3x+1}{2^{a_n}} = f_{n-1}(a_1, \dots, a_{n-1})$

(2) $\dfrac{3(4x+1)+1}{2^{a_n+2}} = \dfrac{12x+4}{2^{a_n+2}} = \dfrac{4(3x+1)}{2^{a_n+2}} = f_{n-1}(a_1, \dots, a_{n-1})$

Corollary: If $x$ is a positive odd integer and $4x+1$ is a solution in $n$ steps, then $x$ is also a solution in $n$ steps:

By assumption, there exist $a_1, a_2, \dots, a_n$ such that:

$$\frac{3(4x+1)+1}{2^{a_n}} = \frac{4(3x+1)}{2^{a_n}} = \frac{3x+1}{2^{a_n-2}}= f_{n-1}(a_1, a_2, \dots, a_{n-1})$$

Update 1: corrected an ambiguity -- my definitions only apply to odd numbers.

Update 2: I wrote a simple java app to validate.  I have been able to validate this result up to $90\times10^6$.

My code iterates through odd numbers $x$ and validates that the count of steps for $x$ matches the count of steps for $4x+1$.

答: If $x$ is odd then after one (×3+1) step we have $3x+1$. We then complete the op's step by applying $k$ (/2) steps. Where $k$ is the largest integer such that $2^k|3x+1$. $4x+1$ is also odd so after one (×3+1) step we have $12x+4$ then if we apply two (/2) steps we get $3x+1$ then we divide by $2^k$ and get to the same number. So after one of the op's steps $x$ and $4x+1$ arrive at the same number $\frac{3x+1}{2^k}$.


问:
Proving that $f$ verifies $f(1-x)=-f(x)$
I'm trying to show an identity verified by a function. I have this function which is defined for all real numbers as

$ f(x) = \sum_{k=0}^{p-1} \frac{(x+k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!}$

I would like to show that $f(1-x) = -f(x)$.

First thing that tried:

$
\begin{align*}
f(1-x)
&= \sum_{k=0}^{p-1} \frac{(1-x+k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!} \\
&= \sum_{k=0}^{p-1} \frac{(-1)^{2m-1}(x-1-k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!}
\quad \text{factoring the numerator by }(-1)^{2m-1}\\
&= -\sum_{k=0}^{p-1} \frac{(x-1-k)^{2m-1}}{(-1)^{p+k} (p-1-k)!(p+k)!}
\quad \text{reducing }(-1)^{2m-1}=-1\\
&= -\sum_{i=-p}^{-1} \frac{(x+i)^{2m-1}}{(-1)^{p-i-1} (p+i)!(p-i-1)!}
\quad \text{doing a change of variable } i = -k-1 \\
\end{align*}
$

I'm kind of stuck at this point and I don't see how to move forward with it. I think I'm not going to the optimal direction to prove that relation.

Second thing that I am investigating:

Since $f(x)$ is a polynomial of degree $2m-1$, I could write it as a finite power series centered at $0$. So with Taylor's theorem I would have

$f(x) = \sum_{i=0}^{2m-1}  \frac{f^{(i)}(0) }{i!} x^{i}$.

and

$
\begin{align*}
f(1-x)
&= \sum_{i=0}^{2m-1}  \frac{ f^{(i)}(0) }{i!} (1-x)^{i} \\
&= \sum_{i=0}^{2m-1}  \frac{ f^{(i)}(0) }{i!} \Bigg( \sum_{j=0}^{i} \binom{i}{j} (-x)^{j} \Bigg)
\quad \text{binomial expansion applied to } (1-x)^{i} \\
&= \sum_{j=0}^{2m-1}  \Bigg((-1)^{j}\sum_{i=j}^{2m-1} \binom{i}{j} \frac{ f^{(i)}(0) }{i!} \Bigg) x^{j}
\quad \text{changing the summation order} \sum_{i=0}^{2m-1}\sum_{j=0}^{i} \text{ to }   \sum_{j=0}^{2m-1}\sum_{i=j}^{2m-1}  \\
\end{align*}
$

So showing that $f(1-x) = -f(x)$ is equivalent to showing that
$(-1)^{j}\sum_{i=j}^{2m-1} \binom{i}{j} \frac{ f^{(i)}(0) }{i!} = - \frac{f^{(j)}(0) }{j!} $.
I think this is the more elegant way to go but I don't know if it is going to make things worse.

Any feedback ?

EDIT : clarification of the parameters m and p :
I forgot to mention hypothesis on the parameters $m$ and $p$. So we consider that $ 1 \leq m < p$

答:
We can rewrite the sum as
$$
\eqalign{
  & f\left( x \right) = \sum\limits_{k = 0}^{p - 1} {{{\left( {x + k} \right)^{\,2m - 1} }
 \over {\left( { - 1} \right)^{\,p + k} \left( {p - 1 - k} \right)!\left( {p + k} \right)!}}}  =   \cr
  &  = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k} {{\left( {x + k} \right)^{\,2m - 1} }
 \over {\left( {p - 1 - k} \right)!\left( {p + k} \right)!}}} \quad  \Rightarrow   \cr
  &  \Rightarrow \quad g\left( x \right) = \left( {2p - 1} \right)!f\left( x \right) =   \cr
  &  = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
 \left( \matrix{ 2p - 1 \cr p + k \cr}  \right)\left( {x + k} \right)^{\,2m - 1} }  \cr}
$$

Then for the sum $g(1-x)+g(x)$ we get
$$
\eqalign{
  & g\left( {1 - x} \right) + g(x) =   \cr
&  = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
 \left( \matrix{2p - 1 \cr p + k \cr}  \right)\left( {1 - x + k} \right)^{\,2m - 1} }  +   \cr
  &  + \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
 \left( \matrix{2p - 1 \cr p + k \cr}  \right)\left( {x + k} \right)^{\,2m - 1} }  =   \cr
  &  = \left( { - 1} \right)^{\,2m - 1} \sum\limits_{k = 0}^{p - 1}
 {\left( { - 1} \right)^{\,p + k} \left( \matrix{ 2p - 1 \cr  p + k \cr}  \right)
\left( {x - k - 1} \right)^{\,2m - 1} }  +   \cr
  &  + \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
 \left( \matrix{ 2p - 1 \cr   p + k \cr}  \right)\left( {x + k} \right)^{\,2m - 1} }  =   \cr
  &  =  - \left( { - 1} \right)^{\,2m - 1} \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p - 1 - k}
 \left( \matrix{2p - 1 \cr p - 1 - k \cr}  \right)\left( {x - p + p - 1 - k} \right)^{\,2m - 1} }  +   \cr
  &  + \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k}
 \left( \matrix{2p - 1 \cr p + k \cr}  \right)\left( {x - p + p + k} \right)^{\,2m - 1} }  =   \cr
  &  = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,k}
 \left( \matrix{2p - 1 \cr k \cr}  \right)\left( {x - p + k} \right)^{\,2m - 1} }  +   \cr
  &  + \sum\limits_{k = p}^{2p - 1} {\left( { - 1} \right)^{\,k}
 \left( \matrix{2p - 1 \cr k \cr}  \right)\left( {x - p + k} \right)^{\,2m - 1} }  =   \cr
  &  = \sum\limits_{k = 0}^{2p - 1} {\left( { - 1} \right)^{\,k}
 \left( \matrix{2p - 1 \cr k \cr}  \right)\left( {x - p + k} \right)^{\,2m - 1} }  =   \cr
  &  = \left( { - 1} \right)^{\,2p - 1} \Delta ^{\,2p - 1} \left( {x - p} \right)^{\,2m - 1}  = 0
\quad \;\left| {\,1 \le m < p} \right. \cr}
$$
and that is the $2p-1$-th finite difference of a polynomial of degree $2m-1$
which is null if $ m < p$

ContentLicense: CC BY-SA 4.0


问:
When is a stopped Brownian motion not a markov process?
Let $W_t,t\geq 0$  be a Brownian motion start from zero and $\tau$ a stopping time. For what kinds of $\tau$, $W_{t \wedge \tau}$ is not a markov process?

I have considered choices of constant stopping time, like $\tau=1$; and the hitting time, but they do not work well.

答:
Example: How about $\tau:=\inf\{t>T_1: W_t=0\}$, where $T_1:=\inf\{t>0: W_t = 1\}$. In words, $\tau$ is the time of first return to state $0$ after the first visit to  state 1.

The process $X_t:=W_{t\wedge\tau}$ is certainly not a strong Markov process because state $0$ turns into a trap after the first visit to state $1$. It's not even a simple Markov process because for a bounded measurable $f$ and $0<s<t$,
$$
E[f(X_t)\mid\mathcal F^X_s]=1_{\{\tau\le s\}} f(0)+1_{\{\tau>s\}}h(X_s,t-s),
$$
where $h(x,u):=E^x[f(W_{u\wedge T_0})]$, $T_0$ is the hitting time of $0$ by $W$, and $E^x$ is expectation relative to the Brownian motion started at $x$. The above display shows that the conditional distribution of $X_t$, given the history up to time $s$, depends not just on $X_s$ but also on the event $\{\tau\le s\}$.

A stopping time $T$ of $W$ is called a terminal time if $T(\omega)=t+T(\theta_t\omega)$ for all $\omega$ in the event $\{t<T\}$, for each $t\ge 0$. Here $\theta_t$ is the shift operator acting on the sample space of the Brownian motion by $W_u(\theta_t\omega) = W_{u+t}(\omega)$. A hitting time like $T_A:=\inf\{t>0:W_t\in A\}$ is a terminal time. If $T$ is a terminal time then the stopped process $W_{t\wedge T}$ will be a Markov process.


问:
Change of variables in a convolution..
I'm in trouble with change of variables in a convolution:

Definition: The convolution of two $2\pi$-periodic functions $f$ and $g$ is defined as $$(f*g)(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}
f(y)g(x-y)\ dy.$$

I wan't to show $f*(g*h)=(f*g)*h$. Well, $$[f*(g*h)](x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\tau)(g*h)(x-\tau)\ d\tau\\ =\frac{1}{2\pi}\int_{-\pi}^{\pi}f(\tau)\left(\frac{1}{2\pi}\int_{-\pi}^{\pi}g(\rho)h(x-\tau -\rho)\ d\rho\right)\ d\tau.$$ Now I wan't to make the change of variables $\sigma=\tau+\rho$ but I don't know how to proceed with $d\rho$ and with the extremals $-\pi$, $\pi$. Can anyone explain me that? How can I know what is fixed in $h(x-\tau-\rho)$ inside the integral?

答:
First, use change of variables and the periodicity of $f$ and $g$ to deduce four useful facts about the convolution integral.

$(f\star g)(x)$ is a $2\pi$-periodic function of $x$.

$\displaystyle (f\star g)(x)=\frac{1}{2\pi}\int_{-\pi+a}^{\pi+a}
f(y)g(x-y)\,\mathrm dy$ for each choice of real number $a$.

$\displaystyle (f\star g)(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}
f(y)g(x-y)\,\mathrm dy = \frac{1}{2\pi}\int_{-\pi}^{\pi}
f(x-y)g(y)\,\mathrm dy.$

$\displaystyle \frac{1}{2\pi}\int_{-\pi}^{\pi}
f(a+y)g(b-y)\,\mathrm dy = (f\star g)(a+b \bmod 2\pi)$.

So,
$$\begin{align}
(f \star (g\star h))(x) &= \frac{1}{2\pi}\int_{-\pi}^{\pi}
f(x-y)\left[(g\star h)(y)\right]\ \mathrm dy\\
&= \frac{1}{2\pi}\int_{-\pi}^{\pi}
f(x-y)\left[\frac{1}{2\pi}\int_{-\pi}^{\pi} g(y-t)h(t)\,\mathrm dt\right]\ \mathrm dy\\
&= \frac{1}{2\pi}\int_{-\pi}^{\pi}\left[\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-y)g(y-t)\,\mathrm dy\right]h(t)\,\mathrm dt\\
&= \frac{1}{2\pi}\int_{-\pi}^{\pi} \left[(f\star g)(x-t)\right]h(t)\,\mathrm dt\\
&= ((f\star g)\star h)(x).
\end{align}$$

ContentLicense: CC BY-SA 3.0

Title: Change of variables in a convolution..


问:
Interpreting english EITHER-OR when approaching logical deductions doubt
The following question has been asked in the Indian entrance examination "GATE CS 2021", and it has stirred up a lot of confusion regarding how one should interpret "either-or" when coming to logical questions where we are given certain premises and have to determine if something is true via deduction.

Q. Given below are two statements 1 and 2, and two conclusions I and II

Statement 1: All bacteria are microorganisms.
Statement 2: All pathogens are microorganisms.

Conclusion I: Some pathogens are bacteria.
Conclusion II: All pathogens are not bacteria.

Based on the above statements and conclusions, which one of the following options is logically CORRECT?

Only conclusion I is correct

Only conclusion II is correct '

Either conclusion I or II is correct

Neither conclusion I nor II is correct

refer this link for the debate - https://gateoverflow.in/357468/gate-cse-2021-set-1-ga-question-9
refer this link for my reasoning - https://gateoverflow.in/357468/gate-cse-2021-set-1-ga-question-9?show=358167#a358167

In short -
I feel option D is the correct choice, because neither conclusions follow from the premises, but on the other side a lot of people feel that option C is right because according to them option c is a tautology.

Also there is another confusion regarding the statement of conclusion 2 "All pathogens are not bacteria".
Does this mean that there is no pathogen which is also a bacteria, or does it mean there exists some pathogens who are definitely not bacteria?
Because when we say a similar statement like "All students are not geniuses", we generally mean that there are only few students who are geniuses, and most of the students are not geniuses. We don't mean to say that there isn't even a single student who is not a genius.

Can anybody please provide an unambiguous, mathematically backed argument for choosing the correct answer, or is the given question and options inherently ambiguous?

答:
Intro This type of questions tends to be problematic because you can't be really sure at what level of precision is the author of the question working. So sometimes what looks like a purely logical exercise (which should have a clear answer) ends up being rather vague or weird.

