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

# Fourier series: 2x − x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2080 Baishakh, 2+3 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
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

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

## Short answer

A function f is periodic with period T>0 if f(x+T)=f(x) for all x. Using shift u=x-1, the Fourier series on (0,2) with period 2 is: 2x − x² = 2/3 − (4/π²) Σ_{n=1}∞ (cos nπx)/n².

## Answer

#### Periodic function
A function f(x) is periodic with period T > 0 if f(x + T) = f(x) for all x. The smallest such T is the (fundamental) period. Example: sin x has period 2π.

#### Fourier series of f(x) = 2x − x² on (0, 2)
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. A function is periodic with period T if:
   A. f(x + T) = f(x) for all x
   B. f(x + T) = −f(x) for all x
   C. f(T) = 0
   D. f(−x) = f(x)
   Answer: A. Periodicity means the values repeat after T; the smallest such T is the fundamental period. (Key point: Periodic function)

2. For the interval (0, 2) the series uses:
   A. l = 2, so terms cos nπx/2
   B. l = 1, so terms cos nπx and sin nπx
   C. l = π, so terms cos nx
   D. l = 1/2, so terms cos 2nπx
   Answer: B. The interval length is 2l = 2, hence l = 1. (Key point: Half-period)

3. Putting x = 1 + u turns f into 1 − u^2, which is even, so:
   A. an = 0
   B. a0 = 0
   C. bn = 0
   D. all coefficients vanish
   Answer: C. An even function of u has an odd integrand for the sine integral, so bn = 0. (Key point: Shift trick)

4. The series for 2x − x^2 on (0, 2) is:
   A. 2/3 + (4/π^2) Σ cos nπx/n^2
   B. 1/3 − (4/π^2) Σ cos nπx/n^2
   C. 2/3 − (4/π^2) Σ sin nπx/n^2
   D. 2/3 − (4/π^2) Σ cos nπx/n^2
   Answer: D. a0/2 = 2/3 and an = −4/(n^2π^2) with no sine terms. (Key point: Result)

## 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: (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×

## Related in Mathematics III

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- [Fourier series: x³](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x3) — Mathematics III, chapter 4, asked 6×

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