Math 121, DA Syllabus and Assignments Fall Term, 2005

(subject to change)

This document is also available on-line at http://www.people.carleton.edu/~dappleya.

Text : Stewart, James, Calculus, Fifth Edition, Thomson, 2003.

Instructor : Dave Appleyard

Office : CMC 223

Telephones : 646-4450 (office), 645-8323 (home, before 10 p.m.)

e-mail : dappleya@carleton.edu

Office Hours : 10-11:30 a.m. MW; 2:30-4 p.m. TuTh; 9:30-10:45 F; many other times (come on by!); and by appointment.

Course grade determination: Hour Exams - 20% each

Mathematica Problem Sets - 20%

Final Exam - 30%

Daily Text Assignments - 10%

Late daily assignments will not be accepted - no exceptions. Your two lowest daily assignment scores will not be included in the

computation of your grade.

CONCERNING HONESTY:

The person who reads any work of yours (whether it is an in-class exam, a daily assignment, or, in this course, a

Mathematica problem set) assumes the work is your own and not someone else's (except in the case of a Mathematica problem set,

where you will work with a partner). If you mislead the reader in this regard, you are being dishonest.

Every Carleton student should be familiar with the booklet Academic Honesty in the Writing of Essays and Other

Papers (available from the Dean of the College's office; you received a copy during New Student Week your first year here), which

speaks to these matters in general terms. In mathematics and in this course, there are some special considerations. It is perfectly

legitimate to work with other people on a daily problem assignment. Discussions with other students about daily problem

assignments can be very helpful. If these discussions are fruitful, they should result in your being able to work the problem on your

own and you, therefore, need not refer to your fellow discussants when you write up your work. On the other hand, if you solve a

difficult problem by making use of a clever or unusual idea suggested to you by someone (or by some book different from the text),

it is appropriate to refer to this when you write up your work. A solution to a difficult problem in mathematics may actually be the

result of joint efforts. If so, your written report should indicate that fact and identify your collaborators. Intellectual integrity

demands that we give credit where it is due.

Assignment #1, for Wednesday, September 14

First, send me e-mail (dappleya@carleton.edu) telling me when and where you had the material in Calculus I (at Carleton, that

material is Chapters 1-6 of Stewart’s Calculus) and anything else you want me to know about you and math. Then:

Read:

-The handout “Succeeding in the College Mathematics Classroom.”

-Mathematica Handout #1.

-Section 7.1 of the text.

-Section 7.2 of the text, up to the subsection “Derivatives of Exponential Functions.”

-At the end of Section 7.1, Exercises ## 1, 3, 6, 8, 15, 19 and 33 and their answers.

-At the end of Section 7.2, Exercise #23 and its answer.

NOTE: I’ll talk about Theorem 7 in Section 7.1 next time but you won’t have to do much with it.

Remark on the text:

-In the equation by Figure 11 in Section 7.1: Remember tan (π/2 – θ) = sin (π/2 – θ)/ cos(π/2 – θ) = cos θ/sin θ = 1/tan θ , using

formulas you probably remember.

Problems to turn in:

-From Section 7.1: 9, 10, 11, 13 (in those first four, use graphs and the horizontal line test), 21, 25, 26, 27, 28, 29;

-From Section 7.2: 7, 9, 11, 12, 13, 25, 27, 28.

Some answers:

-Section 7.1. #6. No. #8. Yes. #10. No. #26. f^--1(x) = (1+3x)/(4-2x). #28. f^--1(x) = ((x-3)/2)^1/3.

-Section 7.2. #12. #28. ∞.

Assignment #2, for Friday, September 16

If you have not already done so, please send me e-mail (dappleya@carleton.edu) telling me when and where you had the material in

Calculus I (at Carleton, that material is Chapters 1-6 of Stewart’s Calculus) and anything else you want me to know about you and

math. Then:

Read:

-Mathematica Handout #2.

-The remainder of Section 7.2 of the text.

-Section 7.3 of the text.

-At the end of Section 7.3, Exercises ## 3, 5, 9, 23, 25, 27 and their answers.

Problems to turn in:

-From Section 7.2: 29, 35, 37, 42, 43, 45, 47, 51, 61, 63, 71, 75, 77, 79, 82;

-From Section 7.3: 29, 30, 31, 35, 43.

Some answers:

-Section 7.2. #42. – 4/(e^x – e^-x)² . #82. π (1 – 1/e).

