I wish to ask you guys to fill in a few steps for the derivation of complex form of Fourier series.
This is taken from "Fourier series" of Tolstov (Dover publication).
$$f(x)\sim c_0+\sum_{n=1}^{m}(c_ne^{inx}+c_{-n}e^{-inx})=\sum_{\color{red}{n=-m}}^{m} c_ne^{inx}{\tag1}$$
The author said: "Therefore it is natural to write:
$$f(x)\sim\sum_{-\infty}^{+\infty}c_ne^{inx}\tag{2}$$
$c_n=\dfrac{a_0}{2}$, $c_n=\dfrac{a_n-ib_n}{2}$, $c_{-n}=\dfrac{a_n+ib_n}{2}$
I am totally confused with the summation index. Why does $n=-m$ lead to the exclusion of $c_{-n}e^{-inx}$. I have watched some youtube videos on the complex form of Fourier series but they don't write the standard Fourier series as a partial sum. I think the author skips some steps already.
Isn't $\displaystyle\sum_{n=-m}^{m}c_ne^{inx}$ means $f(x)\sim c_0+\displaystyle\sum_{-\infty}^{+\infty}(c_ne^{inx}+c_{m}e^{imx})$
Also, how can you transform $(1)$ into $(2)$
Best Answer
\begin{eqnarray*} c_0+\sum_{n=1}^{m}(c_ne^{inx}+c_{-n}e^{-inx}) &=& c_0+(c_1 e^{ix}+c_{-1} e^{-ix}) +\cdots + (c_me^{imx}+c_{-m}e^{-imx}) \end{eqnarray*} Now rearrange the order of the terms to \begin{eqnarray*} c_0+\sum_{n=1}^{m}(c_ne^{inx}+c_{-n}e^{-inx}) &=& c_{-m}e^{-imx} +\cdots +c_{-1} e^{-ix} + c_0+c_1 e^{ix} +\cdots + c_me^{imx} \\ &=& \sum_{\color{red}{n=-m}}^{m} c_ne^{inx}. \end{eqnarray*} How do we get from $(1)$ to $(2)$ ? ... What does the symbol $ \sim$ mean ?