Calculus II and Analytic Geometry
Study Guide
1 Section 8.1: Sequences
1.1 Limit Rules
When:
Sum Rule:
Dierence Rule:
Product Rule:
Constant Multiple:
Quotient Rule:
Sandwich Thm:
Check:
Why?
Link:
Given convergent sequences fan g; fbn g with lim an = A; lim bn = B .
lim(an + bn ) = A + B ,
lim(an bn ) = A B ,
lim an bn = AB ,
lim kan = kA,
A when B 6= 0.
lim abnn = B
If an cn bn for some sequence fcn g and A = B , then A = lim cn = B .
This does not hold for divergent sequences.
Sequences are easier when considering these rules.
http://en.wikipedia.org/wiki/Limit of a sequence
1.2 Common Sequence Limits
1.
2.
3.
4.
5.
6.
Link:
limn!1 pnn = 0.
limn!1 n n = 1.
limn!1 x =n = 1 (x > 0).
limn!1 xn = 0(jxj < 1).
n
limn!1 1n + nx = ex .
x
limn!1 n = 0.
http://en.wikipedia.org/wiki/Limit of a sequence
ln
1
!
2 Section 8.2: Innite Series
2.1 Denition
When:
How:
Why?
Link:
1
Given
P1 an innite sum,Pmn an .
n an := limm!1
n an .
There are lots of convergence tests, but this denition will be important for telescoping
sums.
http://en.wikipedia.org/wiki/Innite series
P
=0
=0
=0
2.2 nth Term Test
When:
How:
Check:
Why?
Link:
P
A sum an , but an 6! P
0.
If limn!1 an 6= 0, then an diverges.
This only works to show divergence, not convergence!
Lots of series can be eliminated quickly this way! Always try this rst!
http://en.wikipedia.org/wiki/Term test
2.3 Geometric Series
When:
Necessary:
How:
Check:
Why?
Link:
Given a sum in the form 1
arn .
n
Be
of index requirements, and jrj < 1.
P1careful
n= a .
ar
n
r
Try a few partial sums to see if it converges to the calculated result.
This is a common example.
http://en.wikipedia.org/wiki/Geometric series
P
=0
=0
1
1
Calculus II and Analytic Geometry
Study Guide
2.4 Series Combinations
When:
How:
Why?
P
P
Given two convergent
series Pan = A;
P
P bn = B , then
1. Sum Rule: (anP
+ bn ) = an P
+ bn P
= A + B.
2. Dierence Rule: (an P
bn ) = anP
bn = A B .
3. Constant Multiple Rule: kan = k an = kA.
Knowing parts
P of sums
P rst may help occasionally. Note that this does NOT work when
the either an or bn diverge!
3 Section 8.3: Integral Test
3.1 Integral Test
Consider the sum 1
n A an and there exists a function f (x) with f (n) = an for n A.
fP(x) 0, decreasing.
1 a and R 1 f (x)dx behave the same. Use convergence tricks from 7.7.
n A n
A
Some series can be checked using improper integrals easier than other tests.
http://en.wikipedia.org/wiki/Integral test
When:
Necessary:
How:
Why?
Link:
=
=
3.2 p-Series
When:
How:
Why?
Link:
P
a
A sum can be organized as 1
n
np .
1. If p > 1, converges.
2. If p 1, diverges.
If p 1, then direct comparison test shows n np , so diverges. It turns out that any larger
power will converge.
http://en.wikipedia.org/wiki/Harmonic series (mathematics)#P-series
P
=1
1
1
4 Section 8.4: Comparison Tests
4.1 Direct Comparison Test (DCT)
When:
Necessary:
How:
Link:
P
Showing convergence or divergence of a sum an when an can be compared to bn and
is known.
an 0.
P
P
1. If an bn and P bn converges, thenP an converges.
2. If bn an and bn diverges, then an diverges.
http://en.wikipedia.org/wiki/Comparison test
4.2 Limit Comparison Test (LCT)
When:
How:
Why?
