M257-316Notes_Lecture15

M257-316Notes_Lecture15 - Chapter 11 Lecture 15 -...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 11 Lecture 15 - Convergence of Fourier Series Example 11.1 (Completion of problem illustrating Half-range Expansions) Periodic Extension: Assume that f ( x )= x ,0 <x< 2 represents one full period of the function so that f ( x +2)= f ( x ). 2 L =2 L =1. a 0 = 1 L L Z L f ( x ) dx = 1 Z 1 f ( x ) dx = 2 Z 0 xdx = x 2 2 ¯ ¯ ¯ ¯ 2 0 (11.1) since f ( x f (2). (11.2) 73
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Lecture 15 - Convergence of Fourier Series n 1: a n = 1 L L Z L f ( x ) cos ³ nπx L ´ dx = 1 Z 1 f ( x ) cos( nπx ) dx L =1 = 2 Z 0 x cos( nπx ) dx = x sin( nπx ) ¯ ¯ ¯ ¯ 2 % 0 & µ 1 2 Z 0 sin( nπx ) dx = 1 ( ) 2 cos( nπx ) ¯ ¯ ¯ ¯ 2 0 = 1 ( ) 2 £ cos(2 ) 1 ± =0 (11.3) b n = 1 L L Z L f ( x ) sin ³ nπx L ´ dx = 1 Z 1 f ( x ) sin( nπx ) dx = 2 Z 0 x sin( nπx ) dx = x cos( nπx ) ¯ ¯ ¯ 2 0 & + 1 ( ) 2 Z 0 cos( nπx ) dx = 2 + sin( nπx ) ( ) 2 ¯ ¯ ¯ ¯ 2 % 0 & = µ 2 (11.4) Therefore f ( x )= 2 2 2 π X n =1 sin( nπx ) n (11.5) 2 π X n =1 sin( nπx ) n (11.6) 74
Background image of page 2
-4 -2 0 2 4 -1 0 1 2 3 x S(x) Figure 11.1: Full Range Expansion S N ( x )=1 2 π N =20 n =1 sin( nπx ) n -4 -2 0 2 4 -2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/04/2011 for the course MATH 25 taught by Professor Lo during the Spring '11 term at BC.

Page1 / 8

M257-316Notes_Lecture15 - Chapter 11 Lecture 15 -...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online