Review / Part V — Fourier Transform / Ch 23

Exponential Fourier Series

Ch 23 — Part V — Fourier Transform

Instead of sines and cosines, Fourier series can be written compactly with complex exponentials — the cleaner starting point for transforms.

Exponential form

For \(f\) periodic of period \(2L\): $$f(x) = \sum_{n = -\infty}^\infty c_n e^{i n \pi x / L},\quad c_n = \frac{1}{2L}\int_{-L}^L f(x) e^{-i n \pi x / L}\,dx.$$

Relation to real form

\(c_n = \tfrac{1}{2}(a_n - i b_n),\; c_{-n} = \overline{c_n}\) for real \(f\).

Why this form

Complex exponentials are the simultaneous eigenfunctions of translation and differentiation — key to derivative properties, convolutions, and the transform limit (as \(L\to\infty\)).

Practice Problems

Problem 1easy
Find the exponential Fourier coefficients of \(f(x) = e^{ix}\) on \([-\pi,\pi]\).
Hint
Already a single complex exponential.
Solution

\(c_1 = 1\); all other \(c_n = 0\).

Answer: \(c_1 = 1.\)

Problem 2easy
Express \(\cos(\pi x/L)\) in exponential form.
Hint
Euler formula.
Solution

\(\cos(\pi x/L) = \tfrac{1}{2} e^{i\pi x/L} + \tfrac{1}{2} e^{-i\pi x/L}\), so \(c_{\pm 1} = 1/2\), others zero.

Answer: \(c_{\pm 1} = 1/2.\)

Problem 3medium
If \(f\) is real, show \(c_{-n} = \overline{c_n}\).
Hint
Conjugate the defining integral.
Solution

\(\overline{c_n} = \frac{1}{2L}\int \overline{f} e^{i n\pi x / L} dx = \frac{1}{2L}\int f e^{-i(-n)\pi x/L} dx = c_{-n}\).

Answer: See solution.