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Unformatted text preview: Mathematics 185 – Intro to Complex Analysis Fall 2005 – M. Christ Solutions to Selecta from Problem Set 8 10.5 Let ( F n ) n ≥ be the sequence of coefficients in the power series expansion of (1 z z 2 ) 1 about 0. Show that F n = F n 1 + F n 2 for all n ≥ 2. Show that F n = 1 √ 5 ( γ n +1 + γ n +1 ) where γ ± = 1 2 (1 ± √ 5). Solution. Let f ( z ) = (1 z z 2 ) 1 . Since f is the reciprocal of a differentiable function which does not vanish at 0, f is differentiable in some nonempty open disk D centered at 0, and hence there exists a unique power series expansion f ( z ) = ∑ ∞ n =0 F n z n in D . Then z 2 f ( z ) = ∑ ∞ n =2 F n 2 z n and zf ( z ) = ∑ ∞ n =1 F n 1 z n . Adding/subtracting these three power series gives 1 ≡ (1 z z 2 ) f ( z ) = F z + ( F 1 F ) z 1 + ∞ X n =2 [ F n F n 1 F n 2 ] z n for all z ∈ D . Since the differentiable function 1 has only one power series expansion in D , we must have F = 1, F 1 = F , and F n F n 1 F n 2 = 0 for all n ≥ 2. Another power series expansion can be obtained by first writing (1 z z 2 ) 1 = [( z + γ + )( z + γ )] 1 = 1 √ 5 1 z + γ + 1 √ 5 1 z + γ = 1 √ 5 γ + 1 1 + z/γ + 1 √ 5 γ 1 1 + z/γ = 1 √ 5 γ + ∞ X n =0 ( 1) n γ n + z n 1 √ 5 γ ∞ X n =0 ( 1) n γ n z n = ∞ X n =0 a n z n where a n = ( 1) n h 1 γ + √ 5 γ n + 1 γ √ 5 γ n i . Since any two representations of a function as sums of convergent power series in z must be identical, a n = F n for all n ≥ 0. A bit of algebra, which I leave to the reader, gives eventually F n = 1 √ 5 ( γ n +1 + γ n +1 ) = 1 √ 5 1 + √ 5 2 ! n +1 1 √ 5 1 √ 5 2 ! n +1 ; our authors are off by a factor of √ 5. 10.7 Let ( a n ) be the sequence of coefficients in the Taylor series expansion of f ( z ) = exp( 1 1 z ) about 0. Show that a = 1 and for n ≥ 1, a n = ∑ n s =1 1 s ! ( n 1 s 1 ) . 1 Solution. f ( z ) = exp( z + z 2 + z 3 + ··· ), so the coefficient a n of z n in the Taylor series for f will be ∑ ∞ k =0 c n,k /k ! where c n,k is the coefficient of z n in ( z + z 2 + z 3 + ··· ) k . This is not promising. So let’s try to use more of what we’ve learned. Choose any any r ∈ (0 , 1), and define γ r ( t ) = re it for t ∈ [0 , 2 π ]. Then since f is differentiable in the disk { z :  z  < 1 } , a n = (2 πi ) 1 R γ r f ( z ) z n 1 dz , Substituting z = 1 /w , this becomes a n = (2 πi ) 1 Z γ 1 /r e 1 /w w n 1 dw....
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This note was uploaded on 07/31/2009 for the course MATH 185 taught by Professor Lim during the Spring '07 term at Berkeley.
 Spring '07
 Lim
 Power Series

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