问:
Is number theory $K$ $\aleph_0$-categorical?
Let $K$ be the number theory. The language of $K$ contains the following symbols: $\{0, 1, +, \cdot, <, =\}$. Furthermore, the proper axioms are Peano axioms.

My question is, is $K$ $\aleph_0$-catergorical?

From what I learned, a theory with equality $K$ is $\aleph_0$-categorical if and only if

$K$ has at least one countably infinite normal model

Any two countably infinite normal models of $K$ are isomorphic.

I believe $K$ has at least one countably infinite normal model. We can just take the standard natural number model, right?

What's left is showing that (2) is true, assuming that $K$ is indeed $\aleph_0$-categorical. But we have to take two arbitrary countably infinite normal models (let's call them $M_1$ and $M_2$) of $K$ and show that they are isomorphic. There is not much information about $M_1$ and $M_2$ I believe, so I was wondering if there is a model $M$ of $K$ that has a denumerable normal model, but $M$ is not isomorphic to a standard natural number model (for example).
Any hints/explanation would be appreciated.

答:
This is not even close to $\aleph_0$-categorical.

First of all, per Godel's incompleteness theorem Peano arithmetic isn't even complete. Consequently, per Godel's completeness theorem there are countable models which are not elementarily equivalent to the standard one, so a fortiori not isomorphic to it.

Strictly speaking, the mere statement of the completeness theorem doesn't say that a countable model exists. However, we can get this by looking at the Henkinization proof of that theorem, which explicitly builds a countable model, or by applying downwards Lowenheim-Skolem (see below) after the fact.

What if we replace Peano arithmetic with the full theory $\mathsf{TA}$ (true arithmetic) of the structure $(\mathbb{N}; 0,1,+,\cdot,<,=)$? Well, $\mathsf{TA}$ is by definition complete so the above doesn't work. However, via the Lowenheim-Skolem theorem we can show that even $\mathsf{TA}$ is not $\aleph_0$-categorical.

Via the upwards part of the theorem (which is really just a corollary of compactness), let $\mathfrak{A}$ be an uncountable - hence a fortiori nonstandard - model of $\mathsf{TA}$. Now via the downwards part of the theorem (this is the really interesting half of the theorem), fix some infinite element $a\in\mathfrak{A}$ and let $\mathfrak{B}$ be a countable elementary substructure of $\mathfrak{A}$ with $a\in \mathfrak{B}$. This $\mathfrak{B}$ is a countable nonstandard model of $\mathsf{TA}$.


问:
Epsilon Delta continuity proof
This should be a basic epsilon delta practice but I don't know how to prove it.
Define $H: X\times[0,1]\to Y$ by
$
     f(x) = \begin{cases}
          F(x,2t) & t\in[0,\dfrac1 2] \\
          G(x,2t-1) & t\in[\dfrac1 2, 1] \\
       \end{cases}
    $

where F and G are continuous functions from $X\times[0,1]$ to $Y$ and $F(x,1)=G(x,0)$. I want to show that $H$ is continuous. It suffices to show that H is continuous at $t=\dfrac1 2$ since it is already continuous in $X\times[0,1]\setminus\{\frac1 2\}$. I wanted to state that F and G both converges to the same function at $t=\frac 1 2$ but it is not enough because it has to be continuous at every path around $(x,\frac1 2)$, I assume.

答:
Well, just roll up your sleeves and do it.

Consider the point $(w,\frac 12)$ and let $\epsilon  > 0$.  As $F,G$ are continuous there are $\delta_F, \delta_G$ so that if $D((x,y) - (w,\frac 12)||< \delta_{F,G}$ then $||F(x,2y)-F(w,1)|| < \epsilon$ and $||G(x,2y-1)-G(w,0)|| < \epsilon$.  Let $\delta = \min(\delta_F, \delta_G)$.

Now consider a point $(u,v)$ so that $||(u,v)-(w,\frac 12)|| < \delta$.  If $v < \frac 12$ then $||F(u,2v)-F(w,1)||=|f(u)-f(w)| < \epsilon$.  If $v> \frac 12$ then $||G(u,2v-1)-G(w,0)||=|f(u)-f(w)| < \epsilon$.  If $v = \frac 12$ then $F(x,2v)=G(x,2v-1)$ and $||F(u,2v)-F(w,1)||=||G(u,2v-1)-G(w,0)||=|f(u)-f(w)|< \epsilon$.

And we are done.

ContentLicense: CC BY-SA 4.0

Title: Epsilon Delta continuity proof


问:
Are non-constant meromorphic functions on compact Riemann surjective?
Let $f$ be meromorphic on compact Riemann surface $X$ (Assume $X$ if connected if need be), i.e. we have $f: X \setminus poles(f) \to \mathbb C$ holomorphic.

This corresponds to holormorphic $F: X \to \mathbb C_{\infty}$, where $F(x):=f(x)$, $x \in Domain(f)$ and $F^{-1}(\infty):=poles(f)$.

We have $f$ non-constant if and only if $F$ non-constant.

If $F$ is non-constant, then $F$ is surjective.

$F$ is surjective if and only if $f$ is surjective.

Therefore, if $f$ is non-constant, then by (2),(3) and (4), $f$ is surjective.

Question 1: Is this correct?

Question 2: Is this incorrect if $X$ is not connected?

Remark: I believe for the case of $X$ compact and connected: If question 1 is answered affirmatively, then I believe $f$ has a pole because '$f$ has a pole' is equivalent to '$F^{-1}(\infty) \ne \emptyset$'.

答:
If $f$ doesn't take the value $a$ then $1/(f-a)$ is holomorphic $X\to \Bbb{C}$.

Holomorphic functions from a compact Riemann surface to $\Bbb{C}$ are constant:
they are continuous, (by compactness of $X$) $1/|f-a|$ takes its maximum at some $p\in X$, locally in some chart $1/(f-a)$ is represented by an analytic function which must be constant because locally a non-constant analytic function is $g(z)=\sum_{n\ge 0} c_n (z-b)^k = c_0+c_k (z-b)^k+O((z-b)^{k+1})$ with $c_k\ne 0$ so $|g(z)|$ is not maximum at $b$.


问:
Events related to almost everywhere convergence
We were taught in class that :

A sequence of random variables $(X_n)_{n\geqslant 1}$ converges to $X$ almost surely (or, almost everywhere), if $~\exists~$ an event $A$ such that $\mathbb{P}(A)=0$ , and $\forall~\omega\in\Omega\setminus A$ , $X_n(\omega) \longrightarrow X(\omega)$ pointwise.

Now, my question is, for a sequence $(X_n)_{n\geqslant 1}$ of random variables, if it's given that $X_n\longrightarrow X$ almost surely, then is it true that for all events $E$ such that $\mathbb{P}(E)=0$ and $\forall~\omega\in\Omega\setminus E$ , $X_n(\omega) \longrightarrow X(\omega)$ pointwise ?

If the answer to my doubt is NO, then, in general, is there a common way to show that a sequence of random variables does NOT converge almost surely ?

答:
Suppose $X_n\to X$ almost surely. By definition you are guaranteed a null event $A$ such that $X_n(\omega)\to X(\omega)$ (pointwise) for all $\omega\in A^c.$ But if you take an event $B\subsetneq A,$ and $\omega\in A\setminus B$ then obviously you are not guaranteed that$X_n(\omega)\to X(\omega).$ As long as your measure space is complete and $|A|>1,$ You can always find such events. So in general, as you expected the answer to your first question is ‘No’.

For the second question, I don’t know of a universal technique to show that $X_n$ does not converge to $X.$ But here are a few tricks, I have come across.

If $X_n\to X$ a.s., then $X_n\to X$ in probability. It is easier usually to show that $X_n$ does not converge in probability. Of course, a sequence can converge in probability but may fail to converge almost surely and this method would not work.

Another way would be to try and identify $X^1:=\limsup X_n$ and $X_{1}:=\liminf X_n.$ The good thing is that they are always guaranteed to exist. If you could show that $X^1\neq X_{1}$ then you are done.

If you have added assumptions (like boundedness) that combined with almost sure convergence would give you the convergence of the expectations, then it might be useful to check that the expectations does not converge (this may itself be very tricky sometimes).

A naive trick is to find two subsequences which converge almost surely but to two different limits. This tells you that the original sequence can not converge almost surely.


问:
$H$ must contain every Sylow $p$-subgroup of $G$
Let $p$ be a prime factor of the order of a finite group $G$. If $H$ is a normal subgroup $G$ whose index is not a multiple of $p$, show that $H$ must contain every Sylow $p$-subgroup of $G$.</p>

答:
Hints:

Show that $H$ must contain a Sylow $p$-subgroup $P$ of $G$.

Show that $H$ contains all the subgroups of $G$ that are conjugates of $P$.

ContentLicense: CC BY-SA 3.0


问:
Divergence of an Expression Involving the Gamma Function
I want to show that
$$
\lim_{Re(t) \to \infty} \frac{\Gamma(1+2t)\Gamma(-t)}{\Gamma(1+t)} (-z)^t
$$
diverges if $|z| > \frac14$.
I recognized that the expression involving the Gamma function is just the Beta function, $B(2t+1,-t)$, but I don't know much about the Beta function. Are there some properties of the Gamma function that I can use to prove this divergence? Thanks!

答:
Note that some restriction has to be put on $s$ for this limit to make sense: for example, if $s\to\infty$ through positive real numbers, then the $\Gamma(-s)$ term has a pole at every integer, making the limit undefined.

As long as one is willing to restrict, say, the argument of $s$ to an interval like $[\varepsilon,\pi-\varepsilon]$, then this problem goes away and one can use Stirling's formula to asymptotically evaluate each term.


问:
If $a_{n}$ is an increasing sequence that is not bounded above, show that $a_{n}\to\infty$
My thoughts on this are: assume $a_{n}$ doesn't diverge to infinity, it is increasing so instead must converge to some a but then this a would be an upper bound, yet we said an had no upper bound. This contradiction shows the sequence must diverge to infinity.

Is this enough to show the divergence, what am I missing?

答:
If $a_{n}$ is not bounded above, then for every $M\in\mathbb{R}_{\geq 0}$ there corresponds a natural number $n_{0}$ s.t $a_{n_{0}}\geq M$. Since $a_{n}$ is increasing, this means that:
\begin{align*}
(\forall M\geq 0)(\exists n_{0}\in\mathbb{N})\,\,\text{such that}\,\,(n\geq n_{0} \Rightarrow a_{n}\geq a_{n_{0}} \geq M)
\end{align*}
and we are done.

Hopefully this helps.


答: That strategy works, but you don't need to appeal to proof by contradiction. You showed that if an increasing sequence does not diverge, then it is bounded above. By contrapositive, if an increasing sequence is not bounded above, then it diverges.</p>

Title: If $a_{n}$ is an increasing sequence that is not bounded above, show that $a_{n}\to\infty$

Body:
My thoughts on this are: assume $a_{n}$ doesn't diverge to infinity, it is increasing so instead must converge to some a but then this a would be an upper bound, yet we said an had no upper bound. This contradiction shows the sequence must diverge to infinity.

Is this enough to show the divergence, what am I missing?


问:
Is the limit of uniformly integrable functions integrable?
If $\left\{f_n\right\}$ are uniformly integrable and $f_n\overset{a.e.}{\rightarrow}f$ ($f$ measurable), is $f$ integrable? Can "uniformly integrable" be weakened to "integrable"?</p>

答:
Notes on Additional Assumptions

It should be noted that for this to be true, we need the hypothesis that $|f|<\infty$ a.e..

For example, if $(X,\mathfrak{m},\mu)$ is defined by $X=\{x\},\mathfrak{m}=\{\emptyset,X\},\mu(X)=1$, take $f_n:=n$.

It also should be noted that we also need the hypothesis that $\mu(X)<\infty$.

For example, if the measure space $X$ is $\mathbb{R}$ with Lebesgue measure $m$, take $f_n:=\chi_{[0,n]}$.

A proof assuming the following hypothesis :

$|f|<\infty$ (or only a.e. is also ok)

$\mu(X)<\infty$

can be given as follow.

Proof

By uniform integratability, find $\delta>0$ with $\mu(E)<\delta$ implies $\int_E|f_n|<1$. By Fatou, $\chi_Ef\in L^1(\mu)$.

Since $\mu(X)<\infty$, by successively cutting away these (can be taken to be finitely many) $E$, we may just assume now that every non-empty measurable subset of $X$ has measure $\geq\delta$.

In this case, notice for $x\in X$, $\mu(f^{-1}(f(x)))\geq\delta$, which shows that $f(X)$ is a finite set since $\mu(X)<\delta$, then the condition $|f|<\infty$ applied to each fiber of $f$ shows that $f\in L^1(\mu)$. $\blacksquare$


问:
$\operatorname{Aut_{I}(R)} = \operatorname{End_{I}(R) \cap Aut(R)}$ is a group or not?
I found an answer for part $(c)$ of this question:

Let $R$ be a ring and $I \subset R$ a two-sided ideal, with quotient homomorphism $\pi : R \rightarrow R/I.$ Let $\operatorname{End_{I}(R)}$ be the set of $\varphi \in \operatorname{End(R)}$ such that $\varphi(I) \subset I,$ and let $\operatorname{Aut_{I}(R)} = \operatorname{End_{I}(R) \cap Aut(R)}.$

$(a)$ Given $\varphi \in \operatorname{End_{I}(R)},$ show that there exists $\bar{\varphi} \in \operatorname{End(R/I)}$ such that $\bar{\varphi} \pi = \pi \varphi.$

$(b)$ Given $\varphi, \psi \in \operatorname{End_{I}(R)},$ show that $\overline{\varphi \psi} = \bar{\varphi} \bar{\psi}.$

$(c)$ Given $\varphi \in \operatorname{Aut_{I}(R)},$ show that $\varphi(I) = I$ and $\bar{\varphi} \in \operatorname{Aut(R/I)}.$\ Hint: Use part $(b).$

But as can be seen from this link why $(\varphi \circ \varphi^{-1})(I) = I$ implies that $\varphi^{-1}(I) \subset I$? it seems like the answer is only correct iff $\operatorname{Aut_{I}(R)} = \operatorname{End_{I}(R) \cap Aut(R)}$ is a group and I was told that in general it is not a group and it is a group only if $R$ is a Noetherian ring, is this claim correct? can anyone give me an example showing that it is not a group in general?

