问:
Does the 80-20 Rule apply to the Buffon process?
Whether the needle crosses a line depends greatly on the angle the needle makes with the line, 90 degrees being of course the most favorable for a line-crossing. Does the 80-20 Rule apply to these angles? That is, since 20% of 90 degrees is 18 degrees, are about 80% of the line-crossings (for a large number of needle-tosses) accounted for by about 20% instances, in which the angle of the needle relative to the line was between 72 and 90 degrees? Has anyone done a statistical study or simulation of this?</p>

答:
The angle $\Theta$ is uniformly distributed betwen $0$ and $\pi/2$ radians.

It actually took me a while to figure out that the question is this: If the number $c$ is so chosen that $\Pr(\Theta>c\mid\text{crossing})=0.8$ then is it true that $\Pr(\Theta >c)=0.2$?

I.e. do $80\%$ of crossings occur for only the $20\%$ of values of $\Theta$ most favorable to crossings?  Short answer: no.

The prior density of $\Theta$ is
$$
f_\Theta(\theta) = 2/\pi\quad\text{for }\theta\text{ between }0\text{ and }\pi/2
$$
and $=0$ for $\theta$ not in that space.

The likelihood function is
$$
L(\theta\mid\text{crossing}) = \Pr(\text{crossing}\mid\Theta=\theta) = \sin\theta.
$$

Bayes tells us to multiply the prior by the likelihood function and then normalize to get the posterior probability density $f_{\Theta\mid\text{crossing}} (\theta)$.

Thus the posterior is
$$
f_{\Theta\mid\text{crossing}} (\theta) = 1\cdot\sin\theta,
$$
since $\int_0^{\pi/2}\sin\theta\,d\theta=1$.

Then we have, for example, $\Pr(\Theta>\pi/4)=1/2$ and
$$
\Pr(\Theta>\pi/4\mid\text{crossing}) = \int_{pi/4}^{\pi/2}\sin\theta\,d\theta = \cos\frac\pi4=\frac{\sqrt{2}}{2} \approx 0.707\ldots.
$$

So the most crossing-productive half of the population produces about $70.7\%$ of all crossings.

Plug in numbers to get other results.


问:
How to find sup and inf of a multivariable function
if i'm given that $f(x,y)=(x+y)^2$ with $A=B=[-1,1]$ I want to show that

$$\sup_{x\in A}\inf_{y\in B}f(x,y)\neq\inf_{y\in B}\sup_{x\in A}f(x,y)$$

I can show this by doing it directly, but I'm not sure how can I find $\sup_{x\in A}\inf_{y\in B}f(x,y)$ and the opposite? (inf first then sup). Here's what I tried.

$$\sup_{x\in A}\left(\inf_{y\in B}f(x,y)\right)$$

gives me $-1-x\leq y \leq 1-x$, taking the supremum of this gives me $-2\leq y \leq 0$ giving me $0$. while doing it the with first sup then if gave me $-1-y\leq x\leq 1-y$, taking the infimum gave me $0\leq x\leq 2$ which gives me $0$ too?

答:
I agree that
$$\sup_{x\in A}\left(\inf_{y\in B}f(x,y)\right) = 0.$$
Specifically, we have, for any $x \in [-1, 1]$,
$$\inf_{y \in B} f(x, y) = \inf_{y \in [-1, 1]} (x + y)^2 = 0,$$
as regardless of $x \in [-1, 1]$, we get a minimum of $y = -x \in [-1, 1]$. Then, we are taking the supremum over the constantly $0$ function, which is indeed $0$.

On the other hand, let's consider
$$\sup_{x \in A} f(x, y) = \sup_{x \in [-1, 1]} (x + y)^2.$$
Regardless of the value of $y$, the map $x \mapsto (x + y)^2$ is a parabola with a minimum, and will achieve the maximum in $[-1, 1]$ at an endpoint, i.e. at $x = \pm 1$. The vertex of the parabola occurs at $-y$ and the maximum value occurs at the furthest endpoint from $-y$. So, the maximum occurs at $x = 1$ if $y \ge 0$, and at $x = -1$ if $y \le 0$ (yes, at $y = 0$, it occurs at both endpoints). Therefore,
$$\sup_{x \in [-1, 1]} (x + y)^2 = \begin{cases} (y + 1)^2 & \text{if } y \ge 0 \\ (y - 1)^2 & \text{if } y < 0. \end{cases}$$
If $y \ge 0$, then $y + 1 \ge 1$, and hence $(y + 1)^2 \ge 1$. If $y < 0$, then $y - 1 < -1$, and hence $(y + 1)^2 > 1$. Either way, $1$ is a lower bound for this function, which is achieved at $y = 0$. Therefore,
$$\inf_{y \in [-1, 1]} \sup_{x \in [-1, 1]} (x + y)^2 = 1 \neq 0.$$

ContentLicense: CC BY-SA 4.0


问:
Trying to understand covering maps and local homeomorphisms
I am struggling mightily with the concept of a "covering map".  I think the root cause of my problem is that I thought I understood what a local homeomorphism is, but now I'm not so sure.  An example that illustrates my confusion is the projection of a circle from a point onto a line segment:

My question is: is the projection of the circle from O onto the segment L1L2 a local homeomorphism?  I say no, because open neighborhoods around C1 and C2 will not map to open neighborhoods on the line segment.  They will map to neighborhoods at the end points of the segment.  So each of those neighborhoods will have one "hard" boundary and will, therefore, not be open.  We can therefore see two points in the circle that are not locally homeomorphic to the segment.  Therefore the projection is not a local homeomorphism.

But as I mentioned, I am completely confused, so I am open to correction.

答:
It is indeed not a local homeomorphism but the reason is not exactly what you are saying.

When you are thinking of this projection, what happens around the point $C_1$ is that, as you say, you hit a boundary of [L1, L2], and then move back, but this is not an issue with openess because in the topology of a closed segment (if you are using the subspace topology from $\mathbb{R}$) the sets $[L1, L1 + \epsilon)$ are open. In other words, your map here is an open map.

The issue is about injectivity. If it is a local homemorphism then around every point there must be a neighborhood in which it is injective, but this doesn't happen around the tangency points from O, i.e. in $C_1$ and $C_2$. In these points, no matter how small open set you put around them, there are always two points, corresponding to a line from $O$ very close to the tangent line, that intersect the neighborhood twice destroying the possible injectivity since these points have the same image in $[L1, L2]$.


问:
Chain rule in DARTS – Differentiable Architecture Search
For https://arxiv.org/pdf/1806.09055.pdf#page=4 , could anyone help to see how equation (7) is the chain rule result of equation (6) ?

答:
In order to show that $\color{green}{\text{equation (7)}}$ is the chain rule result of $\color{red}{\text{equation (6)}}$, let's denote $w' = w-\xi \nabla_w \mathcal{L}_{train}(w, \alpha)$

$$\begin{equation}
\begin{split}
\color{red}{\nabla_\alpha \mathcal{L}_{val}(w-\xi \nabla_w \mathcal{L}_{train}(w, \alpha), \alpha)}
&= \nabla_\alpha \mathcal{L}_{val}(w', \alpha) \\
&= \frac{\partial \mathcal{L}_{val}(w', \alpha)}{\partial \alpha} \\
&= \frac{\partial \mathcal{L}_{val}(w', \alpha)}{\partial w'} \frac{\partial w'}{\partial \alpha} + \frac{\partial \mathcal{L}_{val}(w', \alpha)}{\partial \alpha} \\
&= -\xi \color{blue}{\frac{\partial (\nabla_w \mathcal{L}_{train} (w, \alpha))}{\partial \alpha}} \nabla_{w'} \mathcal{L}_{val} (w', \alpha) + \nabla_\alpha \mathcal{L}_{val}(w', \alpha) \\
\end{split}
\end{equation}$$

$$\begin{equation}
\begin{split}
\color{blue}{\frac{\partial (\nabla_w \mathcal{L}_{train} (w, \alpha))}{\partial \alpha}}
= \frac{\partial \frac{\partial \mathcal{L}_{train} (w, \alpha)}{\partial w}}{\partial \alpha}
= \frac{\partial^2 \mathcal{L}_{train} (w, \alpha)}{\partial \alpha \; \partial w}
= \nabla_\alpha (\nabla_w \mathcal{L}_{train} (w, \alpha) )
= \nabla_{\alpha, w}^2 \mathcal{L}_{train} (w, \alpha)
\end{split}
\end{equation}$$

$$\begin{equation}
\begin{split}
\color{red}{\nabla_\alpha \mathcal{L}_{val}(w-\xi \nabla_w \mathcal{L}_{train}(w, \alpha), \alpha)}
&= -\xi \color{blue}{\frac{\partial (\nabla_w \mathcal{L}_{train} (w, \alpha))}{\partial \alpha}} \nabla_{w'} \mathcal{L}_{val} (w', \alpha) + \nabla_\alpha \mathcal{L}_{val}(w', \alpha) \\
&= -\xi \nabla_{\alpha, w}^2 \mathcal{L}_{train} (w, \alpha) \nabla_{w'} \mathcal{L}_{val} (w', \alpha) + \nabla_\alpha \mathcal{L}_{val}(w', \alpha) \\
&= \color{green}{\nabla_\alpha \mathcal{L}_{val}(w', \alpha) - \xi \nabla_{\alpha, w}^2 \mathcal{L}_{train} (w, \alpha) \nabla_{w'} \mathcal{L}_{val} (w', \alpha)}
\end{split}
\end{equation}$$


问:
Expected value of highest card in 4 card draw from standard deck
Suppose you draw 4 cards from a standard 52 cards deck, what is the expected value of the highest card among the 4 cards? (Ace = 1, ..., King = 13)
An accurate result is not necessary, I am more interested in a fast method to approximate it without using a calculator (if the accurate result is not simple enough).

I thought about (falsely) assuming that, by denoting the values of the cards as $X_1,\dots,X_4$,
$$
P(\max(X_1,\dots,X_4)\leq x) = P(X_1\leq x)^4
$$

Then, for simplicity calculating the maximum of 4 i.i.d. uniforms on $[0,1]$ gives
$$
P(\max(U_1,\dots,U_4)\leq x) = \int_0^1x4x^3dx = \frac{4}{5}
$$
and finally,
$$
12\cdot\frac{4}{5}+1 = 10.6
$$
was my approximation for the max of 4 i.i.d. uniforms on $[1,13]$.

How bad is that?

As the 4 draws are in fact not independent because, loosely speaking, drawing a small card reduces the probability of drawing another small card, we have
$$
P(\max(X_1,\dots,X_4)\leq x) < P(X_1\leq x)^4
$$
and thus the expected value should be a little higher, and of course the draws are not uniform from $[1,13]$, does that make a big difference?

In general, how would you approach these kind of questions, for example finding the expected value of the square of the number of red cards in that 4 card draw?
I somehow always assume that the draws are independent, e.g. approximate the probability for a red card by $1/2$ to make the computation of the expected value as simple as possible and then try to correct the answer in the right direction, but I am sure there are better approaches.

答:
$$P[\max\le k]=\frac{{4 k}\choose{4}}{52\choose4}=\frac{4k(4k-1)(4k-2)(4k-3)}{52(51)(50)(49)}$$
$$E[\max]=\sum_{k=1}^{13}kP[\max=k]=P[max\le 1]+\sum_{k=2}^{13}k(P[max\le k]-P[max\le k-1])$$
$$=\frac{32577}{2975}\approx 10.95$$

For an approximation, I would use the fact that the maximum of 4 independent uniform random variables has a Beta distribution with parameters 4 and 1 see here. The mean of that Beta distribution is 0.8. These are not Uniform(0,1) and not independent, but they are Unif(0.5,13.5) rounded to the nearest integer. Thus, if I took Unif(0,1) random variables, multiplied them by 13 and added 0.5, I would have Unif(0.5, 13.5) random variables. The maximum of those would have mean approximately 0.8*13+0.5=10.9. This approximation works very well in this case.


问:
Continuity of Linear Operator
Let $V$, $W$ be normed vector spaces and $A: V \to W$ a linear operator. Show that the following statements are equivalent:

a) $A$ is continuous in $0$,

b) sup$||x||v \leq 1$  ||Ax||W < ∞,

c) $A$ is bounded.

For a) I know that I have to show a neighborhood of $A$ at $0$  shows that $Ax \subset W$ for some neighborhood $x$ of $0$ (in $V$).

b) I am not quite sure how to do this.

c) To prove this I can take any unit set(ball) in V and show that the mapping of points of this set is bounded in W by some $M>0$.

Can someone please help me write these proofs in precise Mathematical format? I don't have enough experience writing proofs.
Thanks

答:
$$ a \implies b$$

$(a)$ means that

$$\forall \epsilon>0,\, \exists \delta > 0,\, \forall x\in V \text{ with } {\lVert x\rVert}_V < \delta, {\lVert Ax\rVert}_W < \epsilon.\tag{1}$$

Now, suppose $(b)$ did not hold but $(a)$ did.
We will show that this is a contradiction.

Since $(b)$ does not hold, we can find a sequence $(x_n)_{n\in\Bbb N}$ in $V$ with ${\lVert x_n\rVert}_V = 1$ for all $n$ and ${\lVert Ax_n\rVert}_W > n$.

Pick any $\epsilon > 0$ and let $\delta > 0$ be the one from $(1)$.
Consider $(y_n)_{n\in\Bbb N}$ given by $y_n = \frac \delta{2}x_n$.
Notice that ${\lVert y_n\rVert}_V = \delta/2 < \delta$ for all $n$, so by $(1)$ we should have ${\lVert Ay_n\rVert}_W < \epsilon$ for all $n$.

