475 answers 5 - ANSWERS PAMPHLET 201 Solutions are A3.8 1,...

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

View Full Document Right Arrow Icon
ANSWERS PAMPHLET 201 Solutions are 2 1 , 1 2 1 i p 3 2 , 1 2 2 i p 3 2 . A3.8 e 1 1 i 5 e 1 ? e i 5 e (cos 1 1 i sin 1) . e p i 6 2 5 cos 2 1 i sin 2 5 i. e 2 2 i 5 e 2 ? e 2 i 5 e 2 (cos( 2 ) 1 i sin( 2 )) 52 e 2 . A3.9 e z 1 ? e z 2 5 ˆ 1 1 z 1 1 z 2 1 2! 1 z 3 1 3! 1??? 1 1 z 2 1 z 2 2 2! 1 z 3 3 3! ! 5 1 1 ( z 1 1 z 2 ) 1 ˆ z 2 1 2! 1 z 1 z 2 1 z 2 2 2! ! 1 1 3! ± z 3 1 1 3! 1! 2! z 1 z 2 2 1 3! 2! 1! z 2 1 z 2 1 z 3 2 1 1 4! ± z 4 1 1 4! 3! 1! z 3 1 z 2 1 4! 2! 2! z 2 1 z 2 2 1 4! 1! 3! z 1 z 3 2 1 z 4 1 5 1 1 ( z 1 1 z 2 ) 1 1 2! ( z 1 z 2 ) 2 1 1 3! ( z 1 1 z 2 ) 3 1 1 4! ( z 1 1 z 2 ) 4 5 e z 1 1 z 2 . The coefficient of z j 1 z k 2 in the big expansion above is 1 j ! ? 1 k ! 5 1 ( j 1 k )! ? ( j 1 k )! j ! k ! . But, ( j 1 k )! 6 ( j ! k !) is precisely the coefficient of z j 1 z k 2 in the expansion of ( z 1 1 z 2 ) j 1 k . A3.10 If x n 5 c 1 3 n 1 c 2 ? 1 n ,then x n 1 1 5 3 c 1 3 n 1 1 ? c 2 ? 1 n and x n 1 2 5 9 c 1 3 n 1 1 ? c 2 ? 1 n . 4 x n 1 1 2 3 x n 5 (12 c 1 3 n 1 4 c 2 ? 1 n ) 2 (3 c 1 3 n 1 3 c 2 ? 1 n ) 5 9 c 1 3 n 1 c 2 ? 1 n 5 x n 1 2 . Let m 5 n 1 2 . Since x n 1 2 5 4 x n 1 1 2 3 x n , x m 5 4 x m 2 1 2 3 x m 2 2 .
Background image of page 1

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

View Full DocumentRight Arrow Icon
202 MATHEMATICS FOR ECONOMISTS A3.11 a )I f x n 5 k 1 (2 1 3 i ) n 1 k 2 (2 2 3 i ) n , x n 1 1 5 (2 1 3 i ) k 1 (2 1 3 i ) n 1 (2 2 3 i ) k 2 (2 2 3 i ) n x n 1 2 5 (2 1 3 i ) 2 k 1 (2 1 3 i ) n 1 (2 2 3 i ) 2 k 2 (2 2 3 i ) n 5 ( 2 5 1 12 i ) k 1 (2 1 3 i ) n 1 ( 2 5 2 12 i ) k 2 (2 2 3 i ) n . 4 x n 1 1 2 13 x n 5 [4(2 1 3 i ) 2 13] k 1 (2 1 3 i ) n 1 [4(2 2 3 i ) 2 13] k 2 (2 2 3 i ) n 5 ( 2 5 1 12 i ) k 1 (2 1 3 i ) n 1 ( 2 5 2 12 i ) k 2 (2 2 3 i ) n 5 x n 1 2 . b, c ) By DeMoivre’s formula, (2 6 3 i ) n 5 (5 n cos n u 0 6 i 5 n sin n 0 ), where tan 0 5 3 6 2. Now, ( c 1 1 ic 2 )(5 n cos n 0 1 i 5 n sin n 0 ) 1 ( c 1 2 ic 2 )(5 n cos n 0 2 i 5 n sin n 0 ) 5 ( c 1 5 n cos n 0 2 c 2 5 n sin n 0 ) 1 i ( c 2 5 n cos n 0 1 c 1 5 n sin n 0 ) 1 ( c 1 5 n cos n 0 2 c 2 5 n sin n 0 ) 2 i ( c 2 5 n cos n 0 1 c 1 5 n sin n 0 ) 5 2( c 1 5 n cos n 0 2 c 2 5 n sin n 0 ) 5 5 n ( C 1 cos n 0 1 C 2 sin n 0 ) , where C 1 5 2 c 1 and C 2 52 2 c 2 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/26/2010 for the course ECON 475 taught by Professor Voyvoda during the Spring '10 term at Middle East Technical University.

Page1 / 43

475 answers 5 - ANSWERS PAMPHLET 201 Solutions are A3.8 1,...

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