---
title: "Half range sine series: x"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:half-range-sine-series-x-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/half-range-sine-series-x-two
site: BCT Engineering
---

# Half range sine series: x

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2069 Chaitra, 5 marks
- Video: [Half range sine series/ Fourier series/ PU 2018/ Engineering mathematics 3/ Third semester](https://www.youtube.com/watch?v=s__Bkco7Upg) — Rupak Deo
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/half-range-sine-series-x-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Express f(x) = x as a half-range sine series in 0 < x < 2.

## Short answer

With L=2, bₙ = ∫₀² x sin(nπx/2) dx = 4(-1)^{n+1}/(nπ). The half-range sine series is x = (4/π) Σ_{n=1}∞ [(-1)^{n+1}/n] sin(nπx/2), 0<x<2.

## Answer

For a function on (0, L) the half range sine series is f(x) = Σ bₙ sin(nπx/L), with bₙ = (2/L)∫₀ᴸ f(x) sin(nπx/L) dx. Here L = 2.

**Coefficient** for f(x) = x:

```text
bₙ = (2/2)∫₀² x sin(nπx/2) dx
   = [ −x cos(nπx/2)·(2/nπ) ]₀² + (2/nπ)∫₀² cos(nπx/2) dx
   = −(4/nπ) cos nπ + (2/nπ)·(2/nπ)[sin(nπx/2)]₀²
   = −4(−1)ⁿ/(nπ) + 0
   = 4(−1)ⁿ⁺¹/(nπ)
```

**Series**

x = (4/π) Σ (n=1 to ∞) [(−1)ⁿ⁺¹/n] sin(nπx/2),  0 < x < 2

= (4/π)[ sin(πx/2) − (1/2) sin(πx) + (1/3) sin(3πx/2) − … ]

Check: at x = 1 the series gives (4/π)(1 − 1/3 + 1/5 − …) = (4/π)(π/4) = 1 = x.

## Check yourself

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

1. On (0, L) with L = 2, bn is:
   A. ∫0^2 x sin(nπx/2) dx
   B. ∫0^2 x cos(nπx/2) dx
   C. (1/2)∫ x sin(nπx/2) dx
   D. ∫0^2 x sin nx dx
   Answer: A. 2/L = 1. (Key point: Formula)

2. The boundary term gives:
   A. (4/nπ) sin nπ
   B. −(4/nπ)cos nπ
   C. −(2/nπ)
   D. 0
   Answer: B. At x = 2, x·(2/nπ) = 4/nπ. (Key point: Parts)

3. bn equals:
   A. 4(−1)^n/(nπ)
   B. 8/(n^2π^2)
   C. 4(−1)^(n+1)/(nπ)
   D. 0
   Answer: C. −4(−1)^n/(nπ). (Key point: bn)

4. At x = 1 the series reduces to:
   A. (4/π)(1 + 1/3 + 1/5 + ...)
   B. (2/π)(1 − 1/2 + 1/3 ...)
   C. 0
   D. (4/π)(1 − 1/3 + 1/5 − ...) = 1
   Answer: D. sin(nπ/2) picks odd n with alternating signs. (Key point: Check)

## Close variants set in other exams

- [Half range sine series: x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-x2) — Mathematics III, chapter 4, asked 3×
- [Half range sine series: x](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Ahalf-range-sine-series-x-four) — Mathematics III, chapter 4, asked 3×

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