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

# Half range cosine series: eˣ

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2074 Chaitra, 5 marks
- 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-e
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Express f(x) = eˣ as a half range Fourier cosine series in 0 < x < 1.

## Short answer

Compute a₀ = 2∫₀¹ eˣ dx = 2(e-1) and aₙ = 2∫₀¹ eˣ cos nπx dx = 2[(-1)ⁿe - 1]/(1+n²π²). The half-range cosine series is eˣ = (e-1) + 2 Σ_{n=1}∞ {[(-1)ⁿe - 1]/(1+n²π²)} cos nπx, 0<x<1.

## Answer

Half-range cosine series on (0, 1) (l = 1): f(x) = a₀/2 + Σ aₙ cos nπx, with a₀ = 2∫₀¹ f dx and aₙ = 2∫₀¹ f cos nπx dx.

#### Coefficients
```text
a₀ = 2 ∫₀¹ eˣ dx = 2(e − 1)
```

Use ∫ eˣ cos nπx dx = eˣ (cos nπx + nπ sin nπx)/(1 + n²π²):

```text
aₙ = 2 [ eˣ (cos nπx + nπ sin nπx)/(1 + n²π²) ]₀¹
   = 2 [ e(−1)ⁿ − 1 ]/(1 + n²π²)
```

**Answer:**

```text
eˣ = (e − 1) + 2 Σ_{n=1}^∞ { [ (−1)ⁿ e − 1 ] / (1 + n²π²) } cos nπx,   0 < x < 1
```

First terms: a₁ = −2(e + 1)/(1 + π²) ≈ −0.68, a₂ = 2(e − 1)/(1 + 4π²) ≈ 0.085, so eˣ ≈ 1.718 − 0.68 cos πx + 0.085 cos 2πx − ...

## Check yourself

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

1. a0 for e^x on (0, 1) is:
   A. 2(e − 1)
   B. e − 1
   C. 2e
   D. (e − 1)/2
   Answer: A. a0 = 2∫0^1 e^x dx = 2(e − 1); a0/2 = e − 1. (Key point: a0)

2. ∫ e^x cos nπx dx equals:
   A. e^x(cos nπx − nπ sin nπx)/(1 + n^2π^2)
   B. e^x(cos nπx + nπ sin nπx)/(1 + n^2π^2)
   C. e^x(sin nπx − nπ cos nπx)/(1 + n^2π^2)
   D. e^x cos nπx/(nπ)
   Answer: B. Standard formula with a = 1, b = nπ. (Key point: Integral formula)

3. an is:
   A. 2[(−1)^n e + 1]/(1 + n^2π^2)
   B. 2(e − 1)/(1 + n^2π^2)
   C. 2[(−1)^n e − 1]/(1 + n^2π^2)
   D. 0
   Answer: C. At x = 1, cos nπ = (−1)^n; at x = 0, 1. (Key point: an)

4. The constant term in the series is:
   A. 2(e − 1)
   B. e
   C. 1
   D. e − 1
   Answer: D. a0/2 = e − 1. (Key point: Constant term)

## Close variants set in other exams

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

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