答:
What we need is an example of a non-Noetherian ring with a ring automorphism that "shifts" an infinite ascending chain of ideals, i.e., maps each ideal to the next in the sequence. Let's take a polynomial ring in infinitely many variables (which is a prototypical example of a non-Noetherian ring) and see if we can find such an ideal and an automorphism. To make "shifting" easier, we'll index the variables by integers.

Let $k$ be a field, and let $R = k[\dots, x_{-2}, x_{-1}, x_0, x_1, x_2, \dots]$. For each $n \in \mathbb{Z}$, let $I_n = (x_n, x_{n-1}, x_{n-2}, \dots)$. We have an infinite chain of ideals
$$\dots \subset I_{-2} \subset I_{-1} \subset I_0 \subset I_1 \subset I_2 \subset \dots$$
where all the inclusions are proper. Can you think of a ring automorphism of $R$ that maps $I_0$ to $I_0$ but whose inverse doesn't have this property?

(Example in spoilers below.)

 It's enough to find an automorphism $\varphi \colon R \to R$ such that $\varphi(I_1) = I_0$, since then $\varphi^{-1}(I_0) = I_1$. Define $\varphi(x_n) = x_{n-1}$ for all $n \in \mathbb{Z}$.


问:
Show that three numbers form an arithmetic progression

The numbers $a,b$ and $c$ form  an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.

We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{   } \forall \text{   }n\ge2$). I am stuck here and I would be very grateful if you could give me a hint.

Body:
$(a^2+ac+c^2)-(a^2+ab+b^2)=a(c-b)+c^2-b^2=(c-b)(a+b+c)\\(b^2+bc+c^2)-(a^2+ac+c^2)=b^2-a^2+(b-a)c=(b-a)(a+b+c)$

Are they equal?


答:
More hint:

\begin{align}&\quad(a^2+ab+b^2) + (b^2+bc + c^2) \\&= a^2+c^2 +2b^2 + b(a+c) \\&= a^2 + c^2 + 2b^2 + b(2b) \\&= a^2 + c^2 + 4b^2\\&=a^2+c^2+(a+c)^2\end{align}

In the worst case you can just substitute $b = \frac {a+c}2$.

Title: Show that three numbers form an arithmetic progression

Body:
The numbers $a,b$ and $c$ form  an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.

We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{   } \forall \text{   }n\ge2$). I am stuck here and I would be very grateful if you could give me a hint.


问:
Graph of $r=4\cos(6\theta)$
I'm learning about polar coordinates. What is the best way to understand $r=4\cos(6\theta)$ without a computer?

I already knew what the graph of $r=\cos(2\theta)$ looks like. Is there a way to get an idea about $r=4\cos(6\theta)$ using this "parent graph"?

What about the graph of $r=4\cos(5\theta)$. Does the fact that $5$ is odd while $6$ is even make a big difference as to the number of "leaves" that appear on the "rose"? Thanks

答:
A particular way to understand how these operations transform the graph is to notice that $f(kx)$ is a horizontal compression of scale factor $\frac{1}{k}$. This idea still applies to polar coordinates, but with the small change that you are now compressing with respect to $\theta$ instead of $x$.

So for the original graph, you could fit 4 pedals from $[0,2\pi]$, but with the factor of 3 in front, you would be now fitting $4 \times 3 = 12$ pedals into $[0,2\pi]$. Formally, this is changing the period of the function from $\pi$ to $\pi/3$.

Similarly we can think of $r=4\cos(5\theta)$ as changing the period to $2\pi/5$, but the difference here is that now the function doubles back on itself from $\pi$ to $2\pi$, so although there should be $4\times\frac{5}{2}=10$ pedals, half of them overlap to create the image of only 5 pedals.

The constant outside of the $\cos$ serves to increase the maximum radius. Originally, the maximum of cosine is 1, but by multiplying the value by 4, all of the r values are increased four times, causing the maximum to increase to 4.

Title: Graph of $r=4\cos(6\theta)$


问:
Re-scaling a vector along a direction
I have a vector $a \in \mathbb{R}^n$ that has the following decomposition:

$$ a = a_1 + a_o $$

where $a_1$ represents the component of $a$ along vector $b \in \mathbb{R}^n$ and $a_o$ is the remainder, i.e., $a - a_1$.

I would like to remove from $a$, some component of $a$ along the direction $b$, i.e.,

$$ a - \gamma . \text{proj}_b (a) $$

for some $\gamma > 0$ using the defn of projection from rmm93's answer. However, when I set $\gamma = 1$, the results were the same as with $a$. So, I am not sure what I might be missing here.

答:
From you equation $\bar a_1^T b = \frac{a_1^T b}{\gamma}$ I read that you want to set
$$
\bar a_1 = \frac{1}{\gamma} a_1.
$$
Since $a_1 = \operatorname{proj}_b(a)$ this does indeed yield
$$
\bar a_1 = \frac{1}{\gamma} \operatorname{proj}_b(a).
$$

I'm not sure if that's all you are asking for. If it isn't, please clarify in your question.

Title: Re-scaling a vector along a direction


问:
A few problems on sup and nested intervals
I've been doing these 3 problems for a `proof´ oriented class, one i have found a solution (in fact has been asked here before but the threads are all closed), and checked a correct solution in the Analysis book from Apostol. The thing is, I think im making a mistake and i cant find it. The other 2 problems are using nested closed intervals and i dont really know where to start to solve them. Thanks for your help in advance, and here they go:

The first one is

If $A$, $B\subseteq\mathbb{R}$ are non empty and bounded above (so sup exists) we consider $A+B =\{a+b:a\in A,b\in B\}$.
Show the relation between $sup(A)$ $sup(B)$ and $sup(A+B)$.

My solution:
First i show $A+B$ is bounded. Consider an arbitrary element of $A+B$ is of the form
$a+b$.

Now  $a \leq sup(A)$  and  $b \leq sup(B)$  from this we get

$a+b \leq sup(A) + sup(B)$. $  $ $ $ So $sup(A+B)$ exists, as $A+B$ is non-empty and bounded (completeness axiom). Also $sup(A+B) \leq sup(A)+sup(B)$ because $sup(A+B)$ is the smallest bound of the set.

Now i want to study the relation between supremums.

I know that for any $a+b$,

$a+b \leq sup(A+B)$ and hence $a \leq sup(A+B) - b$
So $sup(A+B)-b$ is an bound for $a$, and as $sup(A)$ is the supremum for any $a$
$sup(A) \leq sup(A+B) - b$

Doing the same for b, $b \leq sup(A+B) - sup(A)$ and as this is a above bound for $b$ and $sup(B)$ is the smallest one, $sup(B) \leq sup(A+B) - sup(A)$ then $sup(A) + sup(B) \leq sup(A+B)$ and $sup(A+B) \leq sup(A)+sup(B)$

And because of this $sup(A+B) = sup(A)+sup(B)$

And here are the other two excercises that i really dont have idea where to start, specially the second one

Second one:

Be $(I_n)_{n\in\mathbb{N}}$ a sequence of closed intervals and for every
$n\in\mathbb{N}$, be $\lambda_n$ the lenght of $I_n$; lets supose that $I_1\supseteq
I_2\supseteq\dots\supseteq I_n\supseteq I_{n+1}\supseteq\dots$. Show that
$\lim \lambda_n$ exists and that if $\lim\lambda_n>0$ then
$\bigcap_{n\geq1}I_n$ is a closed interval of lenght $\lim\lambda_n$.

My attemp to solution For this one i know that they're not empty and bounded above and below (they're closed intervals) and i know that if two sets $A \subseteq B$ are bounded an non-empty, then $sup A \leq sup(B)$ and as they're closed $max(A) \leq max(B)$ and the same with their minimums so the lenght of $[P_n, Q_n]$ $|Pn-Qn|$ is decreasing.

And the last one is

Be $x_1$, $a\in\mathbb{R}$ Such that $x_1>0$ y $a>x_1^2+x_1$,
and for every $n\in\mathbb{N}$ be $x_{n+1}=a/(1+x_n)$.
Then every interval $[x_1,x_2]$, $[x_3,x_4]$, $\dots$ and so on, form  an intersection of intervals and the only point in common $\xi$ common to every one of them is A root of the equation $x^2+x=a$.

sugestion (this comes with the excercise too)
Show that for every $n\in\mathbb{N}$$x_{n+1}-x_n$ and $x_n-x_{n-1}$ have opposite signs and that exists $\theta\in(0,1)$ such that $|{x_{n+1}-x_n}|\le\theta|{x_n-x_{n-1}}|$ for every $n\in\mathbb{N}$.

答: Put $I_{n}:=[a_{n},b_{n}]$; as the intervals are nested you get that $(a_{n})_{n \in \mathbb{N}}$ is increasing and similarly $(b_{n})_{n \in \mathbb{N}}$ is decreasing. Thus $\lim_{n \to \infty}a_n=sup_{n \in \mathbb{N}}a_{n}$ and $\lim_{n \to \infty}b_n=inf _{n \in \mathbb{N}}b_{n}$. From this we get $I:=\bigcap_{n \in \mathbb{N}} I_{n}=[a,b]$ where $a:=sup_{n \in \mathbb{N}}a_{n}$ and $b:=inf _{n \in \mathbb{N}}b_{n}$ (you have to show that I is in fact an interval and that those are its two extremes). Moreover $ \lambda_{n}:=b_{n}-a_{n}$ and $\lim_{n \to \infty}\lambda_{n}=\lim_{n \to \infty}b_n - \lim_{n \to \infty}a_n=\lambda$ (by additivity of limits) where $\lambda$ is the lenght of I.</p>


Body:
Since $2b=a+c$

$$a^2+ab+b^2+b^2+bc+c^2=a^2+b(a+c)+2b^2+c^2=a^2+\frac{(a+c)^2}{2}+2\times \frac{(a+c)^2}{4}+c^2=a^2+(a+c)^2+c^2=2(a^2+ac+c^2)$$


问:
Solution bounds in SDE, zeros of noise and drift term
Some SDE's even though they are defined on $\mathbb R$ have solutions that are confined to a region. One example with explicit solution could be:

$$
dX_t = -\frac{1}{2} X_t (1-X_t^2) dt + (1-X_t^2) dW_t
$$

with $W_t$ standard Wiener process, and solution:
$$
X_t = \tanh(W_t + \operatorname{arctanh} X_0)
$$

This equation has impermeable barriers at $X=1$ and $X=-1$ where both drift and noise terms are zero.

Is it enough for drift and noise terms to be zero for such barrier to exist?

答:
Even though solutions of SDE's can intersect for different paths of the Wiener process, for one specific path of the Wiener process solutions are unique given initial condition (and thus cannot intersect). With that in mind $X_t = q = const$ for value of $q$ such that it is a root of both drift and noise term is a solution.

Now pick a path of the Wiener process and suppose we have solution that is nonconstant and intersects $X_t = q$ for some $t_0$. Then the remainder of such trajectory ($t>t_0$) would be an alternative (to the constant) solution starting at the same initial condition violating uniqueness of strong solutions.

To summarise: yes, zeros of both drift and noise term guarantee an impermeable barrier for the solutions of the SDE.

On a related note you can see that from another angle by changing SDE into a PDE for the evolution of the $X_t$ distribution w.r.t. $t$ (when possible) and see that at the barrier point probability cannot raise or fall ($\partial_tp(x,t)$), and is initially zero so it shouldn't spill over through such barrier.


答:
Yes, the projection of $a$ onto $b$ is as follows.

$$a_1 = \frac{b^T a}{b^T b} b$$

Hence, if you want to scale $a_1$ by $\gamma$, that would just be

$$\bar{a}_1 = \frac{1}{\gamma} \frac{b^T a}{b^T b} b$$

Title: Re-scaling a vector along a direction

Body:
I have a vector $a \in \mathbb{R}^n$ that has the following decomposition:

$$ a = a_1 + a_o $$

where $a_1$ represents the component of $a$ along vector $b \in \mathbb{R}^n$ and $a_o$ is the remainder, i.e., $a - a_1$.

I would like to remove from $a$, some component of $a$ along the direction $b$, i.e.,

$$ a - \gamma . \text{proj}_b (a) $$

for some $\gamma > 0$ using the defn of projection from rmm93's answer. However, when I set $\gamma = 1$, the results were the same as with $a$. So, I am not sure what I might be missing here.


问:
Power series with cubic factorials
I am looking for a starting point to obtain the closed form of
$$ \sum_{k=0}^{\infty} x^k \frac{(2k)!}{(k!)^3}. $$
I have tried transforming the ratio of factorials into combinations, but this lead me nowhere. I also computed several values of the series on Wolfram and I expect the answer to look like $e^{2x} I_0(2x)$ for $I_0$ the $0$-th modified Bessel function.
Does anyone has an idea on how to properly obtain the result ?

答:
CONJECTURE:
$$\sum_{k=0}^\infty x^k\frac{(2k)!}{k!^3}=e^{2x}I_0(2x)$$
Here's my attempt at showing this using a Cauchy product -

The power series for $e^{2x}$ is:
$$e^{2x}=\sum_{k=0}^\infty \frac{2^k}{k!}x^k=\sum_{k=0}^\infty a_k x^k$$
And a well known expansion for $I_0$ is
$$I_0(z)=\sum_{k=0}^\infty \frac{z^{2k}}{4^k k!^2}$$
So then
$$I_0(2x)=\sum_{k=0}^\infty\frac{x^{2k}}{k!^2}=\sum_{k=0}^\infty b_kx^k$$

Here we have
$$b_k=\frac{1+(-1)^{k}}{2}\frac{1}{(k/2)!^2}$$

Then, the Cauchy Product gives
$$e^{2x}I_0(2x)=\left(\sum_{k=0}^\infty a_k x^k\right)\cdot\left(\sum_{k=0}^\infty b_kx^k\right)=\sum_{k=0}^\infty c_kx^k$$

$$c_k=\sum_{j=0}^k a_jb_{k-j}=\sum_{j=0}^k b_ja_{k-j}$$
$$=\sum_{j=0}^k\frac{1+(-1)^{n-j}}{2}\frac{1}{((n-j)/2)!^2}\frac{2^j}{j!}=\sum_{j=0}^k \frac{1+(-1)^{j}}{2}\frac{1}{(j/2)!^2}\frac{2^{k-j}}{(k-j)!}$$
I suppose all there is left to do now is show the above are equal to $(2k)!/k!^3$. (sum experts, help!) This is confirmed by Mathematica.

