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This comprehensive study guide for math 1522 (calculus ii) at the university of new mexico covers essential topics like power series, taylor series, and complex numbers. It includes worked examples, practice problems with solutions, and unsolved questions to help students prepare for their exam. The guide focuses on sections 11.8 through 11.11 of stewart's calculus and basic complex numbers from appendix h. It assumes familiarity with calculus i and previously covered calculus ii material.
Typology: Lecture notes
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dwhavens@unm.edu
Summary and Disclaimer 2
Methods and Techniques 2
Worked Examples 4
Practice Problems 7
Practice Problem Solutions 8
Unsolved Questions 10
Summary and Disclaimer
This is a study guide for the first exam for math 1522 at the University of New Mexico (Calculus II). The exam covers sections 11.8 through 11.11 of Stewart’s Calculus, as well as basic complex numbers (which can be found in Appendix H of the same text). As such, this study guide is focused on that material. I assume that the student reading this study guide is familiar with the material from a calculus 1 course and the material previously covered in calculus 2. If a you feel that you need to review this material, you can send me an email, or take a look at Paul’s Online Math notes:
https://tutorial.math.lamar.edu/
If you are not in my class, I cannot guarantee how much these notes will help you. With that said, if your TA or instructor has shared these with you, then you will most likely get some use out of them.
Methods and Techniques
The primary focus of this exam is on power series. For this exam, there are four power series that you should have memorized:
Common Power Series
1 − x
= 1 + x + x^2 + x^3 +... =
n=
xn
ex^ = 1 + x +
x^2 2!
x^3 3!
n=
xn n!
cos(x) = 1 −
x^2 2!
x^4 4!
x^6 6!
n=
(−1)n^
x^2 n (2n)!
sin(x) = x −
x^3 3!
x^5 5!
x^7 7!
n=
(−1)n^
x^2 n+ (2n + 1)!
Since we mainly use Taylor series in this course, you should know how they are defined:
Taylor Series
The Taylor series for a function f (x) centered at x = a has the formula
X^ ∞
n=
f (n)(a)
(x − a)n n!
Although it has no explicit name, the identity e(θ+2π)i^ = eθi, which can be derived from the above, is also rather useful.
Worked Examples
We will now work through some examples.
Example: Find the power series for f (x) = ln(1 + x^2 ). We begin by taking the derivative of ln(1 + x^2 ). This gives us (by the chain rule),
f ′(x) =
2 x 1 + x^2
This is a geometric series, since we can write it as
f ′(x) = 2x
1 − (−x^2 ) so f ′(x) = 2x
1 − x^2 + x^4 − x^6 +...
Distribution then gives us that
f ′(x) = 2x − 2 x^3 + 2x^5 − 2 x^7 +....
Integrating this gives us
f (x) = C +
2 x 2
2 x^4 4
2 x^6 6
2 x^8 8
We simplify this down to
f (x) = C + x −
x^4 2
x^6 3
x^8 4
and plug in x = 0 to get C = ln(1 + 0)) = 0. So,
f (x) = x −
x^4 2
x^6 3
x^8 4
Example: Determine the value of
X^ ∞
n=
2 n 32 nn!
We have that 3^2 n^ = 9n, so, we can rewrite the sum as
n=
2 n 9 nn!
n=
2 n 9 n n!
n=
9
n
n!
But this is just the power series of ex^ with 29 where the x should be. So,
X^ ∞
n=
2 n 32 nn!
= e
2 (^9).
Example: Consider the Taylor series of cos(x) centered at a = 0. Find p 2 (x), and determine the error on the interval
−π 2 , π 2
Recall that the Taylor series for cos(x) is given by
cos(x) = 1 −
x^2 2!
x^4 4!
x^6 6!
So, p 2 (x) = 1 −
x^2 2
, since 2! = 2.
For the second part of the question, we have two ways of doing this. The first is the alternating series remainder theorem, and the second is Taylor’s Estimation Theorem. First, we will handle the alternating series remainder theorem, since this is easier. For the alternating series estimation theorem, we know that the value of the estimation is less than the value of the next term. So, if E is the error in the estimation,
x^4 4!
|x^4 | 4!
|x|^4 24
So, we find the maximum value of |x| on our interval. And since the endpoints are where this is greatest, and the absolute value is equal to π 2 at the endpoints, we have that
|x| ≤
pi 2
So,
E ≤
π^4 24 24
π^4 24 · 24
π^4 384
For Taylor’s Estimation Theorem, we are using the formula
M |x − a|n+ (n + 1)!
where E is the error and M is the maximum value of the (n + 1)th derivative of f on the interval. So, we first find the maximum value of |x − a| on the interval
−π 2 , π 2
. Since a = 0, this is just the maximum value of |x|, which we found above to be π 2. Next, we need to find the third derivative of cos(x), since n = 2 and we need to find
Practice Problems
These practice problems are separate from the unsolved problems. They should be used to make sure that you are confident with the material, and are of approximately the same level of difficulty as the unsolved questions. They also include worked solutions, unlike the unsolved questions section.
n=
(−1)n^
3 n 22 nn!
Practice Problem Solutions
Solution: We know that
ex^ = 1 + x +
x^2 2!
So, e^2 x^ = 1 + 2x +
22 x^2 2!
Meaning that p 2 (x) = 1 + 2x + 2x^2 , since this is the polynomial of degree 2 which best estimates the power series. We then need to find something which is bigger than (but not that much bigger than) M (x − a)2+ 2 + 1)!
M |x − a|^3 3! Next, we need to find f ′′′(x). Since f ′(x) = 2e^2 x, f ′′(x) = 4e^2 x, so f ′′′(x) = 8e^2 x. Next, we want to see what the largest value of |f ′′′(x)| is on the interval [− 2 , 1]. Since 8e^2 x^ is strictly increasing (as its derivative is 16e^2 x, which is never negative), we only need to test the values at the endpoints to see what the largest value is. So, note that 8 e^2 ·(−2)^ = 8e−^4 < 8 e^2 = 8e^2 ·^1. So, 8e^2 is the largest value that |f ′′′(x)| attains, so M = 8e^2. Additionally, we know that x − a = x, since a = 0. So,
− 2 ≤ x − a ≤ 1 ,
meaning that |x − a| ≤ 2. So, |x − a|^3 ≤ 23 = 8. This gives us the final error bound: |f (x) − p 2 (x)| ≤
8 e^2 · 8 3!
64 e^2 6
Solution: To find the power series of
(1 − x)^2
, we note that
1 1 − x
(1 − x)^2
So we have that
Unsolved Questions
As I am no longer allowed to distribute practice exams, here is a list of 22 unsolved questions which I believe are of similar difficulty to what might be asked of you on an exam. They are grouped by type of problem, so if you feel like you don’t know how to do one specific type of problem, you should do all of the problems of that type for practice.
n=
xn n
X^ ∞
n=
x^3 n n!
X^ ∞
n=
(x − 5)n 3 n
n=
xn nn
− 1 − x + ex cos(x) − 1
x − sin(x) 1 − (^1) −^1 x 3
n=
2 nπn n!
n=
(−1)n^
4 nπ^2 n (2n)!3^2 n
n=
(−1)n^
π^2 n+ (2n + 1)!
n=
(−1)n^
π^2 n 9 n(2n)!
3, find the polar form of z.
π 4 i , find the standard form of z.
0
ex
2 dx
within 0.^2 5
Approximate (^) Z
0
sin(x) x
dx
within 0.^3
5