241S10 final-ans-1

241S10 final-ans-1 - MATHEMATICS 241 FINAL EXAM MAY 4, 2010...

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Unformatted text preview: MATHEMATICS 241 FINAL EXAM MAY 4, 2010 (RIMMER, SHATZ, ZHU) The examination consists of 20 problems each worth 10 points; answer all of them. They are multiple choice, NO PARTIAL CREDIT given, but NO PENALTY FOR GUESSING. To answer, circle the ENTIRE statement you deem correct in the problem concerned. No work is required to be shown, use your bluebooks for computations and scratch work. DO NOT HAND IN YOUR BLUE BOOKS. No books, tables, notes, calculators, com- puters, phones or electronic equipment allowed; one 8 1 2 inch by 11 inch sheet handwritten both sides allowed. NB. In what follows, we write u x for ∂u ∂x and u xx for ∂ 2 u ∂x 2 , etc. in those prob- lems concerning PDE. YOUR NAME (print please): YOUR PENN ID NUMBER: YOUR SIGNATURE: YOUR INSTRUCTOR’S NAME (CIRCLE): RIMMER SHATZ ZHU YOUR SECTION DAY/TIME: DAY: TIME: ————————————————————————————– SCORE: ————————————————————————————— I) Consider Ω, the region sketched below between the indicated curves. Write γ for the boundary of Ω, so that γ consists of the two curves traced in the directions shown in the picture. PICTURE I Then the integral around γ of the function: f ( z ) = z ( z- 1)( z- 3) , is equal to a) 3 2 πi b)- 3 2 πi c) 3 πi d)- 3 πi 1 2 e) 2 πi You can use either Cauchy’s Integral Formula applied to the function g ( z ) = z z- 1 , which is holomorphic in Ω, or the Residue Theorem applied to the given function f ( z ) in Ω. The answer is 3 πi . II) For the boundary value problem 2 tu xx = u t , < x < 1 , t > 0, u (0 ,t ) = u (1 ,t ) = 0 for t > with the initial condition u ( x, 0) = 2 sin 3 πx, < x < 1, the value of u (1 / 2 , 2) is a)- 2 exp (- 36 π 2 ), (here exp ( s ) = e s ) b) 2 exp (- 36 π 2 ) c) d) 2 exp (- 9 π 2 ( e 4- 1)) e)- 2 exp (- 9 π 2 ( e 4- 1)) You separate the variables, the extra t in the equation gives exp ( at 2 ) as a solution for the t-part. With the initial condition, the Fourier Series for u ( x, 0) collapses and you get the final solution: u ( x,t ) = 2 exp (- 9 π 2 t 2 ) sin (3 πx ). Thus, u (1 / 2 , 2) =- 2 exp (- 36 π 2 ). III) Compute the integral R ∞-∞ cos (3 πx ) ( x 2 +3) dx . a) π √ 3 exp (3 π √ 3) b) π √ 3 exp (- 3 π √ 3) c) π √ 3 exp (3 π ) d) π √ 3 exp (- 3 π ) e) π √ 3 exp (- π √ 3) 3 Here, you use the residue theorem. The region is bounded by the upper half circle of radius R centered at the origin and the x-axis from- R to R ; we trace this combined curve counterclockwise, as usual. Then we integrate the function f ( z ) = exp (3 πiz ) z 2 +3 over our curve, the answer is 2 πiRes | i √ 3 ( f ( z ))....
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This note was uploaded on 04/01/2012 for the course HIST 241 taught by Professor Zhu during the Spring '12 term at UPenn.

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241S10 final-ans-1 - MATHEMATICS 241 FINAL EXAM MAY 4, 2010...

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