I think that there is one more ambiguity that is missing in the discussion of the question and that is the difference between correct and valid conclusion. Where correctness is more about the truth-value of the stamen whereas the validity of a conclusion means that it logically follows from premises (and a valid conclusion can be incorrect in the case of false premises).

So I think that the problem isn't about interpreting either-or which is commonly understood as xor. But the problem lies in understanding what author of the question means by logically correct. I would interpret it as valid...

Argument

Statements 1 and 2 both that both pathogens and bacterias are subsets of the bigger set of microorganism. Nothing more. Namely, you can't tell whether these two (sub)sets intersect (some pathogens are bacteria) or not (All pathogens are not bacteria).

So neither conclusion 1, not conclusion 2is valid. They don't follow from the premises.

So this gives us option 4.

However, when we speak only of the truth value of statements, then indeed, we get options 3 tautologically (with no need to look at the premises, i.e. staments 1 and 2) as mentioned in comments.

About big quantifier  The formulation used by the author of the question is rather weird. I think that reason behind it is, that it tries to mimic the way logical formula looks -with quantifier (all pathogens) at the beginning and the statement about them at the end ( x is not a bacteria).

In normal English, I guess, one would prefer no pathogen is bacteria.  Which is less ambiguous.


问:
finding the number of triplests (a,b,c) for which (G,*) it is a group
Let the set $ G=\lbrace x\in \mathbb{R}|x^3+ax^2+bx+c=0\rbrace $. Find $p$ which is the number of triplets $(a,b,c)$ from $\mathbb{R}^3$ for which $(G,\cdot)$ it is a group. ($\cdot$ is the multiplication of the real numbers).
I do not know how can be $(G, \cdot)$ a group because I learned only groups defined by a binary operation and I do not see this as $x^3+ax^2+bx+c=0$ a binary operation.

答:
Hint :

$G$ can have $0$, $1$, $2$ or $3$ elements. Because $G$ is a subgroup of $\mathbb{R}$, it must be $\lbrace 1 \rbrace$ or $\lbrace -1,1 \rbrace$. Now find all the triplets $(a,b,c)$ that give these two possibilities.


问:
the curve $y=x^2\cos{4x}$ for $0 \leq x \leq \dfrac{\pi}{8}$. The point $P$ is the maximum
Question a) Show that the $x$ coordinates of $P$ satisfies the equation $4x^2\tan{4x}=2x$

b) Show also the $x$ coordinates of $P$ satisfies the equation $x=\dfrac{1}{4}\tan^{-1}(1/2x)$

I have been stuck on this question for the past hour I don't understand it it is the last section of my A level book chapter 6 and I have no professor and the back of my book all that is written is proof. Any help is much appreciated thank you!

答:
The max occurs when $y'(x) = 0$.

The product rule of derivatives says that derivative of $\color{green}{f(x)}\color{red}{g(x)} = \color{green}{f(x)'}\color{red}{g(x)}+\color{green}{f(x)}\color{red}{g(x)'}$

And the chain rule says the derivative $\color{blue}{g({\color{orange}{h(x)}})}$ id $\color{blue}{g'({\color{orange}{h(x)}})}\color{orange}{h'(x)}$

$\color{green}{x^2} \color{red}{\cos 4x} $ is

$\color{green}{(x^2)'}\color{red}{\cos 4x} + \color{green}{x^2}\color{red}{(\cos 4x)'}=$

$\color{green}{(2x)}\color{red}{\cos 4x} + \color{green}{x^2}\color{red}{(\color{blue}{\cos}\color{orange}{4x})'}=$

$\color{green}{(2x)}\color{red}{\cos 4x} + \color{green}{x^2}\color{red}{(\color{blue}{\cos'}\color{orange}{4x})'}\color{orange}{(4x)'}=$

$\color{green}{(2x)}\color{red}{\cos 4x} + \color{green}{x^2}\color{red}{(\color{blue}{-\sin}\color{orange}{4x})'}\color{orange}{4}=$

$\color{green}{2x}\color{red}{\cos 4x} - \color{orange}4\color{green}x^2\color{green}{\sin(\color{orange}{4x})}$

So for $P$ to be the maximum is must occur at $(x, y)$ where $y'(x) = \color{green}{2x}\color{red}{\cos 4x} - \color{orange}4\color{green}x^2\color{green}{\sin(\color{orange}{4x})}=0$.

So that means $2x \cos 4x = 4x^2 \sin 4x$

So $2x = 4x^2 \frac {\sin 4x}{\cos 4x} = 4x^2 \tan 4x$.

That's a)

If we assume $x \ne 0$ then we can

Divide both sides by $4x^2$ to get

$\frac 1{2x} = \tan 4x$.

Take the $\arctan$ of both sides

$\arctan \frac 1{2x} = 4x$.  (Need to note that $-\frac \pi 2 < 4x < \frac \pi 2$ so $\tan $ is invertable)

So $x = \frac 14 \arctan \frac 1{2x}$.

That's $b$.

But what if $x = 0$.  Well, the book is sort of wrong.  $\frac 1{2x}$ is not defined but $\tan {\frac \pi 2} = \infty$ so if we extend $\lim_{x\to 0} \frac 1{2x} = \infty$ and allow for $\arctan \infty = 0$ then we will have

$x =0  = \frac 14 0 = \frac 14\arctan \infty = \frac 14\arctan (\lim_{w\to 0} \frac 1{2w})$, so the book is sort of right if we are allowed to say the scary things like $\frac 1{2\cdot 0} = \infty$ and $\arctan \infty = 0$.


问:
If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$

If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$

This seems clear, but I don't know how to prove this..

I was trying to show this by induction such that if $a^{n+1}$ = $rs$ and $b^{n+1}$ = $rt$, then $s,t$ are divisible by $a,b$ respectively, but i think this is a wrong way..

答:
We can proceed by induction using the associativity of gcd:

$\begin{align}\gcd(a^n,b)
&=\gcd(a^n,b\cdot\gcd(a,1))&\gcd(a,1)=1\\
&=\gcd(a^n,\gcd(ab,b))&\lambda\gcd(x,y)=\gcd(\lambda x,\lambda y)\\
&=\gcd(a^n,ab,b)&\text{associativity}\\
&=\gcd(\gcd(a^n,ab),b)&\text{associativity}\\
&=\gcd(a\cdot\gcd(a^{n-1},b),b)&\text{induction }\gcd(a^{n-1},b)=1\\
&=\gcd(a,b)\\
&=1\end{align}$

By symmetry you get both : $\quad\gcd(a,b^n)=\gcd(a^n,b)=1\quad\forall n\in\mathbb N$

Using exactly the same reduction process we get:

$\begin{align}\gcd(a^n,b^m)
&=\gcd(a^n,b^m\cdot\gcd(a,1))\\
&=\gcd(a^n,\gcd(ab^m,b^m))\\
&=\gcd(a^n,ab^m,b^m)\\
&=\gcd(\gcd(a^n,ab^m),b^m)\\
&=\gcd(a\cdot\gcd(a^{n-1},b^m),b^m)&\text{induction }\gcd(a^{n-1},b^m)=1\\
&=\gcd(a,b^m)\\
&=1\end{align}$

And we can conclude, notice that this is a strong induction because we suppose it verified for all couples $<(n,m)$ with lexicographic ordering. (check this https://math.stackexchange.com/a/7665/399263)


问:
Do the position and velocity vectors live in the same space?
This is a question I've been thinking about for a while. Position vectors are supposed be represented by an arrow from the origin that traces out a path, while one does not need to think of the velocity vector as having any sort of starting point because as long as the length and direction and the same, they represent the same velocity. From a pure math perspective, a vector space is just an object with a dozen or so properties with closure being one of them. But it makes no sense to add a position and a velocity vector right? So that means they do not belong to the same vector space. But at the same time, it is possible to take the dot or cross products of a position and velocity vector. So how do you explain that?

Also when we learn multivariable calculus, why do we need the position vector to originate from the origin? $\Bbb R^3$ or $\Bbb R^2$ satisfies all the properties of a vector space without needing to conjure up any sort arrow.

答:
I think it's important to distinguish between the space of points, which is also known as affine space (let's denote this by $A$), and the space of vectors $V$. The space of points looks a lot like a vector space except there is no point called the origin, so an element in affine space is depicted as just a dot and not an arrow. A vector $v \in V$ is an arrow you can slide around in $A$. The assumption is that given any two points $p, q \in A$, there is a unique vector that starts at $p$ and ends at $q$.

You can indeed add a vector $v$ to a point $p$ by positioning the start of the vector at $p$ and defining $p+v$ to be the point at the end of the vector. You can also subtract two points $p$ and $q$ by defining $q-p$ to be the vector that goes from $p$ to $q$. But it makes no sense to add two points together.

However, if you choose a point $p_0$ in affine space, there is a unique vector from $p_0$ to each point $p$. This is the position vector of a point $p$ relative to $p_0$. So there is a natural isomorphism $I_{p_0}: A \rightarrow V$, where $I(p) = p-p_0$, identifying each point with a vector. The definition of the position vector depends on which point $p_0$ you choose to be effectively the origin of space.

So points and vectors live in different spaces, but the position vector does live in the same vector space as the vector itself.


问:
Sturm Liouville Problem Transcendental Equation
I am trying to solve the following problem

$$ X''(x)+ \lambda X(x)=0$$
$$ X'(0)+2X(0)=0$$
$$ X'(1)=0$$
Show that
$$ \tan\left( \sqrt{ \lambda } \right)= -2/ \sqrt{ \lambda } $$
With the eigenfunctions
$$ X(x) = A\cos( \sqrt{ \lambda } (x-1)) $$

Lambda is positive real number and $x$ is between $0$ and $1$ but not  exactly $1$ or $0$.

答:
The general solution is given by $$X(x)=C_1\cos(x\sqrt{\lambda})+C_2\sin(x\sqrt{\lambda})$$

for arbitrary constants $C_1$ and $C_2$. Then $X'(x)=-C_1\sqrt{\lambda}\sin(x\sqrt{\lambda})+C_2\sqrt{\lambda}\cos(x\sqrt{\lambda})$, so we have

$$X'(0)+2X(0)=C_2\sqrt{\lambda}+2C_1=0\implies\space C_1=-\frac{\sqrt{\lambda}}{2}C_2$$

and

$$X'(1)=-C_1\sqrt{\lambda}\sin(\sqrt{\lambda})+C_2\sqrt{\lambda}\cos(\sqrt\lambda)$$
$$=\frac{\lambda}{2}C_2\sin(\sqrt{\lambda})+C_2\sqrt{\lambda}\cos(\sqrt{\lambda})=0$$

which gives $\tan(\sqrt{\lambda})=-\frac{2}{\sqrt{\lambda}}$ provided $\cos(\sqrt{\lambda})\neq 0.$ Then $$C_2=-\frac{2}{\sqrt{\lambda}}C_1=C_1\tan(\sqrt{\lambda})$$

So the eigenfunctions are given by

$$X(x)=C_1\cos(x\sqrt{\lambda})+C_2\sin(x\sqrt{\lambda})$$
$$C_1\big(\cos(x\sqrt{\lambda})+\tan(\sqrt{\lambda})\sin(x\sqrt{\lambda})\big)$$
$$=C_{1}\big(\frac{\cos(x\sqrt{\lambda})\cos(\sqrt{\lambda})+\sin(\sqrt{\lambda})\sin(x\sqrt{\lambda})}{\cos(\sqrt{\lambda})}\big)$$
$$=A\big(\cos(x\sqrt{\lambda})\cos(\sqrt{\lambda})+\sin(x\sqrt{\lambda})\sin(\sqrt{\lambda})\big)=A\cos(x\sqrt{\lambda}-\sqrt{\lambda})=A\cos(\sqrt{\lambda}(x-1))$$

where we have used the angle addition formula $\cos(\alpha-\beta)=\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)$ and $A=\frac{C_1}{\cos(\sqrt{\lambda})}$ is a constant.


问:
Finding intersection multiplicities of a tangent line and a quartic
I was given the following exercise. Consider the quartic curve $f(x) = xy^3 + yz^3 + zx^3 = 0$ in $P_\mathbb{C}^2$. And let $L$ be the tangent line at $P = [0:0:1]$. I need to find the intersection points of $L$ with the curve $f=0$ and the intersection multiplicities.

My attempt:

I know that by Bezout's theorem, I have that the number of intersections counted with multiplicity would be $4 \times 1 =4$. To find the equation of the line $L$ I have calculated the gradient of the curve:
$$grad f = (y^3 + 3x^2 z, 3y^2x+z^3,3yz^2+x^3)$$

which at the point $P$ is given by $(0,1,0)$. So the equation of the tangent line is $(0,1,0) \cdot (x,y,z) = 0$ or more concisely $y=0$. To find the intersection points I plugged in $y=0$ in the equation of the quartic and obtained: $$zx^3 = 0$$, which is only satisfied by the points $P_1 = [1:0:0]$ and $P_2 = [0:0:1]$. Now I'm not quite sure if what I did is correct and how to proceed to find the intersection multiplicities (which I would guess to be 3 for $P_1$ and 1 for $P_2$).

答:
Your intuition is right on. From your comment, I'm guessing that your definition of intersection multiplicity of affine plane curves $C: F(x,y) = 0$ and $D: G(x,y) = 0$ at a point $P$ with corresponding maximal ideal $m$ is
\begin{align*}
\newcommand{\C}{\mathbb{C}}
I_P(C \cap D) = \dim_{\C} \left(\frac{\C[x,y]}{(F,G)}\right)_m \, .
\end{align*}

I'll compute the multiplicity for $P_2$ and leave $P_1$ to you. Since intersection multiplicity is a local quantity we can compute it in an affine patch. Consider the affine open where $z \neq 0$ with affine coordinates $u = x/z$ and $v = y/z$. Your computations show that $(f,y) = (y, x^3 z)$, which on this affine patch corresponds to $(v,u^3)$. The ring $\C[u,v]/(v,u^3)$ is already local, so localizing at $m = (u,v)$ doesn't change anything, and we have
$$
\frac{\C[u,v]}{(v,u^3)} \cong \frac{\C[u]}{(u^3)} \cong \C \oplus \C u \oplus \C u^2
$$
as $\C$-vector spaces, so its dimension is $3$.


