College of the Holy Cross, Spring Semester, 2005 Math 132 Answers to Review Sheet for Midterm 3

1. The graph at right shows either a pdf or a cdf. Which type of function is it, and why? If it is a pdf, sketch the graph of the corresponding cdf; if it is a cdf, sketch the graph of the corresponding pdf.

Solution The graph encloses more than one unit of area (and is non-decreasing and seems to approach 1 as x → ∞), so it’s a cdf. The corresponding density has one bump, centered just to the right of x = 0 .

2. Let x be the size of an email message in kilobytes (KB). An ISP finds that the fraction of emails between x and x + Δx KB in size is about cxe^{−0.01x}Δx (that is, the density function for x is p(x) = cxe^{−0.01x}).

(a) c = (0.01)^{2} = 10^{−4}

(b) Fraction of email messages that are at most 100 KB in size: 1 − \frac{2}{e} ≃ 0.264

(c) Fraction that are at least 50 KB in size: \frac{3}{2\sqrt{e}} ≃ 0.91.

(d) Median size of an email message: About 167.8 KB.

3. In a certain population, the height x of an individual in inches is normally distributed, with a mean of 68 and a standard deviation of 4. In other words, heights are modeled by the density function p(x) = \frac{1}{4\sqrt{2π}} e^{−(x−68)^{2}/32}.

(a) Write down a definite integral whose value gives the fraction of the population whose height is between 68 and 72 inches, and perform the change of variables z = (x − 68)/4 on your integral.

Solution \frac{1}{4\sqrt{2π}} ∫_{68}^{72} e^{−(x−68)^{2}/32} dx = \frac{1}{\sqrt{2π}} ∫_{0}^{1} e^{−z^{2}/2} dz.

(b) Write out the Taylor series for e^{u} giving both the first four non-zero terms and the summation form.

Solution e^{u} = ∑_{k=0}^{∞} \frac{u^{k}}{k!} = 1 + u + \frac{u^{2}}{2!} + \frac{u^{3}}{3!} + ⋯

(c) Use your answer from (b) to find the Taylor series for

∫_{0}^{x} e^{−z^{2}/2} dz,

giving both the first four non-zero terms and the summation form.

Solution ∫_{0}^{x} e^{−z^{2}/2} dz = ∑_{k=0}^{∞} \frac{(−1)^{k}x^{2k+1}}{(2k + 1) · 2^{k} · k!} = x − \frac{x^{3}}{3 · 2} + \frac{x^{5}}{5 · 2^{2} · 2!} − \frac{x^{7}}{7 · 2^{3} · 3!} + ⋯

(d) Estimate the integral in part (a) using the four-term series from part (c).

Solution \frac{1}{\sqrt{2π}} ∫_{0}^{1} e^{−z^{2}/2} dz ≃ \frac{1}{\sqrt{2π}} ∑_{k=0}^{3} \frac{(−1)^{k}}{(2k + 1) · 2^{k} · k!} = \frac{1}{\sqrt{2π}} (1 − \frac{1}{6} + \frac{1}{40} − \frac{1}{336}) ≃ 0.34

(The absolute error of this estimate is at most \frac{1}{\sqrt{2π}} · \frac{1}{9·2^{4}·4!} ≃ 0.0001

4. (a) Use the Comparison Test to determine whether or not ∑_{n=0}^{∞} \frac{n + 3^{n}}{2^{n}} converges. Solution Diverges by comparison with ∑_{n=0}^{∞} \frac{3^{n}}{2^{n}} = ∑_{n=0}^{∞} (\frac{3}{2})^{n}.

(b) Use the Integral Test to determine whether or not ∑_{k=0}^{∞} \frac{k}{e^{k}} converges. Solution Converges (see also Question 2 above.)

(c) Use the Ratio Test to determine whether or not ∑_{k=0}^{∞} \frac{3^{n}}{n!} converges. Solution Converges, since lim_{n→∞} |\frac{a_{n+1}}{a_{n}} | = lim_{n→∞} |\frac{3}{n + 1}| = 0 < 1.

5. Determine (with justification!) whether or not the following series converge:

∑_{k=1}^{∞} \frac{1}{\sqrt{k}} , ∑_{k=3}^{∞}(−1)^{k−1} \frac{1}{ln k} , ∑_{n=10}^{∞} \frac{n^{2} + n}{3n^{3} − 1000}, ∑_{n=1}^{∞} \frac{1}{n^{1.01}} .

