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**Unformatted text preview: **Hence the Fourier polynomial is F N ( x ) = 1 π ( e π − e- π ) bracketleftbigg 1 2 − 1 2 cos x − 1 2 sin x + 1 5 cos 2 x + 2 5 sin 2 x − 1 10 cos 3 x − 3 10 sin 3 x + 1 17 cos 4 x + 4 17 sin 4 x − ··· + ( − 1) N 1 + N 2 cos Nx + ( − 1) N N 1 + N 2 sin Nx bracketrightbigg . (b) f ( x ) = braceleftBigg − 1 , − π < x ≤ − x , < x ≤ π . a = 1 π integraldisplay π- π f ( x ) dx = 1 π bracketleftbigg integraldisplay- π ( − 1) dx + integraldisplay π ( − x ) dx bracketrightbigg = 1 π parenleftbigg − x vextendsingle vextendsingle vextendsingle vextendsingle- π − x 2 2 vextendsingle vextendsingle vextendsingle vextendsingle π parenrightbigg = 1 π parenleftbigg − π − π 2 2 parenrightbigg = − parenleftbigg 1 + π 2 parenrightbigg . a k = 1 π integraldisplay π- π f ( x ) cos kx dx = 1 π bracketleftbigg integraldisplay- π ( − 1) cos kx dx + integraldisplay π ( − x ) cos kx dx bracketrightbigg = 1 π parenleftbigg a24 a24 a24 a24 a24 a24 a24 a24 a24 a58 = 0 bracketleftbigg − 1 k sin kx bracketrightbigg- π + bracketleftbigg a8 a8 a8 a8 a8 a8 a42 = 0 − x k sin kx − MinusΠ Π MinusΠ-1 y Equal f LParen1 x RParen1 y Equal f LParen1 x RParen1 F F 1 F 2 1 k 2 cos kx bracketrightbigg π parenrightbigg = − 1 π k 2 bracketleftbigg cos kπ − 1 bracketrightbigg = − 1 π k 2 ( ( − 1) k − 1 ) = , k = 2 , 4 , ··· 2 π k 2 , k = 1 , 3 , ··· . b k = 1 π integraldisplay π- π f ( x ) sin kx dx = 1 π bracketleftbigg integraldisplay- π ( − 1) sin kx dx + integraldisplay π ( − x ) sin kx dx bracketrightbigg = 1 π parenleftbiggbracketleftbigg 1 k cos kx bracketrightbigg- π + bracketleftbigg x K cos kx − a8 a8 a8 a8 a8 a8 a42 = 0 1 k 2 sin kx bracketrightbigg π parenrightbigg = 1 π bracketleftbiggparenleftbigg 1 k ( 1 − cos( − k π ) parenrightbigg + π k cos kπ bracketrightbigg = 1 π k ( 1 − ( − 1) k ) + 1 k ( − 1) k = 1 k , k even parenleftbigg 2 π − 1 parenrightbigg 1 k , k odd . Hence the Fourier polynomial is F N ( x ) = 1 2 parenleftbigg π 2 +1 parenrightbigg + 2 π cos x + parenleftbigg 2 π − 1 parenrightbigg sin x + 1 2 sin 2 x + 2 9 π cos 3 x + 1 3 parenleftbigg 2 π − 1 parenrightbigg sin 3 x + 1 4 sin 4 x + ··· + MATB42H Solutions # 1 page 3 2 πN 2 cos Nx + 1 N parenleftbigg 2 π − 1 parenrightbigg sin Nx , N odd 1 N − 1 parenleftbigg 2 π − 1 parenrightbigg sin ( ( N − 1) x ) + 1 N sin Nx , N even ....

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- Winter '10
- EricMoore
- Math, Multivariable Calculus