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T 4/1/08
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Quiz
(possibly a double quiz) on §10-7, including Bob’s method as summarized in
the 3/14 calendar entry.
HW due: Read §11-2, §11-3, and the
green box on p. 585; write §11-2 #1, 4, and the third problem from the 3/11
test.
Please re-do the entire test problem for practice (to build speed, if nothing
else), even if you were able to do it the first time.
In your reading, pay special attention to Example 2 on pp. 559-560. Read
these 2 pages several times and do some pencil-and-paper work as you read so
that you can build mental pathways for the technique. This is a standard
AP-type problem that you must eventually be able to do quickly and efficiently.
Note: We will skip §11-4 and §11-5.
We will do the problems in §11-3 and §11-6, and then we are off to conquer
Chapter 12!
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W 4/2/08
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HW due:
Write §11-3 #2, 6, 12, §11-6 #8, 12.
Hint for §11-3 #12:
Please do not consult this hint unless necessary. Spherical shells is a type
of slicing that we have not previously discussed. However, it is entirely
analogous to plane slicing and cylindrical shell slicing. By way of reminder,
plane slicing involves either the integral of A(x) dx or the integral of A(y)
dy, where A is the appropriate cross-sectional area function; similarly,
cylindrical shells involves an integral of A(r) dr, where A is the appropriate “rolled-out rectangular area” function,
namely  .
Now, extend this idea of A(r) dr
to spherical shells. What is the suitable area function for a thin spherical
shell centered at the earth’s center?
A moment’s thought recalls the surface area of a sphere, namely  . Therefore, if you apply the idea of
A(x) dx, A(y)
dy, and A(r) dr one more time, you get the integral
of dr, where r is a dummy variable. Do not use r to represent the radius of the earth. That radius is  km, which you may round
off to 6375 km or, if you prefer, 6.375 · 108 cm. As in all
density problems, you must figure out an expression for the density function
(taking care that this function depends solely on r!) and merge that
density expression into the integrand.
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Th 4/3/08
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HW due:
Write §11-6 #13 and patch up any lingering work from the last few days. Then
read pp. 598-599 and all of §12-2. I usually do not assign the chapter
overviews as reading material, but the one on p. 598 is crucially important.
You will notice that §12-2 is essentially a review of precalculus.
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F 4/4/08
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HW due:
Make sure that your problem #3 from the most recent test is fully patched up,
because it will be collected at some point. Then read §12-3; write §12-2
#Q1-Q10 all, plus regular exercises #7, 8.
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M 4/7/08
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HW due:
Problem #3 from the most recent test, plus §12-3 #1-11 all. This is an
example of “programmed learning,” i.e., self-paced instruction following a plan
that is designed to introduce the concepts in the context of an ongoing
problem.
Please also read the preamble to §12-3 (top of p. 607) and the silly dialog
that follows.
Students may wonder aloud
why it is worth developing a power series for f (x) = 5e2x. Couldn’t we simply use a calculator and punch in e (approximately 2.718281828459) and
multiply it by itself the appropriate number of times?
Mr. Hansen’s response: What if the
exponent is not an integer? Raising e
to the 4th power is easy (square twice), but how can you raise e to the 3.9 power, for example? What
then? You’re going to need a power series.
Rebuttal from student who has studied
French: Au contraire! I will use my superior technology to raise any base
I wish to any exponent I wish, whether the exponent be positive or negative,
rational or irrational, real or imaginary.
Counter-rebuttal from Mr. Hansen:
But what do you think underlies that fancy technology of yours? Power series,
of course. Your calculator cannot raise an arbitrary base to an arbitrary
exponent by repeated multiplication. Repeated multiplication is a fine way to
define integer exponents, but in previous courses we have generalized to
allow the exponent to be any real number, as long as the base is positive. We
now need a way to compute these exponential functions!