-Section 7.3. #30. (a) ½ (ln 7 – 3). (b) ½ (5 – e^{-3 )}.

Assignment #3, for Monday, September 19

Read:

-Section 7.4 of the text.

-Section 7.5 of the text, just up to the definition of the inverse tangent function.

(We’re skipping the sections called 7.2*, 7.3*, and 7.4*.)

Remarks on the text:

-In Section 7.4, formula (6): Of course, using the Chain Rule, we then also get the companion formula d/dx (log_a u) = (1/(u ln a))

du/dx.

-In Section 7.5, formulas (2): There’s no reason to memorize these formulas. Just think a problem through each time, as in

Example 2(c) that follows.

-In Section 7.5, Example 3(c): The domain of the derivative is more easily seen from the answer to part (b): The denominator there

is 0 if and only if x = 0 or x = ±√(2), so we rule those points out from the domain of the original function when we find the domain

of the derivative.

Problems to turn in:

-From Section 7.3: 55, 56, 67, 70;

-From Section 7.4: 3, 7, 8, 9, 12, 19, 25, 35, 44, 45, 47, 65, 67, 69, 71, 73.

Some answers:

-Section 7.3. #56. 0. #70. y – ½ = - ½ (x – ln 2).

-Section 7.4. #8. 1/(5x) . #12. 1/√(x² -1). #44. x^1/x(1 – ln x)/x².

Assignment #4, for Wednesday, September 21

Read:

- Mathematica Handout #3.

-The remainder of Section 7.5 of the text.

-Problem #66 at the end of Section 7.5 and its answer.

-Section 8.1 of the text.

(We’re skipping Section 7.6 and will do Section 7.7 later.)

Remarks on the text:

-In Section 7.5, Example 4, Solution 2: Remember from right triangle trig that tan y = (length of side opposite y) / (length of side

adjacent to y) and cos y = length of side adjacent to y / length of hypotenuse.

-In Section 7.5, Equation (14): You should memorize this formula.

-In Section 8.1, first line of Solution to Example 6: Then by this clever choice we still have u dv = sinⁿx dx, the original

integrand, which we must.

Problems to turn in:

-From Section 7.5: 23, 24, 25, 26, 27, 31, 39, 49, 59, 61, 62, 63, 64 , 65, 67;

-From Section 8.1: 1, 3, 9, 11, 19, 21, 23, 33.

Some answers:

-Section 7.5.

#24. 1 – x sin^-1x /√(1 – x²).

#26. x / ((1 + x²) tan^-1x) + ln(tan^-1x). (Remember that tan^-1 x and arctan x mean the same thing.)

#62. ½ sin^-1(2t) + C. #64. π/4. #66. ½ (tan^-1x)² + C.

Assignment #5, for Friday, September 23

Read:

-Section 8.2 of the text.

-Section 8.3 of the text, up to Example 3.

Remarks on the text:

-In Section 8.2, the last sentence of the Solution to Example 8: The integral of sec³x on the right side of the equation is taken to the

left side, and then the equation is divided by 2.

-In Section 8.2, the first sentence of the Solution to Example 9: Integration by parts would result in a solution similar to Example 4

in Section 8.1.

Problems to turn in:

-From Section 7.4 (p. 450): 79, 80;

-From Section 7.5 (p. 485): 70;

-From Section 8.1 (pp. 516-17): 14, 15, 34, 46, 59;

-From Section 8.2 (p. 524): 1, 3, 7, 9, 15;

-From Section 8.3 (pp. 530-1): 4, 15.

Some answers and hints:

#79. Hint. Use the disk method. #80. Hint. Use cylindrical shells. The answer is π ln 10.

#70. ½ sin^-1(e^2x) + C. #14. Hint. See Example 3. Answer: (t³ – 3t² + 6t – 6)e^t + C.

#15. Hint. See Example 1. #34. 2e². #4. 40/3.

#15, p. 531. Hint. To integrate tan²θ, use

the identity tan²θ= sec²θ - 1. And to

get the final form for the answer, use

the triangle at the right.

Assignment #6, for Monday, September 26

Read:

-Mathematica Handout #4.

-Examples 3 and 4 of Section 8.3 of the text – skip the rest of the section.

-Section 8.4 of the text, through Example 3; skip everything else in the section except Example 9, which you should read.

-Exercises #1(a) and #3(a) at the end of Section 8.4 and their answers.

Remark on the text:

-In Section 8.4, Figure 1. The red graph is the original function, since its values → 0 as x → ∞ (see why?), while the values of the

derivative → ∞ as x → ∞ (see why?).

