TITLE: Equivalent definitions of a Hodge structure
QUESTION [1 upvotes]: Generally, a Hodge structure of weight $k$ on a finitely generated abelian group $H$ is defined as a decomposition of the complexification:
$$
H\otimes \mathbb C = \bigoplus_{p+q=k} H^{p,q},
$$
where the vector spaces $H^{p,q}$ must satisfy the condition $\overline{H^{p,q}}=H^{q,p}$.
If $H$ carries this structure, then it is possible to define the Hodge filtration as
$$
F^p(H)=\bigoplus_{k-q\geq p} H^{k-q,q} .
$$
However, it seems that if a filtration $F^\bullet(H)$ exists with the property that
$$
F^p(H)\cap \overline{F^q(H)}= 0 \qquad \mbox{ whenever } p+q=k+1
\tag{$\ast$}
$$
then a Hodge structure of weight $k$ exists on $H$ by putting
$$
H^{p,q}=F^p(H)\cap \overline{F^q(H)}.
$$
Now, in Peters and Steenbrink, it is said that $(\ast)$ is equivalent to say that $F^p(H)\oplus F^{k-p+1}(H)=H\otimes\mathbb C$, while on Wikipedia this latter condition is also required (meaning that it might not be equivalent).
Question 1. Is it equivalent or not?
If the answer is yes, notice that the Hodge filtration satisfies much nicer properties than the required on  the existing descending filtration (for the Hodge filtration $F^p(H)\cap\overline{F^q(H)} = 0$ unless $p+q=k$), which makes me think that the Hodge filtration of associated given by the Hodge structure induced by the former filtration would be different. Isn't that absurd?.
Question 2. Either if the answer to question 1 is yes or not, how can one prove that indeed
$$
H\otimes\mathbb C= \bigoplus_{p+q=k} F^p(H)\cap\overline{F^q(H)} ?
$$
I can only see it if the filtration $F^\bullet(H)$ satisfies the same properties that the Hodge filtration.

REPLY [2 votes]: Fortunately for you, I've had a headache because of the conflict between these exact references before.
Wikipedia is correct, Peters and Steenbrink are not.
Question 1. Is it equivalent? No.
Here's an example: Let $k=0$, let $H \cong \mathbb Z$. Let
$$
0= F_0 = F_1=\cdots ;\quad H \otimes \mathbb C= F_{-1}=F_{-2}=\cdots
$$
You can check that the condition in Peters and Steenbrink is satisfied, but this is not a Hodge structure of weight $0$, as all the graded pieces will vanish.
In general, take any Hodge structure of weight $k$. According to the condition in Peters and Steenbrink, it is automatically a Hodge structure of any weight $k'\ge k$, since the Hodge filtration is decreasing, which is never true.


TITLE: Showing the limit is $+\infty$
QUESTION [5 upvotes]: Let $f(x)$ be continuous and positive on $[0,+\infty)$. Suppose
$$\int_0^{+\infty}\frac{1}{f(x)}dx<+\infty. $$
How can one show that
$$\lim_{s\to +\infty}\frac{\int_0^{s}f(x)dx}{s^2}=+\infty?$$

REPLY [6 votes]: For $t>0$, we have
$$t=\int_t^{2t}\frac 1{\sqrt{f(\xi)}}\sqrt{f(\xi)}d\xi\leq \left(\int_t^{2t}\frac 1{f(\xi)}d\xi\right)^{1/2}\left(\int_t^{2t}f(\xi)d\xi\right)^{1/2}$$
so $$s^2\leq \int_s^{2s}\frac{dx}{f(x)}\cdot \int_s^{2s}f(\xi)d\xi$$
and so
$$\frac{\int_0^{2s}f(\xi)d\xi}{s^2}\geq\frac{\int_s^{2s}f(\xi)d\xi}{s^2}\geq \frac 1{\int_s^{2s}\frac{d\xi}{f(\xi)}},$$
and we can conclude, since the fact that $\int_0^{+\infty}\frac{d\xi}{f(\xi)}$ converges implies that $\lim_{s\to \infty}\int_s^{2s}\frac{d\xi}{f(\xi)}=0$.


