M257-316Notes_Lecture15

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

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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 L f ( x ) dx = 1 1 f ( x ) dx = 2 0 x dx = x 2 2 2 0 = 2 (11.1) since f ( x + 2) = f (2). (11.2) 73
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Lecture 15 - Convergence of Fourier Series n 1: a n = 1 L L L f ( x ) cos nπx L dx = 1 1 f ( x ) cos( nπx ) dx L = 1 = 2 0 x cos( nπx ) dx = x sin( nπx ) 2 0 1 2 0 sin( nπx ) dx = 1 ( ) 2 cos( nπx ) 2 0 = 1 ( ) 2 cos(2 ) 1 = 0 (11.3) b n = 1 L L L f ( x ) sin nπx L dx = 1 1 f ( x ) sin( nπx ) dx = 2 0 x sin( nπx ) dx = x cos( nπx ) 2 0 + 1 ( ) 2 0 cos( nπx ) dx = 2 + sin( nπx ) ( ) 2 2 0 = 2 (11.4) Therefore f ( x ) = 2 2 2 π n =1 sin( nπx ) n (11.5) = 1 2 π n =1 sin( nπx ) n (11.6) 74
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-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 -1 0 1 2 x S(x)-1 Figure 11.2: Full Range Expansion S N ( x ) 1 = 2 π N =20 n =1 sin( nπx ) n 75
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