Fourier series invoking real part

[Tolaso J Kos]

We know that \( \displaystyle \ln(1-x) =-\sum_{n=1}^{\infty} \frac{x^n}{n} \) (which is nothing but the MacLaurin series of \( \ln (1-x) \) with center at \( x_0=0 \) ).

Use the above expansion to prove the Fourier series:

$$ \sum_{n=1}^{\infty} \frac{x^n \cos (na)}{n}= -\frac{1}{2} \ln (x^2-2x\cos a +1) $$

[Riemann]

This is not actually a Fourier series but a rather look like Fourier series. Anyway the solution will require us to take a look at the complex exponential series.

Well mapping $x \mapsto xe^{ia}$ we have that:

\begin{equation} \log \left ( 1-xe^{ia} \right ) = -\sum_{n=1}^{\infty} \frac{x^ne^{ina}}{n} \end{equation}

Taking the real part of $(1)$ we have that:

\begin{align*}
\Re \left \{ \sum_{n=1}^{\infty} \frac{x^ne^{ina}}{n} \right \} &= \Re \left \{ \sum_{n=1}^{\infty} \frac{x^n \left ( \cos na + i \sin na \right )}{n} \right \} \\
&= \sum_{n=1}^{\infty} \frac{x^n \cos na}{n}
\end{align*}

and

\begin{align*}
\Re \left \{ \log \left ( 1-xe^{ia} \right ) \right \} &=\Re \left \{ \log \left | 1-xe^{ia} \right |+ i \arg \left ( 1-xe^{ia} \right )\right \} \\
&= \frac{1}{2}\log \left ( x^2 \sin^2 a + \left ( 1- x \cos a \right )^2 \right )\\
&= \frac{1}{2} \log \left ( 1-2x \cos a + x^2 \right )
\end{align*}

Thus:

$$\bbox[blue, 2pt]{{\color{white}\log \left ( x^2-2x \cos a + 1 \right ) = -2 \sum_{n=1}^{\infty} \frac{x^n \cos na}{n}}}$$