This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Since f ( x ) = 2 3 x is piecewise very smooth on [ π, π ], the Fourier series converges to f ( x ) for x ∈ ( π, π ) and to 2 + 3 π + 2 3 π 2 = 2 at x = ± π . (b) f ( x ) = 2 3 x restricted to bracketleftbig , 2 π bracketrightbig . a = 1 π integraldisplay 2 π ( 2 3 x ) dx = 1 π bracketleftbigg 2 x 3 2 x 2 bracketrightbigg 2 π = 2 ( 2 3 π ) . For k > 0, a k = 1 π integraldisplay 2 π ( 2 3 x ) cos kx dx = 1 π parenleftbiggbracketleftbigg 2 sin kx k bracketrightbigg 2 π 3 parenleftbiggbracketleftbigg x sin kx k bracketrightbigg 2 π a24 a24 a24 a24 a24 a24 a24 a24 a24 a58 = 0 1 k integraldisplay 2 π sin kx dx parenrightbiggparenrightbigg = 0. b k = 1 π integraldisplay 2 π ( 2 3 x ) sin kx dx = 1 π parenleftbiggbracketleftbigg 2 cos kx k bracketrightbigg 2 π 3 parenleftbiggbracketleftbigg x cos kx k bracketrightbigg 2 π + a24 a24 a24 a24 a24 a24 a24 a24 a24 a58 = 0 1 k integraldisplay 2 π cos kx dx parenrightbiggparenrightbigg = 3 π bracketleftbigg 2 π k bracketrightbigg = 6 k . Hence, the Fourier series is ( 2 3 π ) + 6 ∞ summationdisplay k =1 sin kx k . Since f ( x ) = 2 3 x is piecewise very smooth on [0 , 2 π ], the Fourier series converges to f ( x ) for x ∈ (0 , 2 π ) and to 2 + 2 6 π 2 = 2 3 π at x = 0 and at x = 2 π . 3. On assignment 1 we discovered that the function f ( x ) (of period 2), which corresponds to y = 1 x 2 on [ 1 , 1], has Fourier series F ( x ) = 2 3 + 4 π 2 ∞ summationdisplay k =1 ( 1) k +1 k 2 cos( kπx ). Since f ( x ) is continuous and piecewise very smooth for all x , F ( x ) converges to f ( x ) for all x . (a) Since 1 x 2 = 2 3 + 4 π 2 ∞ summationdisplay k =1 ( 1) k +1 k 2 cos( kπx ) for all x ∈ [ 1 , 1], substituting x = 1 gives 0 = 2 3 + 4 π 2 ∞ summationdisplay k =1 ( 1) k +1 k 2 cos k π = 2 3 + 4 π 2 bracketleftbigg cos π 1 4 cos 2 π + 1 9 cos 3 π 1 16 cos 4 π + 1 25 cos 5 π + ··· bracketrightbigg = 2 3 + 4 π 2 bracketleftbigg 1 1 4 1 9 1 16 1 25 ··· bracketrightbigg = ⇒ 2 3 = 4 π 2 bracketleftbigg 1 1 4 1 9 1 16 1 25 ··· bracketrightbigg = ⇒ π 2 6 = 1+ 1 4 + 1 9 + ··· 1 n 2 + ··· . (b) Since 1 x 2 = 2 3 + 4 π 2 ∞ summationdisplay k =1 ( 1) k +1 k 2 cos( kπx ) for all x ∈ [ 1 , 1], substituting x = 0 gives 1 = 2 3 + 4 π 2 ∞ summationdisplay k =1 ( 1) k +1 k 2 cos 0 = 2 3 + 4 π 2 bracketleftbigg 1 1 4 + 1 9 1 16 + 1 25 ··· bracketrightbigg = ⇒ 1 3 = 4 π 2 bracketleftbigg 1 1 4 + 1 9 1 16 + 1 25 ··· bracketrightbigg = ⇒ π 2 12 = 1 1 4 + 1 9 1 16 + 1 25 ··· + ( 1) n +1 n 2 + ··· . MATB42H Solutions # 2 page 3 4. f ( x ) = , π ≤ x < π 2 2 , π 2 ≤ x < π 2 , π 2 ≤ x < π ....
View
Full Document
 Winter '10
 EricMoore
 Math, Multivariable Calculus

Click to edit the document details