College of the Holy Cross, Spring 2008 Solutions for Math 132, Midterm Exam 3 (All Sections) Wednesday, April 23, 7 PM

I.

A. [10 points] Circle the number of the plot showing the direction field for each of the following differential equations. (Note that there are only 3 plots, so the correct answer for one is “None.”)

Solution:

(1) y′ = x is plot III (note negative slopes for all x < 0 and positive slopes for all x > 0)

(2) y′ = 1 + y^{2} matches None of the plots (by process of elimination, or by noting that none of the direction fields has positive slope everywhere)

(3) y′ = 1 − y is plot I (note zero slope along the line y = 1)

(4) y′ = x − y is plot II (note zero slope along the line y = x)

I: II:

III:

B. [5 points] On the plot for the equation y′ = 1 − y from (3) of part A, give a qualitative sketch of the graph of the solution satisfying the initial condition y(−1) = 0. Show as much of the graph as you can for both positive and negative x.

Solution: See plot I above.

II. All parts of this problem deal with the differential equation y′ = 7 − y.

A. [4 points] Use 4 steps of Euler’s method to approximate the solution of this equation with the initial condition y(0) = 4 at x = 2.

Solution: With x_{0} = 0 and x_{4} = 2, we will use Δx = \frac{2−0}{4} = \frac{1}{2}. The steps of Euler’s Method are

y_{1} = y_{0} + (7 − y_{0})Δx = 4 + (3)(.5) = 5.5
y_{2} = y_{1} + (7 − y_{1})Δx = 5.5 + (1.5)(.5) = 6.25
y_{3} = y_{2} + (7 − y_{2})Δx = 6.25 + (.75)(.5) = 6.625
y_{4} = y_{3} + (7 − y_{3})Δx = 6.625 + (.375)(.5) = 6.8125

The approximate value of y(2) is y(2) ≐ y_{4} = 6.8125.

B. [6 points] Find the general solution y(x) of the equation by separating variables and integrating.

Solution: We separate variables, integrate, then exponentiate to solve for y:

\frac{dy}{7 − y} = dx
∫ \frac{dy}{7 − y} = ∫ dx

− ln |7 − y| = x + c

7 − y = be^{−x} where b = ±e^{−c} is another arbitrary constant
y = 7 − be^{−x}.

C. [5 points] Find the particular solution y(x) satisfying the initial condition y(0) = 4 and compute the exact value of y(2).

Solution: Substituting x = 0 and y = 4 gives 4 = 7 − b, so b = 3. The particular solution is y = 7 − 3e^{−x}, and y(2) = 7 − 3e^{−2} ≐ 6.59399. The approximation given by Euler’s Method in part A. is an overestimate.

III. All parts of this question deal with the infinite series

\frac{1}{2} + \frac{2}{3} + \frac{3}{4} + \frac{4}{5} + · · · .

A. [4 points] Find a general formula for the nth term of the series as a function of n and write the series in summation (“sigma”) notation. Solution: If we index the first term above with n = 1, the second with n = 2, etc., then the general term is a_{n} = \frac{n}{n+1}, and the series is ∑_{n=1}^{∞} \frac{n}{n + 1}.

B. [4 points] Call the nth term in the series a_{n}. What is lim_{n→∞} a_{n}?

Solution: We have

lim_{n→∞} \frac{n}{n + 1} = lim_{n→∞} \frac{n}{n + 1} · \frac{\frac{1}{n}}{\frac{1}{n}}
= lim_{n→∞} \frac{1}{1 + \frac{1}{n}}
= 1.

(This limit could also be computed using L’Hopital’s Rule.) C. [4 points] Write out the first 3 partial sums of the series.

Solution: The first three partial sums are

s_{1} = \frac{1}{2}
s_{2} = \frac{1}{2} + \frac{2}{3} = \frac{7}{6}
s_{3} = \frac{1}{2} + \frac{2}{3} + \frac{3}{4} = \frac{23}{12}.

D. [3 points] Does this series converge or diverge? Explain your answer.

Solution: Since lim_{n→∞} \frac{n}{n + 1} = 1 ≠ 0, this series diverges by the Test for Divergence. IV.

A. [5 points] Does the series ∑_{n=0}^{∞} \frac{(−1)^{n}e^{n}}{π^{n}} converge or diverge?

Solution: This series:

∑_{n=0}^{∞} \frac{(−1)^{n}e^{n}}{π^{n}} = ∑_{n=0}^{∞} (\frac{−e}{π})^{n}

is a geometric series with first term a = 1 and ratio r = −e/π. Since |r| < 1, the series is convergent and

∑_{n=0}^{∞} (\frac{−e}{π})^{n} = \frac{1}{1 − \frac{−e}{π}} = \frac{π}{π + e}.

B. [10 points] Explain why the Integral Test can be applied to the series ∑_{n=1}^{∞} \frac{n}{e^{3n}} and use

it to determine if the series converges or diverges.

