final_solution - Physics 3513 Final exam solution(1 Let z =...

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Physics 3513, Final exam: solution (1) Let z = x + iy , then we have x 2 - y 2 + 2 xyi = x - iy , or x 2 - y 2 = x, 2 xy = - y . The second equation has two solutions y = 0 and x = - 1 / 2. Substituting them to the frst equation, we get x = 0 , 1 For the frst case and y = ± 3 / 2 For the second one. Answer: 0 , 1 , ( - 1 ± 3) / 2. ( * ) z + iz = ( x - y ) + ( y + x ) i . Since | 4 z 4 | is real, this means y + x = 0, or z = x (1 - i ). Substituting, we get 2 x = 16 x 4 , so x = 0 or x = 1 / 2. Answer: 0 , (1 - i ) / 2. (2) g ( ω ) = 1 2 π i + a a 1 2 a e iωx dx = 1 2 π ( - 2 aiω ) p e iωa - e iωa P = sin 2 πaω . ( * ) lim a 0 f ( x ) = δ ( x ), so lim a 0 i + −∞ f ( x ) e x 2 dx = 1. (3) A 1 e = ± 1 2 3 2 ²± 1 - 1 ² = ± - 1 1 ² = - e , so λ 1 = - 1. Similarly, A 2 e = ± 2 0 1 3 1 - 1 ² = ± 2 - 2 ² = 2 e , so λ 2 = 2. ( * ) ( A 1 + A 2 ) e = A 1 e + A 2 e = - e + 2 e = e , so ( A 1 + A 2 ) 10 e = ( A 1 + A 2 ) 9 e = ··· = e . The corresponding eigenvalue is 1. (4) The characteristic equation D 2 - 1 = 0 has roots ± 1. Trying particular solutions oF the Form k 1 e 2 x and k 2 , we fnd k 1 = 1 / 3 and k 2 = 1, respectively. Answer: y = C 1 e x + C 2 e x + 1 3 e 2 x + 1. ( * ) The equation can be re-written as ( y ′′ - y + 1) 2 = e 2 x , or y ′′ - y = ± e x - 1. Answer: y = C 1 e x + C 2 e x ± 1 2 xe x + 1. (5) The series converges iF lim n →∞ v v v v a n +1 a n v v v v = | x | < 1, and diverges iF | x | > 1. At | x | = 1, the series s n =1 2 n + 1 2 n n > s n =1 1 n diverges, and the series s n =1 2 n + 1 2 n n ( - 1) n = s n =1 1 + 2 n n
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