问:
Circles and lines geometry problem

Here is my task: An arbitrary ray $OE$ crosses the circle at point $D$:
$$x ^ 2 + (y - a / 2) ^ 2 = (a / 2) ^ 2$$
and a tangent to it(at point $E$), passing through point $C$, diametrically opposite to $O$.

Through points $D$ and $E$, straight lines are drawn parallel to the axes $Ox$ and $Oy$, respectively, up to the intersection at point M.

Equate the line formed by the $M$ points and draw it (Agnesi's curl).

My attempt

I think that this problem need using Polar coordinates. But I have no idea how to start. Maybe if we assume that $M$ has coordinates $(x,y)$ we can find this point(using polar coordinates like $x=r\sin(q)$ or something like that) but how to equate the line? And of course I have already drawn it. After an hour i have idea if we find point $M$ can we use $M$-coordinates to get parametric equation of a line? If yes, how to make first step? Ok, after a few time i found $x$ like a coordinate of a point $M$ (i use as a parameter angle between $OE$ and $OC$ (q) so $x=tgqa/cosq$) how can find $OB$?

答:
In conventional polar coordinates $(\rho, \theta)$ we measure the angle $\theta$ counterclockwise from the $x$ axis.
Of course you can instead use the angle $q$ measured clockwise from the $y$ axis, but then you must be careful not to simply copy any formulas that were developed for shapes in conventional polar coordinates.
You have already correctly rederived at least one such formula, which is
$x = r \cos\theta$ when converting conventional polar coordinates to Cartesian
but is $x = r \sin q$ in your coordinates.

There are well-known formulas for a circle tangent to the origin in polar coordinates. You can derive a formula for your circle by substituting expressions with $r$ and $q$ for $x$ and $y$ in
$$ x^2 + (y - a/2)^2 = (a / 2)^2, $$
or perhaps more easily by adding the segment $CD$ to the figure and examining
the right triangle $\triangle ODC$ with right angle at $D.$

You should eventually be able to figure out both the $x$ and $y$ coordinates of $M$ using the angle $q$ as a parameter. That would be one way to describe the curve.

However, the fact that you have the formula for the circle in Cartesian coordinates suggests that someone was thinking of doing the whole problem in Cartesian coordinates.
Using the $x$ coordinate of $M$ -- you might say the $x$ coordinate is $x_M$ --
you can write an equation of the line $OE$ in Cartesian coordinates.
Then the coordinates of $D$ are what you get when you solve for $x$ and $y$
simultaneously in $ x^2 + (y - a/2)^2 = (a / 2)^2 $
and in the equation of the line $OE.$
What you're looking for is an equation relating the $x$ and $y$ coordinates of $M$.

So those are two ways of describing the curve.
There may be other hints in the context of the question that would suggest which one will be more desirable.

Title: Circles and lines geometry problem


问:
computing the limit $\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta)$
I'm trying to compute the following limit and would greatly appreciate your heartening feedback on my solution.

The limit:

$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta)$

My steps in deriving the solution:

Preliminary identities:

$\sec \theta = \frac{1}{\cos \theta}$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

$\frac{1}{\cos \theta}-\frac{\sin\theta}{\cos\theta} = \frac{1-\sin\theta}{\cos\theta} = \frac{1-\sin^2\theta}{(1+\sin\theta)\cos\theta} = \frac{\cos^2\theta}{\cos\theta}\cdot \frac{1}{1+\sin\theta} = \frac{\cos\theta}{1+\sin\theta} = \frac{0}{1+1}$

When $\theta \to \frac{\pi}{2}$ then $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$

The answer being:

$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta) = 0$

答: Your proof is fine. Here's another:$$\sec\theta,\,\tan\theta\to\infty\implies\sec\theta+\tan\theta\to\infty\implies\sec\theta-\tan\theta=\frac{1}{\sec\theta+\tan\theta}\to0.$$


问:
Vector $x$ Solves Least Squares Problem $\iff$ $b - Ax \in \text{Null}(A^*)$
I am preparing for a graduate exam in Numerical Linear Algebra and I have to solve the following question (without calculus, and without any other theorems about Least Squares):

Let $A \in \mathbb{C}^{m \times n}$ with $n < m$ (not necessarily of full rank). For any $x \in \mathbb{C}^n$ let $r(x) = b - Ax$. Show that $x$ solves the least squares problem $\min_{x \in \mathbb{R}^n}||b - Ax||_2$ if and only if $r(x) \in \text{Null}(A^*).$

Below, I wrote down my attempt. Not only is it incomplete (it purports to prove just one direction of the equivalence), but later I realized that it contains a major flaw. I would immensely appreciate your help  in either fixing this solution, or finding a new one.

I have very little time on the exam, so as a bonus, I would be very thankful if you could tell me if you see bad habits in my mathematical writing which make the solution too long.

P.S. I have an inkling that a correct solution might involve a QR decomposition, but I'm not sure.

Flawed Solution:

Lemma: The equation $A^*Ap = A^*b$ has a solution.

Proof: We are trying to show that $A^*b \in \text{Col}(A^*A) = \text{Null}(A^*A)^{\perp}$. Take any $y \in \text{Null}(A^*A).$ We will show that $A^*b \perp y$.

But first, we will show $Ay = 0$. Indeed, $$y \in \text{Null}(A^*A) \implies A^*Ay = 0 \implies y^*A^*Ay = 0 \implies ||Ay|| = 0 \implies Ay = 0.$$

Now, we show perpendicularity. We have
$$(A^*b)^*y = b^*(Ay) = b^*0 = 0.$$
So $A^*b \in \text{Col}(A^*A)$.

Q.E.D.

We proceed to the main proof. Let $p$ be a solution of $A^*Ap = A^*b$. Then $A^*b - A^*Ap = 0$.

We claim $r(p) \perp \text{Col(A)}.$ Indeed, for any $Az \in \text{Col}(A)$, we have
$$(Az)^*(b - Ap) = z^*(A^*b - A^*Ap ) = 0.$$
Now, by the Pythagorean Theorem,
$$||b - Ay||^2 = ||b - Ap + Ap - Ay||^2 = ||b - Ap||^2 + ||Ap - Ay||^2 \ge ||b - Ap||^2$$
with equality $\iff Ay = Ap$.

$\color{red}{\text{The flaw is here. I need to bound the function } ||b - Ax||}$
$\color{red}{\text{ by a constant, not by a function of p.}}$

Now suppose $x$ is a vector that solves the L.S. problem. Then $Ax = Ay$, so $r(x) = b - Ax = b - Ap = r(p)$
and $r(p) \in \text{Null}(A^*)$ because $A^*Ap = A^*b \implies A^*(b - Ap) = 0$.

Now for the converse, suppose that $A^*Ax = A^*b$. I don't yet know how to show that $Ax = Ap$.

答: You're not using "WLOG" correctly. It's when you're making an assumption that might technically not be true but obviously doesn't affect the logic of the proof. It should be pretty obvious how the proof would work without that assumption. See wikipedia for an example. You assuming that $A$ is full rank is a big assumption, on the other hand, so it doesn't make sense to tag "WLOG" at the start because it drastically changes the problem.


问:
Differentiability of convolution
First let me say that I have used the search bar and looked through all the "differentiability of convolution" questions that I saw, but none of them seem to cover this case. (If one of them did and I missed it, of course please post the link and I will look more closely).

Suppose $h\in L^\infty(\mathbb{R})$. How nice does $g$ have to be in order to guarantee that $h * g \in C^1(\mathbb{R})$?

In my particular situation $g$ is a Schwartz function, infinitely differentiable and all of its derivatives decay faster than any polynomial, and $h$ is continuous and bounded.

I know that if $h \in L^p$ for $p<\infty$ then $f *g \in C^\infty$, and I know it works if $g$ is compactly supported, but in my case neither of these two hold.

All I have is that $h$ is continuous and bounded, and $g$ is Schwartz.

EDIT: $h*g$ being differentiable almost everywhere would actually suffice for my purposes.

EDIT2: More specifically,  $g(x) = e^{-x^2}$.

答:
Let us focus on dominating the difference quotient's integrand, if this can be done the result follows by dominated convergence.
By the mean value theorem, for any $i$, we have
$$
\frac{g(x-y+te_i) - g(x-y)}{t} = \partial_i g(\xi(x-y,t))
$$
where $\xi(u,t)$ lies on the segment between $u$ to $u+te_i$.
Without loss we consider $|t|<1$.
Then
$$
\left|\frac{g(x-y+te_i) - g(x-y)}{t} h(y) \right| \leq | \partial_i g(\xi(x-y,t))|\cdot |h(y)| \leq C |\partial_i g(\xi(x-y,t))|
$$
where $|h|\leq C$ since $h$ is bounded, and this is
$$
\leq C \sup_{z \in B(x-y,1)} |\partial_i g(z)|
$$
so it suffices to show
$$\int  \sup_{z \in B(x-y,1)} |\partial_i g(z) | dy < \infty.$$

Since $g \in \mathcal{S}(\mathbb{R}^n)$ we may choose $r>0$ such that $|y|^{n+1}\cdot |\partial_i g(y)| \leq K$ for all $|y|>r$.
Since $\partial_i g$ is bounded it suffices to show
$$\int\limits_{|y|>|x|+r+2}  \sup_{z \in B(x-y,1)} |\partial_i g(z) | dy < \infty.$$

Note that for $|y|>|x|+r+2$ we then have
$$
\sup_{z \in B(x-y,1)} |\partial_i g(z) | \leq\frac{K}{(|x-y|-1)^{n+1}}
$$
and since
$$
\int\limits_{|y|>|x|+r+2}\frac{K}{(|x-y|-1)^{n+1}}dy < \infty
$$
the result follows.

Title: Differentiability of convolution


问:
Need continuous time martingales be predictable?
I know that for discrete time martingales, one can show that if they are predictable, they have to be constant. Now for continuous time martingales, that is not true I suppose, since for example a Brownian motion is predictable as it is continuous (correct?). So I was wondering what the relationship between continuous time martingales and predictability is. I could not come up with an example of a non-predictable martingale. Are all continuous time martingales predictable? If not, what would be a counterexample?</p>

答: The following may be near what you are after: If $M$ is a (cadlag) continuous time martingale, then ${}^pM$ coincides with $M_-$ almost surely. Here ${}^pM$, the predictable projection of $M$, is characterized as the predictable process such that $E[M_T\mid\mathcal F_{T-}] = {}^pM_{T}$ for all bounded predictable stopping times $T$; and $M_-$ is the left limit process $t\mapsto M_{t-}$. In particular, this shows that $M$ is predictable if and only if the paths of $M$ are continuous.


问:
Does the uncomputability of Kolmogorov complexity follows from Rice Theorem?
I was reading the proof of uncomputability of Kolmogorov complexity by Li and Vitányi book and thinking if there isn't another way to do this simpler using Rice theorem. I came up with this argument. Can you say to me if it is correct?

Definition: The set $A ⊆ N$ is an index set if it has the property that if $x ∈ A$ and $\phi_x = \phi_y$ , then $y ∈ A$.

Rice theorem states:

Theorem: Suppose that $A$ is an index set not equal to $∅$ or $\mathbb{N}$. Then A is not computable.

The conditional absolute complexity can be defined as (by Li and Vitányi book's definition):

$$C(x|y) = \min \{ |p| : \phi(\langle y,p \rangle) = x \}$$

We have: $\phi(\langle y,p \rangle) = \phi_y(p)$ by the enumeration theorem and property of the universal additively optimal $\phi$:

$$C(x|y) = \min \{ |p| : \phi_y(p) = x \}$$

But $A = \{ p : \phi_y(p) = x \}$ is a index set. Because if any  $p \in A$ and  $\phi_p= \phi_q$, then (this is the step I have doubts) $\phi_y(p) = \phi_y(q)$, because  $\phi_y$  can be interpreted as a program that receives other programs $p$ and $q$, and if the programs are equal so is $\phi_y(p)$ and $\phi_y(q)$. Then $q \in A$.

So $A$ is a index set and also is not $\emptyset$ or $\mathbb{N}$ (because  there is a  program $p$ such that $\phi_y(p) \not = x$). So, by Rice Theorem, $A$ is not computable, and $C(x|y)$ also. And so $C(x) = C(x|\varepsilon)$ is also uncomputable.

答:
Your argument that $A$ is an index set is not correct: you haven't used anything particular about $y$, so your argument would imply that $$\{p: \phi_y(p)=x\}$$ would be an index set for every choice of $y$. But that's clearly incorrect: for example, take $x=1$ and let $\phi_y(p)=1$ if $p=0$ and $0$ otherwise. By the padding lemma, this is not an index set.

While we may want to think of $\phi_y$ as acting on computable functions in a well-defined way rather than natural numbers, that doesn't mean that it actually does.


问:
Differentiating $\frac{1}{7}\cos^4(x^2+2x+3)$
I'm trying to differentiate this function using the chain rule, but not sure whether I'm doing it right.

I did the following:

I let $u = (x^2 +2x+3)$ and $y = \cos^4u$,

$\frac{du}{dx} = 2x +2$, $\frac{dy}{du} = -\sin^4u$

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ = $-\sin^4 \times 2x+2$ = $(1/7) \times -\sin^4(2x+2)$

答:
I will write $D_x$ instead of $\frac{\mathrm{d}}{\mathrm{d}x}$.So, you want to calculate
$$D_x\left[ \frac{1}{7}\cos^4(x^2+2x+3)\right]=\frac{1}{7}D_x\left[\cos^4(x^2+2x+3)\right].$$
I will leave the $1/7$ and multiply it again at the end. Set $g=\cos(x^2+2x+3)$. Now that's
$$D_g\left[g^4\right]\cdot D_x \left[\cos(x^2+2x+3)\right] = 4g^3 D_x \left[\cos(x^2+2x+3)\right].$$
You can evaluate the derivative by the  normal chain rule. Substitute $g$ back again:
$$4g^3\left(-(2x+2)\sin(x^2+2x+3)\right) = 4\left(\cos(x^2+2x+3)\right)^3\left(-(2x+2)\sin(x^2+2x+3)\right).$$
Don't forget to multiply the $1/7$. You should end up with:
$$-\frac{4\cos^3(x^2+2x+3)\sin(x^2+2x+3)(2x+2)}{7}$$


问:
Understand this probability argument
I've got a pretty specific question here. I've included an image of a probabilistic expression relating Brownian Motion and the stopping times for when the Brownian crosses each integer value. This text is Steele: Stochastic Calculus and Financial Applications. The question is, how is this last inequality derived?