Solution: The function f(x) = xe^{−3x} gives the terms in this series when we substitute x = n for n = 1, 2, 3, . . .. The function f(x) is continuous for all x, and f(x) > 0 for all x > 0. Moreover,

f′(x) = −3xe^{−3x} + e^{−3x} = e^{−3x}(1 − 3x) < 0

for all x ≥ \frac{1}{3}. Therefore f(x) is positive and decreasing for x ≥ 1 and the Integral Test applies. We integrate by parts with u = x, dv = e^{−3x} to compute

∫_{1}^{∞} xe^{−3x} dx = lim_{b→∞} \frac{−1}{3}xe^{−3x} − \frac{1}{9}e^{−3x}|_{1}^{b}
= lim_{b→∞} \frac{−b}{3e^{3b}} − \frac{1}{9e^{3b}} + \frac{1}{3e^{3}} + \frac{1}{9e^{3}}
= \frac{4}{9e^{3}} .

Since the improper integral converges, the series also converges.

V. All parts of this question refer to the power series

∑_{n=1}^{∞} \frac{(x − 3)^{n}}{\sqrt{n}} .

A. [9 points] Use the Ratio Test to determine the radius of convergence.

Solution: Applying the Ratio Test,

lim_{n→∞} |\frac{(x − 3)^{n+1}}{\sqrt{n + 1}} · \frac{\sqrt{n}}{(x − 3)^{n}} | = lim_{n→∞} \sqrt{\frac{n}{n + 1}}|x − 3|
= |x − 3|.

For absolute convergence, we need |x − 3| < 1, so 2 < x < 4. The series is centered at a = 3, so the radius of convergence is 1.

B. [6 points] Test convergence at the endpoints of the interval from part A to determine the interval of convergence. Explain your conclusions.

Solution: At x = 2, we substitute and obtain

∑_{n=1}^{∞} \frac{(−1)^{n}}{\sqrt{n}} .

This is an alternating series and \frac{1}{\sqrt{n+1}} < \frac{1}{\sqrt{n}} with lim_{n→∞} \frac{1}{\sqrt{n}} = 0. The Alternating Series Test implies that this is a convergent series. At x = 4, we have

∑_{n=1}^{∞} \frac{1}{\sqrt{n}} = ∑_{n=1}^{∞} \frac{1}{n^{1/2}} .

This is a p-series with p = 1/2. Since p < 1, the series diverges there. The interval of convergence is [2, 4).

VI. [10 points] Find the Taylor polynomial of degree n = 3 for f(x) = 2 + 3x + x^{3} at a = 1.

Solution: We have

f(1) = 6
f′(x) = 3 + 3x^{2} ⇒ f′(1) = 6
f′′(x) = 6x ⇒ f′′(1) = 6
f′′′(x) = 6 ⇒ f′′′(1) = 6.

Therefore the Taylor polynomial is

p_{3}(x) = 6 + 6(x − 1) + \frac{6}{2!}(x − 1)^{2} + \frac{6}{3!}(x − 1)^{3} = 6 + 6(x − 1) + 3(x − 1)^{2} + (x − 1)^{3}.

(If you expand this out and collect powers of x, you will see that p_{3}(x) = f(x).) VII.

A. [5 points] Starting from the Taylor series for sin(x) at a = 0, find a series representation for f(x) = \frac{sin(x) − x}{x^{3}}. Give the first three nonzero terms in your series.

Solution: Since

sin(x) = x − \frac{x^{3}}{3!} + \frac{x^{5}}{5!} − \frac{x^{7}}{7!} + ⋯,

we have

\frac{sin(x) − x}{x^{3}} = \frac{1}{x^{3}} (−\frac{x^{3}}{3!} + \frac{x^{5}}{5!} − \frac{x^{7}}{7!} + ⋯) = \frac{−1}{3!} + \frac{x^{2}}{5!} − \frac{x^{4}}{7!} + · · · .

B. [5 points] Use your answer in part A to determine lim_{x→0} \frac{sin(x) − x}{x^{3}}.

Solution:

lim_{x→0} \frac{sin(x) − x}{x^{3}} = lim_{x→0} \frac{−1}{3!} + \frac{x^{2}}{5!} − \frac{x^{4}}{7!} = \frac{−1}{6}.

C. [5 points] Use your answer in part A to determine the first three nonzero terms in a series representing the function

F (x) = ∫_{0}^{x} \frac{sin(t) − t}{t^{3}} dt.

Solution: We integrate the series from part A term by term to obtain

∫_{0}^{x} \frac{−1}{3!} + \frac{t^{2}}{5!} − \frac{t^{4}}{7!} + ⋯ dt = \frac{−x}{3!} + \frac{x^{3}}{3 · 5!} − \frac{x^{5}}{5 · 7!} + · · · .

A summation notation form would be

∑_{n=0}^{∞}(−1)^{n+1} \frac{x^{2n+1}}{(2n + 1) · (2n + 3)!}.