Counter-counter-rebuttal from student:
OK, so you can compute e to some
power. Big deal. You still can’t raise arbitrary bases to arbitrary powers by
using power series.
Mr. Hansen (going in for the kill):
Au contraire! Remember that for any positive base b, bx can be
rewritten as ex ln b.
Student (filled with pride): Aha!
But you can’t compute ln b
accurately unless b is something
really simple like 1, e, or e2.
Mr. Hansen (smiling broadly): Oh, you
want to know about natural logs? No problem! There’s a power series to
compute those, too.
Student (starting to lose steam):
So, whatever, you can explain how the transcendental functions can be
computed on a scientific calculator. That brings us up to, what, 1976
technology?
Mr. Hansen: Actually, we can go far
beyond that with power series. Remember, one of the difficult problems of the
calculus is efficient quadrature. If the integral you are trying to compute
must be executed millions (or billions) of times as part of some slick new
software you are writing, the run time becomes an important consideration.
Rarely does the integrand have a closed-form antiderivative that would allow
you to compute an answer by FTC1. However, general quadrature using the
midpoint rule, the trapezoid rule, Simpson’s Rule, etc., may be too slow
and/or inaccurate for what you need. If your integrand is assembled from
functions whose power series are known, you may be able to write a power
series representation of the integrand and then antidifferentiate “term by
term” to get a good approximation for the original definite integral.
Student: Using what? FTC1 applied
to the terms of a polynomial?
Mr. Hansen: Precisely.
Student: Why is that better than
Simpson’s Rule? And does this method always work?
Mr. Hansen: If the Simpson’s Rule
implementation involves a number of expensive evaluations of transcendental
functions, then yes, power series might offer a faster, cheaper alternative.
The other caveat is that power series are valid only on a certain interval of convergence that must be
determined for each problem depending on the specifics.
Student: Sounds like a lot of work.
I think I’ll stick to Simpson’s Rule.
Mr. Hansen: Oh, I forgot to mention
one other big advantage of power series estimation. Often (especially in the
case of series whose signs alternate) it is relatively easy to compute an
error bound that depends on the number of terms. In other words, we can
compute in advance precisely how
many terms of the power series will be needed in order to guarantee accuracy
to within, say, 7 decimal places. With Simpson’s Rule, you may have a lot of
trial and error involved.
Student: Error bounds? Oh my! That
sounds like a lot of work.
Mr. Hansen, comfortingly: It is,
but we will take it a step at a time. Before long, you’ll know more than you
would ever have thought possible!
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T 4/8/08
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HW due:
Read §§12-4 and 12-5; write §12-4 #2, 4, 7. You should be able to do these by
emulating the examples given in the text. (For example, #7 is very similar to
Example 3. The main difficulties are adapting the calculator function
appropriately and pushing the correct buttons.)
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W 4/9/08
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HW due: Read §12-6; write §12-5 #9, 12, 13, 14, 15-33 mo3. The
first four of these should go very
quickly.
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Th 4/10/08
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Quiz (10 pts.) on Taylor’s Theorem. Difficulty will be
comparable to the sample problem we did in class yesterday (proposed by Ned)
in which we wrote the expansion about a
= 1 for ex.
HW due: Write §12-6 #1, 4, 7, 12, 13, 14, 15, 17, 19.
Please correct the typo in #14 on p. 627, where the numerator should be n!(xn).
Also, if you have not already done so, correct the typo we discussed Tuesday.
In the first green box on p. 616, the second line should say “expansion about
x = a” instead of “expansion about x = 1.”
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F 4/11/08
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“Regurgitation”
Quiz (10 pts.) on the 8 standard
Taylor series (p. 616) and their intervals of convergence. The intervals of
convergence are the real line for all except the last three listed:
- The series for ln x converges only on (0, 2].
- The series for 1/(1 – x) converges only on (–1, 1).
- The series for arctan x converges only on [–1, 1].
