How can I compute this limit : $$\lim_{n \to \infty } {1 \over n}\sum\limits_{k = 1}^n {\left\lvert \sin k\right\rvert} $$ Thank you.
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1What are your thoughts and approach? – Jaideep Khare Mar 22 '17 at 17:39
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Think of which techniques you have learned in your course. Does any of those ring a bell? – mathreadler Mar 22 '17 at 17:43
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I tried to apply Stolz theorem but i have not got any result. – Gustave Mar 22 '17 at 17:43
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Equidistributed sequence, ergodic process, Monte-Carlo integration - did you see any of those terms in your course? Is it analysis or statistic course? – A.Γ. Mar 22 '17 at 18:40
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No..it is just an analysis exercise. – Gustave Mar 22 '17 at 18:47
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You need to invoke the equidistribution theorem (https://en.wikipedia.org/wiki/Equidistribution_theorem) because the sequence ${\left|\sin(k)\right|}_{k\geq 0}$ is not regular enough to apply standard manipulations. – Jack D'Aurizio Mar 22 '17 at 20:18
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By Weyl's equidistribution theorem the sequence $\{e^{in}\}_{n\geq 0}$ is dense in the unit circle and much more: it is equidistributed. In particular, since $\sin(k)=\text{Im}\,e^{ik}$, the limit $\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}\left|\sin(k)\right|$ is the average value of the function $\left|\sin(x)\right|$, i.e. $$ \frac{1}{\pi}\int_{0}^{\pi}\sin(x)\,dx = \color{red}{\frac{2}{\pi}}.$$
Jack D'Aurizio
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@Gustav: I believe we cannot. As a matter of fact, we may build sequences that are dense in $[-1,1]$ such that $\lim_{n\to +\infty}\sum_{k=1}^{n}a_k$ does not exist or differs from the expected value given by the associated integral. – Jack D'Aurizio Mar 22 '17 at 21:33
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However, we may prove the above result by exploiting the convergents of the continued fraction of $pi$, Lagrange's theorem and the Lipschitz-continuity of the sine function: $\left|\sin(k)\right|$ is an almost-periodic sequence with respect to the periods $7,106,113,33102,\ldots$. But that is not really different than going through the equidistribution theorem. – Jack D'Aurizio Mar 22 '17 at 21:37
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+1. A 'naive' application of Stolz-Ces$\mathrm{\grave{a}}$ro fails !!!. It "is not regular enough to apply standard manipulations" as you pointed out in the above comments. – Felix Marin Mar 23 '17 at 06:26
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Hint: $|\sin k|=\sin (k \bmod \pi)$ The values of $k \bmod \pi$ will bounce around in the interval $[0,\pi)$ so you are asked for the average value of $\sin x$ over this interval. What integral can you do to get this?
Ross Millikan
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Really Ross Millikan, i didn't understand your question, can you aske it in more simple way? Thanks. – Gustave Mar 22 '17 at 18:22
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2Important to note this is guaranteed because integers are not fractions of any irrational (which $\pi$ is), if the k were a fraction of $\pi$ the answer might be different. – mathreadler Mar 22 '17 at 18:26
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How would you find the average value of $\sin x$ over $[0,\pi)$? – Ross Millikan Mar 22 '17 at 18:33
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@Gustav, your expression can be interpreted as the limit of a Riemann sum. – Martín-Blas Pérez Pinilla Mar 22 '17 at 18:36
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There is a chance that limits are done before integrals are introduced and then it could be confusing. – mathreadler Mar 22 '17 at 18:36
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Martín-Blas Pérez Pinilla.. How it could be interpreted as a limit of Riemann integral ? – Gustave Mar 22 '17 at 18:52
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The area under the sine curve is the same as the average value of the sine times the length of the interval. – Ross Millikan Mar 22 '17 at 18:56
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2Bouncing around or density is not enough: we need equidistribution, that is stronger than density. – Jack D'Aurizio Mar 22 '17 at 20:16
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@JackD'Aurizio But if we had a non-equidistribution and density and could compute the density we could do a weighted integral instead? – mathreadler Mar 23 '17 at 12:39
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1@mathreadler: what does it mean compute the density? If it stands for know the distribution, well, of course, a weighted integral solves the problem. – Jack D'Aurizio Mar 23 '17 at 14:13
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EDIT as Jack points out below in comments I forgot the absolute value. I am sure this answer can be fixed by various means, but it would take some time, won't be pretty and probably add to any confusion already present. I should probably remove it and make a question of how to solve it using a similar technique.
Here is another approach, although probably not easier than the integral: use the addition formula
$$\sin(v+w) = \sin(v)\cos(w)+\cos(v)\sin(w)$$
a lot.
$$\sin(n+1) = \sin(n)\cos(1)+\cos(n)\sin(1)$$ $$\cos(n+1) = \cos(n)\cos(1)+\sin(n)\sin(1)$$
If you have learned to calculate recurrence relations with matrices for example a course in linear algebra, now is the time to put that to use.
mathreadler
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Maybe you can come back and look at it some time after you done some linear algebra. – mathreadler Mar 22 '17 at 19:02
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You are right @JackD'Aurizio and I can't find a way to patch it. Maybe in combination with Fourier series estimation of $|\sin(x)|$ could do it but that would be even more horribly overkill and not make for a very clean answer. – mathreadler Mar 23 '17 at 13:16