Problems to turn in:

-From Section 8.2 (pp. 524-5): 4, 23, 24, 27, 54 (“Explain the difference” by using trig identities to reduce the answers for (b), (c),

(d) to the answer for (a));

-From Section 8.3 (pp. 530-1): 7 , 9, 11, 12, 31(a);

-From Section 8.4 (p. 540): 9, 10, 11, 39.

Some answers and hints:

#4. 8/15. #24. ⅓ tan³x – tan x + x + C.

#54. (a) - ½ cos²x + C. (b) ½ sin²x + C. (c) – ¼ cos 2x + C. (d) ½ sin²x + C or -1/2 cos²x + C.

#7. Hint. Recall that an antiderivative of csc²θ is – cot θ. And to get the final form of the answer, use a right triangle with leg x and

hypotenuse 5.

#9. Hint. Somewhere you may use Equation (1) in Section 8.2. And to get the final form of the answer, use a right triangle with

legs x and 4.

#11. Hint. Let 2x = sin θ. And to get the final form of the answer, use a right triangle with leg 2x and hypotenuse 1.

#12. Hint. Don’t make this too hard! Answer is ⅓ (5√(5) - 8).

#31(a). Hint. Somewhere you may use Equation (1) in Section 8.2.

#10. (1/5) ln | (t – 1)/(t + 4) | + C.

#39. Hint. Make a substitution, then use the result of problem #11 you just did.

Assignment #7, for Wednesday, September 28

Read:

-Section 8.5 of the text.

-Section 8.7 of the text, to “Simpson’s Rule” (approximately the first half of the section).

Remark on the text:

-In the “Table of Integration Formulas” in Section 8.5, you should know formulas ## 1, 2, 3, 5, 6, 7, 13, 17, 18.

Problems to turn in:

-From Section 8.2 (pp. 525): 59, 60, 61;

-From Section 8.4 (pp.540-1): 18, 36, 42, 43, 44, 62(b);

-From Section 8.7 (pp. 563-4): 13(a), 13(b), 15(a), 15(b), 31.

Some answers and hints:

#60. Hint. You may use the answer to problem #24 on p. 524 in Section 8.2 last time. Answer here is π(π/4 – 2/3).

#18. ln|x| - ln|x + 1| + ln|x – 1| + C. #36. No need for partial fractions! Answer is ½ ln 5/2.

#42. Hint. Make a substitution, then do a long division. Answer is 3 (ln 2 – ½). #44. π/3.

#62(b). Hint. Use cylindrical shells. Answer is 2π ln 9/8.

#31(b). Hint. Use one of the error bounds in Formula (3) in the section.

Assignment #8, for Friday, September 30

Read:

-Mathematica Handout #5.

-Section 9.1 of the text.

Remarks on the reading:

- Example 3. Mathematica’s command

NIntegrate[Sqrt[x^4 + 1]/x^2, {x, 1, 2}]

gives an answer of 1.13209.

-Example 4. Study this example carefully! Notice what happens: When 2x – 1/(8x) is squared, we get 4x² – ½ + 1/(64x²); when 1 is

added to this we get 4x² + ½ + 1/(64x²), which is the same as our earlier square, but with a + ½ replacing the – ½. Thus this new

quantity simplifies to 2x + 1/(8x) squared, as you can verify. Then the integrand can actually be antidifferentiated!

-Example 4, fourth line from the end. Typo. The integrand should read 2t + 1/(8t), not 2t – 1/(8t).

Problems to turn in:

-From Section 8.7 (pp. 563-5): 20, 21 (ignore anything about S₁₀ and E_S and (4) and S_n, which involve Simpson’s Rule), 42;

-From Section 9.1 (pp. 588-9): 1, 5, 7, 8, 9, 17, 19, 31.

Some answers and hints:

#42. 20 versus 0.

#20. Show that you can take K (which you are free to choose to be any number that is bigger

than | f '' (x)| in the interval [0, 1]) in the error bound in (3) to be 6, and use that value in parts (b) and (c) of the problem.

Answers: (a) 0.902333; 0.905620. (b) | E_T | ≤ .0078125;

| E_M | ≤ .00390625. (c) 224; 159.

#21. Show that you can take K (which you are free to choose to be any number that is bigger than | f '' (x)| in the interval [0, 1]) in

the error bound in (3) to be e, and use that value in parts (b) and (c) of the problem.