HW due:
Read §12-7 and make extensive reading notes. This is difficult material. If
you have reading notes but they do not appear to be of good quality, then
your score will be determined based on a 1-question oral quiz.
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M 4/14/08
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HW due: Read §12-8 (final section of the book!); write
§12-7 #1, 2, 4, 5-14 all, 19, 25-30 all. Answers for #25-30 are provided
below in case you get stuck. If time is short, focus on #5-14, since those
are the most representative of AP-type problems.
25. Since this is a geometric series with common ratio r = 1/4 satisfying | r
| < 1, the series converges.
26. Much as in #25, this is a geometric series with r = 3/4, which satisfies | r
| < 1. The series converges.
27. The first term, in which n = 0,
is a term having value 1 with no effect on convergence. All other terms are
positive and are bounded above by  . Therefore, the series converges by comparison to a
p-series with p = 2. Or, if you prefer, you could say that the terms are all
bounded above by  , implying convergence by comparison to a geometric series
with r = 1/6.
Loose end for #27: If anyone doubts that the “nth term” (by which we mean the term indexed by n, or technically the (n + 1)st term of the series) is
bounded above by for all n > 0, then here is a sloppy but
valid proof:
28. This is a convergent geometric series with r = –1/3, which satisfies | r
| < 1. Or, if you prefer, use AST instead.
29. Since  ,
each term is bounded below by the corresponding term of the harmonic series.
Since the harmonic series diverges, the given series also diverges.
You could also use the integral test or the limit comparison
test. The limit comparison test is not covered in our textbook, nor is it
a requirement for the AP exam, but it is one of the most useful tests
available, since it can save you from the tedious inequality-crunching that
often accompanies other comparison tests.
30. Series diverges by the nth
term test. (The limit of the nth
term is not 0 as required for convergence.)
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T 4/15/08
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No class (Diversity Day). However, life goes on. The
assignment below should be completed as your HW for today. Show it to me
during Math Lab, 2:05 to 3:30 p.m., in order to earn your homework score for
today. If you cannot make it to Math Lab for some reason (chorale rehearsal,
perhaps, though I cannot reasonably imagine any other excuses), then see me
at table 30 during lunch.
HW due: Complete yesterday’s
assignment, especially problems 5-14 all, and write the following short
additional exercise.
Exercise: As you recall from much
earlier in the year,  .
We can prove this either by the Squeeze Theorem or by L’Hôpital’s Rule.
However, Chapter 12 gives us yet a third way to do this.
(a) Write a power series for the quotient. No work is required. (Use the
“cheat-y” way.)
(b) Show that your power series converges for all x in a neighborhood of 0.
(c) To what value does your power series converge when x = 0?
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W 4/16/08
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Double
Quiz (20 pts.) on “full
regurgitation” (all 8 Taylor series and their intervals of convergence) and “full
thinking” (a problem or two based on recent HW through §12-7).
HW due: §12-8 #2, 3, 4, 13, 14.
Hint for #13: Draw a picture.
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Th 4/17/08
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HW due: Finish all previously assigned problems. Then read
#21 on p. 648 (no need to solve, but take reading notes), and begin writing
§12-8 #5, 6, 11. You may benefit from seeing #6, which is fully worked below.
Copy my work if you must. Problems 5, 6, and 11 will not be collected until
Friday, on account of the late posting of this assignment and the fact that
several confusing typos managed to sneak through in the first version. The
typos did not affect the partial sum cross-check at the bottom, nor did they
affect the answer (11 terms), but they made the inequality incorrect as
originally stated.
6. cosh 3 =  , where the terms are numbered according to the exponent.
“Real” term 1 has exponent 0, “real” term 2 has exponent 2, and so on. As n goes from 0 to  , the real term number (let us call it q) equals half the exponent, plus 1. In other words,  .
Let f (x) = cosh x and a = 0, thus giving Lagrange error term
of form  .
