--- title: "Half range sine series: x − x²" subject: "Mathematics III" chapter: 4 key: "semester_3/mathematics_iii:half-range-sine-series-x-x2-three" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/half-range-sine-series-x-x2-three site: BCT Engineering --- # Half range sine series: x − x² Mathematics III · Chapter 4: Fourier Series - Asked in: 2080 Bhadra, 5 marks - Video: [Half Range Sine Series For F(x) = lx - x^2 in ( o, l ) - Fourier Series - Engineering Mathematics 3](https://www.youtube.com/watch?v=BetZOt2cCgc) — Ekeeda - 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-x2-three - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4 ## Question Find the half-range Fourier sine series for f(x) = x − x² in 0 < x < 1. ## Short answer Half-range sine series on (0,1): f(x) = Σ b_n sin nπx with b_n = 4(1−(−1)^n)/(n³π³), so b_n = 8/(n³π³) for odd n, 0 for even. Thus f(x) = (8/π³) Σ_{m=0}∞ sin((2m+1)πx)/(2m+1)³. ## Answer Half-range sine series on (0, 1) (l = 1): f(x) = Σ bₙ sin nπx with bₙ = 2 ∫₀¹ (x − x²) sin nπx dx. Integrate by parts twice (u = x − x²): ```text ∫₀¹ (x − x²) sin nπx dx = [ −(x − x²) cos nπx/(nπ) ]₀¹ + (1/(nπ)) ∫₀¹ (1 − 2x) cos nπx dx = 0 + (1/(nπ)) { [ (1 − 2x) sin nπx/(nπ) ]₀¹ + (2/(nπ)) ∫₀¹ sin nπx dx } = (2/(nπ)²) · (1 − (−1)ⁿ)/(nπ) = 2 (1 − (−1)ⁿ)/(nπ)³ ``` ```text bₙ = 4 (1 − (−1)ⁿ)/(n³π³) → 8/(n³π³) for odd n, 0 for even n ``` **Answer:** ```text x − x² = (8/π³) [ sin πx/1³ + sin 3πx/3³ + sin 5πx/5³ + ... ], 0 < x < 1 ``` (Check at x = 1/2: 1/4 = (8/π³)(1 − 1/27 + 1/125 − ...) = 0.2580 × 0.9689 ≈ 0.25 ✓.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For f(x) on (0, 1), the half-range sine series coefficient is: A. bn = 2 ∫0^1 f(x) sin nπx dx B. bn = ∫0^1 f(x) sin nπx dx C. bn = 2 ∫0^1 f(x) cos nπx dx D. bn = (1/2) ∫0^1 f(x) sin nπx dx Answer: A. With l = 1 the half-range sine series uses bn = (2/l)∫ f sin(nπx/l) dx. (Key point: Half-range sine formula) 2. Why do the boundary terms vanish in the first integration by parts? A. cos nπx is zero at both limits B. x − x^2 is zero at both x = 0 and x = 1 C. sin nπx is odd so it cancels D. n is always even Answer: B. The factor (x − x^2) is 0 at 0 and 1, so [−(x − x^2)cos nπx/(nπ)] is zero. (Key point: Integration by parts) 3. What is bn for even n? A. 8/(n^3π^3) B. 4/(n^3π^3) C. 0 D. 8/(nπ) Answer: C. bn = 4(1 − (−1)^n)/(n^3π^3); for even n, (−1)^n = 1 so bn = 0. (Key point: Coefficient result) 4. The series is: A. (8/π)[sin πx + sin 3πx/3 + ...] B. (4/π^3)[sin πx + sin 2πx/2^3 + ...] C. (8/π^3)[cos πx + cos 3πx/3^3 + ...] D. (8/π^3)[sin πx + sin 3πx/3^3 + sin 5πx/5^3 + ...] Answer: D. Only odd n survive with bn = 8/(n^3π^3), giving sine terms with odd multiples. 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