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limit of arithmetic mean of $|sin n|$, $ninmathbb{N}$

Mathematics Asked by Jean L. on November 9, 2021

In this posting, it is stated that

it is well known that $$ lim_{nrightarrowinfty}frac{|sin 1|+ldots +|sin n|}{n}=frac{2}{pi} $$ which can be obtained by the
uniform distribution.

I have tried to work that out but I can’t get too far.

I tried instead to look at the problem as a Riemann sum

$$frac{1}{n}sum^n_{k=1}|sin k|=frac{1}{n}sum^n_{k=1}big|sin big(ntfrac{k}{n}big)big|approx int^1_0|sin nt|,dt=frac{1}{n}int^n_0|sin u|,du$$

Tha last integral can be broken down in integrals over $(kpi,(k+1)pi]$, $k=1,ldots big[tfrac{n}{pi}big]-1$, and a residual integral over $Big[big[tfrac{n}{pi}big],nBig]$. The residual is bounded by $pi$ and so, when hit by $1/n$ it goes to $0$. The other integrals amount to $big[tfrac{n}{pi}big]2$ and so,
$$frac{1}{n}int^n_0|sin u|,duxrightarrow{nrightarrowinfty}frac{2}{pi}$$

This however does not answer fully the problem since the Riemann sum I propose is not for an integral over a fixed interval.

Can anybody give another suggestion?

2 Answers

I am not aware thus far if there is a simple Calculus argument for this problem. Here is I present an argument that requires basic knowledge of Ergodic theory.


Consider the unit circle $mathbb{S}^1={zinmathbb{C}:|z|=1}$ equipped with the $sigma$-algebra inherited as a subspace of $mathbb{R}^2$ and the measure $lambda_{mathbb{S}^1}:=frac{1}{2pi}lambda_1$, where $lambda_1$ is the the arch-length measure. (Equivalently, this is the space $[0,1]mod 2pi$ with the Borel $sigma$-algebra and Lebesgue measure restricted to $[0,1]$.)

For any $thetain(0,1)$ define the rotation map

$$ begin{align} R_theta&:mathbb{S}^1rightarrowmathbb{S}^1\ z&mapsto z e^{2pi itheta} end{align} $$

As you can easily see, $R_theta$ is invariant with resect to $lambda_{mathbb{S}^1}$, that is $lambda_{mathbb{S}^1}big(R^{-1}_{theta}(A)big)=lambda_{mathbb{S}^1}(A)$, that is rotating the circle by some angle $theta$ does not change the length of an arch.


A little ergodic theory:

Theorem (Weyl): Suppose $R_theta$ is an irrational rotation. Then, for every bounded Riemann integrable function $f:mathbb{S}^1:rightarrowmathbb{C}^1$, and any $zinmathbb{S}^1$, $$ lim_limits{nrightarrowinfty}frac{1}{n}sum^{n-1}_{k=0}fcirc R^k_theta(z)=int_{mathbb{S}^1} f(w),lambda_{mathbb{S}^1}(dw) $$ where $R^0_theta(z)=z$ and $R^k_theta(z)=R_theta(R^{k-1}_theta(z))$ for all $zinmathbb{S}^1$ and $kgeq1$.


Back to your problem.

Consider the rotation $R_theta$ with $theta=frac{1}{2pi}$. For $z=e^{ix}$, $$ R^k_theta(z)=ze^{2pi itheta k}=e^{i(x + k)} $$

Consider the function $f:mathbb{Z}rightarrowmathbb{R}$ given by $f(z)=|operatorname{Im}(z)|$. Then $$ f(R^k_theta(z))=|sin(x+k)| $$

Since $f$ is continuous and thus, integrable, we can use Weyl's theorem to obtain that

$$ lim_{nrightarrowinfty}frac{1}{n}sum^{n-1}_{k=0}|sin(x+k)| = frac{1}{2pi}int^{2pi}_0|sin x|,dx=frac{2}{pi} $$ for all $x$. In your case, $x=1$.


Edit: Since Weyl's theorem can be proved with only tools from Calculus, I think that It would be worth while to add a short proof:

Proof of Weyl's theorem:

For any function $f$ on $mathbb{S}^1$, let $S_nf(z)=frac{1}{n}sum^{n-1}_{j=1}f(R^j_theta(z))$. Consider polynomials $f_k(z)=z^k$, $kinmathbb{mathbb{Z}}$.

For $k=0$, $$S_nf_0(x)equiv1=frac{1}{2pi}int^{2pi}_0 f_0 $$

For $|k|geq1$, $$S_nf_k(z)equiv1=frac{z^k}{n}sum^{n-1}_{j=0}e^{i2pitheta k j}= frac{z^k}{n}frac{1-e^{in 2pi ktheta}}{1-e^{i2pi ktheta}}xrightarrow{nrightarrowinfty}0=int_{mathbb{S}^1}f_k,lambda_{mathbb{S}^1}=frac{1}{2pi}int^{2pi}_0 e^{ikx},dx $$ since $e^{i2pi ktheta}neq0$ for all $kinmathbb{Z}$. This means that the statement holds for all trigonometric polynomials. By the (complex) Stone-Weierstrass theorem, the result then holds for any $finmathcal{C}(mathbb{S}^1)$.

To extend the result to any Riemann integrable function $f$, it is enough to assume that $f$ is real valued. Given $varepsilon>0$, we can choose continuous functions $g_varepsilon$ and $h_varepsilon$ such that $ g_varepsilon < fleq h_varepsilon$, and $$ int_{mathbb{S}^1}f,lambda_{mathbb{S}^1} -varepsilon <int_{mathbb{S}^1} g_varepsilon, lambda_{mathbb{S}^1}leq int_{mathbb{S}^1}h_varepsilon ,lambda_{mathbb{S}^1}< int_{mathbb{S}^1}f,lambda_{mathbb{S}^1} +varepsilon $$ Then $$ S_ng_varepsilon-int g_varepsilon -varepsilon leq S_nf(z)-int f leq S_nh_varepsilon(z)-int h_varepsilon+varepsilon $$ whence we conclude that $$-varepsilonleq liminf_{nrightarrowinfty}S_nf(z)-int fleqlimsup_n S_nf-int fleqvarepsilon$$ for all $varepsilon>0$ and $zinmathbb{S}^1$. This completes the proof.


Comment: Weyl's result is a little stronger than the plain Ergodic theorem since in the former there are no $mu$-a.s. exceptional points where convergence fails (even though, were there any, they would form a set of $lambda_{mathbb{S}^1}$-measure zero).

Answered by Oliver Diaz on November 9, 2021

Suggestion:

$$|sin(k)|=|sin(kbmodpi)|$$ and you can integrate over $[0,pi)$.

Answered by user65203 on November 9, 2021

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