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

# Half range cosine series: x

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2074 Asoj, 5 marks
- Video: [How To Find The Half Range Cosine Series of f(x)=4x](https://www.youtube.com/watch?v=fMAiHSfr-ik) — DrAvitech
- 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-cosine-series-x
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Obtain the half range cosine series for f(x) = x in the interval 0 ≤ x ≤ π.

## Short answer

a₀ = (2/π)∫₀^π x dx = π. For n≥1, aₙ = (2/π)∫₀^π x cos nx dx gives 0 for even n, -4/(πn²) for odd n. Hence x = π/2 - (4/π)(cos x + cos 3x/3² + cos 5x/5² + ...), 0≤x≤π.

## Answer

Half-range cosine series on (0, π): f(x) = a₀/2 + Σ aₙ cos nx, with a₀ = (2/π)∫₀^π f dx and aₙ = (2/π)∫₀^π f cos nx dx.

For f(x) = x:

```text
a₀ = (2/π) ∫₀^π x dx = (2/π)(π²/2) = π
aₙ = (2/π) ∫₀^π x cos nx dx
   = (2/π) [ x sin nx/n + cos nx/n² ]₀^π
   = (2/(πn²)) [ (−1)ⁿ − 1 ]
```

So aₙ = 0 for even n, and aₙ = −4/(πn²) for odd n.

**Answer:**

```text
x = π/2 − (4/π) [ cos x/1² + cos 3x/3² + cos 5x/5² + ... ],   0 ≤ x ≤ π
```

(Check at x = 0: 0 = π/2 − (4/π)(π²/8) ✓, since Σ 1/(2k+1)² = π²/8.)

## Check yourself

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

1. a0 for f = x on (0, π) is:
   A. π
   B. π/2
   C. 2π
   D. π^2
   Answer: A. (2/π)(π^2/2) = π. (Key point: a0)

2. For even n, an is:
   A. −4/(πn^2)
   B. 0
   C. π
   D. 2/n
   Answer: B. (−1)^n − 1 = 0 for even n. (Key point: an)

3. For odd n, an is:
   A. 4/(πn^2)
   B. −2/(πn)
   C. −4/(πn^2)
   D. −4/(πn)
   Answer: C. (2/(πn^2))(−2) = −4/(πn^2). (Key point: Odd n)

4. The series is:
   A. π/2 + (4/π)[cos x + cos 3x/3^2 + ...]
   B. π − (4/π)[cos x + ...]
   C. π/2 − (4/π)[sin x + sin 3x/3^2 + ...]
   D. π/2 − (4/π)[cos x + cos 3x/3^2 + ...]
   Answer: D. a0/2 = π/2. (Key point: Series)

## Close variants set in other exams

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