College of the Holy Cross, Spring Semester, 2005 Math 132, Midterm 3 Solutions April 27, 2005

1. [5 points each] Suppose f(0) = 0.5, f′(0) = −1 and f′′(0) = −2.

(a) Write down the Taylor polynomial of degree 2 for f near a = 0.

Solution: The polynomial is

p_{2}(x) = f(0) + f′(0)x + \frac{f′′(0)}{2!}x^{2} = 0.5 − x − x^{2}

(b) Use your answer to part (a) to estimate f(0.3).

Solution: Substitute x = 0.3 in p_{2}(x):

p_{2}(0.3) = 0.5 − 0.3 − (0.3)^{2} = .11

(c) Which (if any) of the following could be the graph of f? (Recall that f(0) = 0.5, f′(0) = −1 and f′′(0) = −2. More than one answer may be correct.)

◯ (ii)

(i)^{−1}  ^{1}_{−1}^{1}  (ii)^{−1}  ^{1}_{−1}^{1}  (iii)^{−1}  ^{1}_{−1}^{1}  (iv)^{−1}  ^{1}_{−1}^{1}

Solution: At a = 0, the function should be positive (f(0) > 0), decreasing (f′(0) < 0), and concave down (f′′(0) < 0). Only (ii) is all three.

2. [10 points] Evaluate (find the sum of) the series 5 − \frac{10}{3} + \frac{20}{9} − \frac{40}{27} + \frac{80}{81} − ⋯

Solution: This is a geometric series with a = 5 and r = \frac{−2}{3}. Since −1 < r < 1, the series is convergent and the sum is

\frac{a}{1 − r} = \frac{5}{1 + \frac{2}{3}} = 3.

3. [10 points] Use the power series for e^{x} about a = 0 to find the power series for ze^{−z^{2}} about a = 0. Express your answer both in summation form, and by writing out the first four nonzero terms.

Solution: The Taylor series for e^{x} about a = 0 is

∑_{k=0}^{∞} \frac{x^{k}}{k!}

To get the series for ze^{−z^{2}}, we substitute x = −z^{2}, and multiply the result by z to obtain:

z ∑_{k=0}^{∞} \frac{(−z^{2})^{k}}{k!} = z ∑_{k=0}^{∞}(−1)^{k} \frac{z^{2k}}{k!} = ∑_{k=0}^{∞}(−1)^{k} \frac{z^{2k+1}}{k!}

The first four nonzero terms are

z − z^{3} + \frac{z^{5}}{2!} − \frac{z^{7}}{3!} .

4. The distribution of length in a certain population of inchworms is described by the pdf p(x) = c(3x^{2} − x^{3}) for 0 ≤ x ≤ 3.

(a) [8 points] Find the value of the constant c.

Solution: The total area under the graph of a density must be 1, so

1 = c ∫_{0}^{3} 3x^{2} − x^{3} dx
= c (x^{3} − \frac{x^{4}}{4} |_{0}^{3}
= c\frac{27}{4}
⇒ c = \frac{4}{27}

(b) [7 points] Find the proportion of inchworms with length at most 2 cm.

Solution: This proportion is computed by the integral ∫_{0}^{2} p(x) dx,

which is \frac{4}{27} ∫_{0}^{2} 3x^{2} − x^{3} dx = \frac{16}{27} ≐ .593

(about 59.3% of the inchworms).

5. [5 points each] Use the indicated test to determine whether the given series converges or diverges. Justify your answers.

(a) ∑_{n=1}^{∞} \frac{n^{2}}{n^{3} + 10}; integral test

Solution: We apply the substitution u = x^{3} + 10 to the integral

∫ \frac{x^{2} dx}{x^{3} + 10} = ∫ \frac{\frac{1}{3} du}{u} = \frac{1}{3} ln(u) = \frac{1}{3} ln(x^{3} + 10) + C.

Then ∫_{1}^{b} \frac{x^{2} dx}{x^{3}+10} = ln(b^{3} + 10) − ln(11). Taking lim_{b→∞} ln(b^{3} + 10) − ln(11), we do not get a finite value so the integral diverges, and so does the series.

◯ Diverges

(b) ∑_{n=1}^{∞} \frac{n}{4n^{3} + 3n^{2} + 5}; comparison test

Solution: For all n ≥ 1:

\frac{n}{4n^{3} + 3n^{2} + 5} < \frac{n}{4n^{3}} = \frac{1}{4n^{2}} < \frac{1}{n^{2}}

The series ∑_{n=1}^{∞} \frac{1}{n^{2}} is the p-series with p = 2 > 1. Hence it is convergent. By the comparison test, the given series converges too.

◯ Converges

6. [15 points] Let f(x) = \sqrt{x}. Use the definition to calculate its Taylor polynomial of degree 3 at a = 1.

Solution: We have f(1) = 1. f′(x) = \frac{1}{2}x^{−1/2}, so f′(1) = \frac{1}{2}. f′′(x) = \frac{−1}{4}x^{−3/2}, so f′′(1) = \frac{−1}{4}. Finally, f′′′(x) = \frac{3}{8}x^{−5/2}, so f′′′(1) = \frac{3}{8}. The Taylor polynomial is

f(1)+f′(1)(x−1)+ \frac{f′′(1)}{2}(x−1)^{2}+ \frac{f′′′(1)}{6}(x−1)^{3} = 1+ \frac{1}{2}(x−1)− \frac{1}{8}(x−1)^{2}+ \frac{1}{16}(x−1)^{3}.

7. Both parts refer to the power series ∑_{n=1}^{∞} \frac{(x − 1)^{n}}{n · 5^{n}}.

(a) [10 points] Use the ratio test to find the radius of convergence.

Solution:

lim_{n→∞} |\frac{(x − 1)^{n+1}}{(n + 1) · 5^{n+1}} · \frac{n · 5^{n}}{(x − 1)^{n}} | = lim_{n→∞} \frac{n}{n + 1} · \frac{1}{5} · |x − 1| = \frac{1}{5}|x − 1|.

So the radius of convergence is R = \frac{1}{\frac{1}{5}} = 5.

(b) [5 points] Investigate the endpoint behavior, and determine the interval of convergence.

Solution: At x = 1 + R = 6, ∑_{n=1}^{∞} \frac{(6 − 1)^{n}}{n · 5^{n}} = ∑_{n=1}^{∞} \frac{1}{n}. This is the harmonic series (p

series with p = 1). So it diverges. At x = 1 − R = −4, we get ∑_{n=1}^{∞}(−1)^{n} \frac{1}{n}, which converges by the alternating series test. The interval of convergence is −4 ≤ x < 6.

8. [5 points each] Suppose

p(x) = \frac{4}{π} · \frac{1}{1 + x^{2}}

is the probability density function for the quantity x, for 0 ≤ x ≤ 1.

(a) Find the mean of x. Solution: The mean is given by

∫_{−∞}^{∞} xp(x) dx = \frac{4}{π} ∫_{0}^{1} \frac{x}{1 + x^{2}} dx = \frac{4}{π} · \frac{1}{2} ln(1 + x^{2})|_{0}^{1} = \frac{2 ln(2)}{π} ≐ .4413

(b) Find the median of x. Solution: The median is the t such that

\frac{1}{2} = \frac{4}{π} ∫_{0}^{t} \frac{dx}{1 + x^{2}} = arctan(t)

This says

arctan(t) = \frac{π}{8}

or t = tan (\frac{π}{8}) =≐ .4142.