**Unformatted text preview: **3. f ( x ) = x 3 , restricted to ( − π, π bracketrightbig . Since f ( x ) = x 3 is an odd function, a k = 0, k = , 1 , 2 , ··· . b k = 1 π integraldisplay π- π sin kxdx parts = 1 πk 4 bracketleftbigg − k 3 x 3 cos kx + 3 k 2 x 2 sin kx + 6 kx cos kx − 6 sin kx bracketrightbigg π- π = 6 cos kπ k 3 − π 2 cos kx k + 6 cos( − kπ ) k 3 − π 2 cos( − kπ ) k = 2 k parenleftbigg 6 k 2 − π 2 parenrightbigg cos kπ = ( − 1) k parenleftbigg 2 k parenrightbiggparenleftbigg 6 k 2 − π 2 parenrightbigg . The Fourier series is F ( x ) = 2 ∞ summationdisplay k =1 ( − 1) k k parenleftbigg 6 k 2 − π 2 parenrightbigg sin kx = 2 bracketleftbigg ( − 6 + π 2 ) sin x + parenleftbigg 3 4 − π 2 2 parenrightbigg sin 2 x + parenleftbigg − 2 9 + π 2 3 parenrightbigg sin 3 x + parenleftbigg 3 32 − π 2 4 parenrightbigg sin 4 x + ··· bracketrightbigg . MinusΠ Π-30-20-10 10 20 30 f LParen1 x RParen1 Equal x 3 F 5 4. (a) For k > 0 the energy of the k th harmonic is A 2 k = a 2 k + b 2 k . Since f ( x ) is an even function, b k = 0, k = 0 , 1 , 2 , 3 , ··· . a k = 1 π integraldisplay π- π f ( x ) cos kxdx even = 2 π integraldisplay π 2 π cos kxdx = 2 bracketleftbigg sin kx k bracketrightbigg π 2 = 2 k sin parenleftbigg kπ 2 parenrightbigg = 2 k , k = 1 , 5 , ··· − 2 k , k = 3 , 7 , ··· . Hence the energy of the k th harmonic is A 2 k = 4 k 2 , k odd , k even . (b) For k = 0, a = 1 π integraldisplay π- π f ( x ) dx = 2 π integraldisplay π 2 π dx = π , so the energy of the con- stant term is A 2 k = 1 2 a 2 = π 2 2 . The total energy is E = 1 π integraldisplay π- π ( f ( x ) parenrightbigg 2 dx = 2 π integraldisplay π 2 π 2 dx = π 2 . Now A 2 + A 2 1 + A 2 2 E = π 2 / 2 + 4 + 0 π 2 = π 2 + 8 2 π 2 , so π 2 + 8 2 π 2 of the energy is contained in the first three terms of the Fourier series. (This is about MATB42H Solutions # 1 page 4 9 10 of the energy.) (c) The n th Fourier polynomial F n ( x ) is given by F n ( x ) = 1 2 a + n summationdisplay k =1 ( a k cos kx + b k sin kx ) = 1 2 π + m summationdisplay ℓ =1 ( − 1) ℓ 2 2 ℓ − 1 cos ( (2 ℓ − 1) x ) = 1 2 π + 2 cos x − 2 3 cos 3 x + 2 5 cos 5 x − 2 7 cos 7 x + ··· + ( − 1) m 2 m cos mx , where m = n + 1 2 , n odd n 2 , n even . Minus 3 Π Minus 2 Π MinusΠ Minus Π FractionBarExtFractionBarExtFractionBarExtFractionBarExt 2 Π FractionBarExtFractionBarExtFractionBarExtFractionBarExt 2 Π 2 Π 3 Π Π f LParen1 x RParen1 F 3 5. f ( x ) = 1 − x 2 restricted to [ − 1 , 1] and extended to all of R with period 2....

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- Winter '10
- EricMoore
- Math, Multivariable Calculus