Download MATH 241 Final Exam May 10, 2011 and more Exams Mathematics in PDF only on Docsity!
MATH 241 FINAL EXAM MAY 10, 2011 (CLEE, GUPTA, SHATZ)
The examination consists of 15 problems each worth 10 points; answer all of them. The first 10 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 for these 10 problems, use your bluebooks for computations and scratch work. The next 5 are long answer questions, PARTIAL CREDIT GIVENโYOU MUST SHOW YOUR WORK. SHOW THE WORK FOR THESE IN THE SPACE PROVIDED ON THE EXAM SHEETS. DO NOT USE BLUE BOOKS FOR THESE 5 QUESTIONS; DO NOT HAND IN YOUR BLUE BOOKS. No books, tables, notes, calcula-tors, computers, phones or electronic equipment allowed; one 8 and 1/2 inch by 11 inch paper allowed, handwritten on both sides, in your own handwriting; no substitutions of this aid allowed. NB. In what follows, we write ux for โuโx and uxx for โโx^2 u 2 , etc. in those problems concerning PDE. YOUR NAME (print please): YOUR PENN ID NUMBER: YOUR SIGNATURE: INSTRUCTORโS NAME (CIRCLE ONE): CLEE GUPTA SHATZ
SCORE: Do not write belowโfor grading purposes only. โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ I VI XI II VII XII III VIII XIII IV IX XIV V X XV โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ TOTAL: โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
1
I) โฎ If ฮณ is the ellipse x^2 + y 42 = 1, traced counterclockwise, evaluate ฮณ z^2 (ezzโ2) dz. a) 5 ฯi 2 b) โ^32 ฯi c) 0 d) 2 ฯi e) โ 4 ฯi II) If H(x) = โ1 for โฯ โค x < 0 and H(x) = 1 for 0 โค x < ฯ and
if we extend H to be 2ฯ periodic, then when we expand H(x) in a complex Fourier Series โโ k=โโ ckeikx, we find the sum cโ 2 + cโ 1 + co equals a) 2 ฯi b) 3 ฯi c) (^) ฯi d) (^2) ฯi e) โ 1
III) The complex number (iฯ)i^ may have many values. One of its val-
ues is: a) e โ^2 ฯ^ (cos(logฯ) + i sin(logฯ)) b) logฯ + i logฯ c) eฯ(cos(logฯ) + i sin(logฯ)) d) cos(logฯ) + i sin(logฯ) e) eโฯ(cos(eฯ) + i sin(eฯ)).
VII) For the PDE: ux + 3uy = 0, we are interested only in solutions of the form u(x, y) = X(x)Y (y). For such a solution, suppose we know u(0, 1) = e and u(1, 0) = e^2. Compute u(1, 1): a) exp( 14 ) b) exp( 34 ) c) exp( 54 ) d) exp( 74 ) e) exp( 94 )
VIII) A certain function u(x, y) is defined on the square whose vertices are (0,0), (1,0), (0,1) and (1,1). The function u satisfies the two PDEโs: uxx = 0 and uyy = 0 and has values at the corners of the square given by u(0, 0) = 0, u(1, 0) = 3, u(0, 1) = โ 2 , u(1, 1) = 1. For this function u, the value u( 12 , 12 ) is: a) 1 b) - c) 0 d) -3/ e) 1/
IX) We expand the function f (z) = (^) (zโez1) 2 in a Laurent Series valid in the annulus 1 < |(z โ 1) โ| <โ 2. This series has the form m=1 bm^ (zโ^1 1)m^ +^ โโ k=o ak(z^ โ^ 1)k. Then the value of the product b 2 b 1 aoa 1 a 2 is: a) e^5 b) e^5 / 288 c) e^5 / 120 d) e^5 / 84 e) e^5 /316.
X) For the function f (z) = (^) z(^2 zzโโ5)^12 , find its residue at z = 5.
a) 0 b) i/ c) 2/ d) ฯ/ e) 1/
THE NEXT 5 PROBLEMS ARE LONG ANSWERโWORK IS REQUIRED
TO BE SHOWN.
XII) If u(x, y) is a harmonic function (that is, it satisfies Laplaceโs DE: โu = 0) defined on the whole plane, then it is known that there is another harmonic function, call it v(x, y), so that the complex function f (z) = f (x, y) = u(x, y) + i v(x, y) is an entire function (i.e., is holomorphic (= analytic) on the entire complex plane). Write g(z) for the function g(z) = eโf^ (z). a) If u(x, y) is always โฅ 0, show, with an explicit upper bound, that |g(z)| is bounded.
b) Again assume u(x, y) โฅ 0 for all x, y. Use a) to find all such (every- where non-negative harmonic) functionsโbe careful, clear and explicit in your reasoning.
XIII) Consider the wave equation utt = 4uxx on the interval [0, ฯ] with boundary values u(0, t) = u(ฯ, t) = 0. If the initial conditions are u(x, 0) = 13 sin(3x) + 14 sin(4x) ut(x, 0) = 15 sin(5x) + 16 sin(6x), find (explicitly) the solution, u(x, t), of the problem.
XV) Suppose f (x) = |x| โ ฯ on [โฯ, ฯ] and is made 2ฯ periodic. We compute the Fourier Series for ao f (x):
Find the sum of the series^2 +^ โโ^ k=1^ akcos(kx) +^ bksin(kx). a 1 + a 2 + a 3 + ยท ยท ยท = โโ k=1 ak. Be explicit with your reasoningโa simple numerical answer is not sufficient.
END OF THE EXAM