On the other hand, by linearity and the construction of $(x_n)_{n\in\Bbb N}$, we should have ${\lVert Ay_n\rVert}_W > \frac{\delta n}2$, and for $n$ large enough this is $>\epsilon$, a contradiction.
It follows that $(b)$ must hold whenever $(a)$ does.

$$ b \implies c$$

We need to show that $A$ maps bounded sets to bounded sets.
Letting $s = \sup_{{\lVert x\rVert}_V \leqslant 1} {\lVert Ax\rVert}_W$ $($and note that $(b)$ implies $s< \infty)$, it follows that $A\left(\overline{B_V(0; 1)}\right)\subset \overline{B_W(0; s)}$.

Now, any bounded set  $U\subset V$ is contained in some closed ball $\overline{B_V(0; r)}$ of finite radius $r$.
It's easy to see from the linearity of $A$ that $A(U) \subset \overline{B_W(0; rs)}$.

$$ c \implies a$$

Suppose $(c)$ holds but $(a)$ doesnt.
We will show that this is a contradiction.
The negation of $(a)$, that is, of $(1)$, is

$$\exists \epsilon>0,\, \forall \delta > 0,\, \exists x\in V \text{ with } {\lVert x\rVert}_V < \delta, {\lVert Ax\rVert}_W \geqslant \epsilon.\tag{2}$$

Let $\epsilon>0$ be the value from $(2)$ and consider a sequence of deltas $\delta_n = \frac1n$ for all $n\geqslant 1$.
With $(2)$, this yields a sequence $(x_n)_{n\geqslant 1}$ in $V$ with ${\lVert x_n\rVert}_V < \frac1n$ and ${\lVert Ax_n\rVert}_W \geqslant \epsilon$.

Now, consider $y_n = nx_n$.
Then, for all $n\geqslant 1$, we have ${\lVert y_n\rVert}_V < 1$ and ${\lVert Ay_n\rVert}_W \geqslant n\epsilon$.
In other words, $U = \{y_n\in V:n\geqslant 1\}\subset V$ is bounded, but $A(U)$ isn't, which contradicts $(c)$.

It follows that $(a)$ must hold whenever $(c)$ does, and concludes the proof.

Title: Continuity of Linear Operator


问:
Isomorphic of tensor products
I am working on this problem and need a little guidance. I have 2 $k$-algebras $$\frac{k[x_1, \ldots, x_n]}{I} \quad \text{ and } \quad \frac{k[y_1, \ldots, y_m]}{J}$$
where $k$ is a field and $I$ is an ideal of $k[x_1, \ldots, x_n]$ and $J$ is an ideal of $k[y_1, \ldots, y_m]$

My goal here is to show that $$\frac{k[x_1, \ldots, x_n]}{I} \otimes_k \frac{k[y_1, \ldots, y_m]}{J} \cong_k \frac{k[x_1, \ldots, x_n, y_1, \ldots, y_m]}{(I,J)}.$$

My plan is to define a mapping: $$f: \frac{k[x_1, \ldots, x_n]}{I} \times \frac{k[y_1, \ldots, y_m]}{J} \rightarrow \frac{k[x_1, \ldots, x_n, y_1, \ldots, y_m]}{(I,J)}$$
If I can define a mapping that is bi-additive and k-balanced, then I can use the universal mapping property of tensor products to have a mapping $$\tilde{f} : \frac{k[x_1, \ldots, x_n]}{I} \otimes_k \frac{k[y_1, \ldots, y_m]}{J} \rightarrow \frac{k[x_1, \ldots, x_n, y_1, \ldots, y_m]}{(I,J)}.$$ I would still have to make the argument that $\tilde{f}$ is an isomorphism of $k$-algebras however.

Can anyone help me in defining $f$ explicitly.

答:
See if you can use this in the definition of $\tilde{f}$ to explicitly define a map?

Edit: This can also be seen as a very nice application of Yoneda's lemma. You can find a related post here

Title: Isomorphic of tensor products


问:
The sum of a homomorphism that sends $M$ to $\mathfrak aN$ and a surjection is an isomorphim between finitely generated modules $M$ and $N$

$R$ is a commutative ring, $M$ and $N$ finitely generated $R$-modules, $\alpha, \beta\in \operatorname{Hom}_{R}(M,N)$, $\mathfrak a\subset \operatorname{rad}(R)$ and $\alpha $ is surjective while $\beta(M) \subseteq\mathfrak aN$, prove that $\gamma=\alpha + \beta $ is an isomorphism between $M$ and $N$.

I understand why it is a surjection according to Nakayama's lemma but I don't know how to prove injectivity since $M$ might be way bigger than $N$ with regard to their generating sets.

答: For $\beta=0$, the statement is that every surjective homomorphism between f.g. modules is an isomorphism. Of course, this is wrong. However, every surjective endomorphism of a f.g. module is an isomorphism, this is an application of the general form of Nakayama's lemma applied to the corresponding $R[x]$-module where $x$ acs as the endomorphism (details). This shows that the claim is true for $M=N$, and hence also when $M \cong N$.</p>

ContentLicense: CC BY-SA 3.0


问:
An Einstein summation question
Let $\eta_{ab}$ be the Minkowski metric and $T_{ab}=T_{[ab]}+T_{(ab)}$, where $T_{(ab)}=\frac{1}{2}(T_{ab}+T_{ba})$ and $T_{[ab]}=\frac{1}{2}(T_{ab}-T_{ba})$. Is it true that $\eta_{ab}T_{ab}=\eta_{ab}T_{(ab)}$, assuming $a,b$ range over the same integer values? I think so, since assuming the latter, $\eta_{ab}T_{ab}=\eta_{ab}T_{ba}$, so $\eta_{ab}T_{(ab)}=\eta_{ab}\frac{1}{2}(T_{ab}+T_{ba})=\frac{1}{2}\eta_{ab}T_{ab}+\frac{1}{2}\eta_{ab}T_{ba}=\frac{1}{2}\eta_{ab}T_{ab}+\frac{1}{2}\eta_{ab}T_{ab}=\eta_{ab}T_{ab}$. Is this correct? Thanks in advance!

答: Yes, because $\eta_{ab}=\eta_{ba}\implies\eta_{ab}T_{ba}=\eta_{ba}T_{ba}=\eta_{ab}T_{ab}$, where the last $=$ relabels indices.

Title: An Einstein summation question


问:
Finding the value of $\sqrt{z \sqrt{z \sqrt{z}}}...$
I was working on the following nested square root problem:

Let $a \in \mathbb R ^+$, what is the value of: $$\sqrt{a \sqrt{a \sqrt{a}}}...$$

I concluded that the answer is $a$ and then I thought about generalizing this to the complex numbers. The question became:

Let $z \in \mathbb C$, what is the value of: $$\sqrt{z \sqrt{z \sqrt{z}}}...$$

I did the following:

Let $x \in \mathbb C$ with $x =\sqrt{z \sqrt{z \sqrt{z}}}... $. Then:
$$\sqrt{z \sqrt{z \sqrt{z}}}... = \sqrt{z} \sqrt{\underbrace{\sqrt{z \sqrt{z \sqrt{z}}}...}_x}$$

So we concude that $x = \sqrt{z} \sqrt{x}$ or: $$x^2 = |z| |x| \ \ (1)$$.

Let $x = r e^{i \phi}$ and $ z = p e^{i \alpha}$. Then, eq. (1) becomes:

$$r^2e^{i 2 \phi} = pr$$

$$r = 0\vee r = p e^{- i 2 \phi}$$

We know that $r \neq 0$ so we have that $r = p e^{- i 2 \phi}$. because $x = r e^{i \phi}$ if we substitute this we get:

$$x = p e^{- i \phi}$$

So this concludes that:

$$\left| \sqrt{z \sqrt{z \sqrt{z}}}... \right| = |z|$$

My question is: I was able to find the absolute value of the solution, now to fully have a solution for this problem I need the argument of it. How can find the argument of the solution?

答:
Over the entire complex field, it's not possible to consistently (continuously) define $\sqrt .$ everywhere. It's the same problem as defining a logarithm : you have to cut somewhere.

So your question won't have a definite answer. My guess is : for every phase $\varphi$, there is a definition of $\sqrt .$ such that your answer is $|z|e^{i\varphi}$.

The idea is that the series $\sum \frac {x_i} 2$ for $(x_i) \in \{0,1\}^{\mathbb N}$ can converge to anything in $[0,1]$ so by defining the cut to realise the right series $(x_i)$ we can add any phase we want.


问:
Sequence/List Formula
Consider the following sequence $\{1,1,2,1,2,3,4,1,2,3,4,5,6,7,8,1,...,16,1,...,32,1...\}$; it is the list of integers $1$ to $2^0$, then $1$ to $2^1$, then $1$ to $2^2$, etc. Let $f(n)$ return the $n$th entry in the sequence.

What is a closed form for $f(n)$? Currently I've just been building the sequence as a list and using the $n$th entry... but an analytical expression would be quite helpful. So far what I have is:
\begin{equation}
f(n)=j\quad \text{ if }\quad n=2^{m+\lceil \log_2 j\rceil}+j-1
\end{equation}
where $m$ is a positive integer or $0$. It's a little small but we're taking the ceiling of $\log_2 j$ in the above equation. This question has been bothering me for a while because it seems like I should've gotten this immediately... but I couldn't do it!

答:
$$f(n)=n-2^{\lfloor log_2 n \rfloor}+1$$

where the sequence starts $a_1=1, a_2=1,a_3=2, a_4=1, \dots$

 Add $\{0,1,1,3,3,3,3,7,7,7,\dots\}$ to your existing sequence.

Title: Sequence/List Formula


问:
Is it possible to solve a recurrence in which the biggest factor is negative (thus alternating)?
Apologies if the question is ambiguous; English is not my native language. My partial solution to the problem may convey the issue more accurately: the recurrence to solve is

$$g(n) - g(n-1) - 2g(n-2) + 2g(n-3) = 0.$$

Its polynomial is thus

$$x^3 - x^2 - 2x + 2$$

and the roots I obtain are

$$(x - 1)(x - \sqrt{2})(x + \sqrt{2}).$$

With the initial conditions being

$$g(0) = 106$$
$$g(1) = 100$$
$$g(2) = 112$$

the general form is therefore

$$g(n) = 3\sqrt{2}^n + 3(-\sqrt{2})^n + 100.$$

The issue arises when I try to prove the exact order of the recurrence ($\Theta(\sqrt{2}^n)$) using the limit test. My attempt goes as such:

$$
\begin{align*}
\lim_{n\to\infty} \frac{3\sqrt{2}^n + 3(-\sqrt{2})^n + 100}{\sqrt{2}^n} & = \lim_{n\to\infty} \frac{3\sqrt{2}^n}{\sqrt{2}^n} + \lim_{n\to\infty} \frac{3(-\sqrt{2})^n}{\sqrt{2}^n} + \lim_{n\to\infty} \frac{100}{\sqrt{2}^n}\\
& = 3 + \lim_{n\to\infty} \frac{3(-\sqrt{2})^n}{\sqrt{2}^n} + 0\\
& = 3 + \lim_{n\to\infty} \frac{3(-\sqrt{2})^n}{\sqrt{2}^n}
\end{align*}
$$

The remaining limit diverges and I am confused at this point. I expect the explanation will be something relatively simple in retrospect and also that it will be more closely related to the notion of limits than anything else, but I nonetheless phrased my question in terms of recurrence because my uncertainty partially stems from the fact that I am now unsure whether it is possible or not to give a Big-Theta for every recurrence out there.

Thank you very much for your time and hopefully your help as well. :)

答:
The function $g$ isn’t $\Theta(f(n))$ for any $f$. This is especially clear when you write the function like this:

$$g(n)=\begin{cases}
6\cdot 2^{n/2},&\text{if }n\text{ is even}\\
100,&\text{if }n\text{ is odd:}
\end{cases}$$

this makes it clear that $g(n)$ is $O(\sqrt2^n)$ and $\Omega(1)$, and that neither estimate can be improved.

In fact if $f(n)=g(2n)$ and $h(n)=g(2n+1)$, then $f(n)$ is $\Theta(2^n)$ and $h(n)$ is $\Theta(1)$. The function $g$ restricted to even arguments grows at an entirely different rate from the function $g$ restricted to odd arguments.


问:
Solving $n\log{n}=10^9$
The logarithm here is of base $2$. I tried to solve the following equation to find out $n$, where $n \in \mathbb{R}$,

$$n\log{n}=10^9$$

Edited: My naive approach was to raise both sides to exponential $2$,

$$2^{n\log{n}}=2^{10^9}$$
$$2^{\log{n^n}}=2^{10^9}$$
$$n^n=2^{10^9}$$

I got stuck here. I saw another solution suggesting to use Lambert $W$ function,

$$n=\frac{10^9\times \log{2}}{W(10^9\times \log{2})}$$

Question 1: Can you please tell me if it's possible to solve it without Lambert function as I did? But I got stuck at $n^n=2^{10^9}$, so how to proceed from here?

Question 2: How the solution that used Lambert $W$ function get to that factor $n=\frac{10^9\times \log{2}}{W(10^9\times \log{2})}$?

