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

# Fourier series: eˣ

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2076 Chaitra, 5 marks
- 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
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Obtain the Fourier series to represent the function f(x) = eˣ for −π ≤ x ≤ π.

## Short answer

Coefficients: a0 = 2 sinh π/π, a_n = 2(-1)^n sinh π/(π(1+n²)), b_n = -2n(-1)^n sinh π/(π(1+n²)). The series is e^x = (sinh π/π)[1 + 2 ∑_{n=1}^∞ (-1)^n (cos nx - n sin nx)/(1+n²)].

## Answer

f(x) = eˣ on (−π, π), period 2π. Use ∫ eˣ cos nx dx = eˣ(cos nx + n sin nx)/(1 + n²) and ∫ eˣ sin nx dx = eˣ(sin nx − n cos nx)/(1 + n²).

#### Coefficients
```text
a₀ = (1/π) ∫_{−π}^{π} eˣ dx = (e^π − e^(−π))/π = 2 sinh π / π

aₙ = (1/π) [ eˣ (cos nx + n sin nx)/(1 + n²) ]_{−π}^{π}
   = (−1)ⁿ (e^π − e^(−π)) / (π(1 + n²)) = 2 (−1)ⁿ sinh π / (π (1 + n²))

bₙ = (1/π) [ eˣ (sin nx − n cos nx)/(1 + n²) ]_{−π}^{π}
   = −n (−1)ⁿ (e^π − e^(−π)) / (π(1 + n²)) = −2n (−1)ⁿ sinh π / (π (1 + n²))
```

(since cos nπ = (−1)ⁿ and sin nπ = 0.)

#### Series
```text
eˣ = (sinh π / π) { 1 + 2 Σ_{n=1}^∞ [ (−1)ⁿ/(1 + n²) ] (cos nx − n sin nx) }
```

Expanded:

```text
eˣ = (sinh π/π) [ 1 − cos x + sin x + (2/5)(cos 2x − 2 sin 2x) − (2/10)(cos 3x − 3 sin 3x) + ... ]
```

(sinh π/π ≈ 3.676.) At x = ±π the series gives the mean (e^π + e^(−π))/2 = cosh π.

## Check yourself

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

1. For e^x on (−π, π), the coefficients use:
   A. (1/π)∫ over −π to π
   B. (2/π)∫ over 0 to π only
   C. (1/2π)∫ over 0 to 2π
   D. (1/π)∫ over 0 to π
   Answer: A. The function is neither even nor odd, so the full interval is needed. (Key point: Interval)

2. a0 equals:
   A. 2 cosh π/π
   B. 2 sinh π/π
   C. sinh π/π
   D. 0
   Answer: B. (e^π − e^−π)/π = 2 sinh π/π. (Key point: a0)

3. bn is:
   A. 2n(−1)^n sinh π/(π(1 + n^2))
   B. 0 since e^x is even
   C. −2n(−1)^n sinh π/(π(1 + n^2))
   D. 2(−1)^n sinh π/(π(1 + n^2))
   Answer: C. The sine integral gives −n cos nx in the numerator. (Key point: Sign of bn)

4. At x = ±π the series converges to:
   A. e^π
   B. e^−π
   C. 0
   D. cosh π, the mean of e^π and e^−π
   Answer: D. At the jump the series gives the average of the two end values. (Key point: End points)

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