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

# Fourier series: (π − x)²/4

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2079 Bhadra, 5 marks
- Video: [105) P3) Find the Fourier series of f(x)=1/4(π-x)² || Fourier Series](https://www.youtube.com/watch?v=qgFfJLZEc6w) — Sky Dhoni Production House
- 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-2-4
- 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)²/4 in the interval 0 ≤ x ≤ 2π.

## Short answer

By substituting u=x-π and noting evenness, obtain Fourier cosine series: (π − x)²/4 = π²/12 + Σ_{n=1}∞ (cos nx)/n².

## Answer

f(x) = (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)²/4 = 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. Why substitute u = x − π for f = (π − x)^2/4 on [0, 2π]?
   A. It turns f into u^2/4, an even function on (−π, π)
   B. It makes f odd
   C. It changes the period to π
   D. It removes the need for integration
   Answer: A. f = u^2/4 is even on a symmetric interval, so bn = 0. (Key point: Substitution)

2. The value of a0 is:
   A. π^2/12
   B. π^2/6
   C. π^2/3
   D. 2π^2/3
   Answer: B. a0 = (1/(4π))(2π^3/3) = π^2/6, so a0/2 = π^2/12. (Key point: a0)

3. The coefficients an and bn are:
   A. an = 4/n^2, bn = 0
   B. an = 0, bn = 1/n
   C. an = 1/n^2, bn = 0
   D. an = (−1)^n/n^2, bn = 0
   Answer: C. The (−1)^n from the shift cancels the (−1)^n from the integral, leaving 1/n^2. (Key point: an and bn)

4. Putting x = 0 in the series gives which known sum?
   A. Σ (−1)^n/n^2 = −π^2/12
   B. Σ 1/n = ∞
   C. Σ 1/n^4 = π^4/90
   D. Σ 1/n^2 = π^2/6
   Answer: D. π^2/4 = π^2/12 + Σ1/n^2 gives Σ1/n^2 = π^2/6. (Key point: Check at x = 0)

## Close variants set in other exams

- [Fourier series: ((π − x)/2)²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x-2-2) — 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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