And of course the half integer Gamma is well known. We then have
$$=\frac{4^k(2k)!\sqrt{\pi}}{\sqrt{\pi}4^k k!(k!)^2}=\frac{(2k)!}{k!^3}$$
However I think I'm a bit out of my depth to show this using "pencil and paper" methods.

Title: Power series with cubic factorials


问:
Finding the useable area of a triangular lot.
I have purchase a lot to build a cabin. the lot is triangle shaped. There are streets on 2 sides and a neighboring cabin on the other. The lot dimensions are 103.43' at the neighbors property line, 93.53' along the left side and 99.10' along the right side (facing from the neighbors lot). My setbacks are 15' from my neighbors lot and 20' from each street. Can anyone help? Thanks in advance!! David</p>

答:
A construction to scale using Cinderella shows the area to be approximately 531.88. The edge $AB$ in the following image is the one to your neighbors.

If this approximation wasn't good enough, you'd have to start computations. Using the law of cosines you could compute the angles of your triangles. From the intercept theorem you could conclude the distance between the points where the line $DE$ intersects the outer triangle: that length relates to the length $AB$ like the original height of the plot relates to the height minus 15'. That original height can in turn be computed from one angle and the adjacent edge. Now you still need the distances between the points $D$ and $E$ and the corresponding endpoints on the outer triangle. You can obtain these by dividing 20' by the sine of the angle at that corner. I'm pretty sure that this is correct, but giving a sound proof would take a bit of time which I don't have just now. Once you have the length $DE$, you can obtain the other two inner lengths as well, since the inner and outer triangle are similar. You can therefore compute the area from these lengths using Heron's formula.

Computed with perfect precision (treating your decimal numbers as exact), the area computed as outlined above will be

$$A = \frac{33217}{200}\cdot\sqrt{\frac{7496394733369083}{5868607457510}} -
\frac{108081}{20} \approx 531.88454926$$