I am not seeing exactly how we get the connection between $|B_{\tau_n}-B_n| \ge \epsilon\sqrt{n}$ and $\sup_{s:|s-n| \le \delta n} |B_s - B_n| \ge \epsilon\sqrt{n}$. There isn't really any more explanation in the book about this. I guess it should be obvious but I've spent a long time on it and can't figure it out. Thanks!

答:
Define $S:=\{s:|s-n|\le \delta n\}$ for simplicity. We want to show that
$$\left \{|B_{\tau_n}-B_n|\ge \epsilon \sqrt{n}, |\tau_n-n|<\delta n\right \}\subset\left \{\sup_{s\in S}|B_s-B_n|\ge \epsilon\sqrt{n}\right \},$$
namely that if $$\omega\in\left\{|B_{\tau_n}-B_n|\ge \epsilon \sqrt{n}, |\tau_n-n|<\delta n\right\},$$ then $$\omega\in \left\{\sup_{s\in S}|B_s-B_n|\ge \epsilon\sqrt{n}\right\}.$$

Now, let us fix an $\omega\in\{|B_{\tau_n}-B_n|\ge \epsilon \sqrt{n}, |\tau_n-n|<\delta n\}.$ Then $|\tau_n(\omega)-n|<\delta n$, which implies that $\tau_n(\omega)\in S.$ Moreover, $|B_{\tau_n}(\omega)-B_n(\omega)|\ge \epsilon \sqrt{n},$ but $\tau_n\in S$ and hence clearly $\sup_{s\in S}|B_s(\omega)-B_n(\omega)|\ge \epsilon\sqrt{n}.$
Consequently, $$\left \{|B_{\tau_n}-B_n|\ge \epsilon \sqrt{n}, |\tau_n-n|<\delta n\right \}\subset\left \{\sup_{s\in S}|B_s-B_n|\ge \epsilon\sqrt{n}\right \}$$ and we have that
$$P\left \{|B_{\tau_n}-B_n|\ge \epsilon \sqrt{n}, |\tau_n-n|<\delta n\right \}\le P\left \{\sup_{s\in S}|B_s-B_n|\ge \epsilon\sqrt{n}\right \}.$$

Title: Understand this probability argument


答:
$\frac {d\cos^4 u}{du}\ne -\sin^4 u$.

Let $v(x) = x^4$ then we have

SO if you have $\cos^4 u = v(\cos u)$ then

$\frac {d\cos^4 u}{du} = \frac {d v(\cos u)}{du} = \frac {d v(\cos u)}{d(\cos u)} \frac d{\cos u}{du}=$

$4(\cos u)^3\cdot (-\sin u)=$

$-4\sin u\cos^3 u$.

Doing it via the chain rule is

$[\frac{1}{7}cos^4(x^2+2x+3)]'=$

$\frac 17[4(\cos^3(x^2+2x+3))(-\sin(x^2 + 2x + 3)(2x + 2)]=$

$-\frac 47\cos^3(x^2 + 2x + 3)(\sin(x^2 + 2x+3)(2x + 2)$

.....

If you want to be really pendantic

Let $f(x) = x^2 +2x + 3$

Let $g(x) = \cos x$

Let $h(x) = x^4$

Let $i(x) = \frac 17 x$

So we have $[i(h(g(f(x)))]' = i'(h(g(f(x)))\cdot h'(g(f(x)))\cdot g'(f(x))\cdot f'(x)$

$i'(w) = [\frac 17 w]' = \frac 17$.

$h'(w) = [w^4]' = 4w^3$.

$g'(w)= [\cos w]' =-\sin w$

$f'(w) = (w^2 +2w + 3)' = 2w+2$

So  $[i(h(g(f(x)))]' = i'(h(g(f(x)))\cdot h'(g(f(x)))\cdot g'(f(x))\cdot f'(x)=$

$\frac 17\cdot (4(g(f(x)))^3) \cdot (-\sin(f(x)) \cdot (2x + 2) =$

$\frac 17 \cdot 4(\cos(x^2 + 2x + 3))^3 \cdot (-\sin(x^2 + 2x + 3))\cdot (2x+2) =$

$-\frac 47 \cos^3(x^2 + 2x + 3)(\sin(x^2 + 2x + 3)(2x+2)$.

Oh... I guess we should factor out the $2$ from $2x + 2$...

$= -\frac 87 \cos^3(x^2 + 2x + 3)(\sin(x^2 + 2x + 3)(x+1)$.

....

And I suppose proving $[i(h(g(f(x)))]' = i'(h(g(f(x)))\cdot h'(g(f(x)))\cdot g'(f(x))\cdot f'(x)$ by multiple applications of the chain rule can be done.

Let $h(g(f(x))) = j(x)$ then $[i(j(x))]' = i'(j(x))j'(x)$.

If $k(x) = g(f(x))$ then $j(x) = k(g(x))$ and $j'(x) = k'(g(x))g'(x)$.

And $k'(x) = g'(f(x))f'(x)$.  SO the result follows.

======

But one can drive oneself NUTS paying that much attention to detail.  The comes a point where sums,constants, and powers should be automatic:

If $h(x) = c\cdot (f(g))^k$

then $h'(x) = ck(f(g(x)))^{k-1}f'(g(x))\cdot g'(x)$,

shouldn't be tooooo difficult a result to accept off the bat.

(which is our case with $c = \frac 17, k = 4, f=\cos; g(x) = x^2 + 2x + 3$.

It shouldn't be difficult for a student to follow that

$[\frac 17\cos^4(x^2 + 2x + 3)]'=$

$\frac 17\cdot 4\cos^3(x^2+2x+3)\cdot(-\sin(x^2 + 2x + 3))\cdot (2x  +2)$.

Title: Differentiating $\frac{1}{7}\cos^4(x^2+2x+3)$

Body:
I'm trying to differentiate this function using the chain rule, but not sure whether I'm doing it right.

I did the following:

I let $u = (x^2 +2x+3)$ and $y = \cos^4u$,

$\frac{du}{dx} = 2x +2$, $\frac{dy}{du} = -\sin^4u$

$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$ = $-\sin^4 \times 2x+2$ = $(1/7) \times -\sin^4(2x+2)$


答:
The function that you hace to derive is the composition of three two function: $f$, $g$, $h$.

In effect $\frac{1}{7}\cos^4(x^2+2x+3)=\frac{1}{7}h(g(f(x)))$, with $f(x)=x^2+2x+3$, $g(x)=cos(x)$, $h(x)=x^4$.

So applying the chain rule of derivation to this composition the derivative will be =$-\frac{4}{7}\cos^3{(x^2+2x+3)}\sin{(x^2+2x+3)}*(2x+2)$


问:
A conditional probability problem where the next day depends on the last 3 days
For many years, Meteorologists have spent long visits (5 days) at the Bigtown.
They have observed that, for three consecutive days, if there are EXACTLY two sunny days, the next day is a sunny day*, while half of all the cloudy days are followed by another cloudy day. Assuming days are either cloudy or sunny, estimate how many days have been cloudy in their last ten visits. Visits are not necessarily following each other.

''half of all the cloudy days are followed by another cloudy day'' so $P(C\mid C)=50\%$

Is this possible? If not why not?

*i.e.
SSC -> S

SCS -> S

CSS -> S

where -> means followed by

答:
Here is an attempt to solve the problem:

Let's define the problem by a Markov chain, where a state of the chain is defined by the state of last (most recent) 3 consecutive days observed at any point in time (e.g., the state represented by CCC can transition to any of the two states, namely, itself, i.e., CCC and CCS). The state transition diagram will look like the following then

Notice that we don't have all the probabilities given to us (those are circled in the above figure), but some of them can be inferred and some of them can't be. For example, given, P(SCC|CCC)=1/2, so, consequently, P(CCS|CCC)=1-1/2=1/2. But we don't know either of P(SSS|SSS) and P(SSC|SSS), these are marked by x in the above figure.

For the sake of completeness (and simplicity), let's assume uniform probabilities for the unknown ones, e.g., let P(SSS|SSS) = P(SSC|SSS) = 1/2.

Then the Markov transition matrix becomes the one as shown below:

Now, the Markov chain being connected and aperiodic, by Perron-Frobenius theorem, it converges to stationary distribution in the long run.