答:
You could try fixed point iteration and the error of $s_n$ to $10^9$ is tiny after just a few iterations:
$$
x_1=2\\
x_{n+1}\gets \log_2(10^9/x_n)\\
s_n=10^9 /x_n\\
s_\infty\log_2(s_\infty)=10^9
$$

```<code>8.66671*10^8, 1.00824*10^9, 9.99531*10^8, 1.00003*10^9, 9.99998*10^8
</code>```

This works because the Lambert function can be written this way if $|W(x)|>e$:
$$
W(x)=\ln \frac{x}{\ln \frac{x}{\ddots}}
$$
...so we're still technically using Lambert but the iterative version might be useful if you don't have a Lambert function on your calculator or in a mathematics software library.

Title: Solving $n\log{n}=10^9$

Body:
You could try fixed point iteration and the error of $s_n$ to $10^9$ is tiny after just a few iterations:
$$
x_1=2\\
x_{n+1}\gets \log_2(10^9/x_n)\\
s_n=10^9 /x_n\\
s_\infty\log_2(s_\infty)=10^9
$$

8.66671*10^8, 1.00824*10^9, 9.99531*10^8, 1.00003*10^9, 9.99998*10^8

This works because the Lambert function can be written this way if $|W(x)|>e$:
$$
W(x)=\ln \frac{x}{\ln \frac{x}{\ddots}}
$$
...so we're still technically using Lambert but the iterative version might be useful if you don't have a Lambert function on your calculator or in a mathematics software library.


问:
Null Hypothesis: Is it being rejected?
I was doing this particular question that is as following:

A drugs manufacturer claims that the amount of paracetamol in tablets is 60 mg. A sample of 10 tablets is taken and the amount of paracetamol in each is recorded:

59.1,  59.7,  61.0,  59.1,60.6,68.9,60.2,58.6,58.7,58.9

Assuming that this sample came from an underlying normal population, test the claim at 5% significance level that the amount of paracetamol is different from 60 mg.

I got my sample mean as 1512/25 and my unbiased variance as 7073/750

I set my null hypothesis as μ= 60 and my alternative hypothesis as μ ≠ 60 making it two tailed test

After doing the calculations, I got my test statistic as 0.494 and according to t-test, the tabulated value should be 2.262 when t(0.975,9).

Since 0.494 doesn't come in the rejection region, shouldn't we accept null hypothesis? But the answer says it is rejected. What am I doing wrong?

答:
Below your data are imported into R, and a two-sided, one-sample t test is done in R. Your test statistic and conclusion not to reject are correct:

```<code>x = c(59.1, 59.7, 61.0, 59.1,60.6,68.9,60.2,58.6,58.7,58.9)
t.test(x, mu=60)

        One Sample t-test

data:  x
t = 0.49428, df = 9, p-value = 0.633
alternative hypothesis: true mean is not equal to 60
95 percent confidence interval:
  58.28318 62.67682
sample estimates:
mean of x
    60.48
</code>```

If your book says you should reject, then either the data
are not correct or this is one case (out of many) where an answer key is wrong.

Title: Null Hypothesis: Is it being rejected?

Body:
Below your data are imported into R, and a two-sided, one-sample t test is done in R. Your test statistic and conclusion not to reject are correct:

x = c(59.1, 59.7, 61.0, 59.1,60.6,68.9,60.2,58.6,58.7,58.9)
t.test(x, mu=60)

        One Sample t-test

data:  x
t = 0.49428, df = 9, p-value = 0.633
alternative hypothesis: true mean is not equal to 60
95 percent confidence interval:
  58.28318 62.67682
sample estimates:
mean of x
    60.48

If your book says you should reject, then either the data
are not correct or this is one case (out of many) where an answer key is wrong.


问:
Formulating a complicated if else statement with linear programming
I'm trying to convert the following statement into «only one» inequality. Is it possible?

If $M[i] < M[j]$ and $M[j] < M[k] $ and $A[i,k] = 1$ then $A[i,j]=1$

$M$ is an array of size N and its elements are in the range of 1..N. $A$ is a 2d array of size $N*N$ and its elements are either 0 or 1.
I've seen a lot of examples of converting problems into lp problems but none of them were this complicated. I've thought of some solutions but they don't seem to work. Any help would be really appreciated.

答:
You can enforce the desired implication via the following linear constraint:
$$\text{$A[i,k] \le A[i,j]$ for all $i,j,k$ such that $M[i] < M[j]$ and $M[j] < M[k]$}$$


问:
Is function $f:\{1,2\}\rightarrow\{1\}$ continuous?
Under "regular" topology in $\mathbb{R}$, is function $f:\{1,2\}\rightarrow\{1\}$ continuous? Here $f$ is a function defined on $\{1,2\}$ and $f(1)=f(2)=1$.

I think it is. According to the definition of continuity, the pre-image of any open set is open. The only open set in this case is $\phi$, the empty set, and the pre-image is also $\phi$. Thus, $f$ is open. Please correct me if I am wrong.

答: You missed the open set $\{1\}$ whose pre-image is the entire domain which is open. So the function is indeed continuous.


问:
Is it always possible to construct an extension of $\mathbb{Q}_p$ as a totally ramified extension first and then an unramified extension after?
Let $K = \mathbb{Q}_p$, and $L/K$ be a finite Galois extension.

Question: Is it possible to find a totally ramified extension $L'/K$ such that $L/L'$ is unramified?

I know that it is always possible to find an unramified $L'/K$ such that $L/L'$ is totally ramified, and I was wondering whether the converse is true or not.

I guess it is, and the following example demonstrates how it may be generalized. If $K= \Bbb{Q}_3$ and $L=K(\alpha)$ with $\min_K(\alpha) = x^4-3x^2 +18$, then $L/K$ has degree $4$ and ramification index $2$. The field $L$ can be written as $L=K'(\alpha)=K(\alpha,\zeta_4)$ where $K' = K(\zeta_4)$ is the maximal unramified subextension of $L/K$.

Now I could choose an automorphism $\sigma \in \operatorname{Gal}(L/K)$ such that $\sigma(\zeta_4) = \zeta_4^3$ (i.e. $\sigma$ is a lift of the Frobenius), and choose $L' = L^{\langle \sigma \rangle}$. By construction, the residue field of $L'$ is $\mathbb{F}_3$, the same residue field as of $K$, so $L'/K$ is totally ramified. This also implies that $L/L'$ is unramified.

I guess one could always construct such a subextension that way but I am not sure if it can be always generalized like that.

答:
The answer is no, even for Galois extensions. The standard example is as follows. Suppose that $L/K$ is Galois with $\mathrm{Gal}(L/K) = \mathbf{Z}/4 \mathbf{Z}$ is cyclic of order $4$ but the inertia group is cyclic of order $2$, so $f = e = 2$. By Galois theory, there is only one (proper) intermediate subfield $L/E/K$, and it will be unramified over $K$, so $L/E$ will be ramified. You can construct examples of such extensions quite explicitly. Suppose that $p \equiv 1 \bmod 4$ so that $\mathbf{Q}_p$ contains a $4$th root of unity $i$. Now let $u \in \mathbf{Z}_p$ be any element which is a quadratic non-residue. Then take $K = \mathbf{Q}_p$ and $L = \mathbf{Q}_p((up^2)^{1/4})$.

For Galois extensions, with Galois group $G$, inertia group $I$, and cyclic quotient $G/I = \langle \phi \rangle$, the general condition is that there exists a splitting $\langle \phi \rangle \rightarrow G$, that is, there is a lift of $\phi$ to $G$ which has the same order. Then you can take $E$ to be the fixed field of that lift $\phi \subset G$. You can see the example above is non-split.


问:
Finding all primes of the form $n^{5}-1$
I need to find all primes of the form: $n^{5}-1$.

My attempt: First I have denoted $q=n^5-1$. It appears that $n$ is even, otherwise, if $n$ is odd then $n^5$ is odd too, which means that $n^5-1$ is an even number bigger than 2. Therefore, suppose that
$n=2k,\mathbb{N}\ni k >0$. Thus, we have that: $n^5-1=(2k)^5-1=32k^5-1$. If $k=1$, then we have $32-1=31$ and we get a prime number. Now I know that I can proceed and get all 32's multiples, but I don't see how to negate that there are no more prime of the given form.
Thanks!

答:
Hint: Expand the product
$$(n-1)(n^4+n^3+n^2+n+1).$$

In general, when asked to find all primes of the form $P(n)$ for some polynomial with integer coefficients, it is very likely that there are only finitely many, and that they can be found by factoring the polynomial. It turns out that showing that there exist infinitely many primes of the form $P(n)$ is usually extremely difficult if $\deg P>1$, so you are unlikely to be given such an exercise.


问:
An inflection point where the second derivative doesn't exist?
A point $x=c$ is an inflection point if the function is continuous at that point and the concavity of the graph changes at that point.
And a list of possible inflection points will be those points where the second derivative is zero or doesn't exist.
But if continuity is required in order for a point to be an inflection point, how can we consider points where the second derivative doesn't exist as inflection points?

Also, an inflection point is like a critical point except it isn't an extremum, correct?
So why do we consider points where the second derivative doesn't exist as inflection points?

thanks.

答:
A function can be continuous but fail to have a second derivative. For example, consider $$f(x)=\cases{ -x^2 & $x\le 0$ \\ x^2 & $x>0$ }$$
with second derivative
$$f''(x)=\cases{ -2 & $x< 0$ \\ \text{undefined} & $x=0$ \\ 2 & $x>0$ }$$

The statement you give says only that you need to check points without a second derivative or where it's zero. There are examples where

the second derivative doesn't exist like $$f(x)=\cases{ x^2 & $x\le 0$ \\ 2x^2 & $x>0$ }$$

the second derivative does exist and is zero like $f(x)=x^4$

but the function does not have an inflection point.


问:
Where does Euclidean and non-Euclidean geometry lead to?
I'm thinking of whether I should self-study Euclidean, elliptic, and hyperbolic geometry but I don't know where these subjects lead to.

Are they prerequisites to algebraic geometry, differential geometry, projective geometry, etc.? How are they related to other subjects?

答:
The short answer is that in mathematics there are often profound (and surprising and gratifying) connections between different areas.  You might study topic X for its own intrinsic attractions, or you might get into it only because you're fascinated with another topic or problem Y that has connections to field X.

Beyond that, I'd suggest web searches on the connections between topics.  For example, the search "hyperbolic geometry and number theory" turned up lots of results, including

What are the interesting applications of hyperbolic geometry?

The Oracle of Arithmetic: At 28, Peter Scholze is uncovering deep connections between number theory and geometry.

From the latter:

In the middle of the 20th century, mathematicians discovered an
astonishing link between reciprocity laws and what seemed like an
entirely different subject: the “hyperbolic” geometry of patterns such
as M.C. Escher’s famous angel-devil tilings of a disk. This link
is a core part of the “Langlands program,” a collection of
interconnected conjectures and theorems about the relationship between
number theory, geometry and analysis. When these conjectures can be
proved, they are often enormously powerful: For instance, the proof of
Fermat’s Last Theorem boiled down to solving one small (but highly
nontrivial) section of the Langlands program.

Mathematicians have gradually become aware that the Langlands program
extends far beyond the hyperbolic disk; it can also be studied in
higher-dimensional hyperbolic spaces and a variety of other contexts.
Now, Scholze has shown how to extend the Langlands program to a wide
range of structures in “hyperbolic three-space” — a three-dimensional
analogue of the hyperbolic disk — and beyond. By constructing a
perfectoid version of hyperbolic three-space, Scholze has discovered
an entirely new suite of reciprocity laws.

and, speaking to whether you learn topic X first, or get led to it by topic/problem Y:

“I understood nothing, but it was really fascinating,” he said.

So Scholze worked backward, figuring out what he needed to learn to
make sense of the proof. “To this day, that’s to a large extent how I
learn,” he said. “I never really learned the basic things like linear
algebra, actually — I only assimilated it through learning some other
stuff.”

So by all means start studying geometry.  If you don't like it, don't worry:

you may never need it

the textbook you're reading is not the right one for you

the time when you need it for topic/problem Y may be a better time than right now.

On the other hand, learning it now may make you better prepared to tackle topic Y when the time comes.

Some other interesting reads:

Series, The Geometry of Markoff Numbers

Stilwell, Geometry of Surfaces


问:
Can a series $\sum_{n=1}^{\infty} (\frac{1}2)^n=1$ be infinitely substituted into itself?
Example

$$\Large\sum_{n=1}^{\infty} \left(\frac{\sum_{n=1}^{\infty} \left(\frac{\sum_{n=1}^{\infty} \left(\frac{\sum_{n=1}^{\infty} \left(\frac{\sum_{n=1}^{\infty} \left(\frac{...}2\right)^n}2\right)^n}2\right)^n}2\right)^n}2\right)^n$$

Motivation

Recently did a screen share on gather.town with itself and was greeted with this image below. The idea popped into my head, "What if this infinite mirror concept was applied to an infinite series, would it be possible to create a series that has a known result but is not calculable?"

答:
What you end up with is the relation

$$x = \sum_{n=1}^\infty \left(\frac{x}{2}\right)^n.$$

If we assume $x$ to be complex we must have $|x| < 2$ or the RHS diverges. Then, summing the geometric series

$$x = -1 + \sum_{n=0}^\infty \left(\frac{x}{2}\right)^n$$
$$x = -1 + \frac{1}{1-\frac{x}{2}}$$
$$-\frac{1}{2}x^2 +\frac{1}{2}x = 0$$

gives $x = 0 \vee x = 1$ as the only solutions.


问:
Suppose that $X$ and $Y$ are well-ordered sets. Then $X + Y$ and $X \times Y$ are both well-ordered.
I have proved the first part in the title. I'd like to know if my reasoning is sound. If it is, then a similar argument for $X \times Y$ can be proved similarly.

Proof. We have
\begin{equation*}
X + Y = \left(X \times \left\{\boldsymbol{0}\right\}\right) \cup \left(Y \times \left\{\boldsymbol{1}\right\}\right).
\end{equation*}
Suppose $S \subseteq X + Y$ and $S \neq \emptyset$. Define the following set:
\begin{equation*}
A = \left\{ \boldsymbol{n} \in \mathbb{N}: \textrm{ there exists } t \in X \cup Y \textrm{ such that } \left(t,\boldsymbol{n}\right) \in S \right\}.
\end{equation*}
From $S \neq \emptyset$, it is clear that $A \subseteq \left\{\boldsymbol{0}, \boldsymbol{1}\right\}$ and $A \neq \emptyset$. As $\mathbb{N}$ is a well-ordered set, and $A \subseteq \mathbb{N}$, there is some $a_{0} \in A$ such that
\begin{equation*}
a_{0} = \min{A}.
\end{equation*}
It is obvious that $a_{0} = \boldsymbol{0}$ or $a_{0} = \boldsymbol{1}$.

