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

# Fourier series: x − x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2068 Chaitra, 5 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-five
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Obtain the Fourier series for f(x) = x − x², where −1 < x < 1, as a Fourier series of period 2.

## Short answer

The period-2 Fourier series for f(x)=x−x² on (−1,1) is f(x)=−1/3 + Σ_{n=1}^∞ [4(−1)^(n+1)/(n²π²) cos nπx + 2(−1)^(n+1)/(nπ) sin nπx]. Coefficients: a₀=−2/3, aₙ=−4(−1)ⁿ/(n²π²), bₙ=2(−1)ⁿ⁺¹/(nπ).

## Answer

A function of period 2 on (−1, 1) has the series f(x) = a₀/2 + Σ (aₙ cos nπx + bₙ sin nπx), where a₀ = ∫₋₁¹ f dx, aₙ = ∫₋₁¹ f cos nπx dx, bₙ = ∫₋₁¹ f sin nπx dx (half-period l = 1).

**Coefficients** for f(x) = x − x² (x is odd, x² is even):

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

aₙ = −∫₋₁¹ x² cos nπx dx = −2∫₀¹ x² cos nπx dx
   = −2[ x² sin nπx/(nπ) + 2x cos nπx/(nπ)² − 2 sin nπx/(nπ)³ ]₀¹
   = −2·2(−1)ⁿ/(n²π²) = −4(−1)ⁿ/(n²π²)

bₙ = ∫₋₁¹ x sin nπx dx = 2∫₀¹ x sin nπx dx
   = 2[ −x cos nπx/(nπ) + sin nπx/(nπ)² ]₀¹
   = −2(−1)ⁿ/(nπ) = 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/2² + cos 3πx/3² − …) + (2/π)(sin πx − sin 2πx/2 + sin 3πx/3 − …)

At x = 0 the function is continuous with f(0) = 0, which gives 1/1² − 1/2² + 1/3² − … = π²/12, a useful check.

## Check yourself

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

1. For period 2 on (−1, 1), a0 is:
   A. ∫₋₁¹ f dx
   B. (1/π)∫ f dx
   C. (1/2)∫ f dx
   D. 2∫ f dx
   Answer: A. l = 1 so a0 = (1/l)∫ f dx. (Key point: Period 2)

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

3. an and bn are:
   A. an = 0, bn = 2(−1)^(n+1)/(nπ)
   B. an = 4(−1)^n/(nπ), bn = 0
   C. an = 4(−1)^(n+1)/(n^2π^2), bn = 2(−1)^(n+1)/(nπ)
   D. an = 2/(n^2π^2), bn = 0
   Answer: C. Cosine from −x^2 and sine from x. (Key point: Coefficients)

4. At x = 0 we get:
   A. Σ1/n^2 = π^2/6
   B. Σ1/(2n−1)^2 = π^2/8
   C. π/4
   D. 1 − 1/2^2 + 1/3^2 − ... = π^2/12
   Answer: D. 0 = −1/3 + (4/π^2)S. (Key point: Deduction)

## Related in Mathematics III

- [Fourier series: x − x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x-x2) — Mathematics III, chapter 4, asked 6×
- [Fourier series: x³](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x3) — Mathematics III, chapter 4, asked 6×
- [Fourier series of a function](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-of-a-function) — Mathematics III, chapter 4, asked 6×
- [Fourier series: eˣ](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-e) — Mathematics III, chapter 4, asked 6×
- [Fourier series: x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x2-two) — Mathematics III, chapter 4, asked 2×
- [Fourier series: x + x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x-x2-three) — Mathematics III, chapter 4

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