This was done using sage and algebraic numbers:

```<code>AB = QQbar(10343/100)
AC = QQbar(9353/100)
BC = QQbar(9910/100)
cosA = (AB^2 + AC^2 - BC^2)/(2*AB*AC) #   0.49746779652302
sinA = sqrt(1 - cosA^2)               #   0.86748244444630
cosB = (AB^2 + BC^2 - AC^2)/(2*AB*BC) #   0.57418604431081
sinB = sqrt(1 - cosB^2)               #   0.81872485397641
h1 = sinA*AC                          #  81.1356330290626
h2 = h1 - 15                          #  66.1356330290626
DE = h2/h1*AB - 20/sinA - 20/sinB     #  36.82486242666910
EF = BC*DE/AB                         #  35.28322407892205
DF = AC*DE/AB                         #  33.30010038447608
s = (DE+EF+DF)/2                      #  52.70409344503361
T = sqrt(s*(s-DE)*(s-EF)*(s-DF))      # 531.8845492639687
T0, T1, T2 = T.minpoly().coefficients()
(-T1+sqrt(T1^2-4*T0*T2))/(2*T2)
</code>```

Body:
A construction to scale using Cinderella shows the area to be approximately 531.88. The edge $AB$ in the following image is the one to your neighbors.

If this approximation wasn't good enough, you'd have to start computations. Using the law of cosines you could compute the angles of your triangles. From the intercept theorem you could conclude the distance between the points where the line $DE$ intersects the outer triangle: that length relates to the length $AB$ like the original height of the plot relates to the height minus 15'. That original height can in turn be computed from one angle and the adjacent edge. Now you still need the distances between the points $D$ and $E$ and the corresponding endpoints on the outer triangle. You can obtain these by dividing 20' by the sine of the angle at that corner. I'm pretty sure that this is correct, but giving a sound proof would take a bit of time which I don't have just now. Once you have the length $DE$, you can obtain the other two inner lengths as well, since the inner and outer triangle are similar. You can therefore compute the area from these lengths using Heron's formula.

Computed with perfect precision (treating your decimal numbers as exact), the area computed as outlined above will be

$$A = \frac{33217}{200}\cdot\sqrt{\frac{7496394733369083}{5868607457510}} -
\frac{108081}{20} \approx 531.88454926$$

This was done using sage and algebraic numbers:

AB = QQbar(10343/100)
AC = QQbar(9353/100)
BC = QQbar(9910/100)
cosA = (AB^2 + AC^2 - BC^2)/(2*AB*AC) #   0.49746779652302
sinA = sqrt(1 - cosA^2)               #   0.86748244444630
cosB = (AB^2 + BC^2 - AC^2)/(2*AB*BC) #   0.57418604431081
sinB = sqrt(1 - cosB^2)               #   0.81872485397641
h1 = sinA*AC                          #  81.1356330290626
h2 = h1 - 15                          #  66.1356330290626
DE = h2/h1*AB - 20/sinA - 20/sinB     #  36.82486242666910
EF = BC*DE/AB                         #  35.28322407892205
DF = AC*DE/AB                         #  33.30010038447608
s = (DE+EF+DF)/2                      #  52.70409344503361
T = sqrt(s*(s-DE)*(s-EF)*(s-DF))      # 531.8845492639687
T0, T1, T2 = T.minpoly().coefficients()
(-T1+sqrt(T1^2-4*T0*T2))/(2*T2)


问:
How do you define “independence” in combinatorics?
I feel like most definitions of “independence” are circular. Consider how we count the number of cards in a standard deck of cards: $|S \times R| = |S||R|$, where $S$ is the set of suits, and $R$ is the set of ranks. That is, $$S = \{\text{Hearts, Diamonds, Spades, Clubs}\}$$ and $$R = \{ 2,3,\dots,\text{King},\text{Ace} \}.$$ We know that $|S| = 4$ and $|R| = 13$, so $|S \times R| = |S||R| = 4 \cdot 13 = 52.$ In such an example, we define that they are independent, because they are disjoint subsets. But do we know that? In this example, it is “obvious.” What about examples where it is not obvious?

答:
Here is another view point. Independence is a property of partitions, not of individual sets. Namely, let $S$ be a finite set, and let $\{X_1,\dots,X_m\}$ and $\{Y_1,\dots,Y_n\}$ be two partitions of $S$ (meaning the $X_i$ are nonempty, disjoint, and have a union of $S$, and same for the $Y_i$). We say that these two partitions are independent if
$$
|X_i\cap Y_j|\text{ is the same for all }i\in \{1,\dots,m\},j\in \{1,\dots,n\}
$$
For example, let $S$ be a deck of cards, and let
$$
X_1=\{\text{the set of spades}\}=\{A\spadesuit,2\spadesuit,\dots,K\spadesuit\},\\X_2=\{\text{the set of hearts}\},\\ X_3=\{\text{the set of clubs}\}\\
X_4=\{\text{the set of diamonds}\}
$$
and
$$
Y_1=\{\text{set of aces}\},Y_2=\{\text{set of twos}\},\dots,Y_{13}=\{\text{set of kings}\}
$$
Since it is true that $|X_i\cap Y_j|=1$ for all $i,j$, we conclude these two partitions are independent.

A consequence of the definition of independence is that
$$
|S|=m\cdot n\cdot |X_1\cap Y_1|
$$


问:
Divergence of a function product with functions in $L^\infty (\Omega)$ and $H^1(\Omega)$
I am doing the weak formulation of the equation:

$\nabla ( k \nabla u) = f \ \text{in } \Omega$

I want to find the correct space for k in order to have $\nabla ( k \nabla u)$ well defined.
At the moment, I am sure that I have to require $u\in H^2(\Omega)$ during the process to be able to use the Green's formula (or even the Stoke's theorem). Also, I think it is needed that $k\nabla u\in H^1(\Omega)$ in order to have the divergence correctly defined. First of all, I impose $k\in L^\infty(\Omega)$ so $k\nabla u\in L^2(\Omega)$. Does this condition also imply that $k\nabla u\in H^1(\Omega)$?

I was thinking on $k\in W^{1,\infty}(\Omega)$ to have this product well defined, but my professor told me that we can assume that the strong formulation allows us to have well defined the divergence of that product.

I am not sure if I need to impose derivatives of $k$ in $L^\infty (\Omega)$ to have everything well defined, or just think this derivative in the way of distributions and impose $k\in L^\infty(\Omega)$.

Thank you for your time.

答:
Let me reach a bit further back.

Your notation is a bit sloppy, which may mislead and thus confuse you. The correct expression for the differential operator is $\nabla \cdot (k \nabla u) = \operatorname{div} (k \nabla u)$, i.e. the divergence of $k \nabla u$, while $\nabla (k \nabla u)$ would denote the gradient of the vector field $k \nabla u$, i.e. the derivative in all directions (the Jacobian). This distinction is important because requiring $\operatorname{div}(k \nabla u)$ to be well-defined is a much weaker condition than requiring $\nabla(k \nabla u)$ to be well-defined. In the following I assume you wanted to ask what conditions are necessary for $\operatorname{div}(k \nabla u)$ to be well-defined.

To be nit-picky (or maybe not), you must specify in which sense that expression should be well-defined. I assume you meant that $\operatorname{div}(k \nabla u) \in L^2(\Omega)$ since then, under the additional assumption that $f \in L^2(\Omega)$, one can say that the differential equation $\operatorname{div}(k \nabla u(x)) = f(x)$ holds for $x \in \Omega$ almost everywhere.

If you don't assume from the get-go that $k \nabla u \in (H^1(\Omega))^d$ then the question arises what $\operatorname{div}(k \nabla u) \in L^2(\Omega)$ should even mean. You stated in your comments that you never found a use for distributional derivatives, so this is a good time as any to make use of those concepts (and maybe clear up some more confusion). For any vector field $v \in (L^2(\Omega))^d$, with the special case $v = k \nabla u$ in mind, the distributional divergence of $v$ is the distribution $\operatorname{div} v \in C^\infty_0(\Omega)'$ defined by
$$ [\operatorname{div} v](\varphi) := -\int_{\Omega} v \cdot \nabla \varphi \, \mathrm dx \quad \forall \varphi \in C^\infty_0(\Omega). $$

This mirrors the definition of the distributional derivatives which you should already know. Indeed, you can define the distributional version of any linear differential operator in a similar way. It's important to realize that while this may look like "just" partial integration / Green's identities, the basic idea of distributional derivatives is that the right-hand side is well-defined even if only $v \in (L^2(\Omega))^d$ and nothing more. Only if $v$ is more regular, partial integration allows us to a-posteriori identify the distributional derivative with more a classical function. This identification of distributional divergence is done (as for distributional derivatives) by saying that $\operatorname{div} v \in C^\infty_0(\Omega)'$ is represented by a function $h \in L^2(\Omega)$ if there holds
$$ \int_{\Omega} h \varphi \,\mathrm dx = v(\varphi) \left( = -\int_{\Omega} v \cdot \nabla \varphi \, \mathrm dx \right)$$
for all $\varphi \in C^\infty_0(\Omega)$. One then, casually one may say, writes $\operatorname{div} v = h$. This is the sense in which $\operatorname{div} v \in L^2(\Omega)$, e.g. $\operatorname{div}(k \nabla u) \in L^2(\Omega)$, must be understood. Of course one could just say that a vector field $v$ has "a divergence
in $L^2(\Omega)$" if there is a $h \in L^2(\Omega)$ satisfying this equation without ever mentioning "distribution" or "distributional derivative", but this hides the generality of the ideas behind the construction.

With this preparation, I claim that $k \in L^\infty(\Omega)$ is sufficient that any solution $u \in H^1(\Omega)$ of your weak equation
$$ \int_\Omega k \nabla u \cdot \nabla \varphi \,\mathrm dx = \int_\Omega f \varphi \,\mathrm dx $$
for all $\varphi \in H^1_0(\Omega)$ with $f \in L^2(\Omega)$ has regularity $\operatorname{div}(k \nabla u) \in L^2(\Omega)$. Comparing the integral equation for the weak solution with the condition I wrote above for $h \in L^2(\Omega)$ to represent the distributional derivative of $\operatorname{div}(k \nabla u)$, you can see that $h = - f$ and if $f \in L^2(\Omega)$ thus $\operatorname{div}(k \nabla u) \in L^2(\Omega)$. The regularity $k \in L^\infty(\Omega)$ is here only used such that $v = k \nabla u \in L^2(\Omega)$ (and of course for less regular $k$ the integral equation for the weak solutions is no longer well-defined for $\varphi \in H^1_0(\Omega))$). This argument of course only works because $u$ is not any function in $H^1(\Omega)$ but one that satisfies the integral equation, which gives you the regularity $\operatorname{div}(k \nabla u) \in L^2(\Omega)$ "for free" if your right-hand side $f$ has that regularity.

As a last note, the space of vector fields $v \in (L^2(\Omega))^d$ with $\operatorname{div} v \in L^2(\Omega)$ is typically denoted by $H(\operatorname{div}; \Omega)$, thus $k \nabla u \in H(\operatorname{div}; \Omega)$ is the minimal regularity for $\operatorname{div}(k \nabla u)$ to be well-defined.


答:
We have that $r=\cos ( \theta)$ makes a loop for any interval $\left[-\frac \pi 2 +k\pi ,\frac \pi 2 +k\pi\right]$ but since it leads to negative values for $k$ odd we have that only the loops with $k$ even can be plotted, which is a circle in this case (note that the same plot is obtained if we allow negative values for $r$ since in this case the plots coincide).

Therefore $r=\cos ( 2\theta)$ makes a loop for any interval $\left[-\frac \pi 4 +k\frac \pi 2 ,\frac \pi 4 +k\frac \pi 2\right]$ which correspond to $2$ loops if we restrict to positive values for $r$ and to $4$ loops if we allow negative values also.

So also for $\cos ( 6\theta)$ we have a loop for any interval $\left[-\frac \pi {12} +k\frac \pi 6 ,\frac \pi {12} +k\frac \pi 6\right]$ which correspond to $6$ loops if we restrict to positive values for $r$ and to $12$ loops if we allow negative values also.

For $r=\cos ( 5\theta)$ we have a loop for any interval $\left[-\frac \pi {10} +k\frac \pi {5} ,\frac \pi {10} +k\frac \pi {5}\right]$ which correspond in any case to $5$ loops since the loops for $k$ even ($r>0$) are the same we obtain for $k$ odd ($r<0$).

Summarizing, the plot for $r= \cos (n\theta)$ has

$n$ loops for $n$ odd

$n$ loops for $n$ even if negative value for $r$ are not allowed

$2n$ loops for $n$ even if negative value for $r$ are allowed

Title: Graph of $r=4\cos(6\theta)$

Body:
I'm learning about polar coordinates. What is the best way to understand $r=4\cos(6\theta)$ without a computer?

I already knew what the graph of $r=\cos(2\theta)$ looks like. Is there a way to get an idea about $r=4\cos(6\theta)$ using this "parent graph"?

What about the graph of $r=4\cos(5\theta)$. Does the fact that $5$ is odd while $6$ is even make a big difference as to the number of "leaves" that appear on the "rose"? Thanks


答:
An efficient method is to find the $\theta$-values where $r(\theta)=0$ and $r(\theta)=a$, where a is the amplitude of the cosine. Combine this with your knowledge that the graph is a rose-type. Now plot the corresponding points $(r,\theta)$, tracing in order of increasing $\theta$.  This should give you a reliable skeleton on which you can draw the rose.

This strategy is generally a useful one when graphing other types of well-know polar curves.

Title: Graph of $r=4\cos(6\theta)$


Body:
Well,  If $a,b,c$ form an arithmetic progression then there is a $d$ so that $a=b-d$ and $c=b+d$.

And to show that $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ form an arithmetic progression we have to show there is a $k$ so that

$k = (a^2+ac+c^2) - (a^2+ab+b^2)$ and $k= (b^2+bc+c^2) -(a^2+ac+c^2)$ or in other words that

$(a^2+ac+c^2) - (a^2+ab+b^2)= (b^2+bc+c^2) -(a^2+ac+c^2)$.  So if we replace $a=b-d$ and $c = b+d$ we must show

$((b-d)^2+(b-d)(b+d)+(b+d)^2) - ((b-d)^2+(b-d)b+b^2)= (b^2+b(b+d)+(b+d)^2) -((b-d)^2+(b-d)(b+d)+(b+d)^2)$

Can we?

$((b-d)^2+(b-d)(b+d)+(b+d)^2) - ((b-d)^2+(b-d)b+b^2)=$

$((b-d)(b+d)+(b+d)^2) - ((b-d)b+b^2)=$

$((b-d)d+(b+d)^2) - (b^2)=$

$(b-d)d + 2bd+d^2=$

$3bd $.

And

$(b^2+b(b+d)+(b+d)^2) -((b-d)^2+(b-d)(b+d)+(b+d)^2)=$

$(b^2+b(b+d)) -((b-d)^2+(b-d)(b+d))=$

$(b^2) -((b-d)^2-d(b+d))=$

$(b^2) -(b^2-2bd-bd)=$

$3bd$.

Yep, they are equal all right.

========

To take your idea of $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{   } \forall \text{   }n\ge2$ into account:

We know  $b = \frac{a+c}2$ and we need to prove

$a^2+ac+c^2= \frac {(a^2+ab+b^2)+(b^2+bc+c^2)}2$ given that $b = \frac {a+c}2$

So

$\frac {(a^2+ab+b^2)+(b^2+bc+c^2)}2=$

$\frac {a^2 + c^2 + b(a+c) + 2b^2}2=$

$\frac {a^2 +c^2 + \frac {a+c}2(a+c) + 2(\frac {a+c}2)^2}2=$

$\frac {a^2 + c^2 + \frac {(a+c)^2}2 + 2\frac {(a+c)^2}4}2=$

$\frac {a^2 + c^2 + (a+c)^2}2=$

$\frac {a^2 + c^2 + a^2 + 2ac + c^2}2 =$

$\frac {2a^2 + 2ac + 2c^2}2 =$