Since the derivatives of sinh x are
all either sinh x or cosh x, an upper bound for |Rn(x)| when 0 < x <
3 would be
 .
The rationale for wanting the expression to be less than 5 · 10–9
is that 8 decimal places of precision will generally be ensured [see note
below] whenever the error is less than half of a unit in the 8th decimal place,
i.e., 0.5 · 10–8 or, equivalently, 5 · 10–9. The
easiest way to solve the inequality is to put the function
(e^3+e^(–3))/2/(X+1)!*3^(X+1)
into the Y1 function on your calculator. Then push 2nd TBLSET and
set TblStart=1,  Tbl=1, Auto, Auto. When you push 2nd TABLE, the values will
all be displayed for you. Scroll down to see the first value for which the Y1
expression is less than 5 · 10–9, namely 20.
[Note: Because there are certain
pathological cases where having error less than 5 · 10–9 does not
guarantee 8 digits of displayed accuracy, the AP exam does not pose questions
in this format. Instead, you will see a much more straightforward demand. For
example, you may be asked to compute the number of terms needed in the partial
sum in order to guarantee an error less than .001. That is a much easier
question, since you simply take whatever value is given to you and put it on
the right side of your “want” inequality.]
What does n = 20 mean? It does not mean 20 terms. What it means is
that 20 is the first n value that
ensures an error term less than 5 · 10–9 in absolute value. By our
rule above, the term number is found by  .
Final answer: The partial sum must have 11 terms.
If you wish, you can cross-check your work by computing partial sums as
follows:
Press MODE and select Seq instead of Func.
Press 2nd QUIT.
Press the Y= key.
Set nMin=0,
u(n)=u(n–1)+3^(2n)/(2n)! by using the 2nd function of 7 to
get u and the “ ” key to get n.
Set u(nMin)={1}.
Press 2nd QUIT.
Key in 2nd CATALOG D uparrow uparrow uparrow uparrow ENTER 3 ENTER to compute
cosh 3, namely 10.067662. (The actual value, which your calculator does not
show you, is 10.06766199578 . . .)
Key in u(1) ENTER to compute the 2nd partial sum, namely 5.5. This is, of
course, nowhere close to cosh 3.
Key in u(2) ENTER to compute the 3rd partial sum. Not much better!
Key in u(2)–cosh(3) ENTER to compute the error after the 3rd partial sum,
namely –1.192661996. No good!
Key in 2nd ENTRY and modify to get u(9)–cosh(3) ENTER. This gives the error
after the 10th partial sum, a very respectable –1.461 · 10–9. Now,
you might think that this would be enough to guarantee 8 places of accuracy,
but unfortunately here it does not. The issue is not whether the actual error
is less than 5 · 10–9, but rather whether the Lagrange error bound
can be guaranteed to be less than 5 · 10–9. The latter does not
occur until after the 11th partial sum, namely u(10).
Key in u(9) ENTER and notice that the 10th partial sum does not exactly equal
cosh(3). In fact, u(9), which is the 10th partial sum, displays as
10.06766199 to 8 places, while cosh(3) displays as 10.06766200 to 8 places.
The difference is caused by the small discrepancy in the 9th decimal place as
computed above.
Finally, key in u(10) ENTER to find the 11th partial sum. Now we have an
answer that agrees with cosh 3 to 8 decimal places. Whew!
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F 4/18/08
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HW due: §12-8 #5, 6, 11.
Double Quiz (10 + 10 pts.) in
preparation for the AP exam. Questions will be drawn from the BC Calculus Cram Sheet.
Next week’s quizzes will involve questions drawn randomly from an AP review
book similar to the one you should be working on every night. All material
from the entire year is fair game beginning next week.
Optional reading assignment: Check
out the Bailey-Borwein-Plouffe information under “Fun Links.”
