---
title: "Fourier series: x − x²"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:fourier-series-x-x2-seven"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-series-x-x2-seven
site: BCT Engineering
---

# Fourier series: x − x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2082 Kartik (new course), 4 marks
- Video: [Fourier Series of (x-x^2): Full Calculation](https://www.youtube.com/watch?v=yMWKgKi0qDk) — Ayan Sarkar
- 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-x-x2-seven
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Find the Fourier series representation of f(x) = x − x² for −1 ≤ x ≤ 1.

## Short answer

a0 = −2/3; a_n = 4(−1)^(n+1)/(n^2 π^2), b_n = 2(−1)^(n+1)/(nπ). Fourier series: x − x^2 = −1/3 + Σ_{n=1}^∞ [4(−1)^(n+1)/(n^2 π^2) cos(nπx) + 2(−1)^(n+1)/(nπ) sin(nπx)].

## Answer

On (−1, 1) the period is 2 (half-period l = 1), so f(x) = a₀/2 + Σ (aₙ cos nπx + bₙ sin nπx) with a₀ = ∫₋₁¹ f dx, aₙ = ∫₋₁¹ f cos nπx dx, bₙ = ∫₋₁¹ f sin nπx dx.

Here x is odd and x² is even.

```text
a₀ = ∫₋₁¹ (x − x²) dx = −2∫₀¹ x² dx = −2/3

aₙ = −2∫₀¹ x² cos nπx dx
   = −2[ x² sin nπx/(nπ) + 2x cos nπx/(nπ)² − 2 sin nπx/(nπ)³ ]₀¹
   = −4(−1)ⁿ/(n²π²)

bₙ = 2∫₀¹ x sin nπx dx = 2[ −x cos nπx/(nπ) + sin nπx/(nπ)² ]₀¹
   = 2(−1)ⁿ⁺¹/(nπ)
```

**Fourier series**

x − x² = −1/3 + Σ (n=1 to ∞) [ 4(−1)ⁿ⁺¹ cos nπx / (n²π²) + 2(−1)ⁿ⁺¹ sin nπx / (nπ) ]

= −1/3 + (4/π²)(cos πx − cos 2πx/4 + cos 3πx/9 − …) + (2/π)(sin πx − sin 2πx/2 + sin 3πx/3 − …)

(At the end points x = ±1 the periodic extension jumps, so the series gives the mean value −1 there.)

## Check yourself

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

1. On (−1, 1):
   A. terms cos nx
   B. a0 = (1/π)∫ f dx
   C. a0 = ∫ f dx, terms cos nπx and sin nπx
   D. period 1
   Answer: C. l = 1. (Key point: Period)

2. a0 =
   A. 2/3
   B. −1/3
   C. 0
   D. −2/3
   Answer: D. −2∫0^1 x^2 dx. (Key point: a0)

3. bn =
   A. 2(−1)^(n+1)/(nπ)
   B. 2(−1)^n/(nπ)
   C. 4/(n^2π^2)
   D. 0
   Answer: A. From x. (Key point: bn)

4. At x = ±1 the series gives:
   A. 0
   B. the mean value −1
   C. 1
   D. −2
   Answer: B. Mean of f(1)=0 and f(−1)=−2. (Key point: Endpoints)

## Close variants set in other exams

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