Supposing I want to find the conjugate of
$$\tag{1}\label{eq1}\sum_{n=0}^{m}n \cos(2\pi x n)$$
If I view (1) as a Fourier series then what would be the conjugate Fourier series and how would I go about doing so?
If I first change the cosine using the identity $\cos (ax) = \frac{1}{2} (e^{iax}+e^{-iax})$ would that make \eqref{eq2}?
$$\sum_{n=0}^{m}\frac{n}{2} (e^{i2\pi x}+e^{-i2\pi x})\tag{2}\label{eq2}$$
then could I simply from $\psi=Ae^{i(kx-\omega t)}+ Be^{-i(kx+\omega t)}$ using identity $\bar{\psi} = \bar{A} \mathrm{e}^{-i (k x - \omega t)} + \bar{B} \mathrm{e}^{i (k x + \omega t)}$
do this?
$$\sum_{n=0}^{m}\frac{n}{2} (e^{-i2\pi x}+e^{i2\pi x})\tag{3}\label{eq3}$$
So I suppose from \eqref{eq3} if my working is correct we end up where we begun?
Using the Wikipedia definition, as $a_n = n$ and $b_n = 0$, the conjugate series will be $$\sum_{n=1}^m n\sin(2\pi x n).$$