# fourier series and exponential

By equating the sine-cosine and exponential Fourier series for the function $f$:

$$f(x) = \sum_{n = 0}^\infty f_n \cdot e^{inx} = a_0 + \sum_{n=1}^{\infty} (a_n\cos(nx) + b_n\sin(nx))$$

and using Euler's formula, how can we find the explicit relation between the coefficients $f_n$ and the $a_n$, $b_n$?

-

Note first that we have in the complex fourier series also negaitve $n$, that is $$f(x) = \sum_{n=-\infty}^\infty f_n \exp(inx)$$ Now, by Euler $\exp(inx) = \cos(nx) + i\sin(nx)$. Now $\cos$ is even and $\sin$ is odd, so we have for $n\in \mathbb N$: \begin{align*} f_n\exp(inx) + f_{-n}\exp(-inx) &= f_n\cos(nx) + f_{-n}\cos(-nx) + i\bigl(f_n\sin(nx) + f_{-n}\sin(-nx)\bigr)\\ &= (f_n + f_{-n})\cos(nx) + i(f_n - f_{-n})\sin(nx) \end{align*} So we have $$f(x) = f_0 + \sum_{n=1}^\infty \bigl((f_n+f_{-n})\cos(nx) + i(f_n-f_{-n})\sin(nx)\bigr)$$ Equating coefficients yields \begin{align*} a_0 &= f_0\\ a_n &= f_n + f_{-n} \qquad n \ge 1\\ b_n &= i\bigl(f_n - f_{-n}\bigr) \end{align*}
We have the following trigonometric identities: $$\cos(nx) = \frac{e^{inx} + e^{-inx}}{2}, \quad \sin(nx) = \frac{e^{inx} - e^{inx}}{2i}$$ From this, it follows that $$a_0 = f_0, \, a_n = f_n + f_{-n}, \, b_n = i(f_n - f_{-n})$$
EDIT: I have noticed that your formula in your question is wrong. In the exponential Fourier series, you must sum $n$ over all integers, not just non-negative.
I think, the main point is that the exponential version sums up from $-\infty$ instead of $0$. – Berci Nov 17 '12 at 11:17