Let's compute the stationary distribution, using the following R code with the Markov transition matrix $T$:

```<code>S <- T
for (i in 1:100)
  S <- S %*% T
</code>```

Which shows that in the long run the Markov chain will be in the states CSS, SSC, SCS and SSS with probabilities 0.2, 0.2, 0.2 and 0.4, respectively. But the proportion of sunny days in these states are 2/3, 2/3, 2/3 and 1, respectively.

Hence, the expected number of sunny days will be = 0.22/3 + 0.22/3 + 0.2*2/3 + 0.4 = 0.8

Body:
Here is an attempt to solve the problem:

Let's define the problem by a Markov chain, where a state of the chain is defined by the state of last (most recent) 3 consecutive days observed at any point in time (e.g., the state represented by CCC can transition to any of the two states, namely, itself, i.e., CCC and CCS). The state transition diagram will look like the following then

Notice that we don't have all the probabilities given to us (those are circled in the above figure), but some of them can be inferred and some of them can't be. For example, given, P(SCC|CCC)=1/2, so, consequently, P(CCS|CCC)=1-1/2=1/2. But we don't know either of P(SSS|SSS) and P(SSC|SSS), these are marked by x in the above figure.

For the sake of completeness (and simplicity), let's assume uniform probabilities for the unknown ones, e.g., let P(SSS|SSS) = P(SSC|SSS) = 1/2.

Then the Markov transition matrix becomes the one as shown below:

Now, the Markov chain being connected and aperiodic, by Perron-Frobenius theorem, it converges to stationary distribution in the long run.

Let's compute the stationary distribution, using the following R code with the Markov transition matrix $T$:

S <- T
for (i in 1:100)
  S <- S %*% T

Which shows that in the long run the Markov chain will be in the states CSS, SSC, SCS and SSS with probabilities 0.2, 0.2, 0.2 and 0.4, respectively. But the proportion of sunny days in these states are 2/3, 2/3, 2/3 and 1, respectively.

Hence, the expected number of sunny days will be = 0.22/3 + 0.22/3 + 0.2*2/3 + 0.4 = 0.8


答:
You should consider
$$u=x^2+2x+3$$
$$y=\cos(u)$$
$$z=y^4$$

and

$$\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}=4y^3\color{red}{\frac{dy}{dx}}$$

$$\color{red}{\frac{dy}{dx}}=\frac{dy}{du}\frac{du}{dx}=-\sin(u)\frac{du}{dx}$$

$$\frac{du}{dx}=(2x+2)$$

you can finish.


问:
Arrow-theoretic proof that cohomology is a functor.
Let $\sf{A}$ be an abelian category and $\sf{C}(\sf{A})$ be its category of complexes. I want to prove that the $i$-th cohomology is an additive functor $\sf{C}(\sf{A})\to\sf{A}$. For that, let $\varphi^\bullet:M^\bullet\to N^\bullet$ be a morphism of complexes. By the universal property of kernels, we have an induced morphism

In order for the universal property of cokernels to induce a morphism $H^i(\varphi^\bullet):H^i(M^\bullet)\to H^i(N^\bullet)$ making the diagram

commute, we have to show that the morphism $I^{i-1}_{M^\bullet}\to K^i_{M^\bullet}\to K^i_{N^\bullet} \to H^i(N^\bullet)$ is zero. That would follow from the existence of an induced map $I^{i-1}_{M^\bullet}\to I^{i-1}_{N^\bullet}$, but I can't seem to figure out why this exists.

Also, I would love to know if there's a simpler way to construct the induced morphism $H^i(\varphi^\bullet):H^i(M^\bullet)\to H^i(N^\bullet)$ since my construction seems a little tough to work with.

答:
Deepening what I said in comments, I will show that for a commutative rectangle

Where $f, f'$ are surjective (means zero cokernel, equivalent to being epimorphism) and $g, g'$ are injective (means zero kernel, equivalent to being monomorphism), there exist an arrow $h_3 : X \to X' $ making the two squares commutative. Indeed let $C, C'$ be the kernels of $f, f'$ with natural maps $i: C \to A, i': C' \to A'$. Note that they are also the kernels of $gf, g'f'$ since $g,g'$ are monomorphisms. Consider the map $h_0 : C \to C'$ induced by the commutative square and the kernel property.

The map $f' h_1 : A \to X' $ is such that $ f'h_1 i = f'i' h_0 = 0$, so that by the universal property of
$$0 \to C \to A \to X \to 0$$
being an exact sequence there exist a map $h_3 : X \to X'$ such that $ h_3 f = f' h_1 $.

To verify that the rightside square is commutative, aka $h_2 g = g' h_3$ , precompose with $f$, then use the fact that $f$ is epi to deduce that the identity must be true.

I have been deliberately pedant and formal, so that you can practice with general diagram chasing. I suggest you to verify the small things that I have left here and there: you will get a lot more confident!


答:
Your $\frac{dy}{du}$ is wrong. $\cos^4(u) = \left((\cos u)^4\right),\ $ so let $v = \cos u\ $ and $\ y = v^4.\ $ Then,

$\frac{dy}{dv}= 4v^3,\ $ and $\frac{dv}{du}= -\sin(u).$ Therefore:

$\frac{dy}{du} = \frac{dy}{dv} \times \frac{dv}{du} = 4v^3  \times -\sin(u) = 4(\cos (u))^3  \times -\sin(u) = -4(\sin (u))\cos^3(u).$


问:
Counterexample to basis of subspace (Axler 2.B.7)
I'm stuck at a problem from the Axler (Linear algebra done right) 2.B.7

Prove or give a counterexample: If $v_1,v_2,v_3,v_4$ is a basis of $V$ and $U$ is a subspace of $V$ such that $v1,v2 \in U$ and $v_3 \notin U$ and $v_4 \notin U$, then $v_1,v_2$ is a basis of $U$.

While I looked up the solution and fully agree with it, I don't know a structured way how to come up with such a solution.

The solution (counterexample) provided:

let $V = \mathbb{R}^4, v_1 = (1,0,0,0), v_2=(0,1,0,0), v_3=(0,0,1,1),v_4=(0,0,0,1)$ and $U =\{(x,y,z,0)|x,y,z\in\mathbb{R}\}.$

Is it correct that for a counter example, I need to find $v_3,v_4$ such that one "contains" the other (like $(0,0,2,1),(0,0,2,0)$ or $(0,0,3,2),(0,0,3,1)$) and then let $U$ to be $\{(x,y,0,z)|x,y,z\in\mathbb{R}\}$ if $v_3-v_4 = \lambda(0,0,0,1)$ or $\{(x,y,z,0)|x,y,z\in\mathbb{R}\}$ if $v_3-v_4 = \lambda(0,0,1,0)$ (for some $\lambda\in\mathbb{R}$).

Is there a more structured / more flexible way of solving this problem?

答:
Prove or give a counterexample: If $v_1,v_2,v_3,v_4$ is a basis of $V$ and $U$ is a subspace of $V$ such that $v_1,v_2 \in U$ and $v_3 \notin U$ and $v_4 \notin U$, then $v_1,v_2$ is a basis of $U$.

Since $v_1, v_2$ are linearly independent we can say that $U$ will be at least 2-dimensional. But $v_1,v_2$ will be a basis for $U$ if and only if $U$ is exactly 2-dimensional. So the question is can we find subspace $U$ of dimension greater than two that which contains $v_1,v_2$ but does not contain $v_3$ and $v_4$. Note that if $U$ were 4-dimensional then $U$ would be all of $V$ so that wouldn't work. So we are specifically looking for a 3-dimensional.

Another way of phrasing it is that we are looking to extend the span of $v_1, v_2$ into a 3-dimensional subspace that avoids including $v_3,v_4$.


问:
Continuity and limits. Please check epsilon delta

Suppose $f$ is continuous at $a$ and $f(a) = 0$. Prove that if $\alpha \neq 0 $, then $f+\alpha$ is nonzero in some open interval containing $a$.

Since $f$ is continuous, we take $\epsilon = |\alpha|$; then we have $|x - a| < \delta \implies -|\alpha| < f(x) < |\alpha| \implies \alpha - |\alpha | <f(x) + \alpha < \alpha + |\alpha |$

We consider two cases. If $\alpha >0$, then $\alpha = |\alpha|$; thus  $|x - a| < \delta \implies 0<f(x) + \alpha < 2|\alpha |$

On the other hand, if $\alpha < 0$ we have $|x - a| < \delta \implies -2|\alpha |< f(x) + \alpha < 0$

EDIT Spivak's answer book referred me back to a theorem in the book. I am unfortunately too lazy to do that and after a quick glimpse of the theorem, it didn't look anything like what I wrote...

答:
Correct. Note that in such situations, people usually give themselves some space and take $\epsilon=|\alpha|/2$. And it is not recommended to take the habit to rely too much on strict inequalities. It does no matter in this case, of course. But in more involved arguments requiring a limiting process, strict inequalities become large...

Also, you can handle both cases at once with the reverse triangular inequality:
$$|t-z|=|z-t|\geq ||z|-|t||\geq |z|-|t|.$$

So take $\delta>0$ such that $|x-a|\leq \delta$ implies $|f(x)|\leq \frac{|\alpha|}{2}$. Then
$$
|f(x)+\alpha|\geq |\alpha|-|f(x)|\geq |\alpha|-\frac{|\alpha|}{2}=\frac{|\alpha|}{2}>0.
$$

Note: the theorem you mention could very well be that if $f$ and $g$ are continuous at $a$, then $f+g$ is continuous at $a$. In this case, take $g$ the constant function equal to $\alpha$, which is obviously continuous. So $h=f+\alpha$ is continuous at $a$ where it takes the value $\alpha\neq 0$. Hence $|h|$ is continuous at $a$ where it takes the value $|\alpha|>0$. Take $\epsilon=\frac{|\alpha|}{2}>0$. Then there is $\delta>0$ such that $|x-a|\leq \delta$ implies
$$
0<\frac{|\alpha|}{2}=|\alpha|-\frac{|\alpha|}{2}\leq |h(x)|=|f(x)+\alpha|\leq |\alpha|+\frac{|\alpha|}{2}
$$
where the rhs is useless for your purposes.


问:
Interpreting english EITHER-OR when approaching logical deductions doubt
The following question has been asked in the Indian entrance examination "GATE CS 2021", and it has stirred up a lot of confusion regarding how one should interpret "either-or" when coming to logical questions where we are given certain premises and have to determine if something is true via deduction.

Q. Given below are two statements 1 and 2, and two conclusions I and II

Statement 1: All bacteria are microorganisms.
Statement 2: All pathogens are microorganisms.

Conclusion I: Some pathogens are bacteria.
Conclusion II: All pathogens are not bacteria.

Based on the above statements and conclusions, which one of the following options is logically CORRECT?

Only conclusion I is correct

Only conclusion II is correct '

Either conclusion I or II is correct

Neither conclusion I nor II is correct

refer this link for the debate - https://gateoverflow.in/357468/gate-cse-2021-set-1-ga-question-9
refer this link for my reasoning - https://gateoverflow.in/357468/gate-cse-2021-set-1-ga-question-9?show=358167#a358167

In short -
I feel option D is the correct choice, because neither conclusions follow from the premises, but on the other side a lot of people feel that option C is right because according to them option c is a tautology.

Also there is another confusion regarding the statement of conclusion 2 "All pathogens are not bacteria".
Does this mean that there is no pathogen which is also a bacteria, or does it mean there exists some pathogens who are definitely not bacteria?
Because when we say a similar statement like "All students are not geniuses", we generally mean that there are only few students who are geniuses, and most of the students are not geniuses. We don't mean to say that there isn't even a single student who is not a genius.

Can anybody please provide an unambiguous, mathematically backed argument for choosing the correct answer, or is the given question and options inherently ambiguous?

答:
Conclusions I and II can be restated as "some pathogens are bacteria" and "no pathogens are bacteria".

Whether option C is a tautology or not depends on the meaning of "some", and whether it is to be interpreted as "at least some" (i.e. some and possibly all) or "only some" (I.e. some but not all).

It is only a tautology if "some" includes the possibility of "all", otherwise it is not a tautology since there is a third possibility that "all (and not merely some) pathogens are bacteria", and both the alternatives expressed in option C could then be false.

And since the question asks us only to resolve the matter by reference to the statements and conclusions, and we cannot infer the exact meaning of "some" from them, then I would say option D is correct, that neither of the conclusions are logically correct.


答:
It's important to remember to $\color{blue}{\text{take the constant out}}$ and $\color{red}{\text{apply the chain rule}}$:

\begin{align}
&\frac{1}{7}\cos^4(x^2+2x+3) \\
&=\color{blue}{\frac{1}{7}\frac{d}{dx}(\cos^4(x^2+2x+3))}\\
&\color{red}{=4(\cos(x^4(x^2+2x+3))^3\frac{d}{dx}(\cos(x^2+2x+3))}\\
&=\frac{1}{7}\cdot4(\cos(x^2(x^2+2x+3))^3(-\sin(x^2+2x+3)(2x+2))\\
&=\boxed{-\frac{4}{7}\cos^3(x^2+2x+3)\sin(x^2+2x+3)(2x+2))}\\
\end{align}

EDIT:

Thanks @fleablood for pointing that out!


问:
Probability of two dependent events
I saw a nice problem on probability online but not sure how to tackle it.

In CS:GO, Bob is using Desert Eagle with 7 bullets in one clip. Bob has a .5 chance of landing the bullet. Of the bullets that landed, there is .2 chance that it is a headshot. If it is a headshot, the target dies instantly, otherwise Bob would need 3 (non-consecutive or consecutive) landed bullets to eliminate the target. What is the probability that Bob eliminate a target with one clip?

So I started out naively with something like, $$P(\text{at least 3 landed shots}) = 1 - 0.5^7 - {7 \choose 1}0.5^7 - {7 \choose 2} 0.5^7 = .77$$
and similarly with $$P(\text{at least 1 headshot}) = 1  - 0.9^7 = .52$$
but not sure how to continue. Any help would be great!

答:
Yes as you started, it is easier to first find probability, say $Q$, that Bob is not able to kill the target in one clip ($7$ bullets).

This can only happen if Bob does not land any bullet on the target or lands either one or two bullets but they are not headshot. The probability that a landed bullet is not a headshot is $0.8$.

$Q = \displaystyle 0.5^7 + {7 \choose 1} (0.5)^7 \times 0.8 + {7 \choose 2} (0.5)^7 \times (0.8)^2 \approx 0.1566$

So the probability that Bob kills the target is $P = 1 - Q \approx 0.8434$.