$a^2 + ac + c^2$.


问:
Proof a function is injective - (intermediary step to proving the rationals are countable)
I am proving the rationals (greater than 0) are countable. I have to prove a function is injective and I just wanted another opinion on whether my proof is correct or if there is an even better method of proof.

I have the following function - f : Q(greater than 0) -> Naturals

f(m/n) = 2^m 3^n (where m and n are coprime)

I have to prove that this function is injective.
Proof:

For all m/n and m'/n' in Q (>0)
WTS: f(m/n) = f(m'/n') implies m/n = m'/n'

f(m/n) = 2^m 3^n

f(m'/n') = 2^m' 3^n'

Suppose m != m' and n != n' for a contradiction.

By the definition of injective functions we have:

2^m 3^n = 2^m' 3^n' which is equivalent to
2^m 3^n (1- 2^(m'-m) 3^(n'-n)) = 0

Since a^b = 0 does not have a solution. We have that:

2^(m'-m) 3^(n'-n) = 1

Since 2 divides the LHS, it must divide the RHS so 2 divides 1. Which is a contradiction as n divides 1 is false for all n > 1. We can do something similar with 3 as well. So our assumption that n != n' and m != m' is incorrect. Yielding m = m' and n = n' as required.

Note: I have a feeling there is a far better way of showing this so if you have a better method please tell me :)

答:
There are problems here. You should not suppose that $m\ne m'$ and that $n\ne n'$; what you should suppose is that $m\ne m'$ or that $n\ne n'$.

And, after writing that $2^{m'-m}3^{n'-n}=1$, you act as if the LHS is a natural number, but it doesn't have to be.

You can simply say that $2^m3^n=2^{m'}3^{n'}\implies m=m'\wedge n=n'$ by the Fundamental Theorem of Arithmetic.


问:
what is the value of this trigonometric expression
I want to find out value of this expression
$$\cos^2 48°-\sin^2 12°$$
Just hint the starting step.Is there any any formula regarding $\cos^2 A-\sin^2 B$?</p>

Body:
$$\cos^{2}(48)-\sin^{2}(12)=\sin^{2}(42)-\sin^{2}(12)=\sin(54)\sin(30)=\dfrac{1}{2(\sqrt{5}-1)}$$

Here we used the formula-$$\sin^{2}(A)-\sin^{2}(B)=\sin(A+B)\sin(A-B)$$ and $\sin(54)=\dfrac{1}{\sqrt{5}-1}$


问:
How to solve this stochastic differential equation? (Ito's formula)
I'm a beginner in stochastic processes and I want to solve the following stochastic equation:

$$X_t = 1 + \int_0^t X_s \sigma_s\, dB_s,$$

where $B_s$ is a standard Brownian motion. I want to apply Ito's formula and it gives me:

$$d X_t = X_t \sigma_s\, dB_t.$$

Is it a proper way to start solving it? How should I continue the calculation to solve this equation?

答:
From here, I think the easiest way is to look at $\ln X_t$.  Ito's formula gives \begin{align*}d\ln X_t &= \frac 1{X_t} dX_t - \frac 1{2X_t^2} dX_t dX_t \\
&= \sigma_t dB_t - \frac 12 \sigma_t^2 dt\end{align*} so we can just integrate to obtain $\ln X_t - \ln X_0 = \int_0^t \sigma_s dB_s - \frac 12 \int_0^t \sigma_s^2 ds$.  Since $X_0 = 1$, by exponentiating we obtain $X_t = \exp(\int_0^t \sigma_s dB_s - \frac 12 \int_0^t \sigma_s^2 ds)$.


问:
Poisson processes introduction - why $\{N(t) \geq n\} = \{S_n \leq t\}$?
I recently got introduced to Poisson Processes and while we were preparing necessary mathematical tools to do it, something along the following lines was in the notes of our professor.

Define 0 < $S_1$ < $S_2$ < $\dots$ < $S_n$ < $\dots$ as a time of arrival of event $S_1$, $S_2$, $\dots, S_n, \dots$ and let $N(t) = \sup\{n > 0: S_n \leq t\}$ (how many events occured until time $t$). Notice that $\{N(t) \geq n\} = \{S_n \leq t\}$.

We then used $\{N(t), t \geq 0\}$ as our first Poisson process (with $S_n$ being $\sum_k T_k$ where $T_k$ i.i.d. exponential).

I am trying to wrap my head around $\{N(t) \geq n\} = \{S_n \leq t\}$. I remember my lecturer saying something along lines "we can infer $t$ from $n$ and $n$ from $t$".

Why are these sets equal? What are their items' "types"? If I understand correctly $\{S_n \leq t\}$ contains all events from 1 to $n$ that occured until time $t$. The second set however, $\{N(t) \geq n\}$ contains just natural numbers.

I appreciate your help.

答:
$\{N(t) \ge n\}$ is not a set of natural numbers, it is an event (a measurable subset of the sample space).  Same with $\{S(n) \le t\}$.

$N(t) \ge n$ means there are at least $n$ arrivals in the time interval $[0,t]$.  That's equivalent to saying that the $n$'th arrival happens by time $t$, i.e. $S_n \le t$.


问:
Normality of product of normal subgroups inside composition series
Assume we have a composition series $\{e\} = G_0 \triangleleft G_1 \triangleleft \dots \triangleleft G_k = G$, i.e. each $G_{i+1}/G_i$ is simple. Now we take some normal subgroup $N$ of $G$. We can create a new (non-strict) composition series as follows:

$$
\{e\} = G_0 \cap N \trianglelefteq G_1 \cap N \trianglelefteq \dots \trianglelefteq G_k \cap N = N = G_0 N \trianglelefteq G_1 N \trianglelefteq \dots \trianglelefteq G_k N = G.
$$

We can then look at $(G_{i+1} \cap N) / (G_i \cap N)$ and $(G_{i+1} N) / (G_i N)$ to determine at which steps the inclusion is strict and at which it is not.

I just didn't fully understand one detail:

Why is $G_i \cap N$ normal in $G_{i+1} \cap N$?

Analogously, why is $G_i N$ normal in $G_{i+1} N$?

I think number 1 is easy:
Because $G_{i}$ is normal in $G_{i+1}$, we have $\pi: G_{i+1} \to G_{i+1} / G_i$ with $\ker(\pi) = G_i$. We can then just construct
$$
f: G_{i+1} \cap N \to G_{i+1} / G_i, g \mapsto \pi(g)
$$
and $\ker(f) = \ker(\pi) \cap N = G_i \cap N$, therefore $G_i \cap N$ is normal in $G_{i+1} \cap N$.

Now I would like to the the same for 2), i.e. construct a homomorphism $f: G_{i+1}N \to X$ with $\ker(f) = G_i N$, but I'm not sure how to construct one. I'd like to reuse $\pi$, but extending it to $h: G_{i+1} N \to G_{i+1}/G_i$ doesn't seem to work, as representing $x \in G_{i+1}N$ via $x = g \cdot n, g \in G_{i+1}, n \in N$ isn't unique. (or is it?)

I'm stuck here, and help would be appreciated.

答:
$G_{i+1} \le N_G(G_i)$ by assumption, and  $G_{i+1} \le N_G(N)$ because $N$ is a normal subgroup of $G$. So $G_{i+1} \le N_G(G_{i}N)$.

Also $N \le N_G(G_{i}N)$ because $ N \le G_{i}N$.

So $G_{i+1}N \le N_G(G_iN)$ and hence $G_iN$ is normal in $G_{i+1}N$.


答:
The probability that $n$ or more events occurred by time $t$ (aka $\{N(t) \geq n\})$is the same as the probability that the nth event arrived before time $t$ (aka $\{S_n \leq t\})$.

Suppose for a moment that the nth event ($S_n$) arrived after time $t$. Then we would have less than n events at time $t$ which is not enough, so we need to have $Sn<t$.

Title: Poisson processes introduction - why $\{N(t) \geq n\} = \{S_n \leq t\}$?

Body:
I recently got introduced to Poisson Processes and while we were preparing necessary mathematical tools to do it, something along the following lines was in the notes of our professor.

Define 0 < $S_1$ < $S_2$ < $\dots$ < $S_n$ < $\dots$ as a time of arrival of event $S_1$, $S_2$, $\dots, S_n, \dots$ and let $N(t) = \sup\{n > 0: S_n \leq t\}$ (how many events occured until time $t$). Notice that $\{N(t) \geq n\} = \{S_n \leq t\}$.

We then used $\{N(t), t \geq 0\}$ as our first Poisson process (with $S_n$ being $\sum_k T_k$ where $T_k$ i.i.d. exponential).

I am trying to wrap my head around $\{N(t) \geq n\} = \{S_n \leq t\}$. I remember my lecturer saying something along lines "we can infer $t$ from $n$ and $n$ from $t$".

Why are these sets equal? What are their items' "types"? If I understand correctly $\{S_n \leq t\}$ contains all events from 1 to $n$ that occured until time $t$. The second set however, $\{N(t) \geq n\}$ contains just natural numbers.

I appreciate your help.


答:
These are events, not sets of numbers.

$N(t)$ is an integer-valued random variable that counts the number of arrivals (these are called "events" in your notes, but I won't use this word to avoid confusion with the above usage of the word "event.") in the time interval $[0,t]$. The event $\{N(t) \ge n\}$ contains all outcomes where "there are at least $n$ arrivals in the time interval $[0,t]$."

$S_n$ is a real-valued random variable that is the time of the $n$th arrival. The event $\{S_n \le t\}$ contains all outcomes where "the $n$th arrival occurs before time $t$."

Do you now see why these two events are equivalent?


问:
Prove by induction that $n^{+}+m=n+m^{+}$ for all $m,n \in \omega$
Just to make sure we use same symbols: $\omega= \mathbb{N}_{0}$, $n^{+}$ is successor of number $n$.
And we define addition by recursion:

$+:\omega \times \omega \rightarrow\omega$

$m+0=m$

$m+n^+=(m+n)^+$

I'm trying to prove: $n^++m=n+m^+$ by induction.

Proof:

Let $m \in \omega$ be fixed and $S=\{n\in \omega| n^++m=n+m^+\}$.

Base: $0\in S$ (proof by induction)

Let $S'=\{m\in \omega| 0^++m=0+m^+\}$ (now I'm using this fixed $m$ as variable in induction (not sure if that's ok?))

$0 \in S$ because  $0^++0=0^+=1$ and $0+0^+=(0+0)^+=0^+=1$

 Let $m\in S$ which means $0^++m=0+m^+$.

 Then $0^++m^+=(0^++m)^+=(0+m^+)^+=0+(m^{+})^{+}$. So, $S'=\omega.$

We proved: $0\in S$.

Hypothesis: $n \in S$ which means  $n^++m=n+m^+$

And now I'm stuck. I don't know how to prove step of induction, and I'm not even sure that begining where I start with fixed $m$ is wnough to prove that statement is true for all $m \in \omega$.

答:
We will prove by induction that the following holds for all $m \in \omega$:
$$n^+ + m = n + m^+ \text{ for all }n \in \omega. \tag{1}$$

For the base case $m = 0$, we just need to show that $n^+ + 0 = n + 0^+$ for all $n \in \omega$.  Note that given any $n \in \omega$ we have
$$\begin{align} n + 0^+
&= ( n + 0)^+ &\text{(definition of addition)} \\
&= n^+ &\text{(definition of addition by }0\text{)} \\
&= n^+ + 0 &\text{(definition of addition by }0\text{)} \end{align}$$

For the induction step suppose that $m \in \omega$ is such that (1) holds.  We now want to show that (1) holds for $m^+$, that is
$$n^+ + m^+ = n + (m^+)^+ \text{ for all }n \in \omega.$$

Given any $n \in \omega$ we then have that
$$\begin{align}
n^+ + m^+
&= ( n^+ + m )^+ &\text{(definition of addition)} \\
&= ( n + m^+ )^+ &\text{((1) for }m\text{)}\\
&= n + (m^+)^+ &\text{(definition of addition)}
\end{align}$$


问:
Find two partitions P,Q of $[a,b]$ such that $||Q||<||P||$ but $\underline{S}(f,P)=\underline{S}(f,Q)$
I'm trying to answer the following exercise:

Give an example of a bounded function $f:[a,b]\rightarrow\mathbb{R}$, with two partitions P,Q of $[a,b]$ such that $||Q||<||P||$ but $\underline{S}(f,P)=\underline{S}(f,Q)$

Where $||Q||$ is the norm of Q and $||P||$ is the norm of P; $\underline{S}(f,P)$ and $\underline{S}(f,Q)$ are the inferior sums.

I've tried to solve it with the following attempts, but with none of them I got that $\underline{S}(f,P)=\underline{S}(f,Q)$:

$f: [0,2]\rightarrow \mathbb{R}$ such that $f(x)=e^x \: \forall x\in [0,2] $; with the partitions $P=\{0, \frac{1}{2}, 1, \frac{3}{2}, 2\}$ and $Q=\{0, \frac{1}{4}, \frac{1}{2},\frac{3}{4}, 1, \frac{5}{4},\frac{3}{2}, \frac{7}{4}, 2\}$, where it's clear that $||Q||=\frac{1}{4}<\frac{1}{2}=||P||$, but $\underline{S}(f,P)<\underline{S}(f,Q)$

And I got to the same result with this attempt:
$f: [0,2]\rightarrow \mathbb{R}$ such that $f(x)=x^2 \: \forall x\in [0,2] $; with the partitions $P=\{0, \frac{1}{2}, 1, \frac{3}{2}, 2\}$ and $Q=\{0, \frac{1}{4}, \frac{1}{2},\frac{3}{4}, 1, \frac{5}{4},\frac{3}{2}, \frac{7}{4}, 2\}$, where it's clear that $||Q||=\frac{1}{4}<\frac{1}{2}=||P||$, but $\underline{S}(f,P)\neq\underline{S}(f,Q)$

I know these are really simple functions but I was trying to see if with those I could get to some function that worked (which it's clear it didn't), and I'm guessing that only certain functions are capable of doing so.

If you could help me with an example I'd appreciate it.
Thank you in advance.

答:
Generally $\underline{S}(f, P)\le \underline{S}(f,Q)$ if $Q$ is a refinement of $P$. But for some special partitions $P$ and $Q$ and some function $f$, $\underline{S}(f, P)$ and $\underline{S}(f,Q)$ may be the same. For example, let
$$ f(x)=\left\{\begin{array}{ll} \frac12 \text{ if }x\in[0,\frac12]\\
\frac32 \text{ if }x\in(\frac12,1]\\
1 \text{ if }x\in(1,\frac32]\\
2 \text{ if }x\in(\frac12,2]
\end{array}\right. $$
be a piecewise constant function. Then for the partitions $P=\{0, \frac{1}{2}, 1, \frac{3}{2}, 2\}$ and $Q=\{0, \frac{1}{4}, \frac{1}{2},\frac{3}{4}, 1, \frac{5}{4},\frac{3}{2}, \frac{7}{4}, 2\}$, we have
$$ \underline{S}(f,P)=\frac12\cdot\frac12+\frac32\cdot\frac12+1\cdot\frac12+2\cdot\frac12=\frac52 $$
and
$$ \underline{S}(f,Q)=\frac12\cdot\frac14+\frac12\cdot\frac14+\frac32\cdot\frac14+\frac32\cdot\frac14+1\cdot\frac14+1\cdot\frac14+2\cdot\frac14+2\cdot\frac14=\frac52. $$


问:
How can I derive the magnitude of the fourier transform of sine wave?
building on this answer https://math.stackexchange.com/a/2152573/845257 . I was deriving the Fourier transform of sine wave  y(t) = sin(2.pi.f.t) and I got it right. Now, I want to derive the magnitude I can't find how he got the magnitude of the sine Fourier transform like that.

答:
It is well known that if $\,z\,$ is a complex number with real
part $\,a\,$ and imaginary part $\,b\,$ then the magnitude of
$\,z\,$ is $\,\sqrt{a^2+b^2}.\,$ so far so good. Note that if
you have a complex function $\,f(\omega)\,$ then to get its
magnitude you have to find the real and imaginary part for each value of $\,\omega.\,$ In the case of interest here we have
$$ f(\omega) := -\frac{i}{2}\delta(\omega -\omega_0) + \frac{i}{2}\delta(\omega +  \omega_0). $$ For each value of $\,\omega\,$ we have to find the imaginary and real parts of $\,f(\omega).\,$
Notice the Dirac delta function $\,\delta()\,$ which means
that the function is zero everywhere except at the two
values $\,\omega=\omega_0\,$ and $\,\omega=-\omega_0\,$, which implies the magnitude is zero also
except at the two values. Now
evaluate the functional expression for those two
exceptional values of $\,\omega\,$ and you will get
the expression given in the answer to the question
you refer to.


问:
What prevents the Expectation conditional on a sub sigma-algebra to be the the variable being conditioned?
I have come to a very simple issue which I have not figured out yet.
In fact there are two questions.
Could you please help me out on this?