HINT for ##7, 8, and 9: See the similar algebra in Example 4 in the section, discussed in the second “Remark” above in this

assignment. Answer to #8 is 6 + ¼ ln 2.

Assignment #9, for Monday, October 3

Read:

-Section 8.6 of the text.

-The handout on arc length.

Remark on the reading:

-Example 8 in the section: This one is a lot like stuff you need to do in your Mathematica Problem #5. Here we note F has a local

max when F ' = f (the red curve) goes from positive to negative (First Derivative Test) and a local min when f goes from negative to

positive. There are inflection points where f ' (the yellow curve) crosses the x-axis.

Problems to do for some review but not to turn in:

-From Section 8.1 (p. 517): 57;

-From Section 8.2 (p. 524-5): 33, 55;

-From Section 8.4 (p. 541): 60, 61;

-From Section 8.5 (p. 546-7): 4, 61;

-From Section 9.1 (p. 589): 29.

Problems to turn in:

-Your first Mathematica assignment! It’s due at 5 p.m. in my mailbox just inside the double glass doors on the second floor of the

CMC.

Some answers and hints:

#60. ln (3/2). #4. ½ sin^-1(x²/√3 ) + C.

#29. Hint. Save yourself some work by using this symmetry of the graph: the length of the whole curve is 4 times its length in the

first quadrant.

Wednesday, October 5: FIRST EXAM (one 4'' by 6'' note card allowed)

For Friday, October 7: No assignment.

Assignment #10, for Monday, October 10

Read:

-Section 7.7 of the text. (Cauchy’s Mean Value Theorem and the proof of l’Hospital’s Rule are optional.)

-The handout on miscellaneous topics.

Remarks on the reading:

-Last line of “NOTE 3” following the statement of l’Hospital’s Rule: this follows from the preceding expression because f (a) =

g(a) = 0 by hypothesis.

-Sixth line on p. 498: Typo. Read “ f ' changes” not “ f changes”.

-Example 9 in the section: This example shows that 1^∞ is not necessarily equal to 1!

Problems to turn in:

-From Section 7.7: 5, 7, 9, 11, 15, 17, 18, 20, 21, 23, 25, 27, 28, 37, 38, 41, 47, 51, 53, 57, 88, 90.

Some answers and hints:

#18. 0. #20. – 1/π . #28. 0. #38. 0.

#47. Hint. Multiply the numerator and denominator by √(x² + x) + x.

#90. Hint. Use l’Hospital’s Rule and the Fundamental Theorem of Calculus, Part I. Answer is 1/3.

Assignment #11, for Wednesday, October 12

Read:

-Section 8.8 of the text.

-Exercise #1 at the end of Section 8.8, and its answer.

Remarks on the reading:

-Solution to Example 7, last line: We don’t need to consider the integral from 1 to 3 because, by definition (3) on the preceding

page, the original integral (from 0 to 3) converges if and only if both the integral from 0 to 1 and the integral from 1 to 3 converge.

-The “Warning” after Example 7: The integral is improper because the integrand is not defined at x = 1, since the denominator is 0

there.

-The paragraph after Example 9: If you take Calc III, you’ll see a really neat way to show that the integral’s value is √π / 2.

Problems to turn in:

-From Section 7.7: 32, 55, 59.

-From Section 8.8: 5, 7, 9, 11, 15, 21, 23, 27, 28, 29, 30, 32.

-NOTE: For the problems in Section 8.8, if an integral is divergent, be sure to argue why: See, for example, the arguments in

Examples 1, 6, and 7 in the section. If you find a value for the integral, by a careful evaluation like those in Examples 2, 3, 5, and 8

in the section, then the integral is convergent to that value. In your arguments, be sure to use limit notation properly.

Some answers:

#32 (from Section 7.7). ¼. #11. Hint. See (c) of Def. 1 in the section.

#23. Hint. See Example 3 in the section. #28. Divergent.

#30. Converges to -6. #32 (from Section 8.8). Converges to π/2.

Assignment #12, for Friday, October 14

Read:

-The last two pages of Section 10.1 (from “General Differential Equations” to the end of the section).

-Section 10.3.

-The handout on indeterminate forms.

Remarks on the reading:

-Section 10.1, Example 1. Of course, here t, not x, is the independent variable. Usually in physical applications the independent

variable is t, for time.