Title: Probability of two dependent events


问:
Non-atomic measure on $\mathbb{Z}\cup\{\pm\infty\}$?
I have a rather naive question?

Is there a non-atomic measure on $\bar{Z}:=\mathbb{Z}\cup\{\pm\infty\}$?

For example, the counting measure is atomic.

答:
In fact, there is no non-atomic measure on $\bar{Z}:=\mathbb{Z}\cup\{\pm\infty\}$, except for the identically null measure.

Let us prove it.

Recall that if $(X,\Sigma)$ is a measurable space, then $A\in \Sigma$ is an atom in $\Sigma$ if and only if $A\neq \emptyset$ and for all $B\in \Sigma$ if $B \subseteq A$ then either $B=\emptyset$ or $B=A$.

Now let us prove the main result:

Let $(X,\Sigma)$ be a measurable space. If $X$ is countable, then, for each $x \in X$, there is $A_x$, an atom  of $\Sigma$, such that $x\in A_x$

Proof:
Consider the the following relation defined in $X$:  $x\sim y$  if and only if, for all $E \in \Sigma$, if $x \in E$ then $y\in E$.

Claim: $\sim$ is an equivalence relation.

$x \sim x$.

It is trivial: for all $E \in \Sigma$, if $x \in E$ then $x\in E$.

if $x \sim y$ then   $y \sim x$ .

If $y \nsim x$ then there is $F \in \Sigma$ such that $y\in F$ and $x\notin F$. Let $E= F^c \in \Sigma$. Then $x\in E$ and $y\notin E$. So $x \nsim y$

if $x \sim y$ and $y \sim z$ the $x \sim z$

It is trivial: for all $E \in \Sigma$, if $x \in E$ then $y\in E$ and, for all $E \in \Sigma$, if $y \in E$ then $z\in E$. So, for all $E \in \Sigma$,  if $x \in E$ then $z\in E$.

This proves the claim.

Now, for each $x\in X$, for each $y\in X$, such that $y \nsim x$, take one $E_y \in \Sigma$ such that $x \in E_y$ and $y\notin E_y$.

Let us define $A_x=\bigcap_{y\in X} E_y$. Since $X$ is countable, we have that $A_x \in \Sigma$. We also have that $x\in A_x$.

It is easy to see that $A_x$ is an atom in $\Sigma$. In fact, since $x\in A_x$ so $A_x \neq \emptyset$. Now suppose that $A_x$ is not an atom in $\Sigma$, then there is $B\in \Sigma$ such that $B \subseteq A_x$, $B \neq \emptyset$ and $B \neq A_x$. Then there are $b, c \in A_x$, such that $b \in B$ and $c \notin B$.

If $x\in B$ then $x\nsim c$, so $A_x \subseteq E_c$ and $c\notin E_c$. So $c\notin A_x$. Contradiction.

If $x\notin B$ then $x \in B^c$ and $b \notin B^c$ so $x\nsim b$, so $A_x \subseteq E_b$ and $b\notin E_b$. So $b\notin A_x$. Contradiction.

So, we have proved that, for all $x\in X$, $A_x$ is an atom in $\Sigma$  and $x \in A_x$. $\square$

Our next result:

Let $(X,\Sigma)$ be a measurable space. If $X$ is countable, then $\mu$ is non-atomic if and only if $\mu$ is identically zero.

Proof:

From the previous result, for all $x\in X$, $A_x$ is an atom in $\Sigma$  and $x \in A_x$.

The only way that an atom $A_x$ in $\Sigma$ will not become an atom of the measure is if $\mu(A_x)=0$.  So $\mu$ is non-atomic if and only if, for all $x\in X$, $\mu(A_x)=0$

But, clearly $X= \bigcup_{x\in X} A_x$  and this union is countable , because $X$ is countable. So

$$\mu(X)\leqslant \sum_{x\in X} \mu(A_x) = 0$$

So, $\mu$ is non-atomic if and only if $\mu$ is identically zero. $\square$

Remark 1: Using the fact that atoms of $\Sigma$ are disjoint, we can show that there is a partition of $X$ made of the atoms of $\Sigma$.

Remark 2: The result

Let $(X,\Sigma)$ be a measurable space. If $X$ is countable, then, for each $x \in X$ there is $A_x$, an atom  of $\Sigma$, such that $x\in A_x$

can be proved using the Zorn lemma

Proof:
Let $\Gamma =\{ E \in \Sigma : x\in E \}$. Let us define the order $\leqslant$ in $\Gamma$ by $E \leqslant F$ if $F \subseteq E$. Suppose $\{E_\lambda\}_{\lambda \in L}$ be a totally ordered family of elements in $\Gamma$. Since $X$ is countable and $\{E_\lambda\}_{\lambda \in L}$ is a totally ordered, we have that $\{E_\lambda\}_{\lambda \in L}$ has only countable distinct elements.
So,
$$ E = \bigcap_{\lambda \in L} E_\lambda$$ is a countable intersection and so $E \in \Sigma$. $E$ is an upper bound of $\{E_\lambda\}_{\lambda \in L}$.

So, by Zorn lemma, there is $A_x \in \Gamma$ such that $A_x$ is a maximal element of $\Gamma$. Clearly, from the definition of $\Gamma$, $x \in A_x$

Now, $A_x$ is an atom of $\Sigma$. In fact, since $x \in A_x$, $A_x \neq \emptyset$. So, if $A_x$ is not an atom of $\Sigma$, there is $B\in \Sigma$ such that $B \subseteq A_x$, $B \neq \emptyset$ and $B \neq A_x$.

if $x \in B$, then $B \in \Gamma$ and $A_x \leqslant B$ and  $B \neq A_x$. Contradiction to the maximality of $A_x$ in $\Gamma$.

if $x \notin B$, then $x \in A_x \cap B^c$. So  $A_x \cap B^c \in \Gamma$ and $A_x \leqslant A_x \cap B^c$ and  $A_x \cap B^c \neq A_x$. Contradiction to the maximality of $A_x$ in $\Gamma$.

So $A_x$ is an atom of $\Sigma$ and $x \in A_x$.


问:
How to calculate $\mathbb{E}[X]\mathbb{E}[Y]$ given $\mathbb{P}(X=k, Y=m)={\frac{1}{36}}\cdot\left( {\frac{5}{6}} \right)^{k-2}$
Let $X,Y$ be random variables, s.t for $1\leq m<k$ $$\mathbb{P}(X=k, Y=m)={\frac{1}{36}}\cdot\left( {\frac{5}{6}} \right)^{k-2}$$
Calculate $\mathbb{E}[X]\mathbb{E}[Y]$

I am not sure even how to approach it, so far I wrote the definition of $\mathbb{E}$ but I am not sure how to write the indexes of $k$ and $m$

答:
I would like for a small hint how do I find $P(X=x)$...

It is very very easy...if you sum the joint pmf summing the y's you get

$$P(X=x)=\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{x-2}\cdot\sum_{y=1}^{x-1}1=(x-1)\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{x-2}$$

That is

$$P(X=x)=\binom{x-1}{2-1}\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)^{x-2}$$

where $x=2,3,4,5,...$

this is a Negative Binomial, counting the number of trial to get 2 successes, thus

$$\mathbb{E}[X]=2\times6=12$$

As an exercise, before getting in calculations, you could prove that your $p(X,Y)$ is a nice joint pmf. This exercise is very useful because it exactly lets you know which is the support. Once you understand where your joint probability function is defined, you are immediately able to understand the problem and thus the solution.

I suggest you to change the index letter in $X=x$ and $Y=y$ because inserting to many letters in the equation can get you in confusion.

$$\sum_{x=2}^{\infty}\sum_{y=1}^{x-1}\frac{1}{36}\left(\frac{5}{6}\right)^{x-2}=\sum_{x=2}^{\infty}\frac{x-1}{36}\left(\frac{5}{6}\right)^{x-2}=1$$


问:
Is there a simpler closed form for $\sum_{n=1}^\infty\frac{(2n-1)!!\ (2n+1)!!}{4^n\ (n+2)\ (n+2)!^2}$
I have the following infinite sum that can be expressed in terms of the generalized hypergeometric function:
$$\sum_{n=1}^\infty\frac{(2n-1)!!\ (2n+1)!!}{4^n\ (n+2)\ (n+2)!^2}=\frac{31}8-4\times{_4F_3}\left(-\frac12,\frac12,1,1;\ 2,2,2;\ 1\right)\\\ \\\approx0.008749644047541935203478962326551903908774780849356243615274...$$
I wonder if it can be expressed in terms of simpler functions and well-known mathematical constants.</p>

答:
Ookay... Consider just the hypergeometric function and its closed form.

Take 16.5.2 from DLMF (with $a_0=1/2$, $b_0=2$) and write (I used Mathematica to substitute the special form for the hypergeometric function in the integrand; I don't really know how to do it by hand):
$$ F(-1/2,1/2,1,1;2,2,2;1) = \int_0^1 \frac{8 \sqrt{1-t} \left(4-4 \sqrt{1-t}+t\sqrt{1-t}-\log8+3 \log(1+\sqrt{1-t})\right)}{9 \pi  t^{3/2}}\,dt. $$

Mathematica can then do the integral in closed form to give
$$\frac{1}{9 \pi }8 \left(\frac{70}{3}+\frac{11 i \pi ^2}{4}+\pi  \left(-7+\log 512-6\log\left(1+\frac{1+i}{\sqrt{2}}\right)\right)+24 i \mathrm{Li}_2\left(-\frac{1+i}{\sqrt{2}}\right)-24 i \mathrm{Li}_2\left(1-\frac{1+i}{\sqrt{2}}\right)\right),$$
where $\mathrm{Li}_2$ is the polylogarithm.

Of this we take the real part only and do some more FunctionExpand:
$$-\frac{56}{9}+\frac{560}{27 \pi }-\frac{16 C}{3 \pi }+\frac{64 \Im\left(\mathrm{Li}_2(1-(-1)^{1/4})\right)}{3 \pi }+\frac{8 \log512}{9}-\frac{8}{3} \log\left(\frac{1}{2}+\left(1+\frac{1}{\sqrt{2}}\right)^2\right)+\frac{\psi_1\left(\frac{1}{8}\right)}{3 \sqrt{2} \pi }+\frac{\psi_1\left(\frac{3}{8}\right)}{3 \sqrt{2} \pi }-\frac{\psi_1\left(\frac{5}{8}\right)}{3 \sqrt{2} \pi }-\frac{\psi_1\left(\frac{7}{8}\right)}{3 \sqrt{2} \pi }, $$
where $\psi_1$ is the polygamma function and $C$ is the Catalan constant.

Now, the polylogarithm term there can be simplified using the identity (DLMF 25.12.6)
$$ \mathrm{Li}_2(x)+\mathrm{Li}_2(1-x)=\frac{\pi^2}{6}-\log x\log(1-x), $$
because $\mathrm{Li}_2((-1)^{1/4})$ is simpler.

After a further FunctionExpand (which gets rid of the polygamma functions also), ComplexExpand to get the real part, and FullSimplify to simplify the expression, the answer is
$$ -\frac{56}{9}+\frac{560}{27 \pi }-\frac{32C}{3 \pi }+\frac{16}{3} \log2$$


问:
How to simplify the summation of a recurrence relation
After solving the recurrence relation

$$T(n) = 3T(\frac{n}{3}) + n\log(n)$$

I get following equation

$$T(n)=3kT(\frac{n}{3k})+ n\log(n) + n\log(\frac{n}{3}) + n\log(\frac{n}{3^2})+\dots+n\log(\frac{n}{3^k})$$

I don't know how to simplify the summation and how to know the asymptotic function?

Body:
Let $n=3^k$. We have,

$T(n)=3^kT(\frac{n}{3^k})+ n\log(n) + n\log(\frac{n}{3}) + n\log(\frac{n}{3^2})+\dots+n\log(\frac{n}{3^k})$

$=3^k.T(1)+n\log\left(n.\frac{n}{3^1}.\frac{n}{3^2}\ldots.\frac{n}{3^k}\right)$

$=3^k.1+n\log\left(n.\frac{n^k}{3^{1+2+\ldots+k}}\right)$ (since $T(1)=1$)

$=3^k+n\log\left(\frac{n^{k+1}}{3^{1+2+\ldots+k}}\right)$

$=3^k+n\log\left(\frac{(3^k)^{k+1}}{3^{k(k+1)/2}}\right)$

$=3^k+n\log\left(\frac{3^{k(k+1)}}{3^{k(k+1)/2}}\right)$

$=3^k+3^k\log\left(3^{k(k+1)/2}\right)$

$=3^k+3^k.k(k+1)/2.\log3$

$=3^k+\Theta(3^k.k^2)$

$=\Theta(3^k.k^2)$

$=\Theta(n.(\log n)^2)$ (since $n=3^k$)

Or use Master theorem:

$T(n) = 3T(\frac{n}{3}) + n\log(n)$, here $c_{crit}=\log_b a = \log_3 3 = 1$, $k=1$, hence, we have,

$T(n) = \Theta(n^{c_{crit}}\log^{k+1}n)=\Theta(n\log^2 n)$


答:
You have a commutative square $$\require{AMScd}\begin{CD}X @>f>> Y \\
@VVV @VVV \\
Z@>g>> W\end{CD}$$
From this you can extract a commutative diagram $$\begin{CD}X @>>> \mathrm{im}(f) @>>> Y \\
@VVV @VVV @VVV \\
Z@>>> \mathrm{im}(g) @>>> W\end{CD}$$

by using the definition of the image either as the kernel of the cokernel, or as the cokernel of the kernel.

Let me do the first one  : you get an induced diagram $$\begin{CD}X @>f>> Y @>>> \mathrm{coker}(f) \\
@VVV @VVV @VVV\\
Z@>g>> W @>>> \mathrm{coker}(g)\end{CD}$$ by universal property of the cokernel, and now it's clear that you also get a diagram $$\begin{CD}\mathrm{im}(f) @>>> Y @>>> \mathrm{coker}(f)\\
@VVV @VVV @VVV \\
\mathrm{im}(g) @>>> W @>>> \mathrm{coker}(g)\end{CD}$$ by universal property of the kernel.

Moreover, by uniqueness in the universal property of the kernel, you also get a commutative diagram $$\begin{CD}X @>>> \mathrm{im}(f) \\
@VVV @VVV \\
Z@>>> \mathrm{im}(g)\end{CD}$$ and pasting them gives the one I claimed.

Since everything here is defined in terms of universal properties this obviously gets you all the functoriality properties you might want

Title: Arrow-theoretic proof that cohomology is a functor.

Body:
Let $\sf{A}$ be an abelian category and $\sf{C}(\sf{A})$ be its category of complexes. I want to prove that the $i$-th cohomology is an additive functor $\sf{C}(\sf{A})\to\sf{A}$. For that, let $\varphi^\bullet:M^\bullet\to N^\bullet$ be a morphism of complexes. By the universal property of kernels, we have an induced morphism

In order for the universal property of cokernels to induce a morphism $H^i(\varphi^\bullet):H^i(M^\bullet)\to H^i(N^\bullet)$ making the diagram

commute, we have to show that the morphism $I^{i-1}_{M^\bullet}\to K^i_{M^\bullet}\to K^i_{N^\bullet} \to H^i(N^\bullet)$ is zero. That would follow from the existence of an induced map $I^{i-1}_{M^\bullet}\to I^{i-1}_{N^\bullet}$, but I can't seem to figure out why this exists.

Also, I would love to know if there's a simpler way to construct the induced morphism $H^i(\varphi^\bullet):H^i(M^\bullet)\to H^i(N^\bullet)$ since my construction seems a little tough to work with.


答:
This is a nice question. Let $\Gamma$ be the two premises (statement 1 and 2) and let us write $\Gamma\models A$ for "$A$ is a valid conclusion from $\Gamma$".

Then for example the formalization of "Only conclusion $1$ is correct" is the (false) statement $$(\Gamma\models\text{conclusion 1})\text{ and }(\Gamma\nvDash\text{conclusion 2}).$$

The ambiguity arises because there are two different readings of the option "3. Either conclusion I or II is correct." This could either be the false statement
$$
(a)\quad\text{either }(\Gamma\models\text{conclusion 1})\text{ or }(\Gamma\models\text{conclusion 2})$$
or it could be the true statement
$$
(b)\quad\Gamma\models\text{either conclusion 1 or conclusion 2}$$

I think $(a)$ is more natural, so that option 3 is false (but I am not a native speaker).

Title: Interpreting english EITHER-OR when approaching logical deductions doubt

Body:
The following question has been asked in the Indian entrance examination "GATE CS 2021", and it has stirred up a lot of confusion regarding how one should interpret "either-or" when coming to logical questions where we are given certain premises and have to determine if something is true via deduction.

Q. Given below are two statements 1 and 2, and two conclusions I and II

Statement 1: All bacteria are microorganisms.
Statement 2: All pathogens are microorganisms.

Conclusion I: Some pathogens are bacteria.
