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Page 1Math 217Exam 3
Name:ID:Section:This exam has 16 questions:
14 multiple choice questions worth 5 points each. 2 hand graded questions worth 15 points each.
Important:
No graphing calculators! Any non-graphing scientific calculator is fine. For the multiple choice questions, mark your answer on the answer card. Show all your work for the written problems. You will be graded on the ease of reading your
solution as well as for your work. You are allowed both sides of 3 5 note cheat card for the exam.
Power Series Representations
The following power series converge to the indicated functions for all x.
ex =n=0
xn
n!= 1 + x+
x2
2!+x3
3!+
cosx =n=0
(1)nx2n(2n)!
= 1 x2
2!+x4
4!+
sinx =n=0
(1)nx2n+1(2n+ 1)!
= x x3
3!+x5
5!
coshx =n=0
x2n
(2n)!= 1 +
x2
2!+x4
4!+
sinhx =n=0
x2n+1
(2n+ 1)!= x+
x3
3!+x5
5!+
ln(1 + x) =n=1
(1)n+1xnn
= x x2
2+x3
3
Geometric series : converges if |x| < 1 but diverges if |x| > 1.1
1 x =n=0
xn = 1 + x+ x2 + x3 +
Binomial series : if is a nonnegative integer, the series terminates. Otherwise, the series convergesif |x| < 1 but diverges if |x| > 1.
(1 + x) = 1 + x+( 1)
2!x2 +
( 1)( 2)3!
x3 +
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Page 2Math 217Exam 3
1. Simple Harmonic Motion: Mechanical VibrationsDetermine the period and frequency of the simple harmonic motion of a body of mass 0.75 kgon the end of a spring with spring constant 48 N/m.
a. frequency = pi3
s, period= 3pi
Hz
b. frequency = pi3
Hz, period= 3pi
s
c. frequency = 3pi
Hz, period=pi3
s
d. frequency = 4pi
Hz, period=pi4
s
e. frequency = 4pi
s, period=pi4
Hz
f. frequency = pi4
Hz, period= 4pi
s
g. frequency = 8pi
Hz, period=pi8
s
h. None of the above
Solution. Ans.: d.
We are given m = 0.75 kg and k = 48 N/m. The circular frequency is 0 =
km
=
480.75
= 8
rad/s. The period is T = 2pi0
= 2pi8
= pi4s. The frequency is equal to the reciprocal of the
period (in Hz), so = 1T
= 4pi
Hz.
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Page 3Math 217Exam 3
2. Simple Harmonic Motion: Mechanical VibrationsA body with mass m = 1
2kg is attached to the end of a spring that is stretched 2m by a force
of 100N. It is set in motion with initial position x0 = 1 m and initial velocity v0 = 5 m/s.Find the position function of the body.
a. x(t) 52
cos(10t 0.4636)b. x(t) 5
2cos(10t 5.8195)
c. x(t) 12
5 cos(10t 6.7468)
d. x(t) 12
5 cos(10t 0.4636)
e. x(t) 12
5 cos(10t 5.8195)
f. None of the above.
Solution. Ans.: e.This is Example 1, page 139. Note that 0 < < 2pi, so (c) and (d) cannot be correct.
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Page 4Math 217Exam 3
3. Simple Harmonic Motion: Mechanical VibrationsA mass m = 4 kg is attached to both a spring (with spring constant k = 169) and a dashpot(with damping constant c = 20). The mass is set in motion with initial position x0 = 4 m andinitial velocity v0 = 16 m/s. Find the position function x(t) and determine whether the motionis overdamped, critically damped, or underdamped.
a. Underdamped. x(t) 13
313e5t/2 cos(6t 0.8254)
b. Critically damped. x(t) 13
313e5t/2 cos(6t 0.8254)
c. Overdamped. x(t) 13
313e5t/2 cos(6t 0.8254)
d. Underdamped. x(t) 13
313e5t/2 sin(6t 0.8254)
e. Underdamped. x(t) 13
313e5t/2 sin(6t+ 0.8254)
f. None of the above.
Solution. Ans.: a.This is 2.4 #19
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Page 5Math 217Exam 3
4. Simple Harmonic Motion: Simple PendulumTwo pendulums are of lengths L1 and L2 and when located at the respective distances R1and R2 from the center of the earth have periods p1 and p2, respectively. Which of thefollowing statements is true?
a. p1p2
= R1L1
R2L2
b. p1p2
= R2L1
R1L2
c. p1p2
= R1L2
R2L1
d. p1p2
= R2L2
R1L1
e. All the above
f. None of the above
Solution. Ans.: a.This is 2.4 #9.
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Page 6Math 217Exam 3
5. Simple Harmonic Motion: Simple PendulumMost grandfather clocks have pendulums with adjustable lengths. One such clock loses 10minutes per day when the length of its pendulum is 30 inches. With what length pendulumwill this clock keep perfect time?
a. 28.41 inches
b. 28.59 inches
c. 29.41 inches
d. 29.59 inches
e. 30.41 inches
f. 30.59 inches
Solution. Ans: d.2.4 #8.
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Page 7Math 217Exam 3
6. Elementary Power SeriesThe power series 1 + x
2+ x
2
4+ x
3
8+ represents which of the following functions?
a. cosh x2
b. sinh x2
c. 22x
d. 2 11x
e. (1 + x)1/2
f. None of the above
Solution. Ans.: c.This is a geometric series:
1 +x
2+x2
4+x3
8+ = 1 + x
2+(x
2
)2+(x
2
)3+ = 1
1 x2
=2
2 x.