If $a_{0} = \boldsymbol{0}$, define the following set:
\begin{equation*}
B_{0} = \left\{x \in X: \left(x,\boldsymbol{0}\right) \in S\right\}.
\end{equation*}
As $A \neq \emptyset$, we have $B_{0} \neq \emptyset$. As $B \subseteq X$ and $X$ is well-ordered, there is some $x_{0} \in X$ such that
\begin{equation*}
x_{0} = \min{B_{0}}.
\end{equation*}
We argue that $\left(x_{0},\boldsymbol{0}\right) = \min{S}$. If it is not, then there is some $x_{1} \in X$ such that $\left(x_{1},\boldsymbol{0}\right) \in S$ and $\left(x_{1},\boldsymbol{0}\right) < \left(x_{0},\boldsymbol{0}\right)$. This indicates that $x_{1} \in B_{0}$ and $x_{1} <_{X} x_{0}$, which contradicts that $x_{0} = \min{B}$. Thus, we have $\left(x_{0},\boldsymbol{0}\right) = \min{S}$.

If $a_{0} = \boldsymbol{1}$, then there is no $x \in X$ such that $\left(x,\boldsymbol{0}\right) \in S$ (otherwise, $a_{0} = \boldsymbol{0}$). Define the following set:
\begin{equation*}
B_{1} = \left\{y \in Y: \left(y,\boldsymbol{1}\right) \in S\right\}.
\end{equation*}
Again, $B_{1} \neq \emptyset$. As $Y$ is well-ordered and $B_{1} \subseteq Y$, there exists some $y_{0} \in Y$ such that
\begin{equation*}
y_{0} = \min{B_{1}}.
\end{equation*}
We argue that $\left(y_{0},\boldsymbol{1}\right) = \min{S}$. If not, then there exists $y_{1} \in Y$ such that $\left(y_{1},\boldsymbol{1}\right) \in S$ and $\left(y_{1},\boldsymbol{1}\right) < \left(y_{0},\boldsymbol{1}\right)$. However, this indicates that $y_{1} \in B_{1}$ and $y_{1} <_{Y} y_{0}$, contradicting that $y_{0} = \min{B_{1}}$. Thus, $\left(y_{0},\boldsymbol{1}\right) = \min{S}$.

As a result, $S$ is indeed of a minimum element. Due to the arbitrariness of $S$, we may conclude that $X + Y$ is well-ordered.

答:
Your argument looks OK, but I’d put it more succinctly: Let $X’ = X \times \{0\} \subseteq X+Y$ and $Y’ = Y \times \{1\} \subseteq X + Y$, and denote by $\pi$ the obvious maps from $X’$ onto $X$ and similarly for $Y’$.

If then $S$ is a non-empty subset of $X+Y$, there are two cases:

$S \cap X’$ is non-empty and let $x_0 = \min_X \pi[X’ \cap S]$ and check that $(x_0,0) = \min S$ in $X+Y$.

Not 1 implies $S \subseteq Y’$ and then let $y_0 = \min_Y \pi[ S]$ and check that $(y_0,1) = \min S$ in $X+Y$.

In either case $S$ has a minimum in $X+Y$ and we’re done.


答:
We have
$$η_{ab}T_{ab}=η_{ab}\left(T_{[ab]}+T_{(ab)}\right)$$
$$=η_{ab}T_{[ab]}+η_{ab}T_{(ab)}$$
$$=0+η_{ab}T_{(ab)}$$
$$=η_{ab}T_{(ab)}$$

because the inner product of a symmetric and an antisymmetric tensor is zero.

Title: An Einstein summation question

Body: Let $\eta_{ab}$ be the Minkowski metric and $T_{ab}=T_{[ab]}+T_{(ab)}$, where $T_{(ab)}=\frac{1}{2}(T_{ab}+T_{ba})$ and $T_{[ab]}=\frac{1}{2}(T_{ab}-T_{ba})$. Is it true that $\eta_{ab}T_{ab}=\eta_{ab}T_{(ab)}$, assuming $a,b$ range over the same integer values? I think so, since assuming the latter, $\eta_{ab}T_{ab}=\eta_{ab}T_{ba}$, so $\eta_{ab}T_{(ab)}=\eta_{ab}\frac{1}{2}(T_{ab}+T_{ba})=\frac{1}{2}\eta_{ab}T_{ab}+\frac{1}{2}\eta_{ab}T_{ba}=\frac{1}{2}\eta_{ab}T_{ab}+\frac{1}{2}\eta_{ab}T_{ab}=\eta_{ab}T_{ab}$. Is this correct? Thanks in advance!


问:
Is $e^{\mathbb{E}\left[X\right]}$ equal to $\mathbb{E}\left[e^X\right]$?
Could anyone please tell me if $e^{\mathbb{E}\left[X\right]}$ is equal to $\mathbb{E}\left[e^X\right]$?

I have:

$\prod^n_{i=1} e^{\mathbb{E}\left[X_i\right]}$

and I'm wondering if I can rearrange this to:

$\prod^n_{i=1} e^{\mathbb{E}\left[X_i\right]} = \prod^n_{i=1} \mathbb{E}\left[e^{X_i}\right] = \mathbb{E}\left[\prod^n_{i=1} e^{X_i}\right]$

答:
No, expectation is a linear operator but it does not have such a property as that. We can do: Jensen's inequality says if f is convex, then $f(E(X))\le E(f(X))$. In fact because $e^x$ is convex, we have:

$$ e^{E(X)}\le E(e^X)$$

Thus

$$\prod e^{E(X)}\le \prod E(e^X)$$


问:
find matrix such that $ Ax=(1,1,1)^t$ has exactly three distinct solutions
Does there exist a matrix $3\times 3$ order such that $ Ax=(1,1,1)^t$ has exactly three distinct solutions? If so, find $A$.

I have no idea in this question please help.

答:
With a very liberal interpretation of the question, let $A$ be the matrix over $\mathbb{F}_3$, such that
$$A = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
2 & 2 & 0
\end{bmatrix}.$$
The solution of $Ax = [ 1,1,1 ]^T$ is $x=
[1,1,t]^T$, where $t\in\mathbb{F}_3$ is arbitrary. In other words, the equation has exactly three solutions.

Over other fields, the answer is no as already stated.


问:
Probability of a specific binary string sub-sequence occurring
I've been wracking my brains for a while to try to come up with a non-brute-force solution for this problem.

If you have have a random binary sequence of length $N$, what is the probability that some sub-sequence $X$ is present? My first thought was that it only depends on $N$ and the length of $X$, but from experimentation it's clear that it depends on the contents of $X$.

It seems similar to the problem where you want to know the probability of at least $A$ heads and $B$ tails when flipping a coin $C$ times, but I'm not sure how to then impose the restriction of a specific sequence (Not that I know how to solve that problem either).

I've computed a few results using brute force:

$N=5$ , $X = "111"$, $P(N,X) = 0.25$

$N=5$, $X = "101"$, $P(N,X) = 0.34375$

$N=5$, $X = "100"$, $P(N,X) = 0.375$

答:
This is expanding on the solution from @hgmath

For the sequence "100", define the generating function $A(x)=\sum_{n=3}^{\infty}a_n x^{n-3}$. Similarly, define $B(x)=\sum_{n=3}^{\infty}b_n x^{n-3}$ and $C(x)=\sum_{n=3}^{\infty}c_n x^{n-3}$.

Then, these can be solved like this in Mathematica (easy to solve by hand, too):

```<code>Solve[{
   A - a3 == x (A + B + C),
   B - b3 == x A,
   C - c3 == x C},
  {A, B, C}] /. {a3 -> 4, b3 -> 2, c3 -> 1}

Simplify[(A + B + C) /. (%[[1]])]
</code>```

The solution is $A+B+C=\frac{-4 x^2-2 x+1}{x^3-2 x+1}$.

Then, you can find the probabilities by differentiating the generating function.

```<code>P[n_] := (2^n - (D[(7 - 2 x - 4 x^2)/(1 - 2 x + x^3), {x, n - 3}] /. x -> 0)/(n - 3)!)/2^n
P[5]
</code>```

P[5] is 3/8 for example.

Unfortunately, there will be a different generating function for every given binary sequence, but the general approach works for any of them.

Body:
This is expanding on the solution from @hgmath

For the sequence "100", define the generating function $A(x)=\sum_{n=3}^{\infty}a_n x^{n-3}$. Similarly, define $B(x)=\sum_{n=3}^{\infty}b_n x^{n-3}$ and $C(x)=\sum_{n=3}^{\infty}c_n x^{n-3}$.

Then, these can be solved like this in Mathematica (easy to solve by hand, too):

Solve[{
   A - a3 == x (A + B + C),
   B - b3 == x A,
   C - c3 == x C},
  {A, B, C}] /. {a3 -> 4, b3 -> 2, c3 -> 1}

Simplify[(A + B + C) /. (%[[1]])]

The solution is $A+B+C=\frac{-4 x^2-2 x+1}{x^3-2 x+1}$.

Then, you can find the probabilities by differentiating the generating function.

P[n_] := (2^n - (D[(7 - 2 x - 4 x^2)/(1 - 2 x + x^3), {x, n - 3}] /. x -> 0)/(n - 3)!)/2^n
P[5]

P[5] is 3/8 for example.

Unfortunately, there will be a different generating function for every given binary sequence, but the general approach works for any of them.


问:
Inverse transform sampling of $f(x)=\lambda e^{-\lambda (x-a)}$
I have the following problem:

Given the PDF $$f(x)=\lambda e^{-\lambda (x-a)}$$ find the CDF through inverse transform sampling.

My question is, with respect to what variable should I integrate. I've seen that it can be from some $a$ to $x$ or from $0$ to $x$. And, if it is from $a$ to $x$, do I need something like an $a'$ ? The answers say it's $F(x)=\frac{\log(1-x)-\lambda a}{-\lambda}$ but I think this is not correct.

Thanks for the help.

答:
The first rule about densities is that they have to integrate to $1$. So in order for $\int f(x)dx$ to equal $1$, the density will be $\lambda e^{-\lambda(x-a)}$ for $x>a$, and is zero otherwise. The CDF is then, for $x>a$,
$$F(x):=\int_{-\infty}^x f(t)\,dt=\int_a^x \lambda e^{-\lambda(t-a)}\,dt$$
which you can evaluate using the change of variables $u:=t-a$. You then invert the equation $y=F(x)$ to obtain $x=F^{-1}(y)$ as a function of $y$. This gives the inverse CDF. BTW, the answer is correct: I get $x$ as the following function of $y$:
$$F^{-1}(y) = a - \frac{\log(1-y)}\lambda$$


答:
Not necessarily. But from Jensen's inequality, due to convexity of $e^x$,
$$\mathbb{E}[e^X] \ge e^{\mathbb{E}[X]}$$.

Verify with $X$ distributed as bernoulli $p$ example. Equality above will happen when $X$ is deterministic.

Title: Is $e^{\mathbb{E}\left[X\right]}$ equal to $\mathbb{E}\left[e^X\right]$?

Body:
Could anyone please tell me if $e^{\mathbb{E}\left[X\right]}$ is equal to $\mathbb{E}\left[e^X\right]$?

I have:

$\prod^n_{i=1} e^{\mathbb{E}\left[X_i\right]}$

and I'm wondering if I can rearrange this to:

$\prod^n_{i=1} e^{\mathbb{E}\left[X_i\right]} = \prod^n_{i=1} \mathbb{E}\left[e^{X_i}\right] = \mathbb{E}\left[\prod^n_{i=1} e^{X_i}\right]$


问:
Expected value of highest card in 4 card draw from standard deck
Suppose you draw 4 cards from a standard 52 cards deck, what is the expected value of the highest card among the 4 cards? (Ace = 1, ..., King = 13)
An accurate result is not necessary, I am more interested in a fast method to approximate it without using a calculator (if the accurate result is not simple enough).

I thought about (falsely) assuming that, by denoting the values of the cards as $X_1,\dots,X_4$,
$$
P(\max(X_1,\dots,X_4)\leq x) = P(X_1\leq x)^4
$$

Then, for simplicity calculating the maximum of 4 i.i.d. uniforms on $[0,1]$ gives
$$
P(\max(U_1,\dots,U_4)\leq x) = \int_0^1x4x^3dx = \frac{4}{5}
$$
and finally,
$$
12\cdot\frac{4}{5}+1 = 10.6
$$
was my approximation for the max of 4 i.i.d. uniforms on $[1,13]$.

How bad is that?

As the 4 draws are in fact not independent because, loosely speaking, drawing a small card reduces the probability of drawing another small card, we have
$$
P(\max(X_1,\dots,X_4)\leq x) < P(X_1\leq x)^4
$$
and thus the expected value should be a little higher, and of course the draws are not uniform from $[1,13]$, does that make a big difference?

In general, how would you approach these kind of questions, for example finding the expected value of the square of the number of red cards in that 4 card draw?
I somehow always assume that the draws are independent, e.g. approximate the probability for a red card by $1/2$ to make the computation of the expected value as simple as possible and then try to correct the answer in the right direction, but I am sure there are better approaches.

