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Unformatted text preview: MTH 122 Calculus II Essex County College Division of Mathematics and Physics 1 Lecture Notes #22 Sakai Web Project Material 1 Finding Power Series We certainly know by now that some functions can be written as power series. For example, we used the geometric series 1 1- x = X n =0 x n , | x | < 1 to generate 2 many related power series. We also used curve fitting and derivatives to find other power series. For example, we now know that e x = X n =0 x n n ! , and ln(1- x ) =- x + x 2 2 + x 3 3 + x 4 4 + =- X n =1 x n n . Not that we need to remember these, or even generate them quickly, but we do need to realize that we were able to find power series for some functions. However, we also had painfully difficult examples 3 that didnt work out so nicely. So lets suppose we have a function f ( x ) that has a power series expansion that is centered at x = c and is valid for all x in an interval. It will be of this form f ( x ) = X n =0 a n ( x- c ) n = a + a 1 ( x- c ) + a 2 ( x- c ) 2 + + a 3 ( x- c ) 3 + . And we can differentiate this series term-by-term to get f ( x ) = = a + a 1 ( x- c ) + a 2 ( x- c ) 2 + + a 3 ( x- c ) 3 + f ( x ) = = a 1 + 2 a 2 ( x- c ) + 3 a 3 ( x- c ) 2 + 4 a 4 ( x- c ) 3 + f 00 ( x ) = = 2 a 2 + 6 a 3 ( x- c ) + 12 a 4 ( x- c ) 2 + 20 a 5 ( x- c ) 3 + . . . = . . . f ( n ) ( x ) = = n ! a n + all terms that follow will have a factor of ( x- c ). 1 This document was prepared by Ron Bannon ( firstname.lastname@example.org ) using L A T E X2 . Last revised January 10, 2009. 2 Examples involved integration, differention, and substitution. 3 The tangent for example. 1 Now if we evaluate these functions at x = c well get: f ( c ) = = a f ( c ) = = a 1 f 00 ( c ) = = 2 a 2 . . . = . . . f ( n ) ( c ) = = n ! a n Here we have a fairly simple way to generate the coefficients, that is, as long as the c is easy to evaluate in f and the derivatives of f are easy to find. a n = f ( n ) ( c ) n ! Lets take a really simple example, let f ( x ) = e x and c = 0. Work: Here we have f ( x ) = e x = f ( n ) ( x ) , and if c = 0 we have a n = f ( n ) (0) n ! = e n ! = 1 n ! . So our power series for e x centered at zero is e x = X n =0 x n n ! The interval of convergence here is all reals. The reason this was so easy to do is that e x is easy to differentiate and evaluate at zero. But what about finding the power series for f ( x ) = sin x and c = 0....
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