Link:
P
bn
1 a when a behaves similarly to b and
Show
n
n
n
n
P1 convergence or divergence of a sum
bn is known to converge or
diverge.
n
P
P
1. If limn!1 abnn = c > 0, then
an and
bn behave the same.
P
P
2. If limn!1 abnn = 0, and bP
an converges.
n converges, then P
3. If limn!1 abnn = +1, and bn diverges, then an diverges.
Sometimes this is an easier comparison than the direct route.
http://en.wikipedia.org/wiki/Limit comparison test
P
=0
=0
2
Calculus II and Analytic Geometry
Study Guide
5 Section 8.5: Ratio and Root Tests
5.1 Ratio Test
When:
Necessary:
Why?
Link:
Consider a sum 1
an where consecutive terms an
n
(nth powers, factorials, etc).
Let limn!1 aann+1 = .
Then, (a) converges if < 1,
(b) diverges if > 1,
(c) inconclusive when = 1.
This is particularly useful for power series.
http://en.wikipedia.org/wiki/Ratio test
P
5.2 Root Test
When:
Necessary:
Why?
Link:
+1
=0
and an have many shared terms
Consider a sum 1
n an where an has nth-exponents around complicated functions.
1. an 0 forp n N ,
2. limn!1 n an = ,
Then, (a) converges if < 1,
(b) diverges if > 1,
(c) inconclusive when = 1.
Some series can only be solved using Root Test
http://en.wikipedia.org/wiki/Root test
P
=0
6 Section 8.6 Alternating Series, Absolute and Conditional Convergence
6.1 Alternating Series Test
When:
Necessary:
n
Given a series 1
n ( 1) an
1. an 0.
2. an an .
3. limn!1 an = 0.
When all three are true, then the sum converges.
Several series converge conditionally, but not absolutely. See alternating harmonic series.
http://en.wikipedia.org/wiki/Alternating series
P
=0
+1
Why?
Link:
6.2 Absolute/Conditional Convergence
When:
Necessary:
Link:
If a series has positive and negative terms, it can converge absolutely (when all the terms
are taken with absolute value) or conditionally (convergent only when the terms keep their
sign) or always diverge (never a convergent series, no matter what the signs do).
Check conditional convergence with the Alternating Series Test,
http://en.wikipedia.org/wiki/Absolute convergence
7 Section 8.7: Power Series
7.1 Power Series
What:
How:
Why?
Link:
n
A power series about a has the form 1
n cn (x a) for a sequence cn .
Power series absolutely converge for values of x given by the Ratio Test.
Check boundary conditions for conditional convergence.
Power series are important to understand before trying Taylor series.
http://en.wikipedia.org/wiki/Power series
http://en.wikipedia.org/wiki/Radius of convergence
P
=0
3
Calculus II and Analytic Geometry
Study Guide
7.2 Term-by-Term Integration and Derivation
When:
How:
Why?
Link:
Approximate
a function by knowing a power series for derivative or intergral.
d P1 c (x a)n = P1 nc (x a)n :
n
n
n
nP
Rdx P1
cn
n
[ n cn (x a)n ] dx = 1
+ C.
n
n (x a)
d
x
x
Try proving dx e = e using its Taylor series.
http://en.wikipedia.org/wiki/Power series#Dierentiation and integration
1
=0
=1
=0
+1
=0
+1
8 Section 8.8: Taylor and Maclaurin Series
8.1 Taylor/Maclaurin Series
When:
Necessary:
How:
Approximating or simplifying expressions for a function f (x) by using a polynomial expression.
f (x) innitely dierentiable on an interval (a R; a + R), and knowing the values of each
derivative at a.
(k)
P
Pn (x) = nk f k a (x a)k is the nth Taylor Polynomial. The limit of this polynomial
P
f (n) a
as n approaches innity is the Taylor Series: 1
(x a)n .
n
n
The Maclaurin series is the Taylor series with a = 0.
Carefully watch the derivatives and the pattern created.
Polynomials are easier to do a lot of things, including derivate and integrate.
http://en.wikipedia.org/wiki/Taylor series
( )
=0
!
( )
=0
Check:
Why?
Link:
!