答:
You want a rough estimate, so I will calculate in my head and check against the calculator sometimes to see how far off I am getting. P(no kings) = $$\frac {48 \choose 4}{52 \choose 4} = \frac{48\cdot47\cdot46 \cdot45}{52\cdot51\cdot50\cdot49}\approx .92^4 \approx .68 $$
so there is very roughly a 1/3 chance that the max is 13.
P(no kings or queens) = $$\frac {44 \choose 4}{52 \choose 4} = \frac{44\cdot43\cdot42 \cdot41}{52\cdot51\cdot50\cdot49}\approx .85^4 \approx .5 $$
(I am using rough fractions and tricks like $$(1-1/n)^4 \approx 1-4/n$$
So at this point a rough lower bound of the expected value is roughly $1/3*13 + 1/6*12 + 1/2*(1) \approx 13/2 = 6.5$
At this point taking a wild stab at it, I would guesstimate that P(13) = 1/3, P(12) = 1/6, P(10) = 1/12, P(9) = 1/24, which gives roughly 4.3 + 2 + .9 + .4 or about 7.6 and that is probably fairly low. After all, the probability of the max being 1 is $$\frac{1}{52 \choose 4} \approx 1/12500$$ (okay, it's exactly 1/270725, but the point is, ignorably small). So the distribution is probably still weighted more heavily to the high end, so let's try $P(13) = 1/3, P(12) = 1/6, P(11) = 1/6, P(10) = 1/6$, and $P(9) = 1/6$, giving about 4.3 + 2 + 1.83 + 1.67 + 1.5 + 1.33 to finish, or ...well, 12.5 is a bit high. So this is a bit too rough, and taking a while.  Still, I wouldn't be surprised if the true answer is more than 11.
Let's try trimming the low values:  $P$(all cards ≤ 5) = $$\frac{20 \choose 4}{52 \choose 4} \approx \frac{20\cdot 19 \cdot 18 \cdot 17}{52\cdot51\cdot50\cdot49} \approx .4^4 = .0256$$. The true value is .01789 so not bad.  The same trick for all cards less than 8 gives roughly $.6^4 = .1296$ and we won't go too far wrong giving say 7 that probability and only other values over 8 get values. So all the values up to 8 contribute about 1 to the expected value.
That just leaves 9 10 and 11 to estimate.
$$P(no cards above 10) = \frac {40 \choose 4}{52 \choose 4} = \frac{40\cdot 39 \cdot 38 \cdot 37}{52\cdot51\cdot50\cdot49} \approx .75^4 = 81/256 \approx 1/3$$, so I wasn't too far wrong with my guess of 1/6 for P(max = 11).
No cards above 9 would involve 36/52 and the like, or about ${\frac{2}{3}}^4 = 16/81 \approx .2$, meaning $P(10) \approx .13$.
No cards above 8 uses 32/52 so about $.6^4 = .13$, making $P(9) \approx .07$
So my new estimate becomes $$13/3 + 12/6 + 11/6 + 10\cdot .13 + 9 \cdot .07 + 1$$ or $$4.33 + 2 + 1.833 + 1.3 + .63 + 1 = 11.1$$ or so.


问:
Sets that are closed or open in other sets
I fear this is terribly basic, so basic that I was sure there would be an answer somewhere, but I can't one.  Anyway, am working through do Carmo's differential geometry book.  He talks about closed and open sets.  And I think I get what they are.  But then he starts talking about one set being closed "in" another set, or open "in" another set.  I can't see where such concepts are defined, and I can't find a definition anywhere on the internet.  So, again, I think I understand what "open" and "closed" mean in and of themselves.  But what does it mean that some set is "closed in" or "open in" some other set?  Perhaps "A open in B" means that A has no boundary other than whatever boundary B may have?  Perhaps "A closed in B" means that if A has a boundary, that boundary is a subset of B?  Or ... heck if I know ...

答:
Being open/closed is not a property of a set. It is a property of some set with respect to another set (or more precisely: to some topological space).

In order for something to be open/closed there has to be some superspace, some universe. Nowadays we defined open/closed via topology. Given a topological space $(X,\tau)$, a subset $U\subseteq X$ being open in $(X,\tau)$ simply means that $U\in \tau$. Analogusly $F\subseteq X$ is closed in $(X,\tau)$ if $X\backslash F\in\tau$. So the first observation is that we can change topology on $X$ to $(X,\tau_2)$ so that a subset that was previously open in $(X,\tau)$ is no longer open in $(X,\tau_2)$. This can be done for any open subset, except $\emptyset$ and $X$ itself (which are always open and closed).

Now if $A\subseteq X$ then it is well known that $(A,\tau')$ is a topological space, where $\tau'=\{A\cap U\}_{U\in\tau}$. This is also known as the subspace topology and it is well defined for any subset $A\subseteq X$.

And so one may ask a question: given $A\subseteq X$, if $V\subseteq A$ is open in $A$ (meaning $V\in\tau'$) then is it also open in $X$ (meaning $V\in\tau$)? Sometimes it is, sometimes it is not. Few examples:

$X=\mathbb{R}$ with the Euclidean topology, $A=(0,1)$, then any open subset in $A$ is also open in $X$. But $(0,0.5]$ is closed in $A$, while it is not closed in $X$.

More generally: if $A\subseteq X$ is open (resp. closed) in $X$ and $U\subseteq A$ is open (resp. closed) in $A$ then $U$ is open (resp. closed) in $X$.

$X=\mathbb{R}$ with the Euclidean topology, $A=[0,1]$. Here for example $(0.1,0.9)$ is open both in $X$ and in $A$. However $[0,0.5)$ is open in $A$, because $[0,0.5)=A\cap (-1,0.5)$, while it is not open in $X$ (the Euclidean topology on $\mathbb{R}$ does not allow half closed intervals).


问:
Evaluating the indefinite trigonometric integral $\int \frac{dx}{(3+2\sin x)^2}$

$$\int \frac{dx}{(3+2 \sin x)^2}$$

ATTEMPT:-

Re Writing the integral as:

$I=\int \frac{2\cos x \sec x \,dx}{2(3+2 \sin x)^2}$ and using by parts:-

$\int u\,dv= uv-\int v\,du$

Here $u=\sec x \implies du=\sec x \tan x \,dx$

$\quad$ $dv=\frac{2\cos x \,dx}{2(3+2sinx)^2} \implies v=\frac{-1}{2(3+2\sin x)}$

$I=\frac{-\sec x\,dx}{2(3+2\sin x)} +\int \frac{\sec x \tan x \,dx}{2(3+2\sin x)}$

Let $I'=\int \frac{\sec x \tan x \,dx}{2(3+2\sin x)}$

$\implies I'=\int \frac{\sin x \,dx}{2\cos^2x(3+2\sin x)}$

$\implies I'=\int \frac{\sin \,dx}{2(1-\sin^2x)(3+2\sin x)}$

$\implies I'=\int \frac{-dx}{4(1+\sin x)} +\frac{3dx}{5(2\sin x+3)} + \frac{-dx}{20(\sin x-1)}$ which can be easily done by weierstrass's substitution.

But I am not able to modify my answer as given in the text.

Text Ans:-$\frac{2\cos x\,dx}{5(3+2\sin x)^2} + \frac{2}{5\sqrt{5}} \arctan\frac{(3\tan(\frac{x}{2})+2)}{\sqrt{5}} +c.$

My Ans:-$\frac{-\sec x\,dx}{2(3+2\sin x)} +  \frac{6}{5\sqrt{5}} \arctan\frac{(3\tan(\frac{x}{2})+2)}{\sqrt{5}}-10\tan x+15\sec x-15.$

答:
Note $\sin x=\cos (\frac\pi2-x)=\frac{1-\tan^2(\frac\pi4-\frac x2)}{1+\tan^2(\frac\pi4-\frac x2)}$
$$\int \frac{dx}{3+2\sin x}=-2\int \frac{d(\tan(\frac\pi4-\frac x2))}{\tan^2(\frac\pi4-\frac x2) +5}=-\frac2{\sqrt{5}}\tan^{-1}
 \frac{\tan (\frac \pi4-\frac x2) }{ \sqrt{5}}
$$
$$\left( \frac{2\cos x}{3+2\sin x}\right)’
= -\frac{3}{3+2\sin x}+\frac{5}{(3+2\sin x)^2}
$$
Integrate both sides to obtain

\begin{align}
I=\int \frac{dx}{(3+2\sin x)^2}=\frac 2{5}\frac{\cos x}{3+2\sin x}- \frac{6}{5\sqrt5}\tan^{-1}
 \frac{\tan (\frac\pi4-\frac x2 )}{ \sqrt{5}}+C
\end{align}


答:
Take for example $$
  f(t) = \begin{cases}
    -x^2 &\text{if $x < 0$} \\
    x^2 &\text{if $x \geq 0$.}
  \end{cases}
$$

For $x<0$ you have $f''(x) = -2$ while for $x > 0$ you have $f''(x) = 2$. $f$ is continuous as $0$, since $\lim_{t\to0^-} f(t) = \lim_{t\to0^+} f(t) = 0$, but since the second-order left-derivative $-2$ is different from the second-order right-derivative $2$ at zero, the second-order derivative doesn't exist there.

For your second question, maybe things are clearer if stated like this

If the second derivative is greater than zero or less than zero at some point $x$, that point cannot be an inflection point

This is quite reasonable - if the second derivative exists and is positive (negative) at some $x$, than the first derivative is continuous at $x$ and strictly increasing (decreasing) around $x$. In both cases, $x$ cannot be an inflection point, since at such a point the first derivative needs to have a local maximum or minimum.

But if the second derivative doesn't exist, then no such reasoning is possible, i.e. for such points you don't know anything about the possible behaviour of the first derivative.

Title: An inflection point where the second derivative doesn't exist?

Body:
A point $x=c$ is an inflection point if the function is continuous at that point and the concavity of the graph changes at that point.
And a list of possible inflection points will be those points where the second derivative is zero or doesn't exist.
But if continuity is required in order for a point to be an inflection point, how can we consider points where the second derivative doesn't exist as inflection points?

Also, an inflection point is like a critical point except it isn't an extremum, correct?
So why do we consider points where the second derivative doesn't exist as inflection points?

thanks.


问:
Does this functor commute with inverse limits?
Let $F$ a contravariant functor from the category $A$ of pointed connected CW-complexes up to homotopy to the category $B$ of pointed sets, with $F$ sending coproducts of $A$ to products of $B$. Let $X_i$ be an ascending series of objects in $A$. Is it true that $F\left(\underset{\longleftarrow}{\lim}X_i\right) = \underset{\longleftarrow}{\lim}F(X_i)$, where inverse limit is defined here?

To be more precise I am trying to understand why the following application is defined $F(f) : F(X) \to \underset{\longleftarrow}{\lim}F(X_i)$, with $X= \bigcup_{i}X_i$. I was presuming that $f : \underset{\longleftarrow}{\lim}X_i \to X_i \to X$ and, given that $F$ reverses all the arrows, I would get $F(X) \to F\left(\underset{\longleftarrow}{\lim}X_i\right)$, but I am not totally sure $F$ commutes with the inverse limit. Is there a different reason why $F(f)$ is defined?

答:
A contravariant functor can -if anything- send direct limits to inverse limits (this is due to the fact that direct = inverse in the opposite category) or dually inverse limits to direct limits; of course, not every contravariant functor has this property.

Since the terminology "direct" and "inverse" limit is a little bit outdated, let me speak about limits (=inverse, or projective $\varprojlim$) and colimits (=direct, or inductive, $\varinjlim$). Contravariant representable functors send colimits to limits, in the sense that $F(\text{colim})\cong \lim F$, and in general they do not transform limits into colimits, apart in very special cases.

Now, if your spaces $X_i$ are organised as a chain of inclusions

$$ X_0 \subset X_1 \subset X_2 \subset \cdots \subset X_n \subset \cdots X=\text{colim } X_i$$

by applying the contravariant $F$ to this chain one obtains a "opchain"

$$ FX_0 \supset FX_1 \supset FX_2 \supset \cdots \supset FX_n \supset \cdots FX=F\big(\text{colim } X_i\big)$$

The universal property of the limit $\lim X_i$, now, yields a unique map $c : F\big(\text{colim } X_i\big) \to \lim FX_i$. So far, no topology was involved (in the sense that this construction carries over to every category and every functor, provided the co/limits exist), but of course, when this construction is specialised to the/a category of topological spaces, there is a canonical topology on the colimit $X$ in question.

Is it the case that you are interested in this map, and not in a putative $f : \underset{\longleftarrow}{\lim}X_i \to X_i \to X$? It is a very natural question, now, to ask whether $c$ is a bijection, and there are theorems in that direction: as I mentioned, if $F$ is representable, it has this property.

If this is not the case, and you're sure that $f$ is what you want, I will show you that $f$ turns out to be a complicated way to express the CW triple $(X_0, X_i,X)$:

first, observe that the object $\varprojlim X_i$ is just homeomorphic to the head of the chain, $X_0$, by virtue of the fact that the limit of a diagram indexed over a category having an initial object coincides with evaluation on said initial object. The chain of spaces in question can be interpreted as a functor from the partially ordered set $\mathbb N = \{0\le 1 \le 2 \le \cdots\}$, and in such category $0$ is an initial object.

Given this, there is a canonical choice of a map $X_0 \to X_i$ for $i\ge 0$, the inclusion $X_0 \subseteq X_i$ that composes the first $i$ steps of the chain.

Given this, there is also a canonical choice of map $X_i \to X$, since $X$ is the colimit of the chain. Of course, such colimit just amounts to the set-theoretic union $\bigcup X_i$ endowed with the initial topology with respect to the subspaces $X_i$. (This is as far as I would go to say that "there is a topological reason" that defines $f$)

All in all, hence, $f$ is the inclusion of $X_0$ in $X_i$, followed by the inclusion of $X_i$ in $X$.


问:
Can we have x-intercept instead of y-intercept in equation of a line
As I have started learning co-ordinate geometry for ML, so got one question. Line equation

Let say in x,y dimension if a line cross over the x-axis instead of y (as shown in image above, unlike y=mx+c ) then how equation will change or will the equation has any impact apart from the x-intercept in this case? Is this a valid case?
Please help me clearing this doubt.