TITLE: How to calculate return time from a transition probability matrix?
QUESTION [2 upvotes]: Consider the transition probability matrix with state space $\{0,1,2,...5\}$ as follows
$$P= \begin{bmatrix} 1/2 & 1/2 & 0 & 0 & 0 & 0 &\\
1/3 & 2/3   & 0 & 0 & 0 & 0\\
0 & 0 & 1/8 & 0  & 7/8 & 0\\
1/4 & 1/4 & 0 & 0 & 1/4 & 1/4\\
0 & 0 & 3/4 & 0 & 1/4 & 0\\
0 & 1/5 & 0 & 1/5 & 2/5 & 1/5\end{bmatrix}$$
a) Let $C=\{0,1\}$. Then find $\rho_{3}(C)$ i.e., absorption probability to the set $C$ starting from $3$.
b) Find $m_0=\mathbb{E}_0(T_0)$, where $\mathbb{E}_i(T_i) := \mathbb{E}(T_i \mid X_0=i)$

Here is what I have thought of.
I know that $\{0,1\}$ is a recurrent irreducible closed set and $\{3,5\}$ is another irreducible closed set. Hence starting from $3$, it would be impossible to go to any other state except $5$ and so the absorption probability of  $\rho_3(C)$ must be $0$. Coming to the second part, I am not sure how to exactly do it? We haven't explicitly learnt how to calculate hitting time, and so I don't know how to start calculating the conditional expectation using this. I also read that $\mathbb{E}_i [T_i]=m_i=\frac{1}{\pi_i}$ where $\pi_i$ is the stationary distribution. But $P$ isn't an irreducible Markov Chain so it doesnt have a unique stationary distribution. So I don't know how to proceed using that formula either. I am unaware of any other formula to calculate return time, so any help would be appreciated!

REPLY [0 votes]: Your understanding of absorption probailities is slightly incorrect. Here is a formula for absorprtion probabilities-
$$\rho_C(x)= \sum_{y \in C} P(x,y) + \sum_{y \in \mathcal{S}_{T}} P(x,y) \rho_C(y)$$
Given that $C=\{0,1\}$ and $\mathcal{S}_T = \{3,5\}$, we have
\begin{align}
    \rho_C(3) &=P(3,0)+P(3,1)+P(3,3)\rho_C(3) + P(3,5) \rho_C(5)\nonumber\\
    &= 1/4 + 1/4 + 1/4 \rho_C(5)\nonumber\\
    \implies &\rho_c(3)- 1/4 \rho_C(5) = 1/2
\end{align}
and
\begin{align}
    \rho_C(5) &=P(5,0)+P(5,1)+P(5,3)\rho_C(3) + P(5,5) \rho_C(5)\nonumber\\
    &= 1/5 + 1/5 \rho_C(3) + 1/5 \rho_C(5)\nonumber\\
    \implies & 4/5\rho_C(5)- 1/5 \rho_C(3) = 1/5
\end{align}
Then, solving the two equations, we get $\rho_C(3)=3/5$ and $\rho_C(5)=2/5$.
For $m_0$, consider the irreducible closed classes in this M.C. We have $\pi P= \pi \implies P^T \pi^T= \pi^T$. Thus,
$$ \left[\begin{array}{cc}
1/2 & 1/3 \\
1/2 & 2/3
\end{array}\right]
\left[\begin{array}{c}
\pi_0 \\
\pi_1 \\
\end{array}\right]
=
\left[\begin{array}{c}
\pi_0 \\
\pi_1 \\
\end{array}\right]
$$
Now we have the equations
$$1/2 \pi_0 + 1/3\pi_1= \pi_0 \Rightarrow -1/2 \pi_0+ 1/3 \pi_1=0$$
$$ \pi_0 +\pi_1=0$$
Solving this yields $\pi(0) = 2/5$ and $\pi(1) = 3/5$.