-Section 10.3, Example 3, next-to-last sentence. The function y = 0 is a solution because if y = 0 for all x, then dy/dx = 0 too, for all

x. Then dy/dx = x²y for all x since the left side is 0 and the right side is x² times 0 = 0. Thus the function y = 0 satisfies the

differential equation for all x, and that’s what we mean by saying it’s a solution.

Problems to turn in:

-From Section 8.8: 31, 33, 49, 54, 58.

-From Section 10.1: 1, 2.

-From Section 10.3: 1, 2, 3, 4, 8, 11, 19, 20.

Some answers and hints:

#31 and #33. Remember to split the integral into two integrals; see Example 7 in the section.

#54. Convergent.

#58. Let u = ln x and then see Example 4. Answer: Converges if p > 1 and diverges otherwise.

#2 (from Section 10.3). y = ±(½ e^2x + C)^1/4 . #4. y = 1/(cos x - C); also y = 0.

#8. y = e^{±√(x^2 /2 + C)} . #20. y = 2x²/(x² + 1).

Monday, October 17: Mid-Term Break

Assignment #13, for Wednesday, October 19

Read:

-Mathematica Handout #6.

-Section 10.4 of the text.

-The handout on the logistic differential equation.

Remark on the reading:

-First paragraph of the subsection “Newton’s Law of Cooling”: there’s really no need to do this change of variable, as I’ll show in

class on Wednesday.

Problems to turn in:

-From Section 8.8: 50, 71(a).

-From Section 10.3: 13, 27, 37.

-From Section 10.4: 1, 3, 9, 11, 19, 20.

Some answers and hints:

#50. Diverges. #71(a). Treat s as a constant in your antidifferentiation.

#13. You will need an integration by parts. #27. See Example 5 in Section 10.3.

#20. (a) About 11.55 years. (b) Remember the annual interest rate in the formula

A = A₀(1 + r)^t is r. Answer is 6.18%.

Assignment #14, for Friday, October 21

Read:

-The subsection “The Limit of a Sequence” on pp. 6-7 of the text.

-Section 12.1 of the text, through the statement of the Monotonic Sequence Theorem. You may ignore its proof and also the example

that follows.

Remarks on the reading:

- Definition 8 in Section 12.1. The last sentence would be more clear if it said a sequence is called monotonic if (1) it is an

- increasing sequence; or (2) it is a decreasing sequence. Also, some people say a sequence is “increasing” if a_n ≤ a_n+1 for

- all n; others say a sequence is “increasing” if a_n < a_n+1 for all n. (Similarly, some use “decreasing” to mean a_n ≥ a_n+1 for

- all n and others use “decreasing” to mean a_n > a_n+1 for all n.)

Problems to turn in:

-From Section 10.1: 4, 6(a).

-From Section 10.3: 15, 28.

-From Section 10.4: 5(a) - ignore the last sentence, 7, 13, 15.

-From Section 12.1: 3, 5, 9, 13, 15, 17, 18, 19, 24, 27.

Some answers and hints:

#4. Force y = e^rt to be a solution, see what happens. Answer: r = 2 or r = -3.

#28. To start, differentiate the given equation implicitly to get dy/dx = x/y. Then proceed as in Example 5 in the section.

#18. Converges to 1. #24. Converges to 1.

Assignment #15, for Monday, October 24

Read:

-Mathematica Handout #7.

-The subsection “The Sum of a Series” on pp. 7-9 of the text.

-Section 12.2 of the text.

-Exercise #1 at the end of Section 12.2 and its answers.

Remarks on the reading:

-Second paragraph of the Section 12.2. You may remember the formula 1 + 2 + 3 + ^… + n = n(n+1)/2 from high school. If not, no

big deal.

-Second paragraph of the Solution to Example 1 in Section 12.2. Here na → ∞ if a > 0,

na → - ∞ if a < 0.

-Second sentence of the Solution to Example 5 in Section 12.2 (the sentence in parentheses). Remember 0⁰ is an indeterminate form

in general (see the class handout).

-Section 12.2, first sentence after Theorem 8. Typo: Read Section 12.1, not 11.1.

Problems to turn in:

-From Section 10.4: 12, 17(b).

-From Section 12.1: 26, 29, 30, 31, 35, 36, 49(b).

-From Section 12.2: 11, 13, 15, 19, 21, 25, 27, 31, 35, 37.

NOTES: You can do all the exercises in Section 12.2 using either result (4) in the section (on the geometric series) or result (7) in

the section (the Test for Divergence). In #35 and #37, you must use result (4), not the high-school short-cut method! See Example

4 in Section 12.2.

