Is there a closed form solution to the following sum:
$$\sum_{i = 1}^n i \cos(\beta i)?$$
Thank you very much.
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1The answer to the posted question is "Yes, there is a closed-form solution to that sum." – Mark Viola Apr 05 '20 at 23:11
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1Hint: What is $\sum_{k=1}^n k z^k$ ? Substitute $z$ by $e^{\beta i}$ and look at the real part... – achille hui Apr 05 '20 at 23:11
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Duplicate of: Summing $ \sum _{k=1}^{n} k\cos(k\theta) $ and $ \sum _{k=1}^{n} k\sin(k\theta) $. – Anne Bauval May 16 '24 at 15:30
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Hint: We have $$\sum_{k=1}^n k\cos(\beta k) = \Re\sum_{k=1}^nk\,(e^{i\beta})^k,$$ and $$\sum_{k=1}^nk\,z^k = z\,\frac d{dz}\sum_{k=1}^n z^k = z\,\frac d{dz}\Big(\frac{z(1-z^n)}{1-z}\Big).$$
Using the hints above, you should arrive at $$\sum_{k=1}^nk\cos(\beta k) = \frac{(n+1)\cos(n\beta) - n\cos((n+1)\beta)-1}{4\sin^2(\beta/2)}.$$
Luke Collins
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Thank you! Can we use this technique to obtain a closed form for
$$\sum_{k = 1}^n k \cos\left(\alpha n + \beta k\right)?$$
I tired... but didn't obtain anything useful. Thanks again!
– user304347 Apr 06 '20 at 23:36 -
@user304347 Yes, since that would correspond to $$\Re\Big(e^{i\alpha n}\sum_{k=1}^nk (e^{i\beta})^k\Big) = \frac{(n+1)\cos(n(\alpha+\beta))-n \cos(\beta+n(\alpha+\beta))-\cos(n\alpha)}{4 \sin^2(\beta /2)},$$ which you get by the same technique. – Luke Collins Apr 07 '20 at 09:49
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Thanks again! Yes, I've made a little mistake... now I have the same result. – user304347 Apr 07 '20 at 11:23