TITLE: Does this question about the sufficiency of the analyticity of f(z) make sense
QUESTION [0 upvotes]: Before I ask the question, I know for sure this following question makes sense:
Prove that a necessary condition for $ f(z) $ to be analytic is that the Cauchy-Riemann equations are satisfied.
This question makes sense because if the Cauchy-Riemann equations are satisfied, $f(z)$ need not be analytic. Hence, why it's called a necessary condition.
But in order for $ f(z) $ to be analytic , there are 2 sufficient conditions, the Cauchy-Riemann equations must be satisfied and the partial derivatives must be continuous.
Now for the main question I want to ask:
Prove that a sufficient condition for $ f(z) $ to be analytic is that the Cauchy-Riemann equations are satisfied.
I'm asking this because I had it on my exam yesterday, now how could this make sense, the Cauchy-Riemann equations being satisfied should not be enough, the question did not mention the continuity of the partial derivatives, thus I treated this question as if it was asking "Prove that a sufficient condition for the Cauchy-Riemann equations to be satisfied is that $f(z)$ must be analytic".
So am I right and should I argue with my professor about that?
My question is not about when a function is holomorphic, my question is if the following makes sense as a proof question or not:
Prove that a sufficient condition for f(z) to be analytic is that the Cauchy-Riemann equations are satisfied.

REPLY [2 votes]: There is a diffrence between differentiability at a point and differentiability/analyticity in a neighborhood of a point. For example $f(x+iy)=\sqrt {|xy|}$ satisfies C-R equations at $0$ but $f$ is not differentiable at $0$. However, in an open set analyticity is equivalent to validity of C-R equations. Continuity of partial derivatives need not be assumed. In fact C-R equations imply that $f$ has continuous partial derivatives of all orders.


TITLE: Binomial approximation with normal distribution
QUESTION [0 upvotes]: Maybe a stupid question but I was trying to approximate the binomial distribution with a normal distribution and I can't understand where the problem is.
Online I read that a binomial distribution can be approximated like this using the central limit theorem:

If we have i.i.d. binomial distributed random variables $X_{i}$ than we can write
$P(\sum_{0}^{n}X_{i}\leqslant z)=Φ(\frac{z-np}{\sqrt{np(1-p)}})$

I have tried to come to this solution but applying the central limit theorem that is formulated: "$\lim_{n \to \infty} P(\frac{S_{n}-nμ}{\sqrt{nσ2}}\leqslant z)\Rightarrow Φ(z)$" I come to the solution that the normal approximation is $Φ(\frac{z-n^2p}{\sqrt{n^2p(1-p)}})$. This is because of the $n$ in the numerator and the $\sqrt{n}$ in the denominator.
Where do these two go in the formula I found online?

REPLY [0 votes]: If you want to approximate $X_n\sim\mathsf{Bin}(n,p)$ with a normal distribution then you must go for $X_n=\sum_{i=1}^nB_i$ where the $B_i$ are iid and have Bernoulli distribution with parameter $p$.
Then: $$\frac{X_n-\mathbb EX_n}{\mathsf{SD}(X_n)}=\frac{X_n-np}{\sqrt{np(1-p)}}\to U\text{ a.s.}$$where $U$ has standard normal distribution.
Note that here:$$\frac{X_n-np}{\sqrt{np(1-p)}}=\frac{\bar B_n-p}{\sigma/\sqrt{n}}$$for $\sigma=\mathsf{SD}(B_1)=\sqrt{p(1-p)}$, showing the connection with CLT.

Edit concerning questions in comments on this question:
If a random variable $Y$ has a second moment then it has a standardized form: $$Y^*:=\frac{Y-\mu}{\sigma}$$where $\mu:=\mathbb EY$ and $\sigma^2:=\mathsf{Var}Y$.
Characteristic for this form are: $$\mathbb EY^*=0\text{ and }\mathsf{Var}Y^*=1$$
Formulation of CLT: If $X_1,X_2,\dots$ are iid random variables that have a second moment and: $$S_n:=X_1+\cdots+X_n$$ then standard form $S_n^*$ converges to a random variable $Z$ that has standard normal distribution.
If in this context $\mathbb EX_1=\mu$ and $\mathsf{Var}(X_1)=\sigma^2$ then we find $\mathbb ES_n=n\mu$ and $\mathsf{Var}(S_n)=n\sigma^2$ so that we find:$$S_n^*:=\frac{S_n-n\mu}{\sigma\sqrt{n}}$$
Applying this on special case where $X_i$ have Bernoulli distribution with parameter $p$ we get:$$S_n^*:=\frac{S_n-np}{\sqrt{np(1-p)}}$$
In this situation $S_n$ has binomial distribution with parameters $n$ and $p$.
Applying this on special case where $X_i$ have Poisson distribution with parameter $\lambda$ we get:$$S_n^*:=\frac{S_n-n\lambda}{\sqrt{n\lambda}}$$
In this situation $S_n$ has Poisson distribution with parameter $n\lambda$.