Let $(\Omega,\mathcal{F},P)$ be a probability space, $X:\Omega\rightarrow R$ be a random variable, and $\mathcal{A}$ be a sub $\sigma$-algebra in $\mathcal{F}$. The conditional expectation of $X$ given $\mathcal{A}$ is the a.s. unique random variable, denoted as $E[X/\mathcal{A}]:\Omega\rightarrow R$ that satisifies: (i) $E[X/\mathcal{A}]$ is $\mathcal{A}/\mathcal{B}$ measurable and (ii) $\int_BE[X/\mathcal{A}]dP=\int_B XdP \hspace{5mm } \forall B\in\mathcal{A}$.

What prevents $X$ to be this random variable regardless of $\mathcal{A}$?

$X$ seems to satisfy both conditions, (i) $\mathcal{A}/\mathcal{B}$ measurability and (ii) $\int_BE[X/\mathcal{A}]dP=\int_B XdP$.

The second question is:

In the same setting as above, it is defined $P[A/\mathcal{A}]=E[I_A/\mathcal{A}] \hspace{5mm}\forall A\in \mathcal{F}$, and if $\mathcal{A}=\sigma(Y)$ for a random element $Y:\Omega\rightarrow\Lambda$.

How does $P[A/Y]=P[A/\mathcal{A}]$ relate with $P[A/Y=y]$ for $A\in\mathcal{F}$ and $y\in\Lambda$.

Best regards,

Juan Manuel

答:
What prevents $X$ from being its own conditional expectation given $\mathcal{A}$ is indeed the measurability part. It can't really fail to satisfy the other condition, can it? In your setting $X$ is a random variable meaning that $X$ is $(\mathcal{F},\mathcal{B}(\mathbb{R}))$-measurable. But $\mathcal{A}$ is a sub sigma-algebra of $\mathcal{F}$, and thus it can be smaller than $\mathcal{F}$ and hence $X$ need not be $(\mathcal{A},\mathcal{B}(\mathbb{R}))$-measurable. However, if $X$ is $(\mathcal{A},\mathcal{B}(\mathbb{R}))$-measurable then
$$
{\rm E}[X\mid\mathcal{A}]=X.
$$

The relationship between $P(A\mid Y)$ and $P(A\mid Y=y)$ is the following: If $\varphi:y\mapsto P(A\mid Y=y)$ for $y\in\Lambda$, then $P(A\mid Y)=\varphi(Y)$.


问:
Existence of a surjective homomorphism
click here for the question

I cannot have an approach to this question. Any help would be appreciated. Thanks in advance.

答:
Hint:  The only way that the product of two $p$-groups can be written as one cyclic group is if the orders are relatively prime, by the Chinese remainder theorem.

So, if you look at $C_{p^r}\times C_{q^s}\cong C_{p^rq^s}$, the RHS is not a $p$-group.

Title: Existence of a surjective homomorphism


问:
is this a theorem of ZF?
Is the following a theorem of ZF?

For any set $x$: the rank of $x$ is the order type of the natural order on the set of all rankes of elements of $TC(x)$.

Where $TC(x)$ is the transitive closure of $x$, defined in the usual manner.

答:
Here is a fun little exercise to confirm your suspicion:

The following are equivalent for any set $x$:

$\operatorname{rank}(x)=\alpha$,

$\alpha=\min\{\beta\mid\operatorname{tcl}(x)\cap V_{\beta+1}\setminus V_\beta=\varnothing\}$.

In other words, the rank of a set is the smallest ordinal that is not the rank of an element of $\operatorname{tcl}(x)$. I never found this claim in any of the standard textbooks (which is admittedly kind of odd), so you can find it embedded in the proof of Proposition 2.2 of my paper with David Asperó,

David Asperó and Asaf Karagila, Dependent Choice, Properness, and Generic Absoluteness. Review of Symbolic Logic (to appear). Journal, arXiv.

Title: is this a theorem of ZF?


问:
Question regarding infinite intersection
Would this intersection be empty?

$\bigcap_{n=1}^{\infty} (1-1/n^2, 2-1/n) $

I generated the intervals with few n's,

N=1, (0,1)

N=2, (0.75, 1.5)

N=3,  (0.88, 1.67)
.
.

and I think it would be empty, but I am not sure?

答:
Yes, the intersection would be empty.

Elements less than 0 and elements greater than 2 are clearly not in the set. It remains to address the intervals (0,1) and [1,2)

Looking at elements in (0,1): None of these can be in the set because suppose one of them is, now we can set n sufficiently large so that the lower bound does not include it.

Looking at elements in [1,2): None of these can be in the set because the case n=1 creates the interval (0,1).

Therefore, no elements can be in our set.


答:
Note:  For any $x \in R$ if there is any $n$ so that $1-\frac 1{n^2}\ge x$ then $x \not \in (1-\frac 1{n^2},2-\frac 1n)$.  As $\sup \{1-\frac 1{n^2}\}= 1$ that means for every $x < 1$ then $x$ is not an upper bound of $\{1-\frac 1{n^2}\}$ so if $x < 1$ then $x \not \in \cap (1-\frac 1{n^2},2-\frac 1n)$.

Likewise if there is any $n$ so that $2-\frac 1n \le x$ then $x \not \in (1-\frac 1{n^2},2-\frac 1n)$.  As $\inf \{2-\frac 1n\} = 1$ that means for every $x > 1$ then $x$ is not a lower boun of $\{2-\frac 1n\}$ so if $x > 1$  then $x \not \in \cap (1-\frac 1{n^2},2-\frac 1n)$

So the only possible $x$ that could be in the intersection is $x=1$.  For $n = 1$ we have $2-\frac 1n \le 1$ so $1\not \in (1-\frac 1{n^2}, 2-\frac 1n) = (0,1)$.  So $1$ is not in the intersection and so the intersection is empty.

Title: Question regarding infinite intersection

Body:
Would this intersection be empty?

$\bigcap_{n=1}^{\infty} (1-1/n^2, 2-1/n) $

I generated the intervals with few n's,

N=1, (0,1)

N=2, (0.75, 1.5)

N=3,  (0.88, 1.67)
.
.

and I think it would be empty, but I am not sure?


问:
Normal generation of the kernel of a surjection of free groups
Suppose that $\phi : F_n \to F_m$ is a surjective homomorphism between free groups of rank $n$ and $m$, respectively.  Is the kernel necessarily finitely normally generated?  If so, can we say anything about the number of normal generators that are needed in terms of $n$ and $m$?

答:
Let me elaborate on Derek Holt's answer, providing some kind of generalization, where we see exactly how the bound appear. The general statement I propose is the following:

Let $p: G \twoheadrightarrow K$ be a split epimorphism between groups, $s: K \hookrightarrow G$ be a section of $p$, and $H$ be the kernel of $p$. If $(x_i)_{i \in I}$ is a generating family for $G$, then the family $(x_i \cdot sp(x_i)^{-1})_{i \in I}$ normally generates $H$.

Here is a proof: Let $N$ be the subgroup of $G$ normally generated by the $x_i \cdot sp(x_i)^{-1}$. These are elements of $H$, hence $N \subseteq H$. Thus $p$ factors through the quotient $\pi : G \twoheadrightarrow G/N$ : there is a unique $\bar p : G/N \rightarrow K$ such that $\bar p \pi = p$. Now, $\pi s$ is a section of $\bar p$, since $\bar p \pi s = ps = id_K$. This implies that $\pi s$ is injective. Moreover, $\pi s$ is surjective, by definition of $N$. Indeed, $G/N$ is generated by the $\pi(x_i)$, and $\pi(x_i) = \pi(sp(x_i)) = \pi s(p(x_i))$ : the image of $\pi s$ contains the $\pi(x_i)$, whence all of $G/N$. Thus, $\pi s$ is an isomorphism, whose inverse mapping is $\bar p$. As a consequence, $1 = \ker(\bar p) = H/N$, which means that $H = N$, and the statement is proved.

In order to see that the answer you seek is a particular case of this, you need to remark that a surjection onto a free group always splits : any choice of lifts of generators ($y_j \mapsto g_j$ in Derek Holt's answer) extends to a section of your epimorphism. Then
the $w_i(g_j)$ in Derek Holt's answer are exactly my $sp(x_i)$, and you see that our answers are the same in this case.

And from the look of my answer, I am guessing that the best bound is indeed $n$, even in the case of free groups (but I do not have a good example at hand ; I need to think about that).


问:
Why is it that $3\in\{1, 2, 3\}$ but $\{1\}\notin \{1, 2\}$?

Fill in the gap with either $\in$ or $\notin$

b) $\{1\}\;\_\_\; \{1, 2\}$. Answer says $\notin$.

but the other exercise
a) $3\;\_\_\;\{1, 2, 3\}$, says that the answer is $\in$.

Why is it not $\in$ on b)? Is the answer in the book wrong here?

Also, what is the difference between "belong to" and a subset?

答:
The elements of $\lbrace 1, 2 \rbrace$ are $1$ and $2$. Because $\lbrace 1 \rbrace \neq 1$ and $\lbrace 1 \rbrace \neq 2$, then $\lbrace 1 \rbrace$ is not an element of the set $\lbrace 1, 2 \rbrace$. So $\lbrace 1 \rbrace \notin \lbrace 1, 2 \rbrace$.

But the elements of $\lbrace 1, 2, 3 \rbrace$ are $1$, $2$ and $3$, so you see that $3$ is an element of the set $\lbrace 1, 2, 3 \rbrace$. So $3 \in \lbrace 1, 2, 3 \rbrace$.


问:
$f: \mathbb{N}\cup\{0\} \longrightarrow \mathbb{N}\cup\{0\}, \forall x\in\mathbb{N}\cup\{0\}, f(x+1)+1 = f(f(x) + 1)$
I try to solve this question : Find all function $f:\mathbb{N}\cup\{0\}\longrightarrow \mathbb{N}\cup\{0\}$ such that $\forall x\in\mathbb{N}\cup\{0\}, f(x+1)+1 = f(f(x)+1)$.

My attempt : By induction, we obtain $$\tag1\forall x, k\in\mathbb{N}\cup\{0\}, f(x+k)+1 = f^{x}(f(k)+1)$$ and $$\tag2\forall k\in\mathbb{N}\cup\{0\}, f(x+1)+k = f(f^{k}(x)+1).$$

Since $\forall x\in\mathbb{N}\cup\{0\}, f(x+1)+1 = f(f(x)+1)$, if $f(x) = f(y)$ then $f(x+1) = f(y+1)$. From $(2)$, we know the image of $f$ is infinite. So $f$ is injective.

From $(2)$, we have $f(f^{k}(f(x)+1)+1) = f(f(x)+2)+k$. From $(1)$, we also have $f(f^{k}(f(x)+1)+1) = f(f^{x+k}(f(0)+1)+1)$ then form $(2)$, $f(f^{x+k}(f(0)+1)+1) = f(f(0)+2)+x+k$. So we have $$\tag3\forall x, k\in\mathbb{N}\cup\{0\}, f(f(x)+2)+k = f(f(0)+2)+x+k.$$

From $(3)$, we have  $$\tag4\forall x\in\mathbb{N}\cup\{0\}, f(f(x)+2) = x + f(f(0)+2).$$ Substitute $f(x)+2$ for $x$ in $(4)$, we have  $$\tag5\forall x\in\mathbb{N}\cup\{0\}, f(x+m) = f(x)+m,$$ where $m = f(f(0)+2)+2$.

Since $f$ is injective, if $m|f(x)-f(y)$, then $f(x)-f(y) = x-y$.

This question is from concours SMF junior 2020
https://smf.emath.fr/evenements-smf/annonce-concours-smf-junior-2020

答:
Some random thoughts ...

From your finding $(5)$, we see that for $m:=f(f(0)+2)+2$, the map $g\colon \Bbb Z/m\Bbb Z\to \Bbb Z/m\Bbb Z$, $x+m\Bbb Z\mapsto f(x)+m\Bbb Z$ is well-defined and obeys
$$g(x+1)+1=g(g(x)+1) $$
and is bijective. In other words, $g$ is a permutation of the $m$ residue classes.
This permutation must be a single $m$-cycle, for if there is some $x\in\Bbb Z/m\Bbb Z$ and $0<k<m$ with $g^{k}(x)=x$, then by the modulo version of $(2)$, $g(x+1)+k=g(g^{k}(x)+1)=g(x+1)$, contradiction.

Moreover, $f(0),\ldots, f(m-1)$ are the smallest representatives in $\operatorname{im}(f)$ of their respective residue class modulo $m$.
In particular, if $f(0)\ge m-2$, we find that $m-2=f(f(0)+2)=m+f(f(0)+2-m)\ge m$, contradiction. Hence
$$\tag6 m\ge f(0)+3\ge4.$$

If $f(x)\ge m-1$, then
$$ f(x+1)=f(f(x)+1)-1=f(f(x)+1-m)+m-1\ge m-1.$$
If $x\ne f(0)+2$ and $f(x)\ge m-2$, then $f(x)\ne f(f(0)+2)=m-2$ and so the previous applies, i.e., $f(x+1)\ge m-1$. It follows that $f(x)<m-2$ for $x<f(0)+2$, and $f(x)\ge m-1$ for $x>f(0)+2$.
In particular, by injectivity $$\tag7m\ge f(0)+4\ge 5.$$

If $a\in\operatorname{im}(f)$ then either $a=f(0)$ or $a+1\in\operatorname{im}(f)$ because
$a=f(x+1)$ implies $a+1=f(f(x)+1)$.
Hence $\operatorname{im}(f)$ is either of the form $[n,\infty)$ or of the form $[n_1,f(0)]\cup [n_2,\infty)$ with $n_1+m>n_2$.

Assume $\operatorname{im}(f)=[n,\infty)$. By $(7)$, $n\le m-4$.
We know that $\{f(0),\ldots,f(m-1)\}=\{n,\ldots, n+m-1\}$.
Of the $m$ numbers $f(r)$, $0\le r<m$, exactly $n+1$ must be $\ge m-1$.
By the results above, these $n+1$ numbers must be $f(0)+3,\ldots, m$.
Thus
$$n=m-f(0)-3.$$
As $n\ge f(0)$, this makes $$m\ge 2f(0)+3.$$


问:
Calculate Beats per Minute (BPM) without using average
TL;DR: My question is not about programming. Calculating BPM (beats per minute) using average is not working well. Any other ideas?

I'm studying a programming language and decided to develop a simple "Tap Tempo" app. It basically get user mouse clicks and calculate the BPM (beats per minute).

However, it's taking too long to get closer to the real BPM and even so it's an approximation. The more samples (clicks) I have, the more it gets near the real BPM, but more samples means the user has to click too many times to discover the BPM.

I'm basically starting a timer when the user clicks the first time, then divide the number of beats (samples/clicks) per the elapsed time since the last click/beat.

Something like: BPM = beats_count / elapsed_time.

The number of beats increases at every click and elapsed_time is not being reset periodically, so it grows "forever". I tried to reset the timer, but it doesn't have the expected effect.

Do you have any clues on other strategy other than using the average of beats per time?

答:
I think I have a solution to part of your problem: subtract $1$ from your number of beats. Let me explain why:

Imagine I am a perfect metronome. I click at $120$ bpm, and I use your app.

Tap. The time starts. $t=0$. Tap, $t=0.5$sec. Tap, $t=1sec$.

If you're keeping track so far, one second has passed between the first and last clicks. I've tapped $3$ times, so by the formula $\text{bpm}=\frac{\text{clicks}}{\text{time}}$ I have a tempo of $\frac{3\text{ clicks}}{1\text{ sec}}=180\text{ bpm}$.

If I do another $4$ clicks, now there have been $3$ seconds since the beginning, and $7$ total taps. Calculated value: $140$ bpm. After $11$ clicks, there have been $5$ seconds since the start, so calculated value: $132$ bpm. Even if I tap for a full $30$ seconds, I've tapped $61$ times so the calculated value is $122$ bpm!

This is what's known as a Fencepost error; if I tap $n$ times, there have only been $n-1$ "gaps" in between my taps. If we take $n$ taps and divide by $n-1$ gaps, our estimate is always too high. But if we subtract $1$ from our taps, then suddenly the error is resolved; taps=gaps and all is right with the world.

See if that helps your count.

Edit: I see that while I was typing this answer, a conversation about this was happening in the comments. I'll leave this up in case it helps anyone.


问:
If $f(x)=a |\sin x| + b e^{|x|}+c|x|^3$ and if $f(x)$ is differentiable at $x=0$, find the values of $a,b$ and $c$.

If $$f(x)=a |\sin x| + b e^{|x|}+c|x|^3$$ and if $f(x)$ is differentiable at $x=0$, find the values of $a,b$ and $c$.

Please note that if a function is differentiable at a point , it is also continuous there.
 Please prove using $$\lim_{x\to c^+}f(x)=\lim_{x\to c^-}f(x)=f(c)$$
and
$$\lim_{x\to c^+}\frac{f(x)-f(c)}{x-c}=\lim_{x\to c^-}\frac{f(x)-f(c)}{x-c}$$

Thankyou

答:
The differentiability of $f$ in $x=0$ is equivalente to the existence, finiteness and equality of the following limits:
$$
  \lim_{h\to 0^-}
    \frac
    {a |\sin h|  +  be^{|h|}  +  c|h|^3  -  b}
    {h}
=
  \lim_{h\to 0^+}
    \frac
    {a |\sin h|  +  be^{|h|}  +  c|h|^3  -  b}
    {h}
$$
Let us compute each summand separately:

1

$$
\lim_{h\to 0^\pm}
  \frac{a|\sin h|}{h}
=
\pm
\lim_{h\to 0^+}
  \frac{a \sin h}{h}
=
\pm a
$$

2

$$
\lim_{h\to 0^\pm}
  \frac{be^{|h|}-b}{h}
=
\pm
\lim_{h\to 0^+}
  \frac{b(e^h-1)}{h}
=
\pm b
$$

3

$$
