3

In this posting, it is stated that

it is well known that $$ \lim_{n\rightarrow\infty}\frac{|\sin 1|+\ldots +|\sin n|}{n}=\frac{2}{\pi} $$ 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

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

Tha last integral can be broken down in integrals over $(k\pi,(k+1)\pi]$, $k=1,\ldots \big[\tfrac{n}{\pi}\big]-1$, and a residual integral over $\Big[\big[\tfrac{n}{\pi}\big],n\Big]$. 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, $$\frac{1}{n}\int^n_0|\sin u|\,du\xrightarrow{n\rightarrow\infty}\frac{2}{\pi}$$

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?

Mittens
  • 39,145
  • 1
    I don't think this approximation works: why is $\sin k \approx \sin nt$ for large $n$? – Alex Jul 21 '20 at 16:47
  • This may help: https://math.stackexchange.com/questions/650403/which-methods-different-than-the-natural-one-can-one-devise-to-confirm-that-the – VIVID Jul 21 '20 at 16:59
  • 1
    The limit of the sum is equivalent to $\frac1{2\pi}\int_0^{2\pi}|\sin(x)|,dx$. – Mark Viola Jul 21 '20 at 17:06
  • @Mark Viola: that is what I got using a little ergodic theory. Is that how you also obtained the results? If you have a slick trick up your sleeve, I would like to see it. I tried simple calculus and even Abel summation but things did not look pretty. – Mittens Jul 21 '20 at 20:00
  • @MarkViola: Do you know a Calculus based trick for this problem? – Mittens Jul 21 '20 at 23:20

2 Answers2

2

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

For any $\theta\in(0,1)$ define the rotation map

$$ \begin{align} R_\theta&:\mathbb{S}^1\rightarrow\mathbb{S}^1\\ z&\mapsto z e^{2\pi i\theta} \end{align} $$

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 $f:\mathbb{S}^1:\rightarrow\mathbb{C}^1$, and any $z\in\mathbb{S}^1$, $$ \lim_\limits{n\rightarrow\infty}\frac{1}{n}\sum^{n-1}_{k=0}f\circ R^k_\theta(z)=\int_{\mathbb{S}^1} f(w)\,\lambda_{\mathbb{S}^1}(dw) $$ where $R^0_\theta(z)=z$ and $R^k_\theta(z)=R_\theta(R^{k-1}_\theta(z))$ for all $z\in\mathbb{S}^1$ and $k\geq1$.


Back to your problem.

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

Consider the function $f:\mathbb{Z}\rightarrow\mathbb{R}$ 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

$$ \lim_{n\rightarrow\infty}\frac{1}{n}\sum^{n-1}_{k=0}|\sin(x+k)| = \frac{1}{2\pi}\int^{2\pi}_0|\sin x|\,dx=\frac{2}{\pi} $$ 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 $f_k(z)=z^k$, $k\in\mathbb{\mathbb{Z}}$.

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

For $|k|\geq1$, $$S_nf_k(z)\equiv1=\frac{z^k}{n}\sum^{n-1}_{j=0}e^{i2\pi\theta k j}= \frac{z^k}{n}\frac{1-e^{in 2\pi k\theta}}{1-e^{i2\pi k\theta}}\xrightarrow{n\rightarrow\infty}0=\int_{\mathbb{S}^1}f_k\,\lambda_{\mathbb{S}^1}=\frac{1}{2\pi}\int^{2\pi}_0 e^{ikx}\,dx $$ since $e^{i2\pi k\theta}\neq0$ for all $k\in\mathbb{Z}$. This means that the statement holds for all trigonometric polynomials. By the (complex) Stone-Weierstrass theorem, the result then holds for any $f\in\mathcal{C}(\mathbb{S}^1)$.

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 < f\leq 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 $$-\varepsilon\leq \liminf_{n\rightarrow\infty}S_nf(z)-\int f\leq\limsup_n S_nf-\int f\leq\varepsilon$$ for all $\varepsilon>0$ and $z\in\mathbb{S}^1$. 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).

Mittens
  • 39,145
  • Thanks for adding a detail explanation of Weyl's theorem. – Mittens Jul 21 '20 at 23:19
  • Oliver [this reference]( https://math.stackexchange.com/questions/650403/which-methods-different-than-the-natural-one-can-one-devise-to-confirm-that-thewas mentioned in a comment) was mentioned in a comment. Note that in one of the two posted solutions, $|\cos(n)|$ is expanded as a Fourier series and then summed over $n$. It seems to lead to an elementary wa forward. – Mark Viola Jul 22 '20 at 01:07
1

Suggestion:

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

  • That is very informative if one is aware of some results of Ergodic theory. I am afraid the OP may not be familiar with that yet. I any case, your hint help me to find an sort of "probabilistic" answer, as the person who made the claim in the OP stated. – Mittens Jul 21 '20 at 20:05
  • @OliverDiaz: I am jut addressing the OP's concern for "the Riemann sum I propose is not for an integral over a fixed interval.". –  Jul 21 '20 at 20:08