solns8 - Mathematics 185 Intro to Complex Analysis Fall...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ( n +1 +- n +1- ) where = 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 + + )( z + - )]- 1 = 1 5 1 z + +- 1 5 1 z + - = 1 5 + 1 1 + z/ +- 1 5 - 1 1 + z/- = 1 5 + X n =0 (- 1) n - n + z n- 1 5 - X n =0 (- 1) n - n- z n = X n =0 a n z n where a n = (- 1) n h 1 + 5 - n +- 1 - 5 - 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 ( n +1 +- 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 lets try to use more of what weve learned. Choose any any r (0 , 1), and define r ( t ) = re it for t [0 , 2 ]. Then since f is differentiable in the disk { z : | z | < 1 } , a n = (2 i )- 1 R r f ( z ) z- n- 1 dz , Substituting z =- 1 /w , this becomes a n = (2 i )- 1 Z 1 /r e 1 /w w n- 1 dw....
View Full Document

Page1 / 5

solns8 - Mathematics 185 Intro to Complex Analysis Fall...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online