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Unformatted text preview: Then ´ ´ ´ ´ f ( s ) ( sz ) s N ´ ´ ´ ´ ≤ M ( rr ) r N , and so  ρ N ( z )  ≤ r N 2 π · M ( rr ) r N · 2 πr = Mr rr µ r r ¶ N . 2 Since r r < 1, we have lim N →∞ ρ N ( z ) = 0. Finally, if z 6 = 0, let g ( z ) = f ( z + z ). Then g is analytic when  ( z + z )z  < R , that is, when  z  < 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  < R . Thus if  zz  < R , f ( z ) = f (( zz ) + z ) = ∞ X n =0 f ( n ) ( z ) n ! ( zz ) n . 3...
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 Spring '04
 DanSloughter
 Math, Power Series, Taylor Series

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