TITLE: Frame of reference of a photon
QUESTION [0 upvotes]: I'm curious about the fact that it is impossible to consider a frame of reference where a photon is the reference itself (meaning a frame of reference where this photon can't move).
I looked for answers in other SE questions. In the first one:
Frame of reference of the photon? I read:

there is no frame of reference for the photon because there is no frame of reference in which the photon is at rest.

This surprises me. Indeed, the Galilean principle of relativity says:

The laws of mechanics have the same form in all inertial frames.

Looking to an answer of a similar question in SE (which is Would time freeze if you could travel at the speed of light?
), I also read:

Since arguments from positivism can often kill off perfectly interesting and reasonable concepts, we might ask whether there are other reasons not to allow such frames. There are. One of the most basic geometrical ideas is intersection.

Unfortunately, those complete explanations are too complicated for me.
Does anyone have a more simple way to explain why it is impossible to consider a photon as a frame of reference? I'm really curious about how a photon would "see" the universe around him.

REPLY [0 votes]: The tangent vector to the worldline of a photon is orthogonal only to multiples of itself.  Therefore  it cannot be part of an orthogonal frame.


TITLE: If an 8 digit number is randomly generated, what are the odds that all 8 digits are different/distinct?
QUESTION [1 upvotes]: If an 8 digit number is randomly generated, what are the odds that all 8 digits are different/distinct?

Assume that we can use the 10 numbers 0-9.
What I got myself is this:

(*)    $\frac{10!}{(10-8)!}$=$\frac{10!}{2!}$=1814400

The total number of 8 digit numbers is $10^8$=100,000,000.

So dividing those numbers gives:
$\frac{1814400}{100,000,000}$=0,018144
What I'm not sure of is whether all possibilities in the step with the (*) are distinct from each other, so there is no repetition of numbers within an 8-digit number. Could someone explain this to me whether it is or not and why? Then I know that my final answer is correct or incorrect.
Maybe good to know, I am just starting with probability theory...
Thanks in advance for your help!

REPLY [0 votes]: Think of your numerator as $10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3$; that is, you have $10$ choices for the first digit, but then only $9$ for the second, $8$ for the third, and so on down to $3$ for the eighth digit. The decreasing number of choices reflects the fact that you are not repeating digits that have already been used. Alternatively, $10!/2!$ means you permute the ten digits in any of $10!$ ways, drop the final two digits, and group them in pairs ($2!$) according to the two digits that get dropped. In either interpretation there is no repetition of digits.
In comments below the OP, it's pointed out that you are allowing your eight-digit numbers to begin with a $0$, which may or may not be what the problem poser intended. As it turns out, however, it doesn't matter for the final result. If starting $0$'s are allowed, the probability is
$$10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\over10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10$$
whereas if starting $0$'s are not allowed, the probability is
$$9\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\over9\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10$$
In either case, cancellation leaves
$$9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\over10\cdot10\cdot10\cdot10\cdot10\cdot10\cdot10$$


TITLE: Irreducible representation of $\mathfrak{so}(4)$
QUESTION [1 upvotes]: I am supposed to give a 9-dimensional irreducible representation of $\mathfrak{so}(4)$.

I know that $\mathfrak{so}(4)\cong\mathfrak{so}(3)\oplus\mathfrak{so}(3)$ and hence I have a 6-dimensional reducible representation of $\mathfrak{so}(4)$. But how do I get a irreducible 9-dimensional representation?
This may be a stupid question but I couldn't find anything.

REPLY [1 votes]: Almost every treatment of representations of real semisimple Lie algebras explicitly treats the basic case $\mathfrak{su}_2$; the standard result is that for each positive integer $n$, there is up to equivalence exactly one irreducible $n$-dimensional $\mathfrak{su}_2$-representation, call it $\rho_n$.
Now if you already know $\mathfrak{so}_4 \simeq \mathfrak{so}_3 \oplus \mathfrak{so}_3$, and further know $\mathfrak{so}_3 \simeq \mathfrak{su}_2$, then let
$pr_1: \mathfrak{so}_4 \twoheadrightarrow \mathfrak{su}_2, \qquad pr_2: \mathfrak{so}_4 \twoheadrightarrow \mathfrak{su}_2$
be the projections on the respective factors. Then $\rho_9 \circ pr_i$ give two inequivalent (why?), irreducible (why?), $9$-dimensional representations of $\mathfrak{so}_4$.


