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Unformatted text preview: Math 445  Solutions to selected problems from HW 1 Exercise 13, page 486. We need to find the Fourier series of the 2 πperiodic function f ( x ) defined by f ( x ) = 1 if π 2 < x < π 2 , and f ( x ) = 0 if π ≤ x ≤  π 2 or π 2 ≤ x ≤ π . Solution Note that f ( x ) is even ! a = 1 2 π R π π f ( x ) dx = 1 2 π R π 2 π 2 1 dx = 1 2 π π = 1 2 . a n = 1 π R π π f ( x ) cos( nx ) dx = 2 π R π 2 cos( nx ) dx = 1 π sin( nπ 2 ) n sin(0) n = 2 nπ sin( nπ 2 ). b n = 0 for all n ≥ 1 because the function is even. Hence the Fourier series of f is: f ( x ) = 1 2 + 2 π cos( x ) 2 3 π cos(3 x ) + 2 5 π cos(5 x ) 2 7 π cos(7 x ) + ... Exercise 2, page 490. We need to find the Fourier series of the p = 2 Lperiodic function f ( x ) defined by f ( x ) = 0 if 2 < x < 0 and f ( x ) = 4 if 0 < x < 2. Here p = 4. Solution Since p = 4 we have L = 2. Then: a = 1 2 L R L L f ( x ) dx = 1 4 R 2 4 dx = 1 4 8 = 2. a n = 1 L R L L f ( x ) cos( nπx L ) dx = 1 2 R 2 4 cos( nπx 2 ) dx = 4 nπ (sin( nπ ) sin(0)) = 0. b n = 1 L R L L f ( x ) sin( nπx L ) dx = 1 2 R 2 4 sin( nπx 2 ) dx = 4 nπ ( cos( nπ ) + cos(0)) = 4(1 ( 1) n ) nπ ....
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 Fall '07
 Friedlander
 Math, Fourier Series, 1 L, 2L, 2 L, 1 2L, 2 0 L

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