Let $A_n=\sum\limits_{k=1}^n \sin k $ , show that there exists $M>0$ , $|A_n|
Show that $A_n=\sum\limits_{k=1}^n \sin k $ is bounded?
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16What are your thoughts on the problem? What have you tried? Just so you know, it is also considered a bit rude to post in the imperative. It is quite alright to just ask a question if you have one instead of giving orders. – 2011-08-25
3 Answers
A more general formula would be:
$A_n=\sum_{k=1}^{n} \sin k\theta = \frac{\sin\theta+\sin n\theta-\sin(n+1)\theta}{2(1-\cos\theta)}$ So $A_n$ is clearly bounded (just simply check the case where $\theta=1$).
The formula can be proved by induction using the trig identity: $\sin\alpha+\sin\beta=2\sin(\frac{\alpha+\beta}{2})\cos(\frac{\alpha-\beta}{2})$.
Hint: Since $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$ we can rewrite $\sum_{n=1}^{K} \sin n = \frac {1}{2i}\sum_{n=1}^K (e^i)^n-\frac {1}{2i}\sum_{n=1}^K (e^{-i})^n$ and both of these are geometric series.
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0@EricNaslund If I've been so insistent it's because I'm unable to work it out. I would be very grateful if you could complete the answer, when you have time :-) – 2014-04-08
Note that $2(\sin k)(\sin(0.5))=\cos(k-0.5)-\cos(k+0.5).$ This is obtained by using the ordinary expression for the cosine of a sum.
Add up, $k=1$ to $n$. On the right, there is mass cancellation. We get $\cos(0.5)-\cos(n+0.5).$ Thus our sum of sines is $\frac{\cos(0.5)-\cos(n+0.5)}{2\sin(0.5)}.$ We can now obtain the desired bound for $|A_n|$. For example, $2$ works, but not by much.
We could modify the appearance of the above formula by using the fact that $\cos(0.5)-\cos(n+0.5)=2\sin(n/2)\sin(n/2+0.5)$.
Generalization: The same idea can be used to find a closed form for $\sum_{k=0}^{n-1} \sin(\alpha +k\delta).$ Sums of cosines can be handled in a similar way.
Comment: This answer was written up because the OP, in a comment, asked for a solution that only uses real functions. However, summing complex exponentials, as in the solution by @Eric Naslund, is the right way to handle the problem.
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0Ok, got it - I just worked out how to show it, too, which is a nice technique to review. Thanks for your time and have a great night @andrenicolas :-) – 2015-11-25