Final Exam 2014 Solutions - Math 121A solutions to nal exam 1(a Calculate the eigenvalues and eigenvectors of the matrix 1 0 1 A = 0 1 1 1 1 0 Answer

# Final Exam 2014 Solutions - Math 121A solutions to nal exam...

• Test Prep
• PresidentHackerIbex3077
• 12
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 1 - 4 out of 12 pages.

Math 121A: solutions to final exam 1. (a) Calculate the eigenvalues and eigenvectors of the matrix A = 1 0 1 0 1 1 1 1 0 . Answer: The eigenvalues are solutions to 0 = det ( A - λ I ) = 1 - λ 0 1 0 1 - λ 1 1 1 - λ = - λ ( 1 - λ ) 2 - 2 ( 1 - λ ) = ( 1 - λ )( - 2 - λ + λ 2 ) = ( 1 - λ )( 2 - λ )( 1 + λ ) and hence λ = 1, - 1, 2. Write u λ = ( u λ , v λ , w λ ) for the eigenvector corresponding to λ . For λ = 1 the eigenvector satisfies 0 0 1 0 0 1 1 1 - 1 u 1 v 1 w 1 = 0 0 0 and thus u 1 = ( - 1, 1, 0 ) is a solution. The other two eigenvectors are given by 2 0 1 0 2 1 1 1 1 u - 1 v - 1 w - 1 = 0 0 0 , - 1 0 1 0 - 1 1 1 1 - 2 u 2 v 2 w 2 = 0 0 0 and thus u - 1 = ( 1, 1, - 2 ) and u 2 = ( 1, 1, 1 ) are solutions. (b) Solve the linear system A x y z = 0 4 2 . Answer: Note that the right hand side is 2 u 2 + 2 u 1 . This is equal to A ( u 2 + 2 u 1 ) and thus ( x , y , z ) = ( - 1, 3, 1 ) .
2. (a) Calculate the radius of convergence R of the power series n = 1 x n n ( - 3 ) n . By considering the series for x = ± R and using appropriate series tests, determine the exact range of x for which it converges. Answer: If the series is written as a n , then the radius of convergence is given by 1 R = lim n a n + 1 a n = lim n ( - 3 ) n n ( - 3 ) n + 1 ( n + 1 ) = lim n n - 3 ( n + 1 ) = 1 3 and hence R = 3. For x = 3, the series is n = 1 ( - 1 ) n n , which converges by the alternating series test. For x = - 3, the series is n = 1 1 n , which diverges by the integral test, since R n - 1 dn = C + log n which tends to infinity as n . Hence the series converges for - 3 < x 3. (b) For the function f ( t ) = - 5 + 2 t 2 , calculate f ( 0 ) , f ( 1 ) , f ( 2 ) , and f ( 3 ) . Use the results to sketch f over the range - 3 t 3. Answer: The function values are f ( 0 ) = - 5, f ( 1 ) = - 3, f ( 2 ) = 3, f ( 3 ) = 13. A graph is shown in Fig. 1 . (c) Determine the precise set of values of t for which the series n = 1 [ f ( t )] n n ( - 3 ) n will converge. Answer: By reference to part (a), it can be seen that the series will converge for - 3 < f ( t ) 3. By reference to the graph from part (b), this will correspond to - 2 t < - 1 and 1 < t 2.
3. (a) Let a function f ( x ) have the Fourier transform ˜ f ( α ) . Let g ( x ) = f ( - x ) and h ( x ) = x f ( x ) . Show that the Fourier transforms of g and h are given by ˜ g ( α ) = ˜ f ( - α ) and ˜ h ( α ) = i ˜ f 0 ( α ) . Answer: The Fourier transform of g is ˜ g ( α ) = 1 2 π Z - f ( - x ) e - i α x dx = 1 2 π Z - f ( y ) e i α ( - y ) dy = 1 2 π Z - f ( y ) e - i ( - α ) y dy = ˜ f ( - α ) . and the Fourier transform of h is ˜ h ( α ) = 1 2 π Z - x f ( x ) e - i α x dx = 1 2 π Z - f ( x )( xe - i α x ) dx = 1 2 π Z - f ( x ) i ∂α ( e - i α x ) dx = i 2 π d d α Z - f ( x ) e - i α x dx = i d ˜ f d α . (b) Determine the Fourier transform ˜ f ( α ) of the function f ( x ) = e - x for x > 0, 0 for x 0. Answer: The Fourier transform is given by ˜ f ( α ) = 1 2 π Z 0 e - x e - i α x dx = 1 2 π " e - ( 1 + i α ) x - ( 1 + i α ) # 0 = 1 2 π ( 1 + i α ) .