TITLE: There are only finitely many elements $z \in \mathbb{Z}[i]$ with $N(z) \leq K$
QUESTION [1 upvotes]: We know $\mathbb{Z}[i]$ is a Euclidean domain, so the question is given any integer $K > 0$, show that $N(z) \leq K$ for only finitely many $z \in \mathbb{Z}[i]$. Once I have this, I think I see how I can use the division algorithm to prove that the quoteint ring $\mathbb{Z}[i]/I$ is finite for any nonzero ideal $I$ of $\mathbb{Z}[i]$, because the Gaussian integers are a P.I.D.

REPLY [2 votes]: For $z=a+bi$, the equation $N(z)\leq K$ determines the circle $a^2+b^2\leq K$, or $|z|^2\leq K$. This is a ball of radius $\sqrt{K}$ centered on the origin on the complex plane; in particular, it is bounded.
On the other hand, $\mathbb{Z}[i]\subset \mathbb{C}$ is discrete, and hence its intersection with any bounded set must be finite.


TITLE: Can Fermi liquid be obtained by a canonical transformation?
QUESTION [7 upvotes]: The basic assumption of the Ferm-liquid theory is the one-to-one correspondence between the states of an interacting Fermi gas to those of a gas of non-interacting quasiparticles.  The question is then, whether one can perform a canonical transformation to tranform the interacting Hamiltonian to a non-interacting one.
Remarks:

It is understood that Fermi liquid description is approximate, so I expect some kind of approximatate procedure along the lines of the Schrieffer-Wolff transformation or the the genuine guesses used for spin Hamiltonians (Holstein-Primakoff, Jordan-Wigner, etc.)
A close relative of the Fermi liquid is Luttinger liquid, which is exactly mapped onto a collection of non-interacting bosons. In this respect, I would like to stress that I am looking for a canonical transformation, similar to canonical bosonization, as described in the reviews of Haldane, Voit or Giamarchi's book (as opposed to more recent popular bosonization via path integrals).

REPLY [2 votes]: It seems that it can be performed in terms of RPA (random phase approximation). In this set up, one should consider Bose-operators for particle-hole pairs,
$$ c_{p,k}=a^{\dagger}_pa_{p+k},\quad c_{p,k}^{\dagger}=a_{p+k}^{\dagger}a_p,$$
where $|p|<p_F$ and $|p+k|>p_F$. Then particle density operator can be expressed in terms of the introduced Bose-operators,
$$\rho_k=\sum_p(c_{p,k}+c_{-p,-k}^{\dagger}). \tag{*}$$
The exact Hamiltonian looks like
$$H=\sum_{p,\sigma}\zeta(p)a_{p,\sigma}^{\dagger}a_{p,\sigma}+\frac{1}{2}\sum_kV_k\rho_k\rho_{-k}\quad \zeta(p)=\frac{p^2}{2m}-\mu,$$
so in terms of introduced operators $c$ and $c^{\dagger}$ it has "free form". Next step is to consider commutation relations and bla-bla. At the final step it is convenient to rotate operators as
$$\phi_{p,k}\equiv \phi_{-p,-k}^{\dagger}=\frac{1}{\sqrt{2\omega_{p,k}}}(c_{p,k}+c_{-p,-k}^{\dagger}),$$
$$\pi_{p,k}\equiv \pi_{-p,-k}^{\dagger}=\frac{i}{2}\sqrt{2\omega_{p,k}}(c_{p,k}-c_{-p,-k}^{\dagger}),$$
where
$$\omega_{p,k}=\frac{(p+k)^2}{2m}-\frac{p^2}{2m}.$$
In conclusion, the Hamiltonian becomes
$$H_{0}=\frac{1}{2}\sum_{p,k}\left(\pi_{p,k}^{\dagger}\pi_{p,k}+\omega_{p,k}^2\phi_{p,k}^{\dagger}\phi_{p,k}\right),$$
$$H_{\text{int}}=\sum_kV_k\left(\sum_p\sqrt{\omega_{p,k}}\phi^{\dagger}_{p,k}\right)\left(\sum_{p'}\sqrt{\omega_{p',k}}\phi_{p',k}\right).$$
The sum of $H_0$ and $H_{\text{int}}$ can be diagonalized with help of Bogoliubov transformation. Resulting spectrum contains two components: 1) continuous branch $\omega=\omega_{p,k}$ (coincides with non-interacting spectrum), 2) collective branch. The dispersion law for collective branch is given by
$$1=V_k\sum_p\frac{\omega_{p,k}}{\omega^2-\omega_{p,k}^2},$$
which corresponds to plasmon mode.
Where the approximation occurs? In the line $(*)$. The strict denotation is
$$\rho_k = \sum_{p\in R_k}(c_{p,k}+c_{-p,-k}^{\dagger}), \tag{**}$$
where $R_k$ is the sickle-shaped domain that is defined by conditions $|p+k|>p_F$ and $|p|<p_F$. In such region the operator $\rho_k$ is real, which means $\rho_k=\rho^{\dagger}_{-k}$. Such approximation $(**)$ implies that we omit terms $a_p^{\dagger}a_{p+k}$ that raise zero when act on state with single particle-hole pair (and on ground state)