Some answers and hints:

#12. y = 5e^2x. #26. Converges. #30. Diverges. #36. Converges (Squeeze Theorem).

Assignment #16, for Wednesday, October 26

Read:

-Section 12.3 of the text.

-Exercise #1 at the end of Section 12.3 and its answer.

NOTE: Example 6 in Section 12.3 is interesting, but we won’t be using the boxed formula (3). But do know the boxed formula (2),

especially the right-side inequality.

Problems to turn in:

-From Section 12.1: 28, 32.

-From Section 12.2: 16, 23, 28, 30, 49.

-From Section 12.3: 3, 5, 9, 11, 12, 15, 17, 21, 30(a), 30(c), 31(a), 31(c).

NOTES: (1) When you use the Integral Test in these problems, you may assume without proving it that the function f(x) is

decreasing on the appropriate interval. (2) In the problems assigned from #9-#24, you may use the p-test (result (1) in Section

12.3) as well as the Integral Test.

Some answers and hints:

#28. Converges to 1. #32. Converges to 0. #16. Diverges.

#23. Hint. Show that 2/(n² – 1) = 1/(n – 1) – 1/(n + 1). Then see Example 6 in the Section 12.2. This one is a “telescoping” series.

#28. Converges to 32/7. #30. Diverges.

#49. Remember Definition (2) in Section 12.2! Also remember that for n > 1, a_n = s_n – s_n-1, by definition.

#12. Converges. #30. (a) S₁₀ = 1.08204; error is at most 1/3000. (c) n > 32.

Assignment #17, for Friday, October 28

Read:

-Section 12.4 of the text.

-Exercises #1 and #2 at the end of Section 12.4 and their answers.

-The handout on improper integrals and infinite series.

Remark on the text:

-Solution to Example 4 in Section 12.4: To go from line 4 to line 5, divide numerator and denominator by n^5/2 = √(n⁵).

Problems to turn in:

-From Section 12.2: 46.

-From Section 12.3: 19, 20, 25 (don’t bother with the case p = 1; you did that case in Exercise #21, for last time), 32.

NOTE: When you use the Integral Test in these problems, you may assume without proving it that the function f(x) is

decreasing on the appropriate interval.

-From Section 12.4: 3, 5, 6, 7, 8, 9, 13, 15, 16, 17, 21, 22, 24, 25, 26. NOTE: For some of these, use the (Ordinary) Comparison

Test; for others, use the Limit Comparison Test. The p-test (Section 12.3) and the Geometric Series Test (Section 12.2) can be

quoted freely to tell whether your comparison series converge or diverge.

Some answers and hints:

#2. (a) If a_n > b_n, then ∑ a_n diverges.

(b) We cannot say anything; ∑ a_n might diverge, or it might converge.

#46. Hint. It’s a “telescoping” series.

#20. Use integration by parts. Answer: Converges.

#25. Hint. In the integral, let u = ln x.

#32. Show that for any n, R_n ≤ 1/(4n⁴). Then show this means n = 5 is good enough. Answer: 1.037.

#6. Diverges. #8. Diverges. #16. Converges.

#22. Expand the denominator first. Answer: Converges.

#24. Diverges. #26. Converges.

Assignment #18, for Monday, October 31

Read:

-Mathematica Handout #8.

-Section 12.5 of the text.

-Exercise #1 at the end of Section 12.5 and its answers.

-Section 12.7 of the text, but omit items #6 and #7 on the first page and Example 5 on the second.

Remarks on the reading:

-Example 2, Section 12.5, last sentence: The limit doesn’t exist since the terms alternate closer and closer to ± ¾ as n → ∞.

-Example 4, Section 12.5, next-to-last sentence. Note that b_n = 1/n! in this series. Thus b₆ = 1/6! = 1/720 = .00139 is not < .0005,

which we need for three-decimal-place accuracy. But b₇ = 1/7! = 1/5040 = .0001984 is.

Problems to write up but not turn in:

-From Section 12.5: 4, 5, 7, 9, 14, 15, 17, 18, 23, 24, 27, 31.

NOTE: The first eight problems assigned use the Alternating Series Test or the Test for Divergence, both of which are very

easy to apply.

Problems to turn in:

-Your second Mathematica assignment. It’s due at 5 p.m. in my mailbox just inside the double glass doors on the second floor of

the CMC.

Some answers and hints:

#4. Converges.