Conclusion II: All pathogens are not bacteria.

Based on the above statements and conclusions, which one of the following options is logically CORRECT?

Only conclusion I is correct

Only conclusion II is correct '

Either conclusion I or II is correct

Neither conclusion I nor II is correct

refer this link for the debate - https://gateoverflow.in/357468/gate-cse-2021-set-1-ga-question-9
refer this link for my reasoning - https://gateoverflow.in/357468/gate-cse-2021-set-1-ga-question-9?show=358167#a358167

In short -
I feel option D is the correct choice, because neither conclusions follow from the premises, but on the other side a lot of people feel that option C is right because according to them option c is a tautology.

Also there is another confusion regarding the statement of conclusion 2 "All pathogens are not bacteria".
Does this mean that there is no pathogen which is also a bacteria, or does it mean there exists some pathogens who are definitely not bacteria?
Because when we say a similar statement like "All students are not geniuses", we generally mean that there are only few students who are geniuses, and most of the students are not geniuses. We don't mean to say that there isn't even a single student who is not a genius.

Can anybody please provide an unambiguous, mathematically backed argument for choosing the correct answer, or is the given question and options inherently ambiguous?


问:
Strongly convex functions (equivalence and intuition)
I trying to prove and understand the equivalence of definitions for a $\gamma$-strongly convex function. I am aware that a function $f:\mathbb{R}^n\mapsto\mathbb{R}$ is strongly convex of modulus $\gamma$ if $\forall$ ${\bf x}$, ${\bf y}\in\mathbb{R}^n$, we have
\begin{align*}
\frac{\gamma}2||{\bf x}-{\bf y}||^2_2&\leq f({\bf y})-f({\bf x})-\nabla f({\bf x})^\mathrm{T}({\bf y}-{\bf x}),\\
f[(1-\lambda)x+\lambda y]&\leq(1-\lambda)f(x)+\lambda f(y)-\frac12\gamma\lambda(1-\lambda)||{\bf x}-{\bf y}||^2_2.
\end{align*}
However, I do not really understand how to prove that the two are equivalent, even after looking through other MSE questions. I found a presentation that talks about the graph being convex but not being flat, and I am having a hard time relating geometry of the problem to the definitions above as well. Could someone help me out/link me to other MSE posts that may help me out? I’ll delete this if it’s been asked before (I don’t know why I couldn’t find anything that made sense to me). Will really appreciate any help, thanks!

答:
Let me provide the geometric interpretation. To start with the first definition:
$$\frac{\gamma}2||{\bf x}-{\bf y}||^2_2\leq f({\bf y})-f({\bf x})-\nabla f({\bf x})^\mathrm{T}({\bf y}-{\bf x})$$
It can be rewritten as:
$$f({\bf y}) \geq f({\bf x})+\nabla f({\bf x})^\mathrm{T}({\bf y}-{\bf x}) + \frac{\gamma}2||{\bf x}-{\bf y}||^2_2$$
The right hand side is the first order taylor approximation at $\bf{x}$ plus a quadratic function. Without the last term, the inequality means that the graph lies above any tangent line, which is valid for convex functions (see this question for a proof). The quadratic term means that the graph is not just above the tangent line, but that the difference between the tangent line and the graph increases at least quadratically.

In the second definition:
$$f[(1-\lambda)x+\lambda y]\leq(1-\lambda)f(x)+\lambda f(y)-\frac12\gamma\lambda(1-\lambda)||{\bf x}-{\bf y}||^2_2$$
if you omit the final term, you recognize the definition of a convex function: the line that connects $x$ and $y$ should be above the graph on the interval $[x,y]$. The final term provides a lower bound on the distance between the line and the graph. For fixed $x$ and $y$, the last term is quadratic in $\lambda$ with the peak at $\lambda=0.5$.

To show equivalence, I suggest you start with the second definition and try to adapt the proof linked above.


问:
What are some examples of functions that are continous from $[a,b]$, differentiable $(a,b)$ but not at $a$ and $b$.
I am studying the MVT and Rolle's Theorem. I would like some examples of functions that are continuous from $[a,b]$, differentiable from $(a,b)$ but not differentiable at $a$ and $b$. I am aware that the derivative at the point c is defined as:

$ f'(c) = \displaystyle \lim_{x \to c} \frac{f(x) - f(c)}{x-c}$

Basically, I'm wondering why does a function have to be continuous at the endpoints of the interval, but it doesn't have to be differentiable at the endpoints. Why is this important?

My other question is, how significant is the study of open/closed sets important in calculus?

Body: $f(x)=\sqrt{a^2-x^2}$ on $[-a,+a]$ or $f(x)=\sqrt[3]{x}$ on $[0,b],~b>0$


问:
Why can we not write: $\lim_{x \to 0}(-\frac{1}{x})=-\infty$
I was skimming through a few books on calculus, and came across the statement that:

we cannot write out this limit as $+\infty$ or $-\infty$  like so: $\lim_{x \to 0}(-\frac{1}{x})=-\infty$, only as $\lim_{x \to 0}(\frac{1}{x})=\infty$. The argument provided was that its a matter of agreement.

Although, it's considered perfectly fine to write this equation likeso: $\lim_{x \to 0}(-\frac{1}{x^2})=-\infty$

My question is whether anyone else has written it in opposition to the statement above, and what is meant by "matter of agreement", and why it may be as such.

Or might it just be the authors own biased opinion, for whatever reason?

I personally see nothing wrong with introducing $+\infty \space or -\infty$ given that it helps to recognise further meaning in supporting your rationale for proving a theorem.

Here's an image of the page where it's taken from:

Reference:
Fort, T. (1951) Calculus. pg 50.

Your comments are most welcome.

答:
Presumably, the "matter of agreement" means that denoting $f(x)=-\frac{1}{x}$, you have
$$
\lim_{x\to 0+}f(x)=-\infty
$$
and
$$\lim_{x\to 0-}f(x)=+\infty
$$ which do not agree.

Alternatively, the definition of "$\lim_{x\to 0}f(x)=-\infty$" says that

for every $M>0$, there exists $\delta>0$ such that $0<|x-0|<\delta$ implies $f(x)<-M$.

But this is impossible in the case when $f(x)=-\frac{1}{x}$.


问:
Degree of Extension from $\mathbb{Q}$ to $\mathbb{Q}(i, \sqrt[4]2)$
I wanted to find the degree of extension from $\mathbb{Q}$ to $\mathbb{Q}(i, \sqrt[4]2)$. The minimal polynomial of $\sqrt[4]2$ over $\mathbb{Q}$ is $f(x)=x^4-2$. Now the roots of $f(x)$ are $\{\sqrt[4]2, -\sqrt[4]2, \sqrt[4]2i, -\sqrt[4]2i\}$. So we can observe $\frac{\sqrt[4]2i}{\sqrt[4]2}=i$ so $i \in \mathbb{Q}(\sqrt[4]2)$ so we can already see the degree of the extension from $\mathbb{Q}$ to $\mathbb{Q(i, \sqrt[4]2)}$ is 4. But also there is another approach to this problem:

Since we know $[\mathbb{Q}(\sqrt[4]2):\mathbb{Q}]$=4, then we can say $\mathbb{Q}(\sqrt[4]2)=\{a+b\sqrt[4]2+c\sqrt[4]2^2+d\sqrt[4]2^3 | a,b,c,d \in\mathbb{Q}\}$. Now I am not sure if my understanding is correct from here or not, but I think we should be able to express $i$ as an element of this linear combination. I have tried a few variations and I cannot seem to show this is possible. There is no negative under the radical sign. Is there something wrong with this second approach or can we express $i$ as a linear combination of the basis?

答: I don't understand how did you get to $i\in\mathbb{Q}(\sqrt[4]{2})$. It's not true that all the roots of the minimal polynomial of $\sqrt[4]{2}$ must belong to this field. Actually, in this example it is clearly false. By definition, $\mathbb{Q}(\sqrt[4]{2})$ is the intersection of all subfields of $\mathbb{C}$ which contain the element $\sqrt[4]{2}$. In particular, $\mathbb{R}$ is such a field, so it is one of the fields in the intersection. This means $\mathbb{Q}(\sqrt[4]{2})$ is a subfield of $\mathbb{R}$, and so $i$ can't be an element of this field. This tells us that we must have $[\mathbb{Q}({i,\sqrt[4]{2})}:\mathbb{Q}(\sqrt[4]{2})]=2$, and so by the tower rule the extension degree you are looking for is $8$.


问:
Backward representation of the general Markov process
One usually deals with a discrete time Markov process in the following form: given a state space $E$ the Markov process is defined by transition kernel $T(B|x)$ such that
$$
\mathsf{P}(X_1\in B|X_0 = x) = T(B|x)
$$
for all $x\in E$, $B\in\mathcal{B}(E)$.

So, given a current state we have a distribution of the future state.

On the other hand it can be an interesting problem given a current state to find a distribution of the previous state which "fits" with a transition kernel $T$.

答:
This is called time-reversal of Markov chains and is presented in numerous lecture notes available on the web, see here for example.

Remark As said in the comments, the sentence of the post stating that given a state $X_{n+1}$ we know the distribution of $X_n$ signals a deep misunderstanding of the structure of Markov processes. In the example considered, $v_n$ is not independent on $X_{n+1}$ and one should note that, given any nonzero $a$ and $b$ and any random $X$, any random variable $Y$ can be written as $Y=X/a-(b/a)V$ for a well chosen random variable $V$.


答:
$i \in \Bbb Q(\sqrt[4]{2})$ is not correct. Indeed, $\Bbb Q(\sqrt[4]{2})$ is completely contained in $\Bbb R$.

Note that this extension is not a normal extension, it does not contain all the roots of the minimal polynomial. This is why your analysis fails.

Since $i \notin \Bbb Q(\sqrt[4]{2})$, you can conclude that $[\Bbb Q(\sqrt[4]{2})(i):\Bbb Q(\sqrt[4]{2})] \ge 2.$ On the other hand, the polynomial $x^2 + 1 \in \Bbb Q(\sqrt[4]{2})[x]$ shows that the degree is exactly $2$.

Since the degrees multiply, we see that the extension $\Bbb Q(\sqrt[4]{2})(i) = \Bbb Q(\sqrt[4]{2}, i) \supset \Bbb Q$ is of degree $\boxed{8}$.

Title: Degree of Extension from $\mathbb{Q}$ to $\mathbb{Q}(i, \sqrt[4]2)$

Body:
I wanted to find the degree of extension from $\mathbb{Q}$ to $\mathbb{Q}(i, \sqrt[4]2)$. The minimal polynomial of $\sqrt[4]2$ over $\mathbb{Q}$ is $f(x)=x^4-2$. Now the roots of $f(x)$ are $\{\sqrt[4]2, -\sqrt[4]2, \sqrt[4]2i, -\sqrt[4]2i\}$. So we can observe $\frac{\sqrt[4]2i}{\sqrt[4]2}=i$ so $i \in \mathbb{Q}(\sqrt[4]2)$ so we can already see the degree of the extension from $\mathbb{Q}$ to $\mathbb{Q(i, \sqrt[4]2)}$ is 4. But also there is another approach to this problem:

Since we know $[\mathbb{Q}(\sqrt[4]2):\mathbb{Q}]$=4, then we can say $\mathbb{Q}(\sqrt[4]2)=\{a+b\sqrt[4]2+c\sqrt[4]2^2+d\sqrt[4]2^3 | a,b,c,d \in\mathbb{Q}\}$. Now I am not sure if my understanding is correct from here or not, but I think we should be able to express $i$ as an element of this linear combination. I have tried a few variations and I cannot seem to show this is possible. There is no negative under the radical sign. Is there something wrong with this second approach or can we express $i$ as a linear combination of the basis?


答:
If you write, $\ \lim_{x \to 0^+}(-\frac{1}{x})=-\infty,\ $ then I would know what you mean, although I may or may not point out that it is more "proper" to say that the limit " diverges to $\ -\infty $", but some people may say that this is just me being pedantic and I think this is a fair opinion also.

However, saying that $\ \lim_{x \to 0}(-\frac{1}{x})=-\infty,\ $ is just incorrect, because

$\ \lim_{x \to 0^-}(-\frac{1}{x})=\infty \neq -\infty = \lim_{x \to 0^+}(-\frac{1}{x}). $

Title: Why can we not write: $\lim_{x \to 0}(-\frac{1}{x})=-\infty$

Body:
I was skimming through a few books on calculus, and came across the statement that:

we cannot write out this limit as $+\infty$ or $-\infty$  like so: $\lim_{x \to 0}(-\frac{1}{x})=-\infty$, only as $\lim_{x \to 0}(\frac{1}{x})=\infty$. The argument provided was that its a matter of agreement.

Although, it's considered perfectly fine to write this equation likeso: $\lim_{x \to 0}(-\frac{1}{x^2})=-\infty$

My question is whether anyone else has written it in opposition to the statement above, and what is meant by "matter of agreement", and why it may be as such.

Or might it just be the authors own biased opinion, for whatever reason?

I personally see nothing wrong with introducing $+\infty \space or -\infty$ given that it helps to recognise further meaning in supporting your rationale for proving a theorem.

Here's an image of the page where it's taken from:

Reference:
Fort, T. (1951) Calculus. pg 50.

Your comments are most welcome.


问:
An example of a class of structures whose axiomatizability depends on the axiom of choice
Is there an example of a specific class $K$ of structures (in the sense of first-order logic), such that $K$ is first-order axiomatizable if the axiom of choice is assumed, but that its axiomatizability is independent of ZF set theory?

答:
Yes, there is such a class of structures. Working in the empty language, consider the class of well-orderable sets (keeping in mind that a structure in the empty language is just a set, full stop). This is axiomatized by the empty theory if $\mathsf{AC}$ holds, and is non-axiomatizable otherwise.

(To see that the class of well-orderable sets is not first-order axiomatizable if $\mathsf{AC}$ fails, just check that $\mathsf{ZF}$ proves the elementary equivalence of any two infinite pure sets.)

In fact, the question - without further restrictions, anyways - is actually trivial: given an arbitrary sentence $\varphi$ in the language of set theory, fixing your favorite non-axiomatizable class $K$ consider the class $$K_\varphi=\begin{cases}
\emptyset & \mbox{if $\varphi$ holds}\\
K & \mbox{otherwise}.
