solut7 - Answers and comments on homework # 7 IV.2. (a) f...

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Answers and comments on homework # 7 IV.2. (a) f ( x ) = 1 2 - 1 π X n =1 sin(2 πnx ) n . (b) - 2 π X n =1 ( - 1) n sin( πnx ) n . (c) The series for [0 , 1] converges faster to the function f on [0 , 1] than the series for [ - 1 , 1]. On the other hand, the second series approximates the function on a larger interval. IV.10. 3 4 + X n =1 - 1 πn sin( πnx ) + ( - 1) n - 1 π 2 n 2 cos( πnx ) ! . IV.13. The Fourier series of f is f ( t ) = - X p =0 2 aT π 2 (2 p + 1) 2 cos (4 p + 2) πt T ! . With α := - a + a 2 - 4 k 2 2 , β := - a - a 2 - 4 k 2 2 , the solution to the given ODE can be written as Ae αt + Be βt + X n IN , n odd ± a n sin ± 2 πnt T ² + b n cos ± 2 πnt T ²² , a n = 2 aT π 2 n 2 a 2 - 4 k 2 α α 2 + 4 π 2 n 2 /T 2 - β β 2 + 4 π 2 n 2 /T 2 ! , b n = 4 a πn a 2 - 4 k 2 1 α 2 + 4 π 2 n 2 /T 2 - 1 β 2 + 4 π 2 n 2 /T 2 ! . IV.16. y ( x, t ) = X n =1 a n sin ± πnx L ² cos ± πnvt L ² , a n = aL 2 π 3 n 3 (3 cos( πn/ 3) + 2 πn sin( πn/ 3) + cos( πn ) - 4) . IV.19. The Fourier series for
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solut7 - Answers and comments on homework # 7 IV.2. (a) f...

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