taylor_series - (12.10 Taylor Series In this section we...

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(12.10) Taylor Series In this section, we learn a general technique for finding power series that represent a given function. Suppose f ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + · · · . (1) We want to figure out what the coefficients a n are, so that the power series does represent the function. If we let x = 0, then f (0) = a 0 + a 1 0 + a 2 0 2 + a 3 0 3 + a 4 0 4 · · · = a 0 . So at least we know a 0 = f (0). Now take the derivative of equation (1). We get f ( x ) = a 1 + 2 a 2 x + 3 a 3 x 2 + 4 a 4 x 3 + · · · . (2) Now let x = 0. We get f (0) = a 1 + 2 a 2 0 + 3 a 3 0 2 + 4 a 4 0 3 + · · · = a 1 . So a 1 = f (0). Now take the derivative of equation (2). We get f ( x ) = 2 a 2 + 6 a 3 x + 12 a 4 x 2 + · · · . (3) Let x = 0. We get f (0) = 2 a 2 + 6 a 3 0 + 12 a 4 0 2 + · · · = 2 a 2 . So a 2 = f (0) 2 . Now take the derivative of equation (3). We get f ( x ) = 6 a 3 + 24 a 4 x + · · · . Let x = 0. We get f (0) = 6 a 3 , so a 3 = f (0) 6 . Keep going with this. We get a n = f (0) (0) n ! . That is, if f ( x ) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + · · · = n =0 a n x n then a n = f (0) (0) n !
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