pow_series_for_funct - (12.9 Power Series for Functions...

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(12.9) Power Series for Functions Recall the formula for geometric series: n =0 ar n = a 1 - r provided | r | < 1. (We really mean, n =0 ar n converges to a 1 - r provided | r | < 1.) So if we let a = 1 and r = x , we have n =0 x n = 1 + x + x 2 + x 3 + · · · = 1 1 - x provided | x | < 1, or - 1 < x < 1. So we think of f ( x ) = n =0 x n as a function of x . We therefore know is that it is really the function f ( x ) = 1 1 - x , but only for - 1 < x < 1 (for x outside that interval, the function f ( x ) is undefined). We can use this series to find power series for other functions. Example Find a power series for the function f ( x ) = 2 x x - 3 . We start with 1 1 - x = 1 + x + x 2 + x 3 + · · · . Trick: substitute x 3 for x . We have 1 1 - x 3 = 1 + x 3 + x 3 2 + x 3 3 + · · · or 3 3 - x = 1+ x 3 + x 2 9 + x 3 27 + · · · . We adjust this a little (by dividing by 3 then multiplying by - 1): 1 x - 3 = - 1 3 - x 9 - x 2 27 - x 3 81 - · · · . This is 1 x - 3 = n =0 - x n 3 n +1 . Now multiply by 2 x to get the desired result: 2 x x - 3 = - 2 x 3 - 2 x 2 9 - 2 x 3 27 - 2 x 4 81 - · · · = n =0 - 2 x n +1 3 n +1 = n =0 - 2 x n 3 n . So we are done, but we should determine the interval of convergence. The geometric series 3 3 - x = 1 + x 3 + x 2 9 + x 3 27 + · · · converges if x 3 < 1 or - 3 < x < 3. Multiplying by 2 x doesn’t affect convergence. So we conclude that
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