9 Section 8.9: Convergence of Taylor Series
9.1 Taylor's Theorem and Formula
Necessary:
How:
A function f (x) continuous and (n + 1)-dierentiable on an interval (a R; a + R).
The Taylor polynomial Pn (x) and the remainder Rn (x) sum to exactly the function: f (x) =
n+1
Pn (x) + Rn (x) where Rn (x) = f n c (x a)n for some c strictly between x and a.
An elaborate use of the Intermediate Value Theorem, paired with the remainder estimation
theorem proves convergence of Taylor series.
( )
Why?
(
+1
+1)!
9.2 Remainder Estimation Theorem
When:
Necessary:
How:
Computing how close a Taylor polynomial of degree n will estimate the function.
Knowing Pn (x) and the (n + 1)th derivative.
Given a Taylor polynomial Pn (x) centered at a, to be calculated at x, let M be an upper
bound for jf n (t)j for x < t < a. Then, the remainder term Rn (x) satises jRn (x)j (
+1)
j
M j (n+1)!
x a n+1
Check:
Why?
Link:
Calculate the polynomial and the function values at x, and see if they are as close as this
estimate.
Computers can calculate polynomials better than general functions, but it would be good
to know how many computations are required to achieve a certain degree of accuracy.
http://en.wikipedia.org/wiki/Taylor polynomial#Estimates of the remainder
9.3 Frequently Used Taylor Series
4
Calculus II and Analytic Geometry
When:
Necessary:
Geometric:
Alternating Geometric:
Exponential:
Sine:
Cosine:
Natural Log:
Natural Log:
Natural Log:
Inverse Tangent:
Check:
Study Guide
Asking to approximate an answer using \known" Taylor series.
Calculating a value usually hard to compute
by hand.
n + = P1 xn ; jxj < 1
=
1
+
x
+
x
+
+
x
n
x
1 ( x)n ; jxj < 1
3 + ( x)nn + = P
n
x = 1 x + 2x
P
xn ; jxj < 1
ex = 1 + x + x + x + + xn + = 1
n
n
P
3
5
2n+1
n x2n+1 ; jxj < 1
sin x = x x + x + ( 1)n xn
+ = 1
n ( 1)
n
P
2
4
2n
n x2n ; jxj < 1
cos x = 1 x + x + ( 1)n xn + = 1
(
1)
n
n
P
2
3
n
n
ln(1 + x) = x x + x + + ( 1)n xn + = 1
(
1)n xn ; 1 < x 1
n
2
n
3
n
P
x
n
ln(x) = (x 1) x
+x
+ +( 1)n x n + = 1
n ( 1)
n ;0 < x 2
P
3
5
2n+1
x2n+1 ; jxj < 1
ln xx = 2 tanh x = 2 x + x + x + + x n + = 2 1
n
n
P1
5
2n+1
2n+1
3
x
x
x
x
n
n
+ + + ( 1) n + = n ( 1) n ; jxj < 1
tan x = x
Compute the rst few derivatives and check their value. Make sure this agrees with your
rst few values. Note that zero values are removed from these sums.
It is easier to memorize a few examples than to repeatedly generate them.
http://en.wikipedia.org/wiki/Taylor series#List of Taylor series of some common functions
1
1
2
1
2
=0
=0
1+
2!
3!
=0
!
3!
5!
(2
+1)!
2!
4!
(2
)!
!
=0
=0
2
(
)!
1
=1
3
1)
2
(
1)
1(
1(
1)
=1
3
1
1+
1
3
5
2
=0
+1
1
3
5
2
+1
10 Section 8.10: Binomial Series
10.1 Denition
Looking for Taylor series
representation of f (x) = (1 + x)m
m k
k
Note
that f (x) = k x
mm
m
m k
m
k =
P1k m k
Then f (x) = k
k x .
Knowing this expansion makes polynomial expansion easier.
http://en.wikipedia.org/wiki/Binomial series
( )
(
1)(
2)
(
+1)
!
Why?
Link:
(2
+1)!
1
Why?
Link:
When:
How:
(2
=0
5
=0
2
+1
2
+1
1)