---
title: "Fourier series: 2x − x²"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:fourier-series-2x-x2-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-series-2x-x2-two
site: BCT Engineering
---

# Fourier series: 2x − x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2078 Kartik, 5 marks
- Video: [Fourier Series of 2x-x^2 over  (0, 3) || 18mat31 || Dr Prashant Patil](https://www.youtube.com/watch?v=5dScBLo99LY) — Dr Prashant Patil
- 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-2x-x2-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Find the Fourier series of f(x) = 2x − x² in (0, 2).

## Short answer

With period 2 (l=1), compute a0 = 4/3, a_n = -4/(n²π²), b_n = 0. The Fourier series is 2x - x² = 2/3 - (4/π²) ∑_{n=1}^∞ (cos nπx)/n².

## Answer

The interval has length 2l = 2, so l = 1 and the period is 2:

```text
f(x) = a₀/2 + Σ [ aₙ cos nπx + bₙ sin nπx ]
a₀ = ∫₀² f dx,  aₙ = ∫₀² f cos nπx dx,  bₙ = ∫₀² f sin nπx dx
```

Put x = 1 + u. Then f = 1 − u² for u in (−1, 1), an even function of u, and cos nπx = (−1)ⁿ cos nπu, sin nπx = (−1)ⁿ sin nπu.

```text
a₀ = ∫_{−1}^{1} (1 − u²) du = 4/3
aₙ = (−1)ⁿ ∫_{−1}^{1} (1 − u²) cos nπu du
   = (−1)ⁿ [ 0 − 4(−1)ⁿ/(nπ)² ] = −4/(n²π²)
bₙ = (−1)ⁿ ∫_{−1}^{1} (1 − u²) sin nπu du = 0   (odd integrand)
```

(using ∫_{−1}^{1} u² cos nπu du = 4(−1)ⁿ/(nπ)².)

**Answer:**

```text
2x − x² = 2/3 − (4/π²) Σ_{n=1}^∞ (cos nπx)/n²
        = 2/3 − (4/π²) [ cos πx + (cos 2πx)/4 + (cos 3πx)/9 + ... ],   0 < x < 2
```

(Check at x = 1: f = 1; the series gives 2/3 + (4/π²)(π²/12) = 1 ✓.)

## Check yourself

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

1. On (0, 2) the Fourier series uses terms:
   A. cos nπx and sin nπx (l = 1)
   B. cos nx and sin nx
   C. cos nπx/2 and sin nπx/2
   D. cos 2nπx
   Answer: A. Length 2l = 2 gives l = 1. (Key point: Period)

2. With x = 1 + u, f = 2x − x^2 becomes:
   A. u^2 − 1, odd in u
   B. 1 − u^2, even in u
   C. 1 + u, odd in u
   D. u(1 − u), neither
   Answer: B. 2(1+u) − (1+u)^2 = 1 − u^2. (Key point: Shift)

3. bn = 0 because:
   A. sin nπu is zero at the limits only
   B. n is an integer
   C. (1 − u^2) sin nπu is odd over (−1, 1)
   D. a0 is nonzero
   Answer: C. Even times odd is odd, integrating to zero. (Key point: Sine coefficients)

4. At x = 1, f = 1 and the series gives:
   A. 2/3 − 4/π^2 = 0
   B. 2/3 only
   C. 1/3
   D. 2/3 + (4/π^2)(π^2/12) = 1
   Answer: D. At x = 1, cos nπ = (−1)^n and Σ(−1)^(n+1)/n^2 = π^2/12. (Key point: Check at x = 1)

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