答:
One way to write an equation of an arbitrary line in the $x,y$ plane is
$$  Ax + By + C = 0.$$

If $A=0$ you get a line parallel to the $x$ axis;
if $B=0$ you get a line parallel to the $y$ axis.

If you set $A=m,$ $B=-1,$ and $C=b$ then the equation $Ax + By + C = 0$
describes the same line as $y = mx + b.$

But people often are interested in the equation $y = mx + b$
for reasons other than the shape it describes in a plane.
We may have some quantity we can either control or observe taking different values,
which we'll represent by the name $x$,
and some other quantity, which we'll call $y$, whose value has some relationship to the value of $x.$

The relationship $y = mx + b$ is one of the simplest possible kinds of relationship that can occur under these circumstances.
And it happens also to be possible to visualize a relationship like this by plotting a line on a graph.

A vertical line can not be the plot of such a relationship, because the first thing we wanted to see was a variety of different values of $x,$
and the vertical line has only one $x$ value.
The fact that $y = mx + b$ cannot describe a vertical line therefore is irrelevant to the study of these kinds of relationship.


问:
Generalisation of $ \sum \frac {1}{k}-\ln n=\gamma$ to $0 \lt\alpha \lt1 , \sum \frac{1}{k^\alpha}-f(n)= \beta$
looking at Find the value of : $\lim\limits_{n\rightarrow\infty}\left({2\sqrt n}-\sum\limits_{k=1}^n\frac{1}{\sqrt k}\right)$

and knowing that for
$$\alpha=1 ,\lim_{ n \to \infty}  \sum_{k=1}^n \frac {1}{k^\alpha}-\ln n=\gamma$$

Makes one wonder if there are other results for
 $0 \lt \alpha \lt 1 $

$$\lim_{ n \to \infty}  \sum_{k=1}^n \frac {1}{k^\alpha}-f(n)=\beta$$

where $f,\beta$ are determined by the value of $\alpha$

is there a topic that relates to some results similar to above?

答:
Suppose that $\Re \alpha>1$. In terms of the Hurwitz zeta function (http://dlmf.nist.gov/25.11)
$$
\sum\limits_{k = 1}^n {\frac{1}{{k^\alpha }}}  = \zeta (\alpha) - \sum\limits_{k = n + 1}^\infty  {\frac{1}{{k^\alpha }}}  = \zeta (\alpha) + \frac{1}{{n^\alpha }} - \sum\limits_{k = n}^\infty  {\frac{1}{{k^\alpha }}}  = \zeta (\alpha) + \frac{1}{{n^\alpha }} - \zeta (\alpha,n),
$$
i.e.,
$$
\zeta (\alpha) = \sum\limits_{k = 1}^n {\frac{1}{{k^\alpha }}}  - \frac{1}{{n^\alpha }} + \zeta (\alpha,n).
$$
By analytic continuation, this formula is valid for all complex $\alpha\neq 1$. Now you can use, for example, http://dlmf.nist.gov/25.11.E28 to obtain a more explicit expression that is valid in half-planes of the form $\Re \alpha>-(2N+1)$. You can also use the asymptotic expansion
$$
\zeta (\alpha,n) \sim \frac{{n^{1 - \alpha} }}{{\alpha - 1}} + \frac{1}{{2n^\alpha }} + \sum\limits_{m = 1}^\infty  {\frac{{B_{2m} }}{{(2m)!}}\frac{{\Gamma (\alpha + 2m - 1)}}{{\Gamma (\alpha)}}\frac{1}{{n^{\alpha + 2n - 1} }}}
$$
as $n\to +\infty$ with fixed $\alpha \neq 1$ (http://dlmf.nist.gov/25.11.E43). Here $B_m$ denotes the Bernoulli numbers. In particular,
$$
\zeta (\alpha) = \mathop {\lim }\limits_{n \to  + \infty } \left( {\sum\limits_{k = 1}^n {\frac{1}{{k^\alpha }}}  + \frac{{n^{1 - \alpha} }}{{\alpha - 1}} - \frac{1}{{2n^\alpha }} + \sum\limits_{m = 1}^N {\frac{{B_{2m} }}{{(2m)!}}\frac{{\Gamma (\alpha + 2m - 1)}}{{\Gamma (\alpha)}}\frac{1}{{n^{\alpha + 2n - 1} }}} } \right)
$$
for all $\alpha\neq 1$ and $N$ satisfying $\Re \alpha >  - (2N + 1)$. Therefore, with your notation, $\beta =\zeta(\alpha)$ and
$$
f(n)= -\frac{{n^{1 - \alpha} }}{{\alpha - 1}} + \frac{1}{{2n^\alpha }} - \sum\limits_{m = 1}^N {\frac{{B_{2m} }}{{(2m)!}}\frac{{\Gamma (\alpha + 2m - 1)}}{{\Gamma (\alpha)}}\frac{1}{{n^{\alpha + 2n - 1} }}}
$$
with $N = \left\lceil { - \frac{{\Re \alpha  + 1}}{2}} \right\rceil$, $\alpha\neq 1$ and with empty sums being interpreted as $0$.


问:
Parametrization of a knot isotopy
I am working on a computer visualization of a knot isotopy of the standard unknot embedding to an unknot with a Reidemeister I move. Does anyone have a formula?</p>

答:
This is very likely not what you're looking for, but I think it illuminates what a Reidemeister I move "is".

Consider this parameterization of an unknotted circle:

Depending on which direction you're looking at it, the projected diagram has either zero or one crossings:

So, by animating the projections of this loop as we rotate it, we get this animation illustrating a sequence of Reidemeister I moves:

The homotopy of these plane curves is
$$f_t(\theta) = (\cos t \sin 2\theta + \sin t \cos \theta, \sin\theta).$$
(Note: in the above animation the slider is mislabeled and it should read $t$ rather than $\theta$.)

The Reidemeister moves come from examining the kinds of singularities in diagrams that occur generically during an isotopy of the projected knots, and Reidemeister I moves come, basically, from when a portion of a curve like this rotates.

The following is potentially more useful.  If you analyze a Reidemeister I singularity carefully, you can isotope the isotopy to give a nice local model for it.  This is a polynomially defined local model for the Reidemeister I singularity:
$$f_t(s)=(s^2,s,s^3-ts)$$
where $s\in\mathbb{R}$ and $t\in[-1,1]$.

Projecting this onto the XZ axis gives this:

Projecting it onto the YZ plane, we can see that what's happening is that a bit of the knot is being turned over:

Title: Parametrization of a knot isotopy


问:
Help with this double summation
I'm having trouble when the indexes are related in a double summation. For example, this problem:
$\sum^n_{i = 1} \sum^n_{j = i+1} j - i + 2$

How could I sum this and what's a general approach to this type of double/triple summations?

Title: Help with this double summation

Body:
$$\sum^n_{i = 1} \sum^n_{j = i+1} j - i + 2=
\sum^n_{i = 1} \sum^{n-i}_{j =1} j + 2=\sum^n_{i = 1}\frac{(n-i)^2+(n-i)}{2}+2(n-i)$$
$$=\sum_{i=1}^{n-1}\frac{i^2+i}{2}+2i=\sum_{i=1}^{n-1}\frac{i^2+5i}{2}=\frac{1}{2}\sum_{i=1}^{n-1}i^2+\frac{5}{2}\sum_{i=1}^{n-1}i=\frac{n(n-1)(n+7)}{6}$$


问:
Can the idea of a 'function of a variable' be made rigorous?
Suppose that $y=x^2$. Very often people describe this relationship by saying that '$y$ is a function of $x$'. It seems that there are several logical problems with this statement:

A function is simply a set of ordered pairs of numbers. It does not matter how you denote the input of a function, and there is no doubt that $x \mapsto x^2$ and $y \mapsto y^2$ are the same function. Indeed, in both cases $y$ and $x$ are simply dummy variables used to illustrate what happens when you plug in an arbitrary number into the function.

$x$ and $y$ are often described as variables that are related in some way; in this case, with $y$ being the square of $x$. However, in this Math Overflow post  Mike Shulman states that the idea of a variable is 'not a standard part of modern formalizations of mathematics'. Strangely, the idea of a dummy variable makes much more sense to me. For instance, when we say 'consider the function $f$ defined by $f(x)=x^2$ for all $x$', it is clear that the only purpose of the letter $x$ is to declare that the second entry of the ordered pair is the square of the first.

If $y=x^2$ and $x \geq 0$, then we just as well might write $x=\sqrt{y}$. This flexibility is not really allowed when speaking of functions: $x \mapsto x^2$ and $x \mapsto \sqrt{x}$ are certainly not the same function. Perhaps it is possible evade this by treating $x$ as the independent variable and $y$ as the dependent variable.

Some authors get around this ambiguity by saying that 'the function $y=x^2$' is just a shorthand for 'the function $y(x)=x^2$', which in turn is just a shorthand for 'the function $y$ defined by $y(x)=x^2$ for all x'. However, this doesn't seem to align with how people treat the relationship between $y$ and $x$ in practice. Even if we accept that $y=x^2$ is simply a shorthand for $y(x)=x^2$, it still seems that there is a tendency towards treating $x$ as an independent variable representing the input of $y$, as opposed to simply a dummy variable that can be replaced by any other letter.

So I ask, in formal mathematics, is it possible to interpret $y=f(x)$ in such a way that $x$ and $y$ are variables representing the inputs and outputs of a function? And if so, how should the statement '$y$ is a function of $x$' be understood?

答:
A formal notation is

$$f:X\to Y:x\to f(x)$$ where $X,Y$ and the expression of $f$ are specified.

When you write $y=f(x)$, it must be obvious from context that

you are defining a function $y(x)$,

$X$ and $Y$ are known (such as $\mathbb R$).

Otherwise, the identity $$y=f(x)$$ might denote an equation where $y$ is explicitly known in terms of $x$ and you could be interested to express $x$ in terms of $y$. But again, this must be clear from context.


问:
For Which Additional constant a Manifold is guaranteed?
Set is given by $M = (x,y) \in \mathbb{R}^2 \vert \underbrace{e^{x^2+2\,y^2+2}}_{f} = c$.

Instantly I noticed $M$ can be counted as Manifold if $c \in(e^2,\infty)$ because $f \geq e^2$.   Now apparently this is not complete, so I wonder which additional c I have to take into consideration. Is it the boundary $[e,\infty)$ ? Or even other sets.

答:
We have $M=\{(x,y)\in \mathbb{R}^{2} \colon x^{2}+2y^{2}=d\}$, where $d=\operatorname{log}(c)-2$. Let $g=x^{2}+2y^{2}$. We have that $\operatorname{grad}(g)=(2x,4y)$ vanishes at $(0,0)$. Now let's look at some cases:

If $d<0$ (which is the same as $c<e^{2}$), $M=\emptyset$ isn't a manifold.

If $d=0$ (i.e. $c=e^{2}$) then $M=\{(0,0)\}$ is a point, which is a manifold as long as you admit zero-dimensional manifolds.

If $d>0$ (which is the same as $c>e^{2}$), then $\operatorname{grad}(g)(x,y)\neq 0$ for every $(x,y)\in g^{-1}(c)$, so by the regular level set theorem, $g^{-1}(c)$ is a ($1$-dimensional) embedded submanifold of $\mathbb{R}^{2}$.

Hope this helps!


问:
What size sample do I need to find errors with a certain amount of confidence
I work in a corporate audit department. Audits are necessarily limited in time and scope, we cannot investigate everything all the time. My question regards how many records I should examine to get a certain level of comfort, recognizing that I will sometimes miss errors or fraud because I cannot examine everything.

Suppose I have 1,000,000 transactions. Some of them may have errors/fraud, and I do not know how often errors/fraud occur. I want to test enough transactions so that I am 95% sure, or 99% sure, that I am finding at least one erroneous/fraudulent transaction if the incidence of error/fraud is greater than or equal to 1%, or 2%, or N%.

An example. Suppose in those 1,000,000 transactions there are 10,000 erroneous/fraudulent transactions and I care about an incidence rate of more than 1%. 10,000 is more than 1% of 1,000,000, so how many transactions would I need to test to be 95% confident that I am testing at least one of those 10,000 erroneous/fraudulent transactions? To be 99% confident? And same questions if the incidence rate I care about is 2%. (I'm looking for a general formula.)

答:
I will leave it to you to find a general formula. Here is the solution for
one of the specific scenarios you mention in your question.

If the fraud rate is $p = .01$ then you need $n$ large enough
that the probability of getting at least one fraudulent item exceeds 95%.
Let $X$ be the (binomial) count of fraudulent items. You need $n$ such that
$$P(X \ge 1) = 1 - P(X = 0) = 1 - (.99)^n > .95.$$

Using logs, one can see that $n = 299$ suffices, as shown below using R as a calculator.

```<code>log(.05)/log(.99)
[1] 298.0729

1 - .99^299
[1] 0.9504637
</code>```

Or from a 'grid' search in R:

```<code>n = 1:1000
min(n[dbinom(0, n, .01)<.05])
[1] 299
</code>```

Body:
I will leave it to you to find a general formula. Here is the solution for
one of the specific scenarios you mention in your question.

