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

# Fourier coefficient bₙ of e^(−x)

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2082 Chaitra (new course), 2 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-coefficient-b-of-e-x
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

In the series f(x) = a₀/2 + Σ_{n=1}^∞ (aₙ cos nx + bₙ sin nx), find the coefficient bₙ for the function f(x) = e^(−x) where x ∈ (0, 2π).

## Short answer

bₙ = n(1−e^(−2π)) / [π(1+n²)], obtained from bₙ = (1/π)∫₀^{2π} e^(−x) sin nx dx.

## Answer

For f(x) on (0, 2π), bₙ = (1/π)∫₀^{2π} f(x) sin nx dx.

With f(x) = e^(−x), use ∫ e^(−x) sin nx dx = e^(−x)(−sin nx − n cos nx)/(1 + n²):

```text
bₙ = (1/π)[ e^(−x)(−sin nx − n cos nx)/(1 + n²) ]₀^{2π}
   = (1/π)·[ −n e^(−2π) + n ]/(1 + n²)          (sin 2nπ = 0, cos 2nπ = 1)
```

**Answer:** bₙ = n(1 − e^(−2π)) / [π(1 + n²)]

For example, b₁ = (1 − e^(−2π))/(2π) ≈ 0.159.

## Check yourself

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

1. On (0, 2π), bn equals:
   A. (2/π)∫0^π f sin nx dx
   B. (1/2π)∫ f sin nx dx
   C. (1/π)∫ f cos nx dx
   D. (1/π)∫0^2π f(x) sin nx dx
   Answer: D. Full-period formula. (Key point: Formula)

2. At 2π, sin 2nπ and cos 2nπ equal:
   A. 0 and 1
   B. 1 and 0
   C. 0 and −1
   D. both 1
   Answer: A. For integer n. (Key point: Evaluation)

3. bn =
   A. (1 − e^(−2π))/(π(1 + n^2))
   B. n(1 − e^(−2π))/(π(1 + n^2))
   C. n(1 + e^(−2π))/(π(1 + n^2))
   D. −n(1 − e^(−2π))/(π(1 + n^2))
   Answer: B. From [−n e^(−2π) + n]/(π(1+n^2)). (Key point: Result)

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