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Page 8Math 217Exam 3
7. Power Series Operations: MultiplicationThe power series 1 + 1
2x 1
8x2 + 1
16x3 + represents which of the following functions?
a. cosh x2
b. sinh x2
c. 22x
d. 2 11x
e. (1 + x)1/2
f. None of the above
Solution. Ans.: e.Note: (a) only involves even powers, (b) only odd powers. Choice (c) is the answer to the lastquestion. By elimination, only (d) and (e) are left and a check shows that (e) is correct.
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Page 9Math 217Exam 3
8. Analytic functionsWhich of the following statements is true?
a. If the Taylor series of the function f converges to f(x) for all x in some open intervalcontaining a, then we say that the function f is analytic at x = a
b. Every polynomial function is analytic everywhere.
c. Every rational function is analytic wherever its denominator is nonzero.
d. The sum of functions which are analytic at x = a is analytic at x = a.
e. The product of functions which are analytic at x = a is analytic at x = a.
f. All of the above.
g. None of the above.
Solution. Ans.: f.These are all true statements. (See page 196).
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Page 10Math 217Exam 3
9. Analytic FunctionsWhich of the following functions is analytic at x = 2?a. x+2
x24b. x2
x24
c.x+2
x24
d.x2
x24
e.x+2
x2+4
f. None of the above
Solution. Ans.: e.Choices (a)-(f) all have a singularity at x = 2 because the denominator is zero there.I also accepted (f): None of the above as an answer. Several students pointed out that thefunction is not differentiable at 0, thus not analytic there.
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Page 11Math 217Exam 3
10. Radius of ConvergenceSuppose we apply the method of power series to a DE and find out that the recurrence relationfor the coefficients of
n=0 is cn+1 =
n+23(n+1)
cn for n 0. What is the radius of convergence ofthe power series solution?
a. 1
b. 2
c. 3
d. 9
e. 0
f. None of the above
Solution. Ans.: c.This question is based on Example 2, page 202. Note that
= limn
cncn+1 = 3.
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Page 12Math 217Exam 3
11. Manipulating Power SeriesConsider the following power series.
I.
n=0 cnxn
II.
n=3 cnn(n 1)(n 2)xn3III.
n=0 cnn(n 1)(n 2)xn3
IV.
n=1 cnxn
Which of these power series represents the same function?
a. I and IV only
b. II and III only
c. I and II only
d. II and IV only
e. All the above
f. None of the above, or some other combination of I, II, III and IV
Solution. Ans.: b.
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Page 13Math 217Exam 3
12. Recurrence RelationSuppose cn+1 = 2n+1cn for n 0. Which of the following is a closed form expression for cn?
a. cn =2n
n!co, n 0
b. cn = (1)n 2n(n+1)!co, n 0c. cn = (1)n 2nn! c0, n 1d. cn = (1)n 2nn! c1, n 0e. cn = (1)n n2n! co, n 0f. None of the above.
Solution. Ans.: c.3.1 Example 1.I also accepted (f): None of the above. Several students pointed out that the formula is true forn 0, not just n 1.
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Page 14Math 217Exam 3
13. Solving DEs with Power SeriesSolve: y=y. What is the radius of convergence?
a. y = c0 sinx; =b. y = c0 sinx; =c. y = c0e
x; = 1
d. y = c0ex; =
e. No power series solution since = 0
f. None of the above
Solution. Ans.: d.This is 3.1 #1.
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Page 15Math 217Exam 3
14. Solving IVPs with Power SeriesSolve the IVP: y + 4y = 0; y(0) = 0, y(0) = 3
a. 23
sin 2x
b. 32
sin 2x
c. 23
cos 2x
d. 32
cos 2x
e. 23
sin 3x
f. None of the above
Solution. Ans.: b.3.1 #19.
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WRITTEN PROBLEMSHOW YOUR WORKMath 217Exam 3
Name:ID:Section:Note: You will be graded on the readability of your work. Use the back of this sheet, ifnecessary.
15. Power Series SolutionsSolve the equation (2x 1)y + 2y using the power series method.
Solution. Ans. y(x) = c01
12xThis is 3.1 #7.
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WRITTEN PROBLEMSHOW YOUR WORKMath 217Exam 3
Name:ID:Section:Note: You will be graded on the readability of your work. Use the back of this sheet, ifnecessary.
16. Power Series Solutions Show that the equation
x2y + x2y + y = 0
has no nontrivial power series solution of the form y =
n=0 cnxn. (Hint: Assume that the
equation does have a power series and work it out. What happens?)
Solution. Assume that there is a power series of the form y =
n=0 cnxn. Then
0 = x2y + x2y + y
= x2n=2
cnn(n 1)xn2 + x2n=1
cnnxn1 +
n=0
cnxn
=n=2
cnn(n 1)xn +n=1
cnnxn+1 +
n=0
cnxn
=n=2
cnn(n 1)xn +n=1
cnnxn+1 + c0 + c1x+
n=2
cnxn
= c0 + c1x+n=2
[(n(n 1) + 1)cn + (n 1)cn1]xn
= c0 + c1x+n=2
[(n2 n+ 1)cn + (n 1)cn1]xn.
Therefore, c0 = c1 = 0 and cn = n1n2n+1cn1 for n 2. Consequently, cn = 0 for all n 0, andso there is no nontrivial power series solution.