Suppose that we wish to calculate values of the function g, where g(x) = n=1 (2n 1) 1 +

Chapter 10, Problem 19

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Suppose that we wish to calculate values of the function g, where g(x) = n=1 (2n 1) 1 + (2n 1)2 sin(2n 1)x. (i) It is possible to show that this series converges, albeit rather slowly. However, observe that for large n the terms in the series (i) are approximately equal to [sin(2n 1)x]/(2n 1) and that the latter terms are similar to those in the example in the text, Eq. (6). (a) Show that n=1 [sin(2n 1)x]/(2n 1) = (/2) f(x) 1 2 , (ii) where f is the square wave in the example with L = 1. (b) Subtract Eq. (ii) from Eq. (i) and show that g(x) = 2 f(x) 1 2 n=1 sin(2n 1)x (2n 1)[1 + (2n 1)2] . (iii) The series (iii) converges much faster than the series (i) and thus provides a better way to calculate values of g(x).