#7. Use the Test for Divergence.

#14. First show (by showing a derivative is negative) that the terms decrease for n > e, then that b_n = (ln n)/ n → 0 as n → ∞, in

order to apply the Alternating Series Test. Answer: Converges.

#18. Diverges by the Test for Divergence.

#23. b₁₀ = 1/10² = .01, b₁₁ = 1/11² = .008 < .01 so we need n = 10.

#24. b₅ = 1/5⁴ = .0016 > .001, b₆ =1/6⁴ = .0008 < .001, so we need n = 5.

#27. b₇ = 1/7⁵ = .0000595 > .00005, b₈ = 1/8⁵ = .0000305 < .00005, so we take n = 7 and add the first 7 terms, getting 0.9721.

#31. Do this without any calculations! See Figure 1 in the section.

Wednesday, November 2: SECOND EXAM (one 4'' by 6'' note card allowed)

For Friday, November 4: No assignment.

Assignment #19, for Monday, November 7

Read:

-Section 12.6 of the text.

-Exercises #1 and #31 at the end of Section 12.6 and their answers.

Remark on the reading:

-The statement of the Ratio Test: In part (a), note that if we find that L = 0, then since 0 < 1 we have absolute convergence.

Problems to turn in:

-From Section 12.6: 2, 3, 5, 7, 8, 9, 12, 13, 15, 17, 20, 23, 24, 25. (See the NOTE on the next page of this syllabus.)

NOTE: If you use the Ratio Test or Root Test and find that L = 1, then of course you must use some other test to determine

whether the series converges or diverges.

Some answers and hints:

#2. Absolutely convergent.

#8. Hint. Compare the series of absolute values with the harmonic series. Answer to the problem: Conditionally convergent.

#12. Absolutely convergent.

#17. Hint. Compare the series of absolute values with the harmonic series. Answer to the problem: Conditionally convergent.

#20. Absolutely convergent. #24. Absolutely convergent.

Assignment #20, for Wednesday, November 9

Read:

-From Section 12.7: the little stuff (items 6 and 7 on the first page, Example 5 on the second) we skipped earlier.

-Section 12.8.

-Remark on the text:

-Third paragraph of Section 12.8: For these power series, we adopt the convention 0⁰ = 1 just so that we can write the series more

compactly. Remember 0⁰ is actually an indeterminate form – see earlier handout.

Problems to turn in:

-From Section 12.6: 29, 30.

-From Section 12.8: 3, 5, 7, 9, 12, 15, 17, 18, 19, 23, 27, 30.

NOTE: Remember the algorithm for most of these: Use the Ratio Test, then check the endpoints, if any.

Some answers and hints:

Section 12.6:

#29. Hint. Use the Ratio Test: a_n+1/a_n is staring you in the face!

#30. Hint. Use the Ratio Test: a_n+1/a_n is staring you in the face! Answer: Converges absolutely.

Section 12.8:

#12. R = 5; interval of convergence is [-5, 5].

#18. R = ½; interval of convergence is ( -7/2, -5/2].

#19. Hint. Use the Root Test.

#27. Hint. Use the Root Test.

#30. These are real easy! No computations needed. Hint. Note that the information given allows you to conclude that 4 ≤ R ≤ 6,

that is, to conclude that the series converges at least for all x such that | x | < 4 and diverges at least for all x such that | x | > 6.

Answers:

(a) Converges; (b) Diverges; (c) Converges; (d) Diverges.

Assignment #21, for Friday, November 11

Read:

-Mathematica Handout #9.

-Section 12.10 up to the paragraph just before Example 8, and after that read just Example 9. (We’re skipping Section 12.9 only

temporarily.)

-Exercises #1 and #2 at the end of Section 12.10 and their answers.

Remark on the text:

-The boxed formulas just before Example 8 in your reading: The series expansion for tan^-1x was derived in Section 12.9 and we’ll

do that on Monday.

Problems to turn in:

-From Section 12.8: 8, 14, 24, 29.

-From Section 12.10: 4, 5, 6, 7, 11, 13, 15, 17, 47, 49. NOTE: In each of 11, 13, 15, 17, find the radius of convergence.

-From Section 12.6: 18.

Some answers and hints:

#8. Hint. Use the Root Test. Answer: R = 0, “interval of convergence” is just {0}.

#14. R = ∞, interval of convergence is (-∞, ∞). #24. R = ∞, interval of convergence is (-∞, ∞).

