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

# Fourier series: x + x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2082 Chaitra (new course), 4 marks
- Video: [Fourier series of the function f(x)=x+x^2 , Engineering Mathematics ,Diploma Mathematics](https://www.youtube.com/watch?v=z8kpAlULwxc) — GBK Academy 
- 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-six
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

If f(x) = x + x² for −π ≤ x ≤ π, then find the Fourier series of f(x).

## Short answer

Constant term π²/3. Coefficients: a_n = 4(-1)^n/n², b_n = 2(-1)^(n+1)/n. Fourier series: f(x) = π²/3 + sum_{n=1}^∞ [4(-1)^n cos(nx)/n² + 2(-1)^(n+1) sin(nx)/n].

## Answer

For f(x) on (−π, π) of period 2π: f(x) = a₀/2 + Σ (aₙ cos nx + bₙ sin nx), a₀ = (1/π)∫f dx, aₙ = (1/π)∫f cos nx dx, bₙ = (1/π)∫f sin nx dx.

Here x is odd and x² is even, so the odd part contributes only to bₙ and the even part only to a₀ and aₙ.

```text
a₀ = (1/π)∫₋π^π (x + x²) dx = (1/π)(2π³/3) = 2π²/3

aₙ = (1/π)∫₋π^π x² cos nx dx = (2/π)[x² sin nx/n + 2x cos nx/n² − 2 sin nx/n³]₀^π
   = (2/π)(2π cos nπ / n²) = 4(−1)ⁿ/n²

bₙ = (1/π)∫₋π^π x sin nx dx = (2/π)[−x cos nx/n + sin nx/n²]₀^π
   = −2 cos nπ / n = 2(−1)ⁿ⁺¹/n
```

**Fourier series**

x + x² = π²/3 + Σ (n=1 to ∞) [ 4(−1)ⁿ cos nx / n² + 2(−1)ⁿ⁺¹ sin nx / n ]

= π²/3 − 4(cos x − cos 2x/2² + cos 3x/3² − …) + 2(sin x − sin 2x/2 + sin 3x/3 − …)

Check: at x = 0 the function is 0 and the series gives π²/3 − 4(π²/12) = 0.

## Check yourself

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

1. The cosine terms come from:
   A. x (odd part)
   B. both equally
   C. x^2 (even part)
   D. neither
   Answer: C. Even part gives a0, an. (Key point: Parts)

2. a0 equals:
   A. π^2/3
   B. 4π^2/3
   C. 0
   D. 2π^2/3
   Answer: D. (1/π)(2π^3/3). (Key point: a0)

3. bn equals:
   A. 2(−1)^(n+1)/n
   B. 4(−1)^n/n^2
   C. 2(−1)^n/n
   D. 0
   Answer: A. From ∫ x sin nx. (Key point: bn)

4. At x = 0 the series gives π^2/3 − 4(π^2/12) =
   A. π^2/3
   B. 0
   C. π^2
   D. −π^2/3
   Answer: B. Matches f(0) = 0. (Key point: Check)

## Related in Mathematics III

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