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 ² ² ² π 0 = - 2 ³ ( - 1) n - 1 ´ = µ 0i f n is even, 4 if n is odd . Thus the Fourier series is 4 π k =0 1 (2 k +1) sin (2 k 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 ¹ º» ¼ ( x - p/ 2) v ± ¹ º» ¼ cos p = - 4 c p 2 =0 ¹ º» ¼ p ( x - p/ 2) sin p x ² ² ² p x =0 - p ± p 0 sin p = - 4 c p 2 p 2 n 2 π 2 cos p x ² ² ² p x =0 = 4 c n 2 π 2 (1 - cos ) = ½ f 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 ¾ (2 k π p x ¿ (2 k 2 . 9. The function is even; so all the
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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