---
title: "Complex Fourier series; verify convolution"
subject: "Applied Mathematics"
chapter: 4
key: "semester_4/applied_mathematics:fourier-series-complex-convolution"
source: https://www.bctengineer.com/q/semester_4/applied_mathematics/fourier-series-complex-convolution
site: BCT Engineering
---

# Complex Fourier series; verify convolution

Applied Mathematics · Chapter 4: Fourier Transform

- Asked in: 2069 Poush
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/applied_mathematics/fourier-series-complex-convolution
- Whole chapter: https://www.bctengineer.com/md/semester_4/applied_mathematics/chapter-4

## Question

Define Fourier series in complex form. Verify the convolution theorem for f(x) = g(x) = e^{-x²}.

## Short answer

Complex Fourier series of period 2L: f(x)=Σ_{n=-∞}^{∞} c_n e^{inπx/L}, c_n = (1/2L)∫_{-L}^{L} f(x) e^{-inπx/L} dx. For f=g=e^{-x²}, the Fourier transform convolution theorem is verified: left and right sides both equal π e^{-s²/2}.

## Answer

#### Fourier series in complex form

A periodic function f(x) of period 2L (satisfying the Dirichlet conditions) can be written as

**f(x) = Σ_{n=−∞}^{∞} cₙ e^{inπx/L}**

with complex coefficients

**cₙ = (1/2L) ∫_{−L}^{L} f(x) e^{−inπx/L} dx**, n = 0, ±1, ±2, …

It comes from the real form by writing cos θ = (e^{iθ} + e^{−iθ})/2 and sin θ = (e^{iθ} − e^{−iθ})/2i, giving cₙ = (aₙ − ibₙ)/2 and c₋ₙ = (aₙ + ibₙ)/2, c₀ = a₀/2. For L = π: f(x) = Σ cₙ e^{inx}, cₙ = (1/2π) ∫_{−π}^{π} f e^{−inx} dx.

#### Verifying the convolution theorem for f = g = e^{−x²}

For non-periodic functions the convolution theorem belongs to the Fourier transform, F(s) = ∫ f(x) e^{−isx} dx, with (f * g)(x) = ∫ f(u) g(x − u) du and F{f * g} = F(s)G(s).

**Right side.** F(s) = G(s) = √π e^{−s²/4}, so F(s)G(s) = π e^{−s²/2}.

**Left side.** Since u² + (x − u)² = 2(u − x/2)² + x²/2,

```text
(f * g)(x) = e^{−x²/2} ∫ e^{−2(u − x/2)²} du = √(π/2) e^{−x²/2}
```

The transform of e^{−x²/2} is √(2π) e^{−s²/2}, hence

F{f * g} = √(π/2) · √(2π) e^{−s²/2} = π e^{−s²/2}

The two sides are equal, so the theorem is verified. ∎

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What are the complex Fourier series coefficients?
   A. c_n = (1/L) int f e^{i n pi x/L} dx
   B. c_n = (a_n + i b_n)
   C. c_n = (1/2L) int from -L to L f e^{-i n pi x/L} dx
   D. c_n = int f e^{-isx} dx
   Answer: C. With period 2L. (Key point: Complex coefficients)

2. How are c_n related to a_n, b_n?
   A. c_n = a_n + i b_n
   B. c_n = (a_n + i b_n)/2 for n>0
   C. c_n = a_n b_n
   D. c_n = (a_n - i b_n)/2 and c_0 = a_0/2
   Answer: D. Via cos and sin in exponentials. (Key point: Relation to real form)

3. Why is the Fourier transform (not series) used for convolution?
   A. e^{-x^2} is non-periodic
   B. It is periodic
   C. It is odd
   D. It has period 2 pi
   Answer: A. Series apply to periodic functions. (Key point: Which transform)

4. What is (f*g)(x)?
   A. sqrt(pi) e^{-x^2/2}
   B. sqrt(pi/2) e^{-x^2/2}
   C. sqrt(pi/2) e^{-x^2}
   D. pi e^{-x^2/2}
   Answer: B. Completing the square gives a Gaussian. (Key point: Verification result)

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