\lim_{h\to 0^\pm}
  \frac{c|h|^3}{h}
=
\pm
\lim_{h\to 0^+}
  c h^2
=
0
$$
Therefore the above equality turns into
$$
a+b = -a -b
$$
so $f$ is differentiable iff you choose
$$
a,b,c\in\mathbb R ~~\text{with}~~b = -a
$$

Continuity

Concerning "differentiability implies continuity",
$$
\lim_{h\to 0} \big\{f(x+h) - f(x)\big\}
~=~
\lim_{h\to 0} \frac{f(x+h) - f(x)}{h}h
~=~
\lim_{h\to 0} f'(x)h
~=~
0
$$
provided $f$ is differentiable at point $x$.


问:
Proving that If $A$ and $B$ are compact then $\operatorname{dist}(A,B) > 0$
I have seen this question on here before, but I have a different attempt at the proof for it. We are given that $A$ and $B$ are compact and that $\operatorname{dist}(A,B) = \inf\{\Vert x-y \Vert: x \in A, y \in B\}.$

Here is my attempt at it:

So since we have that $A$ and $B$ are compact, we have that by the Heine-Borel theorem that both $A$ and $B$ are closed and bounded. There is a theorem in my text that states that for $E \subseteq \mathbb{R^n}$ and $E$ closed, then $E$ contains all of its limit points. So from this we have that there exists $x_k \in A$ which converges to $x$, and some $y_j \in B$ that converges to $y$. Now, by the Bolzano Weierstrass theorem, each convergent sequence, has a convergent subsequence, namely we have that $y_{j_t}$ converges to $y$ as $t$ goes to infinity, and similarly for $x_{k_p}$ converges to $x$ as $p$ goes to infinity. But from here I'm unsure where to finish. I feel as if the approximation theorem for infimum can be used here, but maybe not.

答:
Here is another proof that RobertShore's. I also assume that $A\cap B=\varnothing$ otherwise the result is false. I also relax the hypothesis and just suppose that $B$ is closed.

Suppose $A$ is compact, $B$ is closed and $A$ and $B$ are disjoint. Let $f : A \to \mathbb{R}$ be defined by $f(x) = d(x,B)$. By the triangle inequality, $f$ is $1$-Lipchitz, and thus is continuous. As $A$ is compact and $f$ continuous, $f$ has a global minimum, say $m\geqslant 0$ attained at $x_0$. That is
\begin{align}
d(A,B) = f(x_0) = d(x_0,B) = m \geqslant 0
\end{align}
Suppose by contradiction that $m = 0$. Then as $B$ is closed, $x_0 \in B$, which is a contradiction because $A\cap B = \varnothing$.
Thus, $f(x_0) >0$ and $d(A,B)>0$.


问:
A finite group $G$ contains a subgroup $H \approx C_p \times C_p $ and a subgroup $K \approx C_{p^2}$
A finite group $G$ contains a subgroup $H \approx C_p \times C_p $ and a subgroup $K \approx C_{p^2}$ show that $p^3 | |G|$.

From the problem we know that $G$ contains elements of order $p$ and at least an element of order $p^2$, since $K$ is cyclic. Also as we can see $H$ is Abelian.

So from this I know that $|G| = p^{\alpha} m$ , where $\alpha \ge 2$.

Also I'm not sure if this is correct but $H \cap K = {1}$ or how this can help me with proving what I need.

答: If $p^3 \nmid |G|$ then $C_{p^2}$ and $C_p\times C_p$ would be Sylow subgroups of $G$. On the other hand every two Sylow subgroups must be conjugate to each other. This is a contradiction since every two conjugate subgroups are necessarily isomorphic groups.


问:
Number of non-negative integer solutions of linear Diophantine equation

Given $n \ge 0$, find the number of non-negative integer solutions of the following equation $$2x_1+x_2+x_3+x_4+x_5=n$$

I tried putting one $x_1$ on the right-hand side, as follows

$$x_1+x_2+x_3+x_4+x_5=n-x_1.$$

Then I am not sure what to do. On the left-hand side, we now have non-negative numbers, but not sure how to write this sum because of the right-hand side.

答:
We choose a value for $x_1$ from $0$ to $\lfloor{\frac{n}{2}}\rfloor$.

Then we use combinations with repetition to distribute $r=n-2x_1$ identical objects into the 4 remaining slots ($x_2,x_3,x_4,x_5$).

So for each choice of $x_1$,

we have $m+r-1 \choose r$ solutions

$m+r-1 \choose r$, where m is 4.

=$n-2x_1+3 \choose n-2x_1$

=$n-2x_1+3 \choose 3$

So the total number of solutions is

$\sum_{x_1=0}^{\lfloor\frac{n}{2}\rfloor} {n-2x_1+3 \choose 3}$

Note sure if there's a simplification of this.


问:
Checking derivative of norm squared of a linear transformation

Given $f: L(\mathbb{R}^n, \mathbb{R}^k) \to \mathbb{R}$ where $f(A)=|Ax|^2$ and fix $x\in \mathbb{R}^n$ then find $\partial_H f(A)$ where $H\in L(\mathbb{R}^n, \mathbb{R}^k)$.

My attempt:
$$
\partial_H f(A) = \lim_{t\to 0}\left(\frac{|(A+t\cdot H)x|^2 - |Ax|^2)}{t}\right)\\
= \lim_{t\to 0}\left(\frac{\sum_{j=1}^k\left(\sum_{i=1}^n (a_{ji} + t\cdot h_{ji})x_i\right)^2-\sum_{j=1}^k\left(\sum_{i=1}^n a_{ji}x_i\right)^2}{t}\right)\\
= \lim_{t\to 0}\left(\frac{\sum_{j=1}^k\left(\sum_{i=1}^n a_{ji}x_i\right)^2 + 2t\sum_{j=1}^k\left(\sum_{i=1}^n (a_{ji}x_i) \sum_{i=1}^n(h_{ji}x_i)\right) + t^2\sum_{j=1}^k\left(\sum_{i=1}^n h_{ji} x_i\right)^2-\sum_{j=1}^k\left(\sum_{i=1}^n a_{ji}x_i\right)^2}{t}\right)\\
= \left(2\sum_{j=1}^k\left(\sum_{i=1}^n (a_{ji}x_i) \sum_{i=1}^n(h_{ji}x_i)\right)\right)\\
$$
however I can't seem to simplify this further.

答:
Write the function in terms of the vector $y=Ax$ and calculate its differential.
$$\eqalign{
 f &= y^Ty \\
df &= 2y^Tdy = 2(Ax)^T(dA\,x) = 2x^TA^TdA\,x \\
}$$
Now substitute $\,dA=H\,dt\,$ to recover the desired derivative.
$$\eqalign{
\partial_Hf &= \left(\frac{df}{dt}\right) = 2x^TA^THx  \\
}$$


问:
Solve the equation: $-u''=f(x), 0<x<1, u(0)=u(1)=0$
Solve the equation using green's function $-u''=f(x), 0<x<1, u(0)=u(1)=0$

I got the solution as being $u(x)=\frac{x^2}{2}(x-1)f(x)+xf(x)(\frac{1}{2}-\frac{x^2}{2}-1+x$.

Which is wrong. Can anyone see where i went wrong?

答:
The next to last formula is wrong, it should have $f(s)$ instead of $f(x)$, that is,
$$
u(x)=\int_0^1G(x,s)f(s)\,ds=-\int_0^x (1-x)sf(s)\,ds-\int_x^1 x(1-s)f(s)\,ds
$$
Without knowing $f$ no further evaluation of the integrals is possible.


问:
Divergence of an Expression Involving the Gamma Function
I want to show that
$$
\lim_{Re(t) \to \infty} \frac{\Gamma(1+2t)\Gamma(-t)}{\Gamma(1+t)} (-z)^t
$$
diverges if $|z| > \frac14$.
I recognized that the expression involving the Gamma function is just the Beta function, $B(2t+1,-t)$, but I don't know much about the Beta function. Are there some properties of the Gamma function that I can use to prove this divergence? Thanks!

答:
The arguments can be simplified using the Gamma duplication identity
$$
\Gamma(2s+1) = \frac{\Gamma(s+1)\Gamma(s+1/2)2^{2s}}{\sqrt{\pi}},
$$
and the Gamma reflection identity
$$
\Gamma(s+1)\Gamma(-s) = -\frac{\pi}{\sin(\pi s)}.
$$
So your expression is equivalent to
$$
-\frac{\sqrt{\pi}}{\sin(\pi s)}\frac{\Gamma(s+1/2)}{\Gamma(s+1)}(-4z)^s
$$

The first term in the product has a magnitude in the interval  $[\sqrt{\pi}\mathrm{sech}(\mathrm{Im}[s]), \sqrt{\pi}|\mathrm{csch}(\mathrm{Im}[s])|]$ (finite for all $\mathrm{Im}[s]\ne 0$) while the second term can be fairly easily shown from Stirling's approximation to scale as $s^{-1/2}$. So the limit diverges if $|z| > 1/4$ or $\mathrm{Im}[s] = 0$.


问:
Examine the continuity of $f(x)=x^2+\frac{x^2}{(1+x^2)}+\frac{x^2}{(1+x^2)^2}+...+ \frac{x^2}{(1+x^2)^n}+....$ at $x=0$

Examine the continuity of $$f(x)=x^2+\frac{x^2}{(1+x^2)}+\frac{x^2}{(1+x^2)^2}+...+ \frac{x^2}{(1+x^2)^n}+....$$ at $x=0$

I tried to solve the problem using $$\lim_{x \to0^+}f(x)=\lim_{x \to0^-}f(x)=f(0)$$ and I got both limits to be $0$. I think there is an error somewhere since my book says it is discontinuous.

Can you help me out?

 Thankyou

答:
If the dots imply an infinite series, then you can first consider
$$
g(x)=\sum_{n\ge0}\frac{1}{(1+x^2)^n}
$$
that converges to
$$
g(x)=\frac{1}{1-\dfrac{1}{1+x^2}}=\frac{1+x^2}{x^2}
$$
whenever
$$
\frac{1}{1+x^2}<1
$$
that is, for all $x\ne0$.

Thus your function is
$$
f(x)=
\begin{cases}
1+x^2 &\text{if $x\ne0$}\\
0&\text{if $x=0$}
\end{cases}
$$


问:
Choosing lottery numbers strategy (based on heuristics such as choose as many odd numbers as even numbers)
So I got this question in brilliant.org's probability fundamentals:

Every week you play a lottery game that requires choosing 4 numbers
between 1 and 80. In this particular game, you win "outright" and do
not have to potentially share the prize with others. Your friends give
you some advice for the next week on which numbers to pick.

Which advice is the most reasonable?

You should choose two even and two odd numbers.

You should not pick 4 consecutive numbers (numbers in a row such as
3, 4, 5, 6).

You should pick some low numbers, high numbers, and numbers in the
middle.

It doesn't matter which four numbers you pick. It's just luck.

brilliant.org says the correct answer is advice #4. I do agree that any particular combination of four numbers is just as likely as any other combination; however, combinations having certain characteristics are more likely than combinations not having those characteristics. For example, a combination having 2 even and 2 odd is about six times as likely as a combination having all even or all odd (the math for this is below), so advice #1. is beneficial. Is brilliant.org wrong or am I not thinking about this correctly? Is the question too vague?

I did the math the for advice #1.:

1. You should choose two even and two odd numbers.

The probability that the winning sequence has two even and two odd numbers is:

$P(2,2) = \binom{4}{2} * \frac{40*39*40*39}{80*79*78*77} \approx 0.385$

whereas the probability that the winning sequence has three even and one odd is:

$P(3,1) = P(1,3) = \binom{4}{3} * \frac{40*39*38*40}{80*79*78*77} \approx 0.250$

Finally, the probability that the winning sequence has four even and zero odd is:

$P(4,0) = P(0,4) = \binom{4}{4} * \frac{40*39*38*437}{80*79*78*77} \approx 0.058$

Given that $P(2,2)$ is greater than $P(3,1)$, $P(1,3)$, $P(4,0)$, and $P(0,4)$, wouldn't we be better off following the advice of option #1. than not following it?

Likewise, if I do the math on advice #2. and #3., I'm very sure that the math would show that the advice is beneficial. It should be possible to follow all three pieces of advice (#1., #2., and #3.) as well.

I'm trying to reconcile this with brilliant.org's correct answer that basically says the advice (#1., #2., and #3.) is irrelevant.

Thanks!

答:
It's more likely for the lottery numbers to be two-odd-two-even; however, there are more two-odd-two-even selections to choose from, for the exact same reason. Not only do you have to hope the lottery ends up in your chosen category, but you also have to hope that the lottery numbers hit your exact choice in the chosen category. When that category is large (two-odd-two-even), the first event is more likely but the second event is less likely; when that category is small (all numbers less than 10), the first event is less likely and the second event is more likely.

Another analogy: I have a two-story warehouse full of $N$ candy bars, and one of the candy bars (uniformly randomly determined) contains a golden ticket. You get one guess to point to a candy bar and hope that's the one with the golden ticket. You could just look at the shelf next to you and point to one of its candy bars. Does it benefit you to follow the advice "the second story is bigger than the shelf, so go point to a candy bar on the second story instead"? No—you still have a $1$ in $N$ chance of finding the golden ticket.


问:
Entire function can be approximated uniformly by polynomials with rational coefficients on every compact set.
I am working on Stein Complex Analysis Chapter 2 Problem 5. The question has a hint that I cannot prove. Basically, the hint states as follows:

Let $p_{1}, p_{2},\cdots$ denote an enumeration of the collection of polynomials whose coefficients have rational real and imaginary parts. Given $h$ entire, there exists a sequence $\{n_{k}\}$ such that $\lim_{k\rightarrow\infty}p_{n_{k}}(z)=h(z)$ uniformly on every compact subset of $\mathbb{C}$.

I understand that this is pretty similar to the Runge theorem:

Any function holomorphic in a neighborhood of a compact set $K$ can be approximated uniformly on $K$ by rational functions whose singularities are in $K^{c}=\mathbb{C}\setminus K$. If $K^{c}$ is connected, any function holomorphic in a neighborhood of $K$ can be approximated uniformly on $K$ by polynomials.

To prove the hint quickly, I tried to prove it using this theorem: as $h$ was entire, it is of course holomorphic in a neighborhood of any compact set $K$. Thus, it can be approximated by rational functions whose singularities are in $K^{c}$.

Then I got stuck. If $K^{c}$ is connected, then I can say $h(z)$ is approximated by a sequence of polynomial, and by density it is clear that this sequence of polynomial can be approximated by a sequence of polynomials with rational coefficients. Then we are done.

But generally speaking we do not have $K^{c}$ to be connected. What should I do?

答:
As per comments the crucial observation is that for every $n \ge 1$ we can find a polynomial $P_{n,h}$ st $|P_{n,h}(z)-h(z)| < 1/(2n), |z| \le n$ (eg some Taylor truncation of $h$).

But then since the closed disc is compact one can tweak the coefficients of $P_{n,h}$ very little and find $Q_{n,h}$ with rational coefficients st $|P_{n,h}(z)-Q_{n,h}(z)| < 1/(2n), |z| \le n$

Putting it together we get the sequence $Q_{n,h}$ satisfying $|h(z)-Q_{n,h}(z)| < 1/n, |z| \le n$. But now any compact $K$ is contained in $|z| \le n$ for $n \ge n_K$ so $Q_{n,h}$ converges uniformly on $K$ to $h$


问:
A simple binomial inequality seeking its proof from the book?
Let $k\in \mathbb{N}$ and $1\leq a \leq k$ be fixed. Imagine we have two groups $G_{1}, G_{2}$ of $k$ indistinguishable elements each forming two sequences with indices $1, \dots , k$. Choose the element with index $a$ in $G_{1}$ and the element with index $b$ in $G_{2}$. Consider the expression

$$f(k,a,b) = \binom{a+b-2}{a-1}\binom{2k-b-a}{k-a},$$

which counts the number of ways to merge the elements before the two chosen elements and after them into two sequences.

Example: $k=5$, $a=3$, $b=4$:

$ - - \circleddash - -$

$ - - - \circleddash -$

The number of ways to merge the set of elements before the two chosen is $\binom{2+3}{2}$ and to merge the set of elements after the two chosen is $\binom{2+1}{1}$. In total, we get $\binom{2+3}{2}.\binom{2+1}{1}$.

Show that

$$\binom{a+b-2}{a-1}\binom{2k-b-a}{k-a}\geq min\left(\binom{2k-1-a}{k-a},\binom{a+k-2}{a-1}\right),$$

i.e.,

$$f(k,a,b)\geq min\left(f(k,a,1),f(k,a,k)\right).$$

