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

# Fourier series: ((π − x)/2)²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2081 Bhadra, 5 marks
- Video: [5. FOURIER SERIES EXPANSION OF  f(x) = ( π - x / 2 )²  ON  (0, 2π).](https://www.youtube.com/watch?v=a8iPxIh8tek) — VIJAY MATH CHANNEL
- 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-2-2
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Find the Fourier series of the function f(x) = ((π − x)/2)² in the interval 0 ≤ x ≤ 2π.

## Short answer

Using symmetry shift u=x-π, the even extension gives Fourier cosine series: ((π − x)/2)² = π²/12 + Σ_{n=1}∞ (cos nx)/n².

## Answer

**Reading:** f(x) = ((π − x)/2)² = (π − x)²/4 on [0, 2π]; the period is 2π.

#### Coefficients
Put u = x − π, so u runs over (−π, π) and f = u²/4 (an even function of u). Also, since sin nπ = 0, cos nx = cos(nu + nπ) = (−1)ⁿ cos nu and sin nx = sin(nu + nπ) = (−1)ⁿ sin nu.

```text
a₀ = (1/π) ∫₀^{2π} (π − x)²/4 dx = (1/(4π)) ∫_{−π}^{π} u² du = (1/(4π))(2π³/3) = π²/6
aₙ = (1/(4π)) ∫_{−π}^{π} u² cos(nu + nπ) du
   = ((−1)ⁿ/(4π)) · 4π(−1)ⁿ/n² = 1/n²
bₙ = (1/(4π)) ∫_{−π}^{π} u² sin(nu + nπ) du = ((−1)ⁿ/(4π)) ∫ u² sin nu du = 0   (odd integrand)
```

(using ∫_{−π}^{π} u² cos nu du = 4π(−1)ⁿ/n².)

**Answer:**

```text
((π − x)/2)² = π²/12 + Σ_{n=1}^∞ (cos nx)/n²
             = π²/12 + cos x + (cos 2x)/4 + (cos 3x)/9 + ...,    0 ≤ x ≤ 2π
```

(Check at x = 0: π²/4 = π²/12 + Σ 1/n² gives Σ 1/n² = π²/6 ✓.)

## Check yourself

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

1. Put u = x - pi so that
   A. u runs over (0,pi)
   B. f = u^2
   C. f = u/2
   D. u runs over (-pi, pi) and f = u^2/4
   Answer: D. (pi - x)^2 = u^2. (Key point: Substitution)

2. a0 =
   A. pi^2/6
   B. pi^2/3
   C. pi^2/12
   D. pi^2/2
   Answer: A. (1/(4pi))(2pi^3/3). (Key point: a0)

3. an and bn are
   A. an = 4/n^2, bn = 0
   B. an = 1/n^2, bn = 0
   C. an = 0, bn = 1/n
   D. an = (-1)^n/n^2, bn = 0
   Answer: B. The (-1)^n factors cancel. (Key point: an and bn)

4. f(x) =
   A. pi^2/6 + sum cos nx/n^2
   B. pi^2/12 + sum (-1)^n cos nx/n^2
   C. pi^2/12 + sum cos nx/n^2
   D. pi^2/12 + sum sin nx/n^2
   Answer: C. Constant a0/2. (Key point: Series)

## Close variants set in other exams

- [Fourier series: (π − x)²/4](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x-2-4) — Mathematics III, chapter 4, asked 5×
- [Fourier series: x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x2) — Mathematics III, chapter 4, asked 9×
- [Fourier series: 2x − x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-2x-x2) — Mathematics III, chapter 4, asked 5×

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