217sppedexam3 sols

Math 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 +

Math 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.

Math 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.

Math 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

Math 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.

Math 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.

Math 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


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.

Math 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


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.

Math 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).

Math 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


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.

Math 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


Solution. Ans.: c.This question is based on Example 2, page 202. Note that

= limn

cncn+1 = 3.

Math 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.

Math 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.

Math 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


Solution. Ans.: d.This is 3.1 #1.

Math 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


Solution. Ans.: b.3.1 #19.

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.

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.

217sppedexam3 sols

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