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I'm supposed to prove the following series expansion:

$$s_n=\sum_{n=1}^\infty r^n\cos n\theta=\frac{r\cos\theta-r^2}{1-2r\cos\theta+r^2}$$

I first note the following:

$$s_n=\sum_{n=1}^\infty \Re\left[re^{i\theta}\right]^n=\Re\left[\sum_{n=1}^\infty z^n\right]$$

Using geometric series definition:

$$s_n=\Re\left[\frac{1}{1-z}\right]$$

$$s_n=\Re\left[\frac{1-z}{1-2z+z^2}\right]$$

Substituting polar coordinates of $z$:

$$s_n=\Re\left[\frac{1-r\exp(i\theta)}{1-2r\exp(i\theta)+r^2\exp(2i\theta)}\right]$$

Taking real component:

$$s_n=\frac{1-r\cos\theta}{1-2r\cos\theta+r^2\cos2\theta}$$

I can't really seems to do anything beyond here, could someone point out where I've gone wrong?

Leonidas Lanier
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  • The first equality does not make sense: the LHS depends on $n$ while the RHS does not. –  Oct 31 '17 at 23:43
  • Moreover, how exactly did you "Taking real component" in the last step? One should not simply take the real part in each single term. –  Oct 31 '17 at 23:45
  • Using $\exp(i\theta)=\cos\theta+i\sin\theta$ – Leonidas Lanier Oct 31 '17 at 23:46
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    In the last step: $$ \Re\left[\frac{1-r\exp(i\theta)}{1-2r\exp(i\theta)+r^2\exp(2i\theta)}\right]\neq\frac{1-r\cos\theta}{1-2r\cos\theta+r^2\cos2\theta}$$

    One should not simply take the real part in each single term.

     –  Nov 01 '17 at 00:01
  • @Jack Why can I not do that with the denominator? – Leonidas Lanier Nov 01 '17 at 00:04
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    In general, $$\Re[\frac{f(z)}{g(z)}]\neq\frac{\Re(f(z))}{\Re(g(z))}.$$ For instance $$\Re[\frac{1-i}{1+i}]\neq 1.$$ –  Nov 01 '17 at 00:11
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    Consider the example $i/i$. Try to take real part of each term. – edm Nov 01 '17 at 00:11

2 Answers2

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Notice that $$\sum_{n=1}^\infty z^n=\frac{z}{1-z}\qquad |z|<1$$since the sum starts at $n=1$. Hence, $$\begin{align}\sum_{n=1}^\infty r^n\cos(n\theta)&=\Re\left( \frac{z}{1-z}\right)\\&=\Re\left(\frac{z-|z|^2}{|1-z|^2}\right)\\&=\frac{\Re(z)-|z|^2}{|1-z|^2}\\&=\frac{r\cos(\theta)-r^2}{(1-r\cos(\theta))^2+(-r\sin(\theta))^2}\\&=\frac{r\cos(\theta)-r^2}{1-2r\cos(\theta)+r^2}\end{align}$$ Note, $$\frac{z}{1-z}=\frac{z(1-\bar{z})}{|1-z|^2}$$

Dave
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  • Where does the $z$ in the numerator come from? From subtracting the $n=0$ case? – Leonidas Lanier Oct 31 '17 at 23:54
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    The geometric series starting at $n=1$ is $$\sum_{n=1}^\infty z^n=\frac{z}{1-z}$$which is not the same as when starting at $n=0$, $$\sum_{n=0}^\infty z^n=\frac{1}{1-z}$$Indeed, it can be shown in the same derivation as the geometric series formula, however we can also show it from the $n=0$ starting case: $$\sum_{n=1}^\infty z^n=\frac{1}{1-z}-1=\frac{1-(1-z)}{1-z}$$ – Dave Oct 31 '17 at 23:57
  • In general, $$\sum_{n=k}^\infty z^n=\frac{z^k}{1-z}$$provided the series converges of course. – Dave Oct 31 '17 at 23:58
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Note that the sum starts at $1$ so the geometric sum will give \begin{eqnarray*} s_n=\Re\left[\frac{z}{1-z}\right] \end{eqnarray*} We now multiply top and bottom by the conjugate of the denominator \begin{eqnarray*} \Re\left[\frac{z}{1-z}\right] &=& \Re\left[\frac{r \cos(\theta)+ir \sin(\theta) }{1-r \cos(\theta)-ir \sin(\theta)}\right] \\ &=& \Re\left[\frac{r \cos(\theta)+ir \sin(\theta) }{1-r \cos(\theta)-ir \sin(\theta)}\frac{1-r \cos(\theta)+ir \sin(\theta) }{1-r \cos(\theta)+ir \sin(\theta)}\right] \\ &=& \Re\left[\frac{r \cos(\theta)(1-r \cos(\theta))-(r \sin(\theta))^2 +i(.)}{(1-r \cos(\theta))^2+(r \sin(\theta))^2}\right] \\ &=& \frac{r \cos(\theta)-r^2 }{1-2r \cos(\theta)+r^2}. \\ \end{eqnarray*}

Donald Splutterwit
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