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What is $\sum\limits_{n>0,\text{ odd}} r^n \sin(nx)$ in terms of $z=re^{ix}$? I tried to write $\sin(nx)={e^{inx}-e^{-inx}\over 2i}$ but then I have a sign problem because the $n$ on the associated $r$ is always $>0$.

Thanks in advance!

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For $r,x\in\mathbb{R}$, $\bar z = r e^{-i x}$ so $r^n \sin(n x)=r^n(e^{inx}-e^{-inx})/2i=(z^n-\bar{z}^n)/2i$. – Heike Nov 2 '11 at 16:32
@QED Careful, that is not quite right unless $r=1$. You need $\overline{z}$ instead of $z^{-1}$. – Eric Naslund Nov 2 '11 at 16:34
This question is probably difficult for high school! – NoChance Nov 2 '11 at 22:49

You're on the right track (or at least on a right track). After expanding the sine, put the $r^n$ on the numerator to get $$\sum_{n>0\text{ odd}} \frac{r^n(e^{ix})^n-r^n(e^{-ix})^n}{2i} = \sum_{n\text{ odd}}\frac{z^n-\overline z^n}{2i}$$ You can then split that into two series $$\frac{1}{2i}\left(\sum_{n\text{ odd}}z^n - \sum_{n\text{ odd}}\overline z^n\right)$$ and then get rid of the oddness restriction on $n$ by pulling out one of the factors in each series: $$\frac{1}{2i}\left(z\sum_{k=0}^\infty (z^2)^k - \overline z\sum_{k=0}^{\infty} (\overline z^2)^k\right)$$ Can you take it from there?

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Thanks loads! :) – Chip Nov 2 '11 at 17:31

Note: This is related to the Poisson Kernel. Knowledge of this can shorten the solution.

Your sum is $$\sum_{n=0}^{\infty}r^{2n+1}\sin((2n+1)x)=\frac{1}{2i}\sum_{n=0}^{\infty}r^{2n+1}\left(e^{i(2n+1)x}-e^{-i(2n+1)x}\right).$$ This is then $$\frac{1}{2i}\sum_{n=0}^{\infty}r^{2n+1}e^{i(2n+1)x}-\frac{1}{2i}\sum_{n=0}^{\infty}r^{2n+1}e^{-i(2n+1)x}=\frac{re^{ix}}{2i}\sum_{n=0}^{\infty}\left(re^{ix}\right)^{2n}-\frac{re^{-ix}}{2i}\sum_{n=0}^{\infty}\left(re^{-ix}\right)^{2n}.$$ Summing the geometric series, this becomes $$\frac{re^{ix}}{2i}\frac{1}{1-r^{2}e^{2ix}}-\frac{re^{-ix}}{2i}\frac{1}{1-r^{2}e^{-2ix}}=\frac{1}{2i}\left(\frac{re^{ix}-re^{-ix}-r^{3}e^{-ix}+r^{3}e^{ix}}{1-2r^{2}\cos\left(2x\right)+r^{4}}\right) $$


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Thanks loads! :) – Chip Nov 2 '11 at 17:31

As commented by Heike $$ r^{n}\sin nx=\frac{z^{n}-\overline{z}^{n}}{2i}. $$ This result can be derived by subtracting the complex conjugates $$ \begin{eqnarray*} z^{n} &=&r^{n}e^{inx}=r^{n}\cos nx+ir^{n}\sin nx \\ \overline{z}^{n} &=&r^{n}e^{-inx}=r^{n}\cos nx-ir^{n}\sin nx. \end{eqnarray*} $$ The given series can be rewritten as $$ \begin{eqnarray*} S &:&=\sum_{n>0\;\text{odd}}r^{n}\sin (nx)=\sum_{k=1}^{\infty }r^{2k-1}\sin (\left( 2k-1\right) x) \\ &=&\sum_{k=1}^{\infty }\frac{z^{2k-1}-\overline{z}^{2k-1}}{2i} \\ &=&\frac{1}{2i}\sum_{k=1}^{\infty }z^{2k-1}-\frac{1}{2i}\sum_{k=1}^{\infty } \overline{z}^{2k-1} \end{eqnarray*} $$ because $$ r^{2k-1}\sin \left( \left( 2k-1\right) x\right) =\frac{z^{2k-1}-\overline{z} ^{2k-1}}{2i}. $$ Since the sums of the two geometric series are $$ \sum_{k=1}^{\infty }z^{2k-1}=\frac{z}{1-z^{2}}\qquad (\text{ratio }z^{2}, \text{ first term }z=re^{ix},\ r<1) $$ and $$ \sum_{k=1}^{\infty }\overline{z}^{2k-1}=\frac{\overline{z}}{1-\overline{z} ^{2}}\qquad (\text{ratio }\overline{z}^{2},\text{ first term }\overline{z}) $$ we get $$ S=\frac{1}{2i}\frac{z}{1-z^{2}}-\frac{1}{2i}\frac{\overline{z}}{1-\overline{z }^{2}}. $$

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