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Mathematics 241 Makeup Final Examination
Tuesday, January 13, 2004
Your name, printed:................................................................
Your signature:......................................................................
Your Penn student ID#:........................................................... Circle the name of your Professor: Shatz Wilf
This is a closed book exam. You may use during the exam, a single 5x7 inch card on which you may have written on both sides, if you wish. No more than one such card, and no larger size, please. No other books, papers, calculators, or other materials may be consulted during an exam. The exam consists of twenty (20) multiple choice questions, and each question offers a choice of five or six answers. Circle the answer that you choose, in each case, on the question sheet itself. The exam booklets are given to you for your convenience in working out the problems, but do not hand in these booklets when the exam is over. Hand in only the exam itself, with your answers to the questions circled.
For grader’s use only:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Do not turn the page and start work until the proctors announce the start of the exam.
- While we are solving a certain heat conduction problem, suppose that the solution which satisfies u(x, 0) = 1 is u(x, t) =
∑^ ∞ n=
Ane−kn^2 π^2 t/L^2 sin nπx L. Then which one of the following is true: (a) ux(0, t) = ux(L, t) = 0 (b) ux(0, t) = ux(L, t) (c) ux(0, t) = u(0, t) = 0 (d) u(0, t) = u(L, t) = 0 (e) ux(L, t) = u(L, t) = 0
- The residue of ze^1 /(2z)^ at the origin is
(a) 1/ (b) 2πi (c) 1/ (d) πi (e) none of the above
- Given that k is a constant and that x^2 − ky^2 is the real part of an analytic function, then (a) k > 0 (b) k = 0 (c) k = 1 (d) no value of k can do this (e) k = − 1
- The integral of ez^ /(z − 1)^3 counterclockwise around a small circle centered at z 0 = 1 is
(a) 2πi (b) πi/ 12 (c) πi/ 3 (d) eπi
- The center, z 0 , and the radius, R, of the circle of convergence of the power series ∑^ ∞ k=
k^2
( (^) 3 + 4i 3 − 4 i
)k (z − 3)k
(a) are z 0 = 3 and R = √ 2 (b) are z 0 = 0 and R =
(c) are z 0 = 3 and R = 1 (d) are z 0 = 0 and R =
(e) are none of the above
- The value of ∫^02 πdt/(10 − 6 cos t) is (a) − 1 / 8 (b) π/ 4 (c) 1/ 2 (d) π/ 2 (e) none of the above
- The value of ∫^ −∞∞ dx/(x^2 − 6 x + 25) is
(a) π/ 4 (b) 1/ 8 (c) π/ 8 (d) 1/ 4 (e) none of the above
- In the Frobenius series solution y(x) = ∑∞ n=0 anxn+r^ of the differential equation 2xy′′^ + (1 + x)y′^ + y = 0, the two possible values of the index r are (a) 0 and 1 (b) 0 and 2 (c) 0 and 1/ (d) 1 and 2 (e) none of the above
- Which of the following functions has a pole of order 2 and a pole of order 1 and a removable singularity? (a) (ez^ − 1)/(z^2 (z − 1)) (b) (ez^ − 1)/(z(z^2 − 9)(z + 3)) (c) sin z/(z(z + 1)(z − 1)) (d) sin z/(z^2 (z − 1)) (e) none of the above
- The value of the integral (1/(2πi)) ∫ γ Re(z)dz, where γ is the unit circle |z| = 1 traversed counterclockwise, is (a) 0 (b) (^12) (c) 1 (d) (^32) (e) 2
- The Laurent series of the function 1/(z(z − 1)^2 ) valid in the annulus 0 < |z − 1 | < 1 is
(a) (^) (z−^1 1) 2 + (^) z−^11 + 1 − (z − 1) + (z − 1)^2 −... (b) (^) (z−−^1 1) 2 + (^) z−^11 + 1 − (z − 1) + (z − 1)^2 −... (c) (^) (z−^1 1) 2 − (^) z−^11 + 1 − (z − 1) + (z − 1)^2 −... (d) (^) (z−^1 1) 2 + (^) z−^11 + 1 + (z − 1) + (z − 1)^2 +... (e) (^) (z−−^1 1) 2 + (^) z−^11 + 1 + (z − 1) + (z − 1)^2 +...
- For the differential equation x^2 y′′^ + (x − 1)y = 0, the result of using the method of Frobenius is (a) Two solutions, both of which take finite values at x = 0 (b) Two solutions, neither of which takes finite values at x = 0 (c) An indicial equation whose roots differ by an integer (d) Two solutions, only one of which takes finite values at x = 0 (e) No Frobenius-type solutions exist.
(b) (eπ^ + e−π^ )/(e^2 π^ − e−^2 π) (c) (eπ^ − e−π^ )/(e^2 π^ + e−^2 π) (d) (eπ^ + e−π^ )/(e^2 π^ + e−^2 π) (e) (e^2 π^ − 1)/(e^4 π^ − 1) (f) (e^2 π^ + 1)/(e^4 π^ + 1)
- The differential equation y′′^ + xy′^ + y = 0 has a series solution y = 1 + a 2 x^2 + a 3 x^3 +.. .. Then (a) all ai’s in which i is even and ≥ 4 are 0. (b) a 2 + a 4 > 0 (c) all ai’s in which i is odd and ≥ 3 are 0. (d) The absolute values of the a’s with even subscripts decrease steadily, and a 6 is the first one with absolute value < 1 /100. (e) a 2 + a 3 + a 4 + a 5 + a 6 > 0. (f) y′(0) = 1