Solution The first and third diverge: Use the integral test for the first, comparison with the harmonic series for the third. The second and fourth converge: The second is alternating, and the terms decrease to 0 in absolute value, while the fourth converges by the integral test.

6. Let f(x) = \sqrt{1 + x} = (1 + x)^{1/2}. Find the 4th degree Taylor polynomial of f centered at a = 0. Find a factorial expression for the general term of the Taylor series.

Solution p_{4}(x) = 1 + \frac{x}{2} − \frac{x^{2}}{8} + \frac{x^{3}}{16} − \frac{15}{16 · 24}x^{4}. The first few derivatives are f′(0) = \frac{1}{2},

f^{(2)}(0) = −\frac{1}{2} · \frac{1}{2}  f^{(3)}(0) = \frac{3}{2} · \frac{1}{2} · \frac{1}{2}  f^{(4)}(0) = −\frac{5}{2} · \frac{3}{2} · \frac{1}{2} · \frac{1}{2}  f^{(5)}(0) = \frac{7}{2} · \frac{5}{2} · \frac{3}{2} · \frac{1}{2} · \frac{1}{2}.

For k ≥ 2, the general expression is

f^{(k)}(0) = (−1)^{k−1} \frac{(2k − 3)}{2} · \frac{(2k − 5)}{2} ⋯ \frac{5}{2} · \frac{3}{2} · \frac{1}{2} · \frac{1}{2} = (−1)^{k−1} \frac{(2k − 2)!}{2^{k}} · \frac{1}{2^{k−1}(k − 1)!}

We do not expect you to construct such formulas, and certainly not under test pressure.

7. Consider the geometric series f(x) = ∑_{k=0}^{∞} x^{k} = \frac{1}{1 − x}.

(a) Differentiate and multiply by x; the Taylor series of xf′(x) = \frac{x}{(1 − x)^{2}} is ∑_{k=0}^{∞} kx^{k}.

(b) Integrate; the Taylor series of − ln(1 − x) is ∑_{k=0}^{∞} \frac{x^{k+1}}{k + 1} = ∑_{k=1}^{∞} \frac{x^{k}}{k}.

(c) Find the radius of convergence of the series in part (b), and investigate convergence at the endpoints.

Solution The radius is 1, the interval of convergence is −1 ≤ x < 1.

(d) Set x = 1/2 to see that ∑_{k=1}^{∞} \frac{1}{k · 2^{k}} = ln 2, ∑_{k=1}^{∞} \frac{k}{2^{k}} = 2.

8. For each of the given power series, find the interval of convergence.

f(x) = ∑_{n=1}^{∞} \frac{(2x)^{n}}{\sqrt{n}} , g(x) = ∑_{n=1}^{∞}(−1)^{n−1} \frac{(x − 5)^{n}}{n · 3^{n}} .

(In particular, give the radius of convergence, and investigate convergence at the endpoints.)

Solution For f(x), −\frac{1}{2} ≤ x < \frac{1}{2}. For g(x), 2 < x ≤ 8. (The first has radius 1/2, the second has radius 3.)

9. The second degree Taylor polynomial of f(x) at a = 0 is p_{2}(x) = C_{0} + C_{1}x + C_{2}x^{2}. What can you say about the signs of C_{0}, C_{1}, and C_{2} if you know the graph of f(x) is:

Solution C_{0} < 0, C_{1} and C_{2} > 0.

10. Use the error bound for Taylor approximations to estimate the number of decimal places of accuracy if the 6th degree Taylor polynomial at a = 0 is used to approximate cos(0.8). Do the same for the nth degree polynomial in general. What happens to the error bound as n → ∞?

Solution With a polynomial of degree 2n, the “standard” error bound is

|E_{2n}(0.8)| ≤ \frac{(0.8)^{2n+2}}{(2n + 2)!} ,

the absolute value of the first term of the Taylor series that is omitted from the approximating polynomial. (This observation rests on the fact that the series for cos(0.8) alternates and the terms decrease in absolute value.)

In particular, the degree-6 polynomial estimate is accurate to ±4.16 × 10^{−6}, or 5 decimal places. The error bounds decrease rapidly to 0 as n → ∞: The numerators decrease exponentially and the denominators grow “super-exponentially” (asymptotically faster than any exponential function).