TITLE: Event of positive probability occurs infinitely often in stationary ergodic sequence
QUESTION [4 upvotes]: Setup:
Suppose $X = \{X_n\}_{n\in\mathbb{Z}}$ is a stationary ergodic proces on the real line and let $A = \prod_{n\in\mathbb{Z}}A_n$ be a Borel measurable set such that
$$
P(X \in A) = P\left(X_n\in A_n \quad\forall n\in\mathbb{Z}\right) > 0.
$$
Finally, let $\tau:\mathbb{R}^\mathbb{Z}\rightarrow \mathbb{R}^\mathbb{Z}$ denote the shift.
Question:
I need the result that, almost surely, there are infinitely many $k\in\mathbb{Z}$ such that $\tau^k(X) \in A$. My question  is two-fold

Is my short proof below correct.
Can the result also be shown by using more basis results on ergodic sequences, like Poincarre's recurrence theorem.

Thank you in advance!
My proof:
This uses Birkhoff's ergodic theorem and the fact that measurable functions of stationary ergodic sequences are also stationary ergodic. Define the sequence $\{Y_n\}_{n\in\mathbb{Z}}$ as $Y_n = 1 $ if $\tau^n(X) \in A$ and zero otherwise. Then $\{Y_n\}_{n\in\mathbb{Z}}$ is stationary ergodic and thus
$$
\lim_{k\rightarrow\infty}\frac{1}{k}\sum_{n=0}^{k-1}Y_n = E(Y) = P(A) > 0,
$$
which implies infinitely many of the $Y_n$ have to be equal to one.

REPLY [7 votes]: Yes it works.
No, this can't be deduced from the Poincare recurrence theorem. If it was possible, the ergodic hypothesis would not be needed.
But without the ergodic hypothesis, it is easy to give a counterexample.
Just take a constant sequence of random variables $X_n = X_0$ and constant targets $A_n =A_0$. Then you can get in $A$ only if you are in $A$ from the start.


TITLE: En route to proving the periodicity of sine and cosine
QUESTION [0 upvotes]: This answer shows that $\cos(\kappa) = \sin(\kappa) = \frac{1}{\sqrt{2}}$ (but whilst $\cos$ and $\sin$ are yet to be defined, $c$ and $s$ are used).
The answer continues that if $0 < x < \kappa$, then $\frac{1}{\sqrt{2}} < x < 1$. Afterwards, it is claimed that $\frac{\kappa}{\sqrt{2}} < s(\kappa) < \kappa$ and $\frac{1}{\sqrt{2}} < \kappa < 1$.
Why do these bounds follow from a point of intersection?