#2. (a) The series says that f '(1) = -0.8 (the second coefficient), but the graph shows that f '(1) is positive. (b) The series says that

f ''(2)/2 = 1.5 (the third coefficient), so f '' (2) = 3, but the graph shows that f ''(2) is negative.

_∞

#4. ∑ (-1)ⁿ2²ⁿ⁺¹x²ⁿ⁺¹/(2n+1)! , R = ∞.

ⁿ⁼⁰

_∞

#6. ∑ (-1)^n-1xⁿ/n , R = 1.

ⁿ⁼¹

#17. The answer in the back of the book is wrong; the first term should be 1/3, not –1/3.

#18. L = 1/e < 1, so the series converges (absolutely) by the Ratio Test.

Assignment #22, for Monday, November 14

Read:

-Mathematica Handout #10.

-The remainder of Section 12.10 that you haven’t read, and then reread Example 6.

-Section 12.12 to “Applications to Physics”.

Remarks on the text:

-Section 12.10, the subsection “Multiplication and Division of Power Series”: This material is good to be aware of but we won’t be

doing any problems like the examples in it.

-Example 1(b) in Section 12.12: If x < 8, then all terms except the first are negative, so it’s not an alternating series.

-Example 2(c) in Section 12.12: Remember π radians = 180^o.

Problems to turn in:

-From Section 12.8: 28.

-From Section 12.10: 23, 26, 27, 29, 31, 39, 40, 42, 55, 57, 58.

NOTE: For none of the series in these problems do you need to find the radius of convergence.

-From Section 12.12: 14(a), 14(b), 19(a), 19(b).

Some answers and hints:

#28. R = 1, interval of convergence is (-1, 1). Hint for the endpoints: Consider using the Test for Divergence.

_∞

#26. ∑ (-1)ⁿx^{8n+ 4}/(2n+1)! .

ⁿ⁼⁰

_{∞ ∞}

#40. C + ∑ (-1)ⁿx^{2n+ 1}/((2n+1)(2n+1)!) . #42. C + ∑ xⁿ/(n n!) .

^{n=0 n=1}

#58. e^3/5.

#14. (a) 1–2(x – 1)+3(x – 1)². (b) Hint. Calculate f '''(x) = -24x ^-5, then show you can take M = 24/(0.9)⁵. Answer is 0.00677404.

#19. (b) Hint. Calculate f ''''(x) = e ^{x^2} (12 + 48x² + 16x⁴), then show you can take M = e ^0.01(12 + 0.48 + 0.0016).

Assignment #23, for Wednesday, November 16

Read:

-Mathematica Handout #11.

-Section 12.9.

Remark on the text:

-Next-to-last sentence of Example 5: Just write out the first few terms of each series to see that they are the same.

Problems to write up but not turn in:

-From Section 12.10: 60.

-From Section 12.12: 13(a), 13(b), 25, 27.

-From Section 12.9: 3, 5, 13.

NOTE: In #13, if you’re not sure your answer is the same as the one in the back of the book, write out, say, the first five terms

of each series and see if they agree.

Problems to turn in:

-Your third Mathematica assignment. It’s due at 5 p.m. in my mailbox just inside the double glass doors on the second floor of the

CMC.

Answers and Hints:

#60. Series converges to e^-ln2 = ½.

#13, Section 12.12. (b) Hint. Calculate f ''' (x) = (3/8)x^-5/2, then show you can take M = (3/8)4^-5/2 = 3/256.

#25. Hint. By Taylor’s Inequality, when 0 ≤ x ≤ 0.1, |R_n(x)| ≤ e ^0.1 (0.1)ⁿ⁺¹/(n+1)! . Use trial and error values for n to determine

that n = 3 is the first one that works.

#27. Hint. Using the Alternating Series Estimation Theorem, put |x⁵/5!| ≤ 0.01, then solve for x.

#3. Write 1/(1+x) = 1/(1-(-x)) and substitute –x for x in the geometric series.

#5. Substitute x³ for x in the geometric series.

#13, Section 12.9. (a) Hint. 1/(1+x)² is minus the derivative of your answer to #3. (b) Hint. 1/(1+x)³ is -1/2 the derivative of your

answer to part (a). (c) Multiply your answer to part (b) by x³.

FINAL EXAM:

8:30-11:00 a.m., Saturday, November 19

or

3:30-6:00 p.m., Saturday, November 19

(Your choice.)

Other times may be available.

Two 4'' by 6'' note cards allowed.