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is/are there a closed form for






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I was writing an answer, but Didier gave a much better one. But I will mention that Wolfram Alpha gives closed form for the first three. – please delete me Mar 27 '11 at 9:40
@Elfvind Sorry about that. Re Wolfram Alpha, I would be at a loss to use in any way whatsoever the formula it provides for the tangent sum... – Did Mar 27 '11 at 16:52
@Eivind: I misspelled your name. Sorry. I guess I linked somewhat unconsciously your name to Scandinavia and Scandinavia to elves... :-) – Did Apr 2 '11 at 10:52
@Didier: Hehe, that's OK. =) – please delete me Apr 4 '11 at 6:58
up vote 11 down vote accepted

The two first sums $S$ and $C$ are the imaginary part and the real part of the complex sum $$ \mathrm{e}^{\mathrm{i}a}+\mathrm{e}^{\mathrm{i}(a+d)}+\cdots+\mathrm{e}^{\mathrm{i}(a+nd)}. $$ Ths is nothing but a geometric sum with first term $\mathrm{e}^{\mathrm{i}a}$ and argument $z=\mathrm{e}^{\mathrm{i}d}$, thus $$ C+\mathrm{i}S=\mathrm{e}^{\mathrm{i}a}\frac{z^{n+1}-1}{z-1}=\mathrm{e}^{\mathrm{i}(a+(nd/2))}\frac{\sin((n+1)d/2)}{\sin(d/2)}, $$ for every $z\ne1$, that is, for every $d$ not in $2\pi\mathbb{Z}$. Using the shorthand $d=2b$ (hence $\sin(b)\ne0$), one gets finally $$ C=\cos(a+nb)\frac{\sin((n+1)b)}{\sin(b)}, \quad S=\sin(a+nb)\frac{\sin((n+1)b)}{\sin(b)}. $$ If $d$ is in $2\pi\mathbb{Z}$, $C=(n+1)\cos(a)$ and $S=(n+1)\sin(a)$ (and these are the limits of the formulas valid when $d$ is not in $2\pi\mathbb{Z}$).

I do not think simple closed form formulas exist for the three other sums you are interested in.

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