In other words, when $k$ and $a$ are fixed, $f$ reaches its minimum at one of the endpoints $b=1$ or $b=k$.

An ugly proof by expansion of the binomial coefficients exist, but
does anyone know an elegant proof?

答:
By expanding the binomials, it can be shown directly that

$$
\frac{f(k,a,b)}{f(k,a,b-1)} = \frac{\binom{a+b-2}{a-1}\binom{2k-b-a}{k-a}}{\binom{a+b-3}{a-1}\binom{2k-b-a+1}{k-a}} \\
= \frac{a+b-2}{b-1} \frac{k-b+1}{2k-a-b+1} = (1 + \frac{a-1}{b-1})(1 - \frac{k-a}{2k-a-b+1})
$$

Interpreting this ratio as a function of a continuous $b$, it is straightforward to calculate that the ratio becomes $1$ for $b^* = \frac{ak - 1}{k-1}$.

Since $ b < 2k - a +1$, both terms in the last product always fall with $b$. It follows that the ratio is smaller than $1$ for  $ \frac{ak - 1}{k-1} < b$ and hence $f(k,a,b)$ is falling in that range of $b$, so  the minimum is obtained at the "endpoint", for the highest $b$. Conversely, for $ \frac{ak - 1}{k-1} > b$ the ratio is larger than $1$ and hence $f(k,a,b)$ is rising in that range of $b$, so  the minimum is obtained at the other "endpoint", for the lowest $b$.

Together, this proves the claim that $f(k,a,b)\geq \min\left(f(k,a,1),f(k,a,k)\right).$ $\qquad \Box$


问:
tensor product and matrix multiplication distributive properties
I am trying to find partial trace of some matrix of the form

$M = (A \otimes B)\times (A^{T*} \otimes B^{T*})$

in which $\otimes$ is tensor product, $\times$ is matrix multiplication, $T*$ is conjugate transpose of the matrix. $A$ and $B$ are vectors (but can be matrices too).

I was wondering if I could use the following relation?

$(A \otimes B)\times (A^{T*} \otimes B^{T*}) = (A \times A^{T*})\otimes (B \times B^{T*})$

It checks out for $A=\left(\begin{matrix} a\\b \end{matrix}\right)$ and $B=\left(\begin{matrix} c\\d \end{matrix}\right)$.

if this relation is general, what about this one (for an arbitrary vectors (or matrices) C and D)

$(A \otimes B)\times (C \otimes D) = (A \times C)\otimes (B \times D)$

although I don't know how to trace this matrix. I would appreciate any help. thank you in advance

答:
The equality in the last part of your question  is true. One can prove it easier if we look at a matrix as a linear map and look at a matrix product as a composition of linear maps. Furthermore we consider the equality $$T\otimes S(v\otimes w)=T(v)\otimes S(w)$$ which is an obvious definition of tensor product of two linear maps.

So your equality becomes

$$(A \otimes B)\circ (C \otimes D) = (A \circ C)\otimes (B \circ D)$$

We can easilly evalute both side of the equation on a typical simple tensor $x\otimes y$.

So the validity of the equality under discussion is a result of the concept of tensor product of vector spaces and tensor product of linear maps.

Remark: I think perhaps a proof of this equality in its matrix setting is not short or straithforward but we should keep in mind that the following fact is necessary for our argument. Some books give it as a theorem and some other books present it as an exercize:

Fact: If matrix $A$ is  matrix representatio  of a linear map $T$  and $B$ is a matrix representation of a linear map $S$ then the tensor product matrix $A\otimes B$ is the matrix representation of $T\otimes S$.


问:
Classify the abelian groups of order 81, 144 and 216
Classify the abelian groups of order n, 2n, 4n</p>

答:
Here, you will want to use the theorem of structure of finite abelian groups, that is: Every abelian finite group is isomorphic to a group of the form:

$$\mathbb{Z}_{m_1}\times \mathbb{Z}_{m_2} \times\mathbb{Z}_{m_3} \times... \times\mathbb{Z}_{m_n}$$

In which $m_i$ divides $m_{i-1}$, and abviously the product of all $m_i$ has to be the order of the group.

So you have to find possible combinations of numbers $m_i$ that meet the above conditions. For example $81=3^4$, just find the combinations.

Now, a further result that will allow you to not repeat any group is this one:

If $G$ is a group of the form $G=G_1\times G_2$, then both of the next things are equivalent

G is cyclic

$\gcd(|G_1|,|G_2|)=1$

This means for example, if we have groups of order 18, then the group $\mathbb{Z}_2 \times \mathbb{Z}_9$ will be the same as $\mathbb{Z}_{18}$

With this results you should be able to classify all the abelian groups of those orders.

Sadly, I've studied this from books (really good books), written in spanish and I can't give references to the proves of these results in english, but I suppose that you will find them in any elementary group theory book. Maybe someone wants to comment with some references.

P.S.: As you're not directly asking anything, I haven't explicitly solved the exercise that looks a lot like homework. If you have any quetion regarding some of the results you don't understand, then ask it.


答:
You're almost going about it backwards. Let's say there are five people: Alfonso follows advice #1: He picks two evens and two odds randomly. (We're assuming order doesn't matter here.) Barbara follows advice #2: She picks randomly but makes sure to not pick $4$ consecutive numbers. Cory follows advice #3: he picks one "high number" (say, $>60$), one "low number" (say, $\leq 20$), and one "in the middle" (in the interval $(20,60]$), plus one randomly. Denise follows advice #4 and just goes random. And just for fun, Ed decides that he's always going to pick $(1,2,3,4)$ because that's his phone passcode and he doesn't want to forget it.

As you calculated, Alfonso has a $\binom 42\frac{40\cdot39\cdot40\cdot39}{80\cdot79\cdot78\cdot77}\approx38.5\%$ chance of being "right," i.e. the winning combination has 2 evens and 2 odds. But that's not his probability of winning. No, because Alfonso picked just one of $\frac{40\cdot39\cdot40\cdot39}{2!2!}$ possible "two even, two odd" tickets, and only one of those possible tickets is a winner. Therefore, Alfonso's probability of winning is $\frac{2!2!}{40\cdot39\cdot40\cdot39}\cdot\binom 42\frac{40\cdot39\cdot40\cdot39}{80\cdot79\cdot78\cdot77}=\frac{4!76!}{80!}$ shot at winning.

Let's move on to Barbara. What are the odds that Advice #2 is correct? Well, the only way that two consecutive numbers can win is if the winning combo is $(1,2,3,4),(2,3,4,5),\dots,(77,78,79,80)$. That's only $77$ possible tickets, out of the $\frac{80!}{4!76!}$ available. So that means that the total chance that advice $2$ is correct is $\frac{\frac{80!}{4!76!}-77}{\frac{80!}{4!76!}}\approx99.995\%$. Pretty good, right? So Barbara is in the money? Well, not quite. You see, Barbara has just one ticket, out of the possible $\frac{80!}{4!76!}-77$ possible "non-consecutive" tickets, so her shot is really $\frac1{\frac{80!}{4!76!}-77}\cdot\frac{\frac{80!}{4!76!}-77}{\frac{80!}{4!76!}}=\frac{4!76!}{80!}$, the same as Alfonso.

Hmm. What about the rest of them? Can you do the math out and show that Cory, Denise, and Ed all have the same chances to win? Or can you convince yourself why the probabilities "cancel out" this way?

Title: Choosing lottery numbers strategy (based on heuristics such as choose as many odd numbers as even numbers)

Body:
So I got this question in brilliant.org's probability fundamentals:

Every week you play a lottery game that requires choosing 4 numbers
between 1 and 80. In this particular game, you win "outright" and do
not have to potentially share the prize with others. Your friends give
you some advice for the next week on which numbers to pick.

Which advice is the most reasonable?

You should choose two even and two odd numbers.

You should not pick 4 consecutive numbers (numbers in a row such as
3, 4, 5, 6).

You should pick some low numbers, high numbers, and numbers in the
middle.

It doesn't matter which four numbers you pick. It's just luck.

brilliant.org says the correct answer is advice #4. I do agree that any particular combination of four numbers is just as likely as any other combination; however, combinations having certain characteristics are more likely than combinations not having those characteristics. For example, a combination having 2 even and 2 odd is about six times as likely as a combination having all even or all odd (the math for this is below), so advice #1. is beneficial. Is brilliant.org wrong or am I not thinking about this correctly? Is the question too vague?

I did the math the for advice #1.:

1. You should choose two even and two odd numbers.

The probability that the winning sequence has two even and two odd numbers is:

$P(2,2) = \binom{4}{2} * \frac{40*39*40*39}{80*79*78*77} \approx 0.385$

whereas the probability that the winning sequence has three even and one odd is:

$P(3,1) = P(1,3) = \binom{4}{3} * \frac{40*39*38*40}{80*79*78*77} \approx 0.250$

Finally, the probability that the winning sequence has four even and zero odd is:

$P(4,0) = P(0,4) = \binom{4}{4} * \frac{40*39*38*437}{80*79*78*77} \approx 0.058$

Given that $P(2,2)$ is greater than $P(3,1)$, $P(1,3)$, $P(4,0)$, and $P(0,4)$, wouldn't we be better off following the advice of option #1. than not following it?

Likewise, if I do the math on advice #2. and #3., I'm very sure that the math would show that the advice is beneficial. It should be possible to follow all three pieces of advice (#1., #2., and #3.) as well.

I'm trying to reconcile this with brilliant.org's correct answer that basically says the advice (#1., #2., and #3.) is irrelevant.

Thanks!


问:
Convex function supremum of countable affine function
Let $I\subset\mathbb{R}$ be an interval and $\varphi:I\to\mathbb{R}$ a convex function. I know that then $\varphi(x)=\sup\{f(x):f\leq\varphi \;\;\text{and}\;\;f(y)=ay+b,\;\; a,b\in\mathbb{R}\}$. In lecture notes im reading it says that we even have (given that $\varphi$ is not affine) $\varphi(x)=\sup\{f(x):f\leq\varphi \;\;\text{and}\;\;f(y)=ay+b,\;\; a,b\in\mathbb{Q}\}$.I was wondering how to show that second statement.

答:
Let
$$\let\phi\varphi \psi(x)=\sup\{f(x):f\leq\varphi \;\;\text{and}\;\;f(y)=ay+b,\;\; a,b\in\mathbb{Q}\}.$$
Clearly, $\psi\le\varphi$ as we take the supremum over fewer functions.
Suppose $\psi(a)<\varphi(a)$.
Let $\ell_a$ be a tangent to $\phi$ at $(a,\phi(a))$.
As $\phi$ is not affine, there is some $b\ne a$ with $\phi(b)>\ell_a(b)$.
Let $\ell_b$ be a tangent to $\phi$ at $(b,\phi(b))$.
Now find a rational line $\ell$ as follows:
Start with $\ell_a$; move it slightly downward to achieve $\ell(a)\in \Bbb Q$ and $\psi(a)<\ell(a)<\phi(a)$.
Now rotate $\ell$ around $(a,\ell(a))$ slightly towards $\ell_b$, i.e., if $b>a$ rotate up and if $b<a$ rotate down. With a small enough rotation, we can achieve that the slope of $\ell$ is rational and $\ell$ still remains outside the shape bounded by $\ell_a$ and $\ell_b$. Now we have a rational line going above $(a,\psi(a))$, contradicting the definition of supremum.


问:
Show that $\lim_{n \to \infty} \frac{1!\cdot2!\cdot...\cdot n!}{(2n)!} = \infty$
Show that
$$\lim_{n \to \infty} \frac{1!\cdot2!\cdot...\cdot n!}{(2n)!}=\infty$$
Intuitively, it's obvious that the sequence above diverges to $\infty$, as a product of factorials will grow much faster than $(2n)!$. However, I haven't been able to prove this rigorously, and I would appreciate it if someone could point me in the right direction.

Body:
HINT

We have that

$$\frac{1!\cdot2!\cdot...\cdot n!}{(2n)!} =\frac{1!}{2}\cdot\frac{2!}{3\cdot 4}\cdot \ldots \cdot\frac{n!}{(2n-1)\cdot (2n)}$$

with

$$\frac{5!}{9\cdot 10}=\frac43$$

Or also for $n\ge 3$

$1!\cdot2!\cdot...\cdot n! \ge 6^{n-4}(n-1)!n!$

and

$(2n)!\le (2n)^2(2n-2)^2\ldots 2^2=(2^nn!)^2$


问:
please,find the limit of $x$
please,find the limit of $x$  when

$x=u^{-\frac{1}{\alpha}}$

where ${0}\le u\le{1}$   and $\alpha>0$

i have found the limit of $x$ is also ${0}\le x\le{1}$

when $u=0$ then $x=0$ since  $x=0^{-\frac{1}{\alpha}}=0$

when $u=1$ then $x=1$ since  $x=1^{-\frac{1}{\alpha}}=1$

but the book which i am reading showing that the range of $x\ge{1}$

where is my wrong?

would you like to help me finding the limit?

please, show your works step by step.

If you have doubt on my question then kindly explain me how the power distribution of statistics ranges from $1 to \inf$
when X follows power distribution & U follows standard uniform distribution & $X=U^{-\frac{1}{\alpha}}$

答:
Let's have proper notations:
$$
\forall (u,a),\; 0  0,\quad f_a(u) = \frac{1}{u^a}
$$

What did I do? I adopted a fraction-notation because of the minus sign in the exponent, and I posed $a = \frac{1}{\alpha}>0$.

Now I suppose I don't need to prove that given any exponent $a>0$, for any $u\in]0,1],\ u^a\in]0,1]$, and so $f_a(u)\geq 1$.

Title: please,find the limit of $x$

Body:
Let's have proper notations:
$$
\forall (u,a),\; 0 < u \leq 1,\; a > 0,\quad f_a(u) = \frac{1}{u^a}
$$

What did I do? I adopted a fraction-notation because of the minus sign in the exponent, and I posed $a = \frac{1}{\alpha}>0$.

Now I suppose I don't need to prove that given any exponent $a>0$, for any $u\in]0,1],\ u^a\in]0,1]$, and so $f_a(u)\geq 1$.


问:
Simplifying $\frac{\sqrt[3]{e^{3i\pi}-e^{i\pi/3}}+\sqrt[3]{e^{3i\pi}+e^{i\pi/3}}}{\sqrt[3]{e^{3i\pi}-e^{i\pi/3}}-\sqrt[3]{e^{3i\pi}+e^{i\pi/3}}}$
How to simplify the expression?
$$\frac{\sqrt[3\,] {e^{3i\pi}-e^{\frac{i\pi}{3}}}+\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}}{\sqrt[3\,] {e^{3i\pi}-e^{\frac{i\pi}{3}}}-\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}}$$

We also have
$$\left(-\frac 32 - i \frac{\sqrt 3}{2}\right)^{1/3}= \sqrt[3\,] {e^{3i\pi}-e^{\frac{i\pi}{3}}} \quad\text{and}\quad (-1)^{2\over 9}=\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}$$

WolframAlpha gives this result:
$$-1 - \frac{2 i}{(3^{1/6} -i)}$$
How can we obtain it?

The problem was initially posed by K.Srinivasa Raghava.

Body:
$\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}=\sqrt[3\,]{e^{\frac {2i\pi}{3}}}$

$-\frac {3}{2} - i \frac{\sqrt 3}{2}=-\sqrt 3 e^{i\pi/6}$

$\sqrt[3\,]{e^{\frac {2i\pi}{3}}} +\sqrt[3\,]{-\sqrt 3 e^{i\pi/6}}= 3^{1/6} e^{-5i\pi/18}+e^{\frac {4i\pi}{18}}$ $=e^{\frac {4i\pi}{18}}(3^{1/6}e^{\frac {-9i\pi}{18}}+1)$ $=e^{\frac {4i\pi}{18}}(-i3^{1/6}+1)$

$\frac{\sqrt[3\,] {e^{3i\pi}-e^{\frac{i\pi}{3}}}+\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}}{\sqrt[3\,] {e^{3i\pi}-e^{\frac{i\pi}{3}}}-\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}}$ $=\frac{(-i3^{1/6}+1)}{(i3^{1/6}+1)}=
-1 - \frac{(2 i)}{(3^{1/6} + -i)} =\frac{-3^{1/6}-i}{3^{1/6}-i}
$