Use symmetry before calculating Fourier coefficients
A Fourier series expresses a periodic signal using sine and cosine terms. On a symmetric interval −L ≤ x ≤ L, an even function satisfies f(−x) = f(x), while an odd function satisfies f(−x) = −f(x). Even functions have only the constant and cosine terms; odd functions have only sine terms. A function can be neither even nor odd. Test the whole function, rather than judging it from only one term.
Worked practice example
Classify f(x) = x² + 3, g(x) = x³ − 2x and h(x) = x + 1 on a symmetric interval.
| Step | Working |
|---|---|
| Test f(−x) | (−x)² + 3 = x² + 3 = f(x): f is even, so bₙ = 0. |
| Test g(−x) | (−x)³ − 2(−x) = −x³ + 2x = −g(x): g is odd, so a₀ = aₙ = 0. |
| Test h(−x) | −x + 1 is neither h(x) nor −h(x): h is neither. |
| Separate h if useful | h(x) = 1 + x, its even part plus its odd part; each can be expanded separately. |
Check your method
For period 2L the terms are cos(nπx/L) and sin(nπx/L). For an even function, aₙ = (2/L)∫ from 0 to L f(x)cos(nπx/L) dx, including n = 0 for a₀; the constant in the series is a₀/2. Forgetting that final half is a common error.
Try it yourself
Which coefficients vanish for f(x) = |x| on −π ≤ x ≤ π? What is its constant term?
Show the answer and method
The function is even, so every bₙ vanishes. a₀ = (2/π)∫ from 0 to π x dx = π, hence the constant term is π/2.
This written practice example was prepared for Jemshah revision. It is additional practice, not an official examination question or a transcript of the video.