If the fraud rate is $p = .01$ then you need $n$ large enough
that the probability of getting at least one fraudulent item exceeds 95%.
Let $X$ be the (binomial) count of fraudulent items. You need $n$ such that
$$P(X \ge 1) = 1 - P(X = 0) = 1 - (.99)^n > .95.$$

Using logs, one can see that $n = 299$ suffices, as shown below using R as a calculator.

log(.05)/log(.99)
[1] 298.0729

1 - .99^299
[1] 0.9504637

Or from a 'grid' search in R:

n = 1:1000
min(n[dbinom(0, n, .01)<.05])
[1] 299


问:
Does the Law of Factor Sparsity apply to factorization?
Does the 80-20 Rule (also known as the Law of Factor Sparsity, the Law of the Vital Few, and the Pareto Principle) apply to factorization? That is, if x is a large positive integer, then are about 20% of the primes not exceeding x sufficient for factorizing about 80% of the positive integers not exceeding x? Has anyone done a statistical study of this?</p>

答: Yes. Hint: The set of primes less than 20% of $x$ constitute 20% of the set of primes less than $x$ for large $x$ (use the prime number theorem; using some explicit bounds you can show that the set constitutes at least 19% of the set of primes less than $x$ for all sufficiently large $x$). If any integer $n\leq x$ has a prime factor $p$ larger than 20% of $x$, then that integer $n$ must be one of $p,2p,3p,4p,5p$, so the number of such integers is less than 6 times the number of primes larger than 20% of $x$, and in particular, less than $6\pi(x)$. Thus since $\pi(x)/x\to0$ as $x\to\infty$, the set of primes less than 20% of $x$ are sufficient for factorizing almost all integers $\leq x$.</p>


问:
Alternative definition of Legendre polynomials
I'm studying Panofsky and Phillips' Classical Electricity and Magnetism. In writing the potential of a linear $2^n$-pole lying along the $x$-axis, they make use of the following definition for the Legendre polynomials.
$$
\frac{P_n(\cos\theta)}{r^{n+1}} = \frac{(-1)^n}{n!}\frac{\partial^n}{\partial x^n} \left(\frac{1}{r}\right).
$$

Here, $r:=|\mathbf{r - r_0}|$ and I guess that $\theta$ is the angle that $(\mathbf{r-r_0})$ makes with the $x$-axis (the authors don't mention abot $\theta$). Further I'm not even sure if the RHS is evaluated at $\mathbf r = 0$.

I'm only barely familiar with the Legendre functions and I am not able to show that the above matches with the standard definitions of the Legendre functions. I'll be happy if you can help me see the equivalence with the Rodrigue's formula.

答:
I think that this is the standard generating-function formula
$$
\frac{1}{\sqrt{1+x^2-2x\cos\theta}}= \sum_{n=0}^\infty x^n P_n(\cos\theta), \quad |x|<1
$$
in disguise as the formula for its  Taylor series coefficients
$$
P_n(\cos \theta) =\left. \frac 1 {n!} \frac{\partial^n}{\partial x^n} \frac{1}{\sqrt{1+x^2-2x\cos\theta}}\right|_{x=0}
$$

OwnerDisplayName: Atom

OwnerDisplayName: mike stone


问:
How to calculate limits of functions rigorously
I am using the following definition of the limit of a function "

Definition:   Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces. Suppose that $E \subseteq X$. Let $f : E \to Y$ and suppose that $p$ is a limit point of $E$. We say that $$\lim_{x \to p} f(x) = q$$ if there exists a point $q \in Y$ with the property that for all $\epsilon > 0$ there exists a $\delta > 0$ such that $$0 < d_X(x, p) < \delta \implies d_Y((f(x), q) < \epsilon$$ for any $x \in B_{(X, d_x)}(p, \delta)$

Suppose now I had the following function $f(x) : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ defined by $$f(x) = \frac{x^2}{x^4 + x^2}$$ and I wanted to prove $$\lim_{x \to 0} \ f(x) = 1.$$  Question 1: What is the easiest way to prove this?

In attempting to prove this on my own I simplified $f(x)$ to get that $$f(x) = \frac{x^2}{x^4 + x^2} = \frac{1}{x^2+1}$$, thus $$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{1}{x^2 + 1}$$

Now I could try and prove it directly using the $\epsilon-\delta$ definition above, but I never see anyone actually calculating limits like this in practice (I'm assuming there's a theorem that allows us to prove the above in a much simpler way).

Indeed there is a theorem that looks like it could be useful here

Theorem: Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces, let $E \subseteq X$ and let $f : E \to Y$ . Assume that $p$ is a limit point of $E$, then $f$ is continuous at $p$ if and only if $$\lim_{x \to p} f(x) = f(p)$$

Now note that the function $g : \mathbb{R} \to \mathbb{R}$ defined by $$g(x) = \frac{1}{x^2+1}$$ is continuous on $\mathbb{R}$ and that $0$ is a limit point of $\mathbb{R}$ hence $$\lim_{x \to 0} g(x) = g(0) = 1$$

and this looks like a very promising result that we could use to prove that $\lim_{x \to 0} f(x) = 1$, however there's one big problem, $f$ is defined on $\mathbb{R} \setminus \{0\}$ and since $0$ is not in the domain of $f$, it follows that $f$ can't be continuous at $0$. So we can't make use of the above theorem directly to prove that limit for $f$.

Question 2: Is there any way I can use the above theorem to prove that $\lim_{x \to 0} \ f(x) = 1$?

答: Regarding your Question 1, you can do a substitution $x=1/u$ to find $ \lim_{u \rightarrow  \infty }  \frac{u^2}{1+u^2} $ instead.


问:
A simple yet hard task for (theoretically) Poisson distribution
Sorry if I don't use the words properly, I haven't learnt these things in English, only some of the words.
Anyway, I'm practicing to one of my exams and sadly this task seemed more challanging for me than it should be. Some kind of explain would help a lot!

10 meters of clothes have 6 holes in it.

a) What kind of distribution does the number of holes per meter follow? (I think it must be Poisson distribution)

b) What's the probability that there's more than 10 holes in 5 meters of clothes?

答:
You are right , it's the Poisson distribution. Think like : 6 trains come every 10 hours

Therefore, in $5$ hours the expected number of trains is $3$. Thus, the $\lambda$ in our distribution is $3$.

Back to the problem, the probability that $k$ holes appear in $5$ meters of cloth is $$P(X=k)=\dfrac{\lambda^k e^{-\lambda}}{k!}=\dfrac{3^ke^{-3}}{k!}$$
Thus, we need the probability: $$P(X>10)=1-P(X\leq10)=$$
$$1-[P(0)+P(1)+\dots+P(9)+P(10)]$$
Now, it is a matter of calculations.


问:
How do I plug this function into the limit?
I've been having some trouble with these questions, specifically with plugging the function in.
Could somebody help me with the step by step on how to plug it in and simplify? I want to get an understanding of this.

ORIGINAL QUESTION:
$$\lim _{Δx\to 0}\left(\frac{f\left(x+Δx\right)-f\left(x\right)}{Δx}\right)$$

For the function

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

Body:
hint

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

$$f(a)=a^2-3a+2$$

$$f(x+b)=(x+b)^2-3(x+b)+2$$
$$=(x^2-3x+2)+b(b+2x-3)$$


问:
Permutation and Combinatorics Problem
A function $G$ is defined on a set $S$ with size $k$ : $G(a_1,a_2,a_3,\ldots,a_k)$.
$G(a_1,a_2,a_3,\ldots,a_k) = 1$ if and only if a convex polygon can be created by taking these $k$ elements as the side lengths. Otherwise $G(a_1,a_2,a_3,\ldots,a_k) = 0$.
You are given the identity permutation over $n$ elements - $I_n(I_n=\{1,2,3,\ldots,n\})$.
Find the sum of function $G$ over all possible distinct $k$-sized subsets of $I_n$.

All I could think of involved checking all subsets of size $k$ (which is not efficient for large size of the set obviously).Checking the $G$ value for a subset is easy , you just check if the largest element is greater than the sum of all other or not. Calculating the number of such subsets doesn't seem trivial :/ I wonder if there is some pattern/theorem/recurrence involved.

Title: Permutation and Combinatorics Problem

Body:
The definitions are important here.

Let's start by assuming $k\ge 1$ here so ignoring a whether a polygon with no sides exists.  Let's also assume that a convex polygon has positive area, so the subset $\{1,2,3\}$ produces a degenerate triangle and is not counted, while the subset $\{1,2,3,4\}$ can produce essentially several convex quadrilaterals but we will count it once. As leonboy commented, it is necessary and sufficient that the largest element a subset is strictly smaller than the sum of the others.  That will not happen if $k <3$ and not  when $n=k=3$.

Here is a table for small $n,k$:

 n \ k [,1] [,2] [,3] [,4] [,5] [,6]  [,7]  [,8]  [,9] [,10] [,11] [,12] [,13] [,14] [,15]
 [1,]    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0
 [2,]    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0
 [3,]    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0
 [4,]    0    0    1    1    0    0     0     0     0     0     0     0     0     0     0
 [5,]    0    0    3    5    1    0     0     0     0     0     0     0     0     0     0
 [6,]    0    0    7   14    6    1     0     0     0     0     0     0     0     0     0
 [7,]    0    0   13   32   21    7     1     0     0     0     0     0     0     0     0
 [8,]    0    0   22   63   56   28     8     1     0     0     0     0     0     0     0
 [9,]    0    0   34  112  126   84    36     9     1     0     0     0     0     0     0
[10,]    0    0   50  185  251  210   120    45    10     1     0     0     0     0     0
[11,]    0    0   70  289  459  462   330   165    55    11     1     0     0     0     0
[12,]    0    0   95  431  785  924   792   495   220    66    12     1     0     0     0
[13,]    0    0  125  620 1273 1716  1716  1287   715   286    78    13     1     0     0
[14,]    0    0  161  865 1976 3003  3432  3003  2002  1001   364    91    14     1     0
[15,]    0    0  203 1176 2959 5004  6435  6435  5005  3003  1365   455   105    15     1

The third column is essentially OEIS A002623 offset and its first difference is, as Alexander Burstein commented, OEIS A002620.

My reading of the question is that you are interested in the numbers for $n$, i.e. the row sums.  This starts $$0,   0,     0,     2,     9,    28,    74,   178,   402,   872,  1842,  3821,  7830, 15913, 32161, 64761,   130091,   260911$$ which does not seem to appear in OEIS but seems to be the difference between two sequences which do: A095941 (which would also count cases where the polygons are of zero area) and OEIS A317910 (which seems to only count cases where the polygons are of zero area).

The various OEIS links suggest a relationship between this sequence and the number of ways of partitioning $n$ into distinct parts OEIS A000009; its generating function is $\prod\limits_{m\ge1} (1+x^m)$ while the generating function of this sequence seems to be $$\frac{1-x-x^2}{(1-x)^2(1-2x)} - \frac{1}{(1-x)^2} \prod\limits_{m\ge1} (1+x^m)$$


问:
Characterization of convex differentiable functions
I am trying to prove the following statement

Let $\phi: \mathbb{R}^p \to \mathbb{R}$ be a convex function whose gradient exists at any point. Then for any $x, y$ we have $f(y) - f(x) \geq \langle \nabla f(x), y - x \rangle$.

without geometric interpretations as hyperplanes and such. I take as definition of convexity of $f$ that $\{(x, r) \ | \ f(x) \leq r\}$ is convex.

Any tips?

答:
Let $g(t)=f((1-t)x+ty)$ for $0\leq t \leq 1$. Since $f$ is convex, $g$ too is convex, so $g'$ exists and is non-decreasing. We have by the Mean Value Theorem:

$$ f(y)-f(x)=g(1)-g(0)=g'(c)\geq g'(0) $$

for some $c\in (0,1)$. If we differentiate $g$ at $0$, we get

$$ g'(0)=df_{x}(y-x)=\langle \nabla f(x),y-x \rangle. $$

As for why $g$ is convex, we have that $h\colon t\in \mathbb{R}\mapsto (1-t)x+ty=x+t(y-x)$ is an affine function (i.e. $h(t)-h(t')=(t-t')(y-x)$ is linear in $t-t'$), so $g=f\circ h$ is convex. This is because for $t,t'\in \mathbb{R}$, $\mu \in [0,1]$, we have $h((1-\mu)t+\mu t')=(1-\mu)h(t)+\mu h(t')$ (by direct computation), so
$$ f(h((1-\mu)t+\mu t'))=f((1-\mu)h(t)+\mu h(t'))\leq (1-\mu)f(h(t))+\mu f(h(t')) $$

Hope this helps!


问:
Does $\sum^{\infty}_{n=1}\left(\frac{p_{n+1}}{p_n}-1\right)^2 $ converge?
There appears to be a "prime constant" $\kappa$, generated from the sequence of primes:

$$\kappa = \sum^{\infty}_{n=1}\left(\frac{p_{n+1}}{p_n}-1\right)^2 \approx 1.653$$

Where $p_n$ is the nth prime.

However, how does one prove that such constant, does in fact, exist, that is, how does one prove that the above series converges?

答:
See my answer to the similar problem here: Does this sum of prime numbers converge?

Note that $(\frac{p_{n+1}}{p_n}-1)^2 = \frac{(p_{n+1}-p_n)^2}{p_n^2}$.
Let $g_n=p_{n+1}-p_n$ be the sequence of prime gaps. Since $p_n\geq n$, the convergence would be proved if we have the convergence of
$$
\sum_{n=1}^{\infty} \frac{g_n^2}{n^2}.
$$
By this result of R. Heath-Brown: Here,
we have
$$
\sum_{n\leq x} g_n^2 \ll x^{\frac{23}{18}+\epsilon}.
$$
Applying the partial summation with $A(x)=\sum_{n\leq x}g_n^2$ and $f(x)=x^{-2}$, we obtain
$$
\sum_{n\leq x}\frac{g_n^2}{n^2}=\int_{1-}^x f(x)dA(x)
$$
$$
=A(x)f(x)-\int_{1-}^x A(t)f'(t)dt. \ \ (1)
$$
Since $3-(23/18) >1$, we obtain the convergence of (1). Hence, the desired sum converges by the comparison test.