\end{cases}$$ This class is axiomatizable iff $\varphi$ holds.

So to make the question non-trivial, we need to restrict attention to "nice" (definitions of) classes of structures. One natural framework for doing so is via abstract model theory. Fixing an "appropriately absolute" logic $\mathcal{L}$ we can ask:

Is there a $\mathcal{L}$-sentence $\sigma$ such that the first-order axiomatizability of the class of models of $\sigma$ is undecidable in $\mathsf{ZF}$ (or $\mathsf{ZFC}$, or ...)?

For example, if we take $\mathcal{L}=\mathsf{FOL}$ the answer is obviously no. On the other hand, the example at the start of this answer shows that the answer is yes if $\mathcal{L}=\mathsf{SOL}$, since well-orderability is characterizable in second-order logic.

In my opinion the most natural choice of $\mathcal{L}$ to consider here is existential second-order logic, due to the latter's relative tameness (it is compact and has the downward Lowenheim-Solem property for individual sentences, avoiding contradiction with Lindstrom's theorem by failing to be closed under negation). It is not immediately clear to me what the situation is here.


问:
Why is the universal quantifier $\forall x \in A : P(x)$ defined as $\forall x (x \in A \implies P(x))$ using an implication?
And the same goes for the existential quantifier: $\exists x \in A : P(x) \; \Leftrightarrow \; \exists x (x \in A \wedge P(x))$. Why couldn’t it be: $\exists x \in A : P(x) \; \Leftrightarrow \; \exists x (x \in A \implies P(x))$ and $\forall x \in A : P(x) \; \Leftrightarrow \; \forall x (x \in A \wedge P(x))$?</p>

答:
I can't answer your first question. It's just the definition of the notation.

For your second question, by definition of '$\rightarrow$', we have

$\exists x (x\in A \rightarrow P(x)) \leftrightarrow \exists x\neg(x\in A \wedge \neg P(x))$

I think you will agree that this is quite different from

$\exists x (x\in A \wedge P(x))$


问:
Description of second moment in terms of 2-dimensional marginal distribution.
I'm studying the book Noise Sensitivity of Boolean Functions and Percolation by C. Garban and J.E. Steif and in Chapter 9, when presenting the Spectral Probability, they state that knowing that the 2-dimensional marginals of two distributions are equal is useful when applying second moment arguments. This and some calculations in the Chapter suggests me that there is some way to describe the second moment in terms of the 2-dimensional marginals, which i'm missing.

More precisely, let $A\subseteq \{1,\dots,n\}$ be a random set which we sample according to $\mathbb{P}$. If $|A|$ denotes the cardinality of the set $A$, then
$$
\mathbb{E}|A|=\mathbb{E}[\sum_{i=1}^n\mathbb{1}_{A}(i)]=\sum_{i=1}^n\mathbb{P}(i\in A),
$$
that is, we can describe the expectation in terms of the 1-dimensional marginals $\mathbb{P}(i\in A)$. So my question is

Is there a similar way to describe $\mathbb{E}[|A|^2]$ in terms of $\mathbb{P}(i,j\in A)$?

答:
just noticed that
$$
\mathbb{E}[|A|^2]=\mathbb{E}[\sum_{i,j}\mathbb{1}_{A\times A}(i,j)]=\sum_{i,j}\mathbb{P}((i,j)\in A\times A)=\sum_{i,j}\mathbb{P}(i,j\in A).
$$


问:
What is the antiderivative of one over an arbitrary polynomial?
Suppose we have an arbitrary polynomial as follows.
$$p(x):=\sum_{i=0}^n a_nx^n$$
And we want to find the antiderivative of $(p(x))^{-1}$ with respect to $x$.
$$\int\frac{dx}{p(x)}$$
Messing around with an online CAS suggests that this would be equal to
$$C+\sum_{z\in\{r\mid p(r)=0\}}\frac{\ln |x-Z|}{\sum^n_{i=1}ia_iz^{i-1}}$$
Which sort of makes sense to me, and I think I can visualize that, but my question is, how do you derive this result and how true is it? Do you have to employ a form of partial fraction decomposition? The website also says that this only applies to cases where each root is distinct. What happens in the case of repeated roots? Can I extend this to a power series?

答:
Well, if you have a polynomial of this form you can write it as a product of factors (although the roots will probably be mostly complex):
$$p_n(x)=\sum_{k=1}^na_kx^k=C_1\prod_{k=1}^n(x-\alpha_k)$$
now if you have:
$$\frac{1}{p_n(x)}=C_2\prod_{k=1}^n\frac{1}{(x-\alpha_k)}=C_2\sum_{k=1}^n\frac{b_k}{x-\alpha_k}$$
where each term can then be integrated individually, I think the hard part would be determining $\alpha_k,b_k$ from $a_k$


问:
$ \int_c^\infty \frac{e^{-a x^2 + bx}}{x} \, dx $
Suppose $a > 0$, $b$ is an arbitrary real number, and $c$ is a positive real number. Is there any closed form solution for the following integral

$$
\int_c^\infty \frac{e^{-a x^2 + bx}}{x} \, dx \text{ ?}
$$

答:
That is not a gaussian integral/integrand. That is:
$$\int\limits_{c}^{\infty}e^{-a(x+b)^2}\,\mathrm{d}x=-\dfrac{\sqrt{{\pi}}\left(\operatorname{erf}\left(\sqrt{a}\left(c+b\right)\right)-1\right)}{2\sqrt{a}}$$
But if that's really what you meant, I don't think there is a closed form for this. If you choose $b=0$, one can write
$$\int\limits_{c}^{\infty}\frac{e^{-ax^2}}{x}\,\mathrm{d}x= \dfrac{\operatorname{\Gamma}\left(0,ac^2\right)}{2}.$$
If we were alloud to set $b=d/x$ we could also look at the integral
$$\int\limits_{c}^{\infty}\frac{e^{-ax^2+b}}{x}\,\mathrm{d}x= \dfrac{\operatorname{\Gamma}\left(0,ac^2\right)\mathrm{e}^d}{2}.$$


问:
Question: For what value of $c$ will $P(x)= -2x^3+cx^2-5x+2$ have the same remainder when devided by $x+1$ or $x-2$?

For what value of $c$ will $P(x)= -2x^3+cx^2-5x+2$ have the same remainder when devided by $x+1$ or $x-2$?

No idea where to start with this one. Can't use the remainder theorem to find the remainder because of the leading coefficient and can't find the value of c when I have no known remainder. Be nice if the examples in my schools lessons actually included the same kind of questions that are on the tests but I guess that would make to much sense for school.

答: Hint: use polynomial long division to divide $P(x)$ by $x+1$, and find the remainder term in terms of $c$. Then, divide $P(x)$ by $x-2$, and also find the remainder term in terms of $c$. Set the two remainder terms equal and solve...


问:
Prove that there exist integers $x, y$ so that $2x^2+3y^2-r$ is divisible by a given prime larger than 3.
Prove that there exist integers $x, y$ so that $2x^2+3y^2-r$ is divisible by some fixed prime larger than 3.

$r$ is also an integer.

答:
Let $p > 3$ be the fixed prime.

We consider the following $p + 1$ numbers:

$2x^2$ for $x = 0, 1, \dots, \frac {p - 1}2$;

$r - 3y^2$ for $y = 0, 1, \dots, \frac {p - 1}2$.

Since there are only $p$ different residues mod $p$, there must be two numbers with the same residue.

It is clear that

$2x_1^2 \neq 2x_2^2$ for any $0 \leq x_1 < x_2 \leq \frac{p - 1}2$;

$r - 3y_1^2 \neq r - 3y_2^2$ for any $0 \leq y_1 < y_2 \leq \frac{p - 1}2$.

Therefore there must exist $x, y$ such that $2x^2 = r - 3y^2$.


问:
Is there a simpler approach to these system of equations?
I recently came across the following system of equations:

$$x + y + z = 1 \\
x^2 + y^2 + z^2 = 2  \\
x^3 + y ^3 + z^3 = 3$$

And I have two questions:

One, is there a way to prove or disprove whether there is a solution for this particular set of equations? Furthermore, is there a way to expand the proof for a more generalized set of equations, that is for this set :

$$x + y + z = 1 \\ x^2 + y^2 + z^2 = 2 \\ ...\\x^n + y^n + z^n = n$$

Two, is there a simpler approach for the prior solution set than substitution? As of now, I'm not getting anywhere with this method. I end up getting into a long-winded series of substitution and isolation that yields something like $z = z$ or $1 = 1$.

Sorry if it's a repeat question; I couldn't exactly find an accurate way to search for the equations.

答:
HINT:

$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$

$(x+y+z)^3=x^3+y^3+z^3+3(xy+yz+zx)(x+y+z)-3xyz$

So, we can find $x+y+z,xy+yz+zx,xyz $

Now  $x,y,z$ are the roots of $t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz=0$

I think the same pattern can be applied for the higher powers


问:
How to find $f^{-1}(22)$ with $f(x,y) = 15x+55y$ for integer $x,y$
I don't even know how to get the inverse of this function

The complete function is described as the following:
$f:\mathbb{Z}^2\to\mathbb{Z}$ such that $f(x,y)=15x+55y$

答:
Here says that $ax+by=c$
has solutions if and only if $c$ is a multiple of $\text{GCD}(a,b)$.

$$15 x+55 y=22$$
has no integer solutions because $\text{GCD}(15,25)=5$ but $5$ doesn't divide $22$.

Therefore $f(x,y)$ has no inverse at $22$.


答:
Without entering the limit definition, ask yourself: how do you approach to $ 0 $?

First

You can take $x=0.00000000...1$, then what do you get?

$$\dfrac 1x=\dfrac {1}{0.00000000...1}=100000000...\longrightarrow \color {red} {\text{positive infinity}}$$

Second

You can take $x=-0.00000000...1$, then what do you get?

$$\dfrac 1x=\dfrac {1}{-0.00000000...1}=-100000000...\longrightarrow \color {blue} {\text{negative infinity}}$$

So, you get two different result:

$$\lim_{x\to 0}\dfrac1x=+ \infty$$

$$\lim_{x\to 0}\dfrac1x=- \infty$$

What will we do now?Which do you think is right?  Unfortunately, none of limits are correct. Because, you get two different results. This is called right and left limits in mathematics.  These must be equal for the limit to exist. So, we must say "limit doesn't exist."

But, note that this limit is correct:

$$\boxed {\lim_{x\to 0}\dfrac{1}{|x|} =\infty}$$


问:
Showing that $\frac{f(x)-f(a)}{x-a} \le \frac{f(b)-f(a)}{b-a}-\frac k 2 (b-x)$ when $f'' \ge k$
Let $f$ be a function of class $C^2$ on $[a,b]$ such that $f'' \ge k$ for $k\in\mathbb{R}$.

The problem is to show that for all $x \in [a,b]$, we have $\frac{f(x)-f(a)}{x-a} \le \frac{f(b)-f(a)}{b-a}-\frac k 2 (b-x)$.

I tried Taylor-Lagrange formula but I didn't succeed because of the signs...

I tried to use convex methods but it fails as the condition doesn't imply convexity or concavity (if $k$ is negative, then $f$ isn't necessarly concave).

Is there any idea ?
Thanks in advance.

NB:
Conversely, if this inequality holds for all $[c,d] \subset [a,b]$, do we have $f'' \ge k$ ?

答:
hint

For $ x\in (a,b]$  put

$$g(x)=\frac{f(x)-f(a)}{x-a}$$

then, by MVT,

$$g(b)=g(x)+(b-x)\color{red}{g'(c)}$$

with

$$g'(c)=\frac{f'(c)(c-a)-f(c)+f(a)}{(c-a)^2}$$

and, by Taylor-Lagrange formula,

$$f(a)=f(c)+f'(c)(a-c)+\frac{(a-c)^2}{2}f"(d)$$

which gives
$$\color{red}{g'(c)}=\frac 12f"(d)\ge \frac k2$$

thus
$$g(b)\ge g(x)+\frac k2(b-x)$$


问:
What to set $u$ as in integration by substitution

$$\int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}}x^2\sqrt{1-x^2}\
 \mathrm{d}x$$

The question asks me to use a trigonometric substitution so I tried $u=\sin x$ but that doesn't give me nice $x$ limit values so I'm not sure. I saw a guy using $u=\sec x$ but I don't understand how or why he knew to use that

答:
Set $x=\sin(u)$. First of all the boundaries become
$$\arcsin \frac{\sqrt 3}{2} = \frac{\pi}{3}, \quad \arcsin \frac{1}{2} = \frac{\pi}{6}.$$
Calculating the differential: $\mathrm{d}x = \cos u\,\mathrm{d}u$:
$$\int\limits_{\pi/6}^{\pi/2}\sin^2 u\cos^2 u\,\mathrm{d}{u}.$$


问:
Convergence of a series where terms involve a convergent sequence
I have been given the following question:

Let $ (x_n)_{n \in \mathbb{N}} $ be a sequence of real, positive numbers such that $ x_n \rightarrow L $ as $ n \rightarrow \infty $.
If $L>0$, prove that the following series converges:

$$\displaystyle\sum_{n=1}^\infty\frac{n^2}{n^4x_n+1}$$

I'm very much at a loss as to how to proceed with this...

I had initially went down the path of saying $$ \frac{n^2}{n^4x_n+1} < \frac{n^2}{n^4x_n}=\frac{1}{n^2x_n} = \frac{1}{n^2} \cdot \frac{1}{x_n} $$

and argued that this sequence converges to $0$, however then remembered that convergence to $0$ does not implicate the convergence of the series...

Then considered using the Ratio Test, but I am a bit unsure how to apply this for the specific question.
Any tips would be massively appreciated!!

答:
The usual way to prove that a series converges is to find an asymptotic equivalent ($\sim$, $O$, ...) of the general term of the series. Here since $L>0$
$$
\frac{n^2}{n^4 x_n+1} \sim \frac{n^2}{n^4L}=\frac{1}{n^2L}.
$$
and using the fact that $\sum \frac{1}{n^2}$ converges, we have the result.
Another way: for large $n$
$$
0 \le \frac{n^2}{n^4x_n+1} \le \frac{1}{n^2 x_n} \le \frac{1}{n^2(L/2)}.
$$
(indeed $x_n \to L$ so taking $\epsilon=L/2 >0$ there exists $n_0$ such that $\forall n \ge n_0, x_n \ge L/2$.)
And using the fact that $\sum \frac{1}{n^2}$ converges, we have the result.


问:
need to prove that the determinant of the inverse of a matrix is equal to 1 over the determinant of that matrix
I need to do some practice exercices and I am struggling with this one.

A and B, inversible square matrixes
prove that the determinant of an inverted matrix is equal to 1 over that matrix

|A^-1| = 1/|A|

(edit) im a french speaker from canada so I have a hard time using the english terms correctly since I use them in french normally. excuse me if there are any syntax/usage errors

答:
By multiplicativity of determinant: $|A||A^{-1}|=|A\cdot A^{-1}|=|Id|=1$, where by $Id$   I mean identity matrix.

From this, we can almost conclude that $|A^{-1}|=1/|A|$. The only thing that we need to check, is that we can divide by $|A|$. In other words, we need to make sure that $|A|$ is not zero.

But we know $A$ is invertible ( or equivalently regular/having full rank), hence the determinant is non-zero.