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MathidiaMathematics Academy
Partial differential equations II
Lesson 26 of 75
Introduction

Fourier series and transforms

Objective. Represent functions and solve linear PDEs spectrally.

  1. 1. Learn
  2. 2. Worked example
  3. 3. Practice
  4. 4. Feedback
  5. 5. Continue

Learn: the key idea

Fourier series and transforms represent a function as a sum or integral of sines and cosines, turning differentiation into multiplication in frequency space. For a 2π-periodic f, write f(x) = a₀/2 + Σ(aₙcos nx + bₙsin nx) with coefficients given by integrals of f against cos nx and sin nx. The Fourier transform extends this to non-periodic functions on the whole line via f̂(k) = ∫f(x)e^(−ikx)dx. Because d/dx becomes multiplication by ik, linear PDEs such as the heat equation uₜ = uₓₓ become ordinary differential equations in time for each mode, which are solved and then summed or inverted back. Example: for f(x) = x on (−π, π), odd symmetry gives bₙ = 2(−1)^(n+1)/n and all aₙ = 0.

Worked example

Find the Fourier sine coefficient b₁ for the odd 2π-periodic extension of f(x) = x on (−π, π).

2
Illustrated card showing 2
  1. 1Formula: bₙ = (1/π)∫₋π^π x sin(nx) dx, doubled to (2/π)∫₀^π x sin(nx) dx by oddness.
  2. 2Integrate by parts: ∫₀^π x sin(nx) dx = [−x cos(nx)/n]₀^π + (1/n)∫₀^π cos(nx) dx.
  3. 3Boundary term gives −π cos(nπ)/n; the remaining integral of cos(nx) over (0,π) vanishes for integer n.
  4. 4For n = 1: bₙ = (2/π)(−π cos(nπ)/n) = −2cos(nπ)/n, so b₁ = −2(−1)/1 = 2.

Answer: b₁ = 2

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Common mistake. Students forget that a Fourier series converges to the midpoint of a jump discontinuity, not to either one-sided value; at a jump from −1 to 1 the series sums to 0 there, not ±1.

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