答:
A function is a subset $f$ of $A\times B$ with$$\forall x_1\in A\forall x_2\in B\forall x_3\in B((x_1,\,x_2)\in f\land(x_1,\,x_3)\in f\to x_2=x_3).$$(Some would say a "function" can be multivalued, but I'll address that in another paragraph.) If some $(x_1,\,x_2)\in f$, we write $x_2=f(x_1)$, thereby forgetting the set theory. But if all we know about values $x,\,y$ is $y=f(x)$ or equivalently $(x,\,y)\in f$, recalling $f$ is a set is again useful: $f$ is just the set of values of $(x,\,y)$ consistent with our knowledge. So "$y$ is a function of $x$" is shorthand for "the set of values of $(x,\,y)$ consistent with our knowledge is a function", which is a nontrivial statement because of the unique-images property of functions.

Even if a "multivalued function" is what you have in mind, knowing which function $y$ is of $x$ means knowing the set of values of $(x,\,y)$ consistent with our knowledge. (Indeed, this is true regardless of how we define "function", although in this more general case we lose the nontriviality observed above.)

Feel free to repeat this analysis for concepts other than knowledge. For example, if $x,\,y$ are related by a physical process, the "function" is the set of physically realizable values of $(x,\,y)$. (There's no need to worry about how other variables may come into play, because we can assume $x,\,y$ are tuples in their own right, thereby folding everything relevant into this discussion. Even randomness/indeterminism doesn't trip us up, if you use what are called random functions, which have a nuanced but ultimately non-ruinous effect on the discussion similar to that of multivalued functions. We might end up saying "there exists a function $f$ for which $y$ is the function $f$ of $x$, but the choice of $f$ is random, so itself has a probability space for its possible values".)

Title: Can the idea of a 'function of a variable' be made rigorous?

Body:
Suppose that $y=x^2$. Very often people describe this relationship by saying that '$y$ is a function of $x$'. It seems that there are several logical problems with this statement:

A function is simply a set of ordered pairs of numbers. It does not matter how you denote the input of a function, and there is no doubt that $x \mapsto x^2$ and $y \mapsto y^2$ are the same function. Indeed, in both cases $y$ and $x$ are simply dummy variables used to illustrate what happens when you plug in an arbitrary number into the function.

$x$ and $y$ are often described as variables that are related in some way; in this case, with $y$ being the square of $x$. However, in this Math Overflow post  Mike Shulman states that the idea of a variable is 'not a standard part of modern formalizations of mathematics'. Strangely, the idea of a dummy variable makes much more sense to me. For instance, when we say 'consider the function $f$ defined by $f(x)=x^2$ for all $x$', it is clear that the only purpose of the letter $x$ is to declare that the second entry of the ordered pair is the square of the first.

If $y=x^2$ and $x \geq 0$, then we just as well might write $x=\sqrt{y}$. This flexibility is not really allowed when speaking of functions: $x \mapsto x^2$ and $x \mapsto \sqrt{x}$ are certainly not the same function. Perhaps it is possible evade this by treating $x$ as the independent variable and $y$ as the dependent variable.

Some authors get around this ambiguity by saying that 'the function $y=x^2$' is just a shorthand for 'the function $y(x)=x^2$', which in turn is just a shorthand for 'the function $y$ defined by $y(x)=x^2$ for all x'. However, this doesn't seem to align with how people treat the relationship between $y$ and $x$ in practice. Even if we accept that $y=x^2$ is simply a shorthand for $y(x)=x^2$, it still seems that there is a tendency towards treating $x$ as an independent variable representing the input of $y$, as opposed to simply a dummy variable that can be replaced by any other letter.

So I ask, in formal mathematics, is it possible to interpret $y=f(x)$ in such a way that $x$ and $y$ are variables representing the inputs and outputs of a function? And if so, how should the statement '$y$ is a function of $x$' be understood?


问:
Set of diameter $\le 1$ contained in set of constant width $1$
I'm reading the paper Minimal universal covers in $E^n$ by H.G. Eggleston and they state that every set $A\subseteq{\bf R}^2$ of diameter at most $1$ (the diameter of $A$ is defined as $\sup_{x,y\in A}|x-y|$) is contained in a set which has width $1$ along any direction.  I can't see how the proof of this should work, though.

EDIT: Maybe the proof is something along the lines of constructing a thing that has width $1$ along any rational direction, then it should has width $1$ along all directions.  For this, it should suffice to show that for any $A$ of diameter $\le 1$ you can add a translate $I$ of the unit interval (embedded into ${\bf R}^2$) such that $A\cup I$ has diameter $1$.  Then you are able to add an infinite amount of intervals at different angles without the diameter exceeding $1$.

答:
Let $X\subseteq\mathbf{R}^2$ be of diameter $d(X)\le 1$, w.l.o.g. we can assume $X$ to be closed.  I attempt to show that there exists $a\in\mathbf{R}^2$ such that with $b:=a+(1,0)$ we have $d(X\subseteq\{a,b\})\le 1$.  This way we can add pairs of points with distance $1$ along every rational angle, which should suffice to get a superset of $X$ with constant width $1$.  In the following, $b$, $b'$, etc. always means $a+(1,0)$, $a'+(1,0)$, etc.

[EDIT] Look in the comments for a shorter approach that doesn't use the following section.

First choose $a$ such that we have $$d(X\cup\{a,b\})=\min_{a'\in\mathbf{R}}d(X\cup\{a',b'\}),$$ call this diameter $D$.  Assume there is some $a'\neq a$ with the same property.  Then all points on the line between $a$ and $a'$ must have this property as well, as when you travel in a direction and the distance to some point grows, it will not shrink again when you travel further in the same direction.  Assume $1<D$.  Then there exists $x\in X$ such that $D$ equals the distance from $x$ to $(a+a')/2$ (or just any other point strictly between $a$ and $a'$).  But this implies $D<d(x,a)$ or $D<d(x,a')$, a contradiction.  So if $1<D$ then our $a$ must be unique.

Now assume w.l.o.g. that $a=(-1/2,0)$, $b=(1/2,0)$.  Consider the set $E$ of points $e$ such that $D=d(\{a,b,e\})$, I called it $E$ because it has the shape of a vertical eye, note that $X$ must live in $E$ together with its enclosed area.  For $s,s'\in\{-,+,*\}$ denote by $E_{s,s'}$ the set of points $e\in E$ such that the sign of the first component is $s$ or $0$ or anything if $s=*$ and similar for the sign of the second component with $s'$.  When we shift $a$ to the right a little, the diameter of $X\cup\{a,b\}$ increases, so $X\cap(E_{-,*})$ cannot be empty.  A similar argument can be made for $E_{+,*}$ and by shifting $a$ vertically also for $E_{*,-}$ and $E_{*,+}$.  So there must be two points of $X$ in two opposite components of $E$ (i.e. $E_{-,+}$ / $E_{+,-}$ or $E_{-,-}$ / $E_{+,+}$), and one can easily see that those have distance greater than $1$, a contradiction.


问:
Inverse of a continuous increasing function is continuous (solution verification)
Show that for a continuous, increasing function $f : \mathbb R \rightarrow [0,1]$ that is onto $(0,1)$ it's inverse $f^{-1}$ is also continuous.

What I've tried

My first approach:

$f^{-1}$ is continuous if $f$ maps open sets to open sets. Since $f$ is increasing it is a one-to-one mapping and it suffices to show that $f$ maps open intervals to open intervals. Let $V \subset [0,1]$ so that $f^{-1}(V)$ is open (since $f$ is continuous) and $f(f^{-1}(V))=V$ which is also open. All that remains to show is that every open interval $U = f^{-1}(V)$ for some open interval $V$. This is true since $f$ is a one-to-one mapping.

My second approach:

$f^{-1}$ is continuous if there exists a $\delta$ such that when:

$$|\alpha - \alpha'| < \delta$$

we have:

$$|f^{-1}(\alpha) - f^{-1}(\alpha')| < \epsilon$$

Call the first event $A$ and the second event $B$. To show $A \implies B$ it is sufficient to show that $B^c \implies A^c$. Let $\alpha = f(x),\, \alpha' = f(x')$ for some $x,x'$ so that we need to show for each $\epsilon$: $|x-x'| \geq \epsilon \implies |f(x) - f(x')| \geq \delta$ for some $\delta$. But this is clear since $f$ is strictly increasing.

I could use some feedback on my reasoning for both approaches.

答:
I think the first argument is broken: it is not true that for a continuous map $m:X\rightarrow Y$ with a retraction function $r:Y \rightarrow X$ the map $r$ is continuous iff $m$ is open. The problem is that if $m$ is not surjective onto $Y$ preimages of open sets $U \subseteq X$ under $r$ might contain elements not in the image of $U$ under $m$. We have canonically $m(U) \subseteq r^{-1}(U)$, but the latter needs not be open.

I can give you a counterexample: Consider the topological spaces $X=\{\circ,\square\}$ and $Y=\{\circ,\square,\bullet\}$, where $\circ,\square$ denote open points, $\bullet$ is a closed point and we consider the topologies generated by this conditions. Then the obvious inclusion is a continous and open injection $m:X\rightarrow Y$ and we can define the retraction $r:Y \rightarrow X$ doing the obvious things and sending $\bullet$ to $\circ$. The latter is not continous though as the preimage $r^{-1}(\{\circ\}) = \{\circ,\bullet\}$ is not open in the topology on $Y$.

Edit I think the statement itself is wrong (if you use $[0,1]$). The function $\operatorname{arctan}:\Bbb R \rightarrow (-\frac{\pi}{2}, \frac{\pi}{2}) \subseteq [-\frac{\pi}{2},\frac{\pi}{2}]$ is a continuous strictly increasing function, which is not surjective. If it had a continuous retraction $r:[-\frac{\pi}{2},\frac{\pi}{2}]\rightarrow \Bbb R$ we would have in particular $r\vert_{(-\frac{\pi}{2},\frac{\pi}{2})} = \tan$ and continuity would force
$$r(\frac{\pi}{2})=\lim \limits_{x \rightarrow \frac{\pi}{2}} r(x) = \lim \limits_{x \rightarrow \frac{\pi}{2}} \tan(x) = \infty \in \Bbb R,$$
which is absurd.


问:
Sets that are closed or open in other sets
I fear this is terribly basic, so basic that I was sure there would be an answer somewhere, but I can't one.  Anyway, am working through do Carmo's differential geometry book.  He talks about closed and open sets.  And I think I get what they are.  But then he starts talking about one set being closed "in" another set, or open "in" another set.  I can't see where such concepts are defined, and I can't find a definition anywhere on the internet.  So, again, I think I understand what "open" and "closed" mean in and of themselves.  But what does it mean that some set is "closed in" or "open in" some other set?  Perhaps "A open in B" means that A has no boundary other than whatever boundary B may have?  Perhaps "A closed in B" means that if A has a boundary, that boundary is a subset of B?  Or ... heck if I know ...

答:
Bear with me.

When we talk of a set $A$ being open and/or closed (i.e. topology) we assume there is a "universe" set in which $A$ "lives" and set $A$ is open and/or closed according to the rules of this universe.

A simple example is the "universe" is $\mathbb R$ and $(1,2)$ is open because.... well, it fits all the definitions.

But there are many possible choices for the "universe".  So long as $A \subseteq X$ then we can consider $X$ to be the universe, and we can talk about whether $A$ is open and/or close within the rules of $X$ being or choice of universe.

In our example $(1,2) \subseteq [0, 15) \subseteq \mathbb R$ so we can consider $[0,15)$ as our universe. And $(1,2)$ is open in $[0,15)$ because is still fits all the definitions.

[Note:  For the statement "$A$ is/is not open/closed in $X$" to even make any sense we must have that $A \subseteq X$.]

An issue might come about when we consider $A \subseteq X \subseteq U$. And $A$ might or might not be open/and or closed according to the rules of one universe but not according to the rules of the other universe.

Consider then set $[13,15)\subseteq [0, 15) \subseteq\mathbb R$.

Clearly $[13,15)$ is not closed in $\mathbb R$ because $15$ is an open end point and closed sets can't have open endpoints.

(Okay, the definition is more formal than that.  $15$ is a limit point of $[13,15)$ and $15\not \in [13,15)$ and "closed" by definition means contains all its limit points and $[13,15)$ does not contain $15$.)

But is $[13,15)$ closed in $[0, 15)$?  You might say "of course not" it still has an open point about $15$".  But.... $15$ does not exist in the universe of $[0,15)$.  So it doesn't make any more sense to talk about an open point at $15$ then it does to talk about an open point at $\text{Barbar the elephant}$.

(More formally the limit points of $[13,15)$ are all the values between $13$ and $15$ inclusive.  As all the points between $[13, 15)$ in our univers are $[13, 15)$ they are all in the set.  So $[13,15)$ is closed.)

Also consider $[0, 1)$.  That is not open in $\mathbb R$.  It is not open because $0 \in [0,1)$ and $0$ is not an interior point.  It isn't an interior point because if we can't find any $(0-\epsilon, 0+\epsilon)\subset [0, 1)$. That is because $(0-\epsilon,0)$ will always have points that are not in $[0,1)$.

But that isn't true if our universe is $[0, 15)$.  If we consider all the $x\in [0, 15$ so that $0 -\frac 12 < x < 0 + \frac 12$ then $0 \le x < \frac 12$.  (We can not have $x < 0$ because then $x$ would not be in our universe.)  Ans as all $[0,\frac 12) \subset [0,1)$ that means $0$ is an interior point.

(And all the other points in $(0,1)$ are interior points.)

So all the points of $[0,1)$ are interior points (!!!!IF!!!!! our universe contains not negative numbers) so $[0,1)$ is open in $[0,15)$.

.....

That's it in a nutshell.  We can have all sorts of different definition of open/closed depending on which "universe" they live in, but to say a set $A$ is "open in $X$"  means $A \subset X$ and according to the definitions of $X$ as our universe, $A$ is open.