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Chem Differential Eq HW Solutions Fall 2011 10

# Chem Differential Eq HW Solutions Fall 2011 10 - 10 Chapter...

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10 Chapter 2 Fourier Series Solutions to Exercises 2.3 1. (a) and (b) Since f is odd, all the a n ’s are zero and b n = 2 p p 0 sin p dx = - 2 cos p π 0 = - 2 ( - 1) n - 1 = 0 if n is even, 4 if n is odd . Thus the Fourier series is 4 π k =0 1 (2 k + 1) sin (2 k + 1) π p x . At the points of discon- tinuity, the Fourier series converges to the average value of the function. In this case, the average value is 0 (as can be seen from the graph. 5. (a) and (b) The function is even. It is also continuous for all x . All the b n s are 0. Also, by computing the area between the graph of f and the x -axis, from x = 0 to x = p , we see that a 0 = 0. Now, using integration by parts, we obtain a n = 2 p p 0 - 2 c p ( x - p/ 2) cos p xdx = - 4 c p 2 p 0 u ( x - p/ 2) v cos p xdx = - 4 c p 2 =0 p ( x - p/ 2) sin p x p x =0 - p p 0 sin p xdx = - 4 c p 2 p 2 n 2 π 2 cos p x p x =0 = 4 c n 2 π 2 (1 - cos ) = 0 if n is even , 8 c n 2 π 2 if n is odd . Thus the Fourier series is f ( x ) = 8 c π 2 k =0 cos (2
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