---
title: "Fourier series: e^(3x)"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:fourier-series-e-3x"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-series-e-3x
site: BCT Engineering
---

# Fourier series: e^(3x)

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2071 Shrawan, 5 marks
- Video: [#mathematics fourier series  f(x)=e^x in [0,2π] #diploma](https://www.youtube.com/watch?v=PMCmRPNEM8M) — MB Uday
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-series-e-3x
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Obtain the Fourier series of f(x) = e^(3x) in 0 < x < 2π.

## Short answer

Fourier series on (0,2π) has a₀ = (e^(6π)−1)/(3π), aₙ = 3(e^(6π)−1)/[π(n²+9)], bₙ = −n(e^(6π)−1)/[π(n²+9)]. Thus e^(3x) = ((e^(6π)−1)/π)[ 1/6 + Σ_{n=1}^∞ (3 cos nx − n sin nx)/(n²+9) ].

## Answer

On (0, 2π) of period 2π: f(x) = a₀/2 + Σ (aₙ cos nx + bₙ sin nx), a₀ = (1/π)∫₀^{2π} f dx, aₙ = (1/π)∫₀^{2π} f cos nx dx, bₙ = (1/π)∫₀^{2π} f sin nx dx.

Use ∫ e^(3x) cos nx dx = e^(3x)(3 cos nx + n sin nx)/(9 + n²) and ∫ e^(3x) sin nx dx = e^(3x)(3 sin nx − n cos nx)/(9 + n²). At x = 2π, sin 2nπ = 0 and cos 2nπ = 1.

```text
a₀ = (1/π)[e^(3x)/3]₀^{2π} = (e^(6π) − 1)/(3π)

aₙ = (1/π)·[ 3e^(6π) − 3 ]/(9 + n²) = 3(e^(6π) − 1)/[π(n² + 9)]

bₙ = (1/π)·[ −n e^(6π) + n ]/(9 + n²) = −n(e^(6π) − 1)/[π(n² + 9)]
```

**Fourier series**

e^(3x) = (e^(6π) − 1)/π · { 1/6 + Σ (n=1 to ∞) [ 3 cos nx − n sin nx ] / (n² + 9) },  0 < x < 2π

Writing out: e^(3x) = ((e^(6π) − 1)/π)[ 1/6 + (3 cos x − sin x)/10 + (3 cos 2x − 2 sin 2x)/13 + … ]

At the ends x = 0 and 2π, the series converges to the mean ½(1 + e^(6π)).

## Check yourself

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

1. What is a0 for e^(3x) on (0, 2pi)?
   A. (e^(6 pi) - 1)/pi
   B. (e^(6 pi) + 1)/(3 pi)
   C. (e^(6 pi) - 1)/(3 pi)
   D. (e^(6 pi) - 1)/(6 pi)
   Answer: C. (1/pi)[e^(3x)/3] from 0 to 2pi. (Key point: a0 formula)

2. Which simplification is used at x = 2 pi?
   A. cos 2n pi = (-1)^n
   B. sin 2n pi = 1, cos 2n pi = 0
   C. both equal 1
   D. sin 2n pi = 0, cos 2n pi = 1
   Answer: D. n is an integer so 2n pi is a full number of turns. (Key point: Using sin and cos at 2 n pi)

3. What is bn?
   A. -3(e^(6 pi) - 1)/[pi(n^2+9)]
   B. n(e^(6 pi) - 1)/[pi(n^2+9)]
   C. -n(e^(6 pi) + 1)/[pi(n^2+9)]
   D. -n(e^(6 pi) - 1)/[pi(n^2+9)]
   Answer: D. The sine integral has -n cos nx term, giving a negative sign. (Key point: bn sign)

4. At x = 0 and 2 pi the series converges to?
   A. (1 + e^(6 pi))/2
   B. 0
   C. 1
   D. e^(6 pi)
   Answer: A. At a jump it converges to the mean of left and right limits. (Key point: Convergence at ends)

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