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M 4/21/08
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HW due: Print out the handy
spreadsheet for recording your review problems, and start filling it in
each day. This is primarily for your benefit, though I will periodically
spot-check to make sure that you are making progress. Work a few problems a
day for about 6 days a week. If there are a few days during which you cannot
devote a full study period, that is OK, but please work at least a couple of
problems. Consistent daily effort is the key. Additional instructions are on
the sheet.
Quest (40 pts.) on the entire year’s material. We will plan for 9 of
these over the next 2 weeks, and your lowest pair of scores will be dropped.
Because answers will be discussed immediately after each quest, there will be
no make-ups. (Missing class for any reason will simply constitute one of your
scores to be dropped.) Questions will be drawn primarily from AP review book
material, although some occasional facts from the cram sheet may be sprinkled in as
well. Most questions will be in one of the following formats:
Part IA: Multiple choice, no calculator, approx. 2 minutes per problem
Part IB: Multiple choice, calculator required, approx. 3 minutes per problem
Part IIA: Free response, calculator required, 15 minutes per multi-part
problem
Part IIB: Free response, no calculator, same pace as Part IIA
Note that in the actual AP exam, you may continue to work on Part IIA
problems (however, without calculator) during your time for Part IIB. Timings
during the actual exam are as follows:
Part IA: 55 minutes
Part IB: 50 minutes
Bathroom break: approx. 10 minutes
Part IIA: 45 minutes
Part IIB: 45 minutes
Total: approx. 3.5 hours
As you prepare for the exam, make sure that you are working on a selection of
problems to cover all of the topic areas of the course. Your handy spreadsheet lists all of the topics
that are required by the AP syllabus.
In Parts IIA and IIB, simplification of algebraic expressions is not required
unless specifically requested. For example, if the question says to show that
f is decreasing when x = 2, you would need to compute and
simplify f '(2) and add the
notation “< 0” at the end to indicate that you knew what you were doing.
However, in general, simplification is not required.
In Part IIB, you need not show work for any of the 4 standard calculator
operations:
1. Expression/function evaluation, including plugging in values to produce a
graph
2. MATH 8 (derivative at a point)
3. MATH 9 (definite integral)
4. MATH 0 or alternative methods for root or intersection finding
The following, however, are not permitted in either Part IIA or Part IIB:
1. Finding maxima or minima without using the calculus
2. Reasoning from a graph (e.g., saying a function has a cusp at x = 3 because there is a sharp point
there, or saying that a function has a local minimum at a certain point “by
inspection”)
3. Leaving expressions unevaluated in final answers (e.g., being vague in
what the limits of integration are)
4. Using calculator notation such as fnInt(X^2,X,0,3) instead of proper
mathematical notation
5. Using equal signs inappropriately
In AP Statistics, you are allowed to use the “=” sign with some abandon,
since statistics involves approximations all over the place, and it would be
a nuisance to be always making a distinction between “equals” and
“approximately equals.” However, in AP Calculus, you must never use an “=”
sign unless the quantities are truly equal. For example, a Taylor series that
uses an “=” sign almost always needs dots at the end ( + . . . ) to show that
the series continues.
** Be sure to circle or box your
answers! **
In Parts IA and IB, your work is not graded, and any method other than
cheating is permitted.
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T 4/22/08
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Quest (40
pts.).
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W 4/23/08
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Quest (40
pts.).
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Th 4/24/08
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Quest (40
pts.).
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F 4/25/08
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Quest (40
pts.).
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M 4/28/08
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Phi Beta Kappa day: no school. However, there will
be a full-length practice AP exam for anyone who wishes to take it, beginning
at 8:00 a.m. in the Mathplex. Your actual AP exam will be either in the
Trophy Room or the Activities Gym, depending on Mr. Andreoli’s decision.
Part of the challenge of the AP exam is the 3.5-hour duration. It is
important to learn how to pace yourself so that you can maintain your focus.
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T 4/29/08
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Quest (40
pts.).
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W 4/30/08
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Quest (40
pts.).
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