lecture-35 - Then f ( s ) ( s-z ) s N M ( r-r ) r N , and...

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Lecture 35: Taylor Series Dan Sloughter Furman University Mathematics 39 May 5, 2004 35.1 Taylor series Definition 35.1. If f is analytic at a point z 0 C , we call the power series X n =0 f ( n ) ( z 0 ) n ! ( z - z 0 ) n the Taylor series of f about z 0 . When z 0 = 0, we call X n =0 f ( n ) (0) n ! z n the Maclaurin series of f . The following fundamental theorem is known as Taylor’s theorem . Theorem 35.1. If R 0 > 0, z 0 C , and f is analytic in the disk D = { z C : | z - z 0 | < R 0 } , then f ( z ) = X n =0 f ( n ) ( z 0 ) n ! ( z - z 0 ) n for all z D . 1
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Proof. We first assume z 0 = 0. Let z D , let r = | z | , let r < r 0 < R 0 , and let C 0 be the positively oriented circle of radius r 0 centered at the origin. By the Cauchy integral formula, we have f ( z ) = 1 2 πi Z C 0 f ( s ) s - z ds. Now 1 s - z = 1 s · 1 1 - z s = 1 s ± N - 1 X n =0 ² z s ³ n + ( z s ) N 1 - z s ! = N - 1 X n =0 z n s n +1 + z N ( s - z ) s N . Hence f ( z ) = 1 2 πi N - 1 X n =0 z n Z C 0 f ( s ) s n +1 ds + z N 2 πi Z C 0 f ( s ) ( s - z ) s N ds = N - 1 X n =0 f ( n ) (0) n ! z n + ρ N ( z ) , where ρ N ( z ) = z N 2 πi Z C 0 f ( s ) ( s - z ) s N ds. It remains to show that lim N →∞ ρ N ( z ) = 0. Let M be the maximum value of | f ( s ) | on C 0 and note that | s - z | ≥ || s | - | z || = r 0 - r.
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Unformatted text preview: Then f ( s ) ( s-z ) s N M ( r-r ) r N , and so | N ( z ) | r N 2 M ( r-r ) r N 2 r = Mr r-r r r N . 2 Since r r &lt; 1, we have lim N N ( z ) = 0. Finally, if z 6 = 0, let g ( z ) = f ( z + z ). Then g is analytic when | ( z + z )-z | &lt; R , that is, when | z | &lt; R . Hence f ( z + z ) = g ( z ) = X n =0 g ( n ) (0) n ! z n = X n =0 f ( n ) ( z ) n ! z n when | z | &lt; R . Thus if | z-z | &lt; R , f ( z ) = f (( z-z ) + z ) = X n =0 f ( n ) ( z ) n ! ( z-z ) n . 3...
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lecture-35 - Then f ( s ) ( s-z ) s N M ( r-r ) r N , and...

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