Chem Differential Eq HW Solutions Fall 2011 15

Chem Differential Eq HW Solutions Fall 2011 15 - cos( n x p...

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

View Full Document Right Arrow Icon
Section 2.4 Half-Range Expansions: The Cosine and Sine Series 15 b k_ 8 Pi^3 1 2k 1 ^3; ss n_, x_ : Sum b k Sin 2k 1 Pi x , k, 0, n ; partialsineseries Table ss n, x , n, 1, 5 ; f x_ x 1 x Plot Evaluate partialsineseries, f x , x, 0, 1 0.2 0.4 0.6 0.8 1 0.05 0.1 0.15 0.2 0.25 Perfect! 13. We have sin πx cos πx = 1 2 sin 2 πx. This yields the desired 2-periodic sine series expansion. 17. (b) Sine series expansion: b n = 2 p ± a 0 h a x sin nπx p dx + 2 p ± p a h a - p ( x - p ) sin nπx p dx = 2 h ap ² - x p cos nπx p ³ ³ ³ a 0 + p ± a 0 cos nπx p dx ´ + 2 h ( a - p ) p ² ( x - p ) ( - p )
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: cos( n x p ) p a + p a p n cos n x p dx = 2 h pa -ap n cos na p + p 2 ( n ) 2 sin na p + 2 h ( a-p ) p p n ( a-p ) cos na p-p 2 ( n ) 2 sin na p = 2 hp ( n ) 2 sin na p 1 a-1 a-p = 2 hp 2 ( n ) 2 ( p-a ) a sin na p . Hence, we obtain the given Fourier series....
View Full Document

Ask a homework question - tutors are online