REPLY [0 votes]: This does not directly answer the question but one can prove periodicity with another argument: for all $a\in{\mathbb R}$, one has
\begin{equation}
m:=\inf_{t\in[a, \infty)}|s(t)| = 0
\end{equation}
Indeed, if $b>a$, the mean value theorem applied to $c(x)$ tells us that for a certain number $t\in (a,b)$, one has
\begin{equation}
m\le |s(t)| = |c'(t)| = \left|\frac{c(b)-c(a)}{b-a}\right|\le \frac{2}{b-a}
\end{equation}
which tends to $0$ when $b\to\infty$.
A consequence of this is that $s$ cannot increase monotonically on $(0,\infty)$ otherwise we would have $0<s(a)\le m$ for any $a>0$ and this is a contradiction. Let $\alpha$ be the largest number such that $s$ increases on $(0, \alpha)$. We have $c(\alpha) = s'(\alpha) = 0$ and $s(\alpha)> 0$ and this implies $s(\alpha) = 1$.
The differential equation $y''+y=0$ now implies that $c(x) = s(x + \alpha)$ because both hand sides of this equality satisfy the same Cauchy problem at $x=0$. As $c$ is even, we conclude that $s(\alpha - x) = s(\alpha + x)$ and it follows that $s(x) = -s(x + 2\alpha)$ by the same Cauchy problem argument. Hence $s(x +  4\alpha) = -(-s(x)) = s(x)$ and $s$ is periodic with period $4\alpha$.


TITLE: A question about billiards
QUESTION [10 upvotes]: This is a question in a rather well investigated subject of which I know very little and I have a hard time "translating" the general results available.  Let me also say that I got interested in this question after a conversation with a friend about placements of cell phone towers. (It is too long a story to include here.)
Here's the problem.  $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bsP}{\boldsymbol{P}}$ $\newcommand{\bR}{\mathbb{R}}$ Fix a positive integer $n$ and let $\bsP_n$ denote the regular $n$-gon centered at the origin and inscribed in the unit circle.
The billiard in the title refers to the motion of a unit speed photon inside $\bsP_n$ assuming the edges of $\bsP_n$ are perfect mirrors.  For every $x$ in the interior of $\bsP_n$, any unit vector identified with a $\theta\in S^1$  and any $T>0$ we denote by $N_T(x,\theta)$ the number of "bounces-off-mirrors"  during the time interval $[0,T]$ of a photon that starts its motion at $x$ in the direction $\theta$.  The question involves the behavior of $N_T(x,\theta)$ for large $T$.
I believe that there exists a measurable function $\beta_n:\bsP_n\times S^1\to [0,\infty)$ such that for almost all $(x,\theta)\in\bsP_n\times S^1$ we have
$$\lim_{T\to\infty}\frac{1}{T}N_T(x,\theta)=\beta_n(x,\theta).$$
The questions involve the proportionality constants $\beta_n(x,\theta)$.

Has anyone computed explicitly the function $\beta_n(x,\theta)$?
Has anyone computed explicitly the average

$$\bar{\beta}_n(x):=\frac{1}{2\pi} \int_0^{2\pi}\beta_n(x,\theta) d\theta ?$$
(I am inclined to believe that $\bar{\beta}_n(x)$ is constant.)
Above, the cases $n=3,6$ are of particular interest in the original cell phone tower problem.
Finally, has anyone investigated the behavior of $\beta_n(x,\theta)$ as $n\to\infty$?
Thanks.

REPLY [11 votes]: The answer is given by the general theory of rational billiard flows (i.e., those on polygons whose angles are rational multiples of $\pi$). On such a polygon $Q$ the tangent vectors to any given orbit are parallel to a finite set of unit vectors, so that the orbits with initial direction $\theta$ lie on an invariant surface $M_\theta$ which consists of a finite number of copies of $Q$, one for each potential direction of an orbit with initial direction $\theta$. According to a theorem of Kerckhoff, Masur and Smillie (http://www.ams.org/mathscinet-getitem?mr=855297 or the announcement http://www.ams.org/mathscinet-getitem?mr=799797), for almost all $\theta$ the flow on $M_\theta$ is uniquely ergodic. It implies that for a.e. $\theta$ the projection to $Q$ of every orbit with initial direction $\theta$ is uniformly distributed in $Q$ (with respect to the Lebesgue measure).
Therefore, $1/\beta_n(x,\theta)$ coincides, for a.e. $\theta$ and any $x\in Q$, with the average length of segments in billiard orbits, i.e., by the ergodic theorem, with the average of the function $L(x,\theta)$ (the distance between the points where the line issued from $x$ in the direction $\theta$ intersects the boundary of $Q$) with respect to the Lebesgue measure on $M_\theta$, which only depends on $\theta$ and should be quite easy to caclulate explicitly in the case of regular polygons.