FourierSeries-1

Example 1 says that for 0 x 1 let x 2 1 2 1 2 k1 2

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: odd F OURIER SERIES ■ 15 19. Example 1 says that, for 0 ≤ x < π, 1 = Let x = π 2 1 2 1 2 + ∞ k=1 + 2 sin((2k − 1)x). (2k − 1)π to obtain 1= ∞ k=1 π 2 sin (2k − 1) (2k − 1)π 2 ∞ π 1 = sin((2k − 1)) 4 k=1 (2k − 1) π =1− 4 1 3 + 1 5 − 1 7 + ··· 1 1 −1.25(−1) + 4(0.25) − √2 + 2(−2.25)(0) + 4(−1.25) √2 1 1 a1 = −1 f (x) cos(πx) dx ≈ 12 + 2(3.5)(1) + 4(3.5) √ + 2(0)(0) + 4(−2.75) − √ − 1.25(−1) 1 1 2 2 √ = 19 1 + 2 24 a2 = = b1 = ≈ 1 −1 f (x) cos(2πx) dx ≈ 3 4 1 −1 1 12 f (x) sin(πx) dx 1 1 −1.25(0) + 4(0.25) − √2 + 2(−2.25)(−1) + 4 (−1.25) − √2 + 2(3.5)(0) + 4(3.5) + 4(−2.75) √ 9+7 2 = 24 b2 = = −1.25(1) + 4(0.25)(0) + 2(−2.25)(−1) + 4(−1.25)(0) 1 12 + 2(3.5)(1) + 4(3.5)(0) + 2(0)(−1) + 4(−2.75)(0) − 1.25(1) 1 −1 1 √ 2 f (x) sin(2x) dx ≈ − 1.25(0) + 4(3.5) 1 12 1 √ 2 + 2(0)(1) + 4(−2.75) 1 √ 2 − 1.25(0) −1.25(0) + 4(0.25)(1) + 2(−2.25)(0) + 4(−1.25)(−1) + 2(3.5)(0) + 4(3.5)(1) + 2(0)(0) + 4(−2.75)(−1) − 1.25(0) 31 12 f (x) ≈ 1 − 12 + 19 24 √ 1 + 2 cos(πx) + √ 9+7 2 24 sin(πx) + 3 4 cos(2πx) + y 4 2 0 -1 -0.5 0 0.5 1 x -2 31 12 sin(2πx) 1 √ 2 + 2(0)(1)...
View Full Document

This note was uploaded on 12/05/2011 for the course MATH 41 taught by Professor Bray,c during the Spring '08 term at Duke.

Ask a homework question - tutors are online