Assigned on:

After introducing examples of a new section, if I assign a range of problems from the current section for you to select work from, you should try the odds in that range, check the answers in back, go to tutoring, etc., and bring questions to class.

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I see how far we get in lecture before I come up with a homework assignment, and will generally post that new assignment by 12pm the day of lecture!

F 05/01

Homework due Monday

11.1 p560 #3, 10, 16, 21, 22

The third Taylor polynomial for e^x is given on p556. Substitute 2x for x and multiply thru by 5 and see if you get the same as in homework prob. #3!

M 05/04

We worked on Taylor series some more and the new part has some examples below on how to tell how accurate your Taylor approximation is. Before working on the Remainder Theorem, try a few in 11.5 like what we talked about with 11.1 #3, where you morph an easier given Taylor series by substitution (read examples 3, 5, 6 pp586-588). Then try one Remainder Theorem problem after reading the below examples.

Homework due Weds. 05/06

11.1 p560 # 29 (it’s like ex. 6 p559, but about x=9, not x=1!)

11.5 p590 #  8, 12, 14, 16 (expansion for sin is in solution to prac. prob #1 p591) 

More details about 11.1 Remainder Formula in ex. 6, #29, #30

W 05/06

We went over 11.1 #29 in detail. Looking at it and the examples above, try the process again on a few more below. We also looked at geometric series use in problems like ex. 2p572

Homework due Friday

11.1 p560 #28 and

     Find the second Taylor polynomial of the squareroot of x at x=16 to estimate the squareroot of 16.7 (some things can be recycled from p560 #21/29 but make sure you see that f(a), f’(a), etc. are different!) and find the maximum error using the Remainder Formula.

11.3 p576 #18, 20, 21, 22

F 05/08

Here are the Taylor series answers from the last homework (it took awhile to type in and I was getting sick of it, so check for errors…I can’t look anymore right now!):

Homework due Monday

11.3 p576 # 4, 6, 10, 12, 14, 36, 40

M 5/11

Homework due Weds,

11.2 p567 #6, 8, 10 and chapter supplementary p592 #9

W 5/13

We got a start on section 11.4 today briefly. Recall that you know how to evaluate improper integrals from section 9.6. The idea behind the Integral test on p580 is that a particular series and the related improper integral both want to measure area (a sum). If an improper integral has a limit, then you can fit the related series under it on a graph and therefore it converges to a limit. If the improper integral does not have a limit, then the series can be shown to diverge also, if you see that it fits outside the bounds of the function.

Read the section and try the following for hmk:

11.4 p583 # 2, 6, 8, 12, 14

On Friday, we will do the last of our material. We will finish out section 11.4 with the comparison test, then probably look at some applications in 11.2, 11.3.

F 5/15

The Remainder Theorem examples were not fitting with the text well, so SCROLL DOWN beyond this box to see them. They are still down there!

Test #5 is your last graded activity and will occur as scheduled on Weds. 5/20. The format is as follows:

1.     Find up to a fourth Taylor polynomial for a function (depending on the difficulty of the function, it might be a second Taylor!) and use it to estimate the value of a function at a certain point, like 11.1 p560 #9, 10, 21, 22

2.     Determine the accuracy of a Taylor approximation using the Remainder Theorem (this may or may not be related to problem #1, I will decide that after I decide what function to use). This would be like 11.1 p560 # 27-30 and examples 5/6 p558/559.

a.     Evaluate the given Taylor series at the given value p(x) and compare it to the calculator answer f(x). Find the remainder f(x)-p(x).

b.     Find an estimate of the remainder using the Remainder Theorem

c.     Possibly a part to use the Newton-Raphson method from 11.2 as in supplementary exercises p592 #9, 10 and example 2 p563

3.     Given the Taylor series for a function, use it to estimate the value of an integral, as in 11.1 p560 #15, 16 (I might throw this question out after I write the test if it looks like too much!)

4.     Use a given Taylor expansion to find a related Taylor expansion (possibly two of these) like 11.5 p590 #5-8, 11-14, 17-20

5.     Find the sum of a geometric series, like 11.3 p576 #1-14

6.     Sum a series to find the rational number for a decimal expansion like 11.3 p576 #15-21

7.     Determine if a series converges using the integral test as in 11.4 p583 #1-16

8.     Use the comparison test to determine if a series is convergent or divergent as in 11.4 p583 #21-26

I am required to hold a final activity for the class during finals week. It will be optional for you, however, if you come on 5/20 to take Test 5! The final activity will be that you can come pick up your last test and grade for the course if you want to. If you did not come to take test 5 on 5/20, you will take a harder longer comprehensive test during finals week to replace it!

M 5/20

We reviewed today (format of test is above in previous notes). I promised to put examples of test question #2 here since we didn’t get to do much with them in class. Here they are:

p592 #9

a. Given: p2(x) = 3+(1/6)(x-9) –(1/216)(x-9)^2   as in #29 p560

(^ means to the power of…so I don’t have to use another pesky pdf to write it!)

By calculator, f(8.7)= 2.949576241 and by plugging into the Taylor series,

p2(8.7) = 3+(1/6)(8.7-9) –(1/216)(8.7-9)^2

= 3 – 0.05 – 0.000416667 = 2.949583333

so the remainder is f(x) – p2(x) = 0.000007092

b. using the Remainder Theorem:

by #29 p560, the third derivative of f(x) is (3/8)c^(-5/2) where 8.7< c <9

Since c > 8.7,   c^(-5/2) < 8.7^(-5/2) < 0.005 (I just picked this cutoff number as a nice number above 0.004479215!)

Then the third derivative at c is < (3/8)(0.005) = 0.001875

so the absolute value of second remainder

< {(0.001875)/3!}*(8.7 – 9)^3 = 0.000008438

so 0.000007092 < 0.000008438

c. Newton Raphson: Let f(x) = x^2 – 8.7 then f’(x) = 2x

x0=3

x1=3 – (0.3/6) = 2.95

x2=2.95 – (2.95^2 – 8.7)/(2*2.95) = 2.949576

calculator, Taylor series, and Newton Raphson all are close!

p592 #10

a. Given: p3(x) = –x – (1/2)x^2 – (1/3)x^3   as in #10 p560

(^ means to the power of)

By calculator, ln(1.3) = f(1.3)= 0.262364264

and plugging into the Taylor series (be careful---1.3 is not x!),

ln(1.3) = ln(1-(-0.3)) = p3(-0.3)

= –(–0.3) – (1/2)( (–0.3) ^2 – (1/3) (–0.3) ^3 =

= 0.3 – 0.045 + 0.009 = 0.264

so the remainder is f(x) – p3(x) = 0.264 – 0.262364264 = 0.001635736

b. using the Remainder Theorem:

by #10 p560, the fourth derivative of f(x) is -6/(1-x)^4 where –0.3 < c < 0

0.3 > –c > 0 so 1.3 > 1-c > 1 so 1.3^(-4) < (1-c)^(-4) < 1

(1-c)^(-4) < 1

Then the absolute value of the fourth derivative at c is < (6)(1) = 6

so the absolute value of third remainder

< {(6)/4!}*(-0.3)^4 = 0.002025

so 0.001635736 < 0.002025 as predicted by the Remainder Theorem!

c. Newton Raphson: Let f(x) = e^x – 1.3 then f’(x) = 2x

x0=3

x1=0.2631

x2=0.2624

x3=0.2624

The Remainder Theorem examples were not fitting with the text well (overwriting the text following), so I put them after the text. SCROLL DOWN beyond this box if there is a space so that you can’t see them!