Chapter 9
AP Calculus BC
9.1 Power Series
1  1  1  1 ... 1
2 4 8 16
1 1  1  1  1 ... 
2 3 4 5
111111...  ???

If the sequence of partial sums has a limit
S, as ninfinity, then we say the series
Converges to S. Otherwise it Diverges.
Infinite Series:
 ak  a1  a2  a3 ... an ...
k 1
a1, a2 are terms
an is the nth term
Partial Sums:
S1  a1
S2  a1  a2
S3  a1  a2  a3
n
Sn   ak
k 1
Geometric Series:
a  ar  ar 2  ar3 ... ar n1 ... 
converges if r 1 and
diverges if r 1

 ar n1
n1
Interval of convergence : -1 < r < 1
Is the IOC for Geom. Series.
converges to sum
a
1 r
Representing functions by series:
If x 1, then the Geometric Series formula assures us that:
1 x  x2  x3  x4 ... xn ...  1
1 x

 xn is a Power Series centered at x=0 and it converges on the interval (-1,1)
n0
Definition of a Power Series: An expression of the form:

 cn xn  c0  c1x1  c2 x2 ... cn xn ... is a Power Series centered at x=0.
n0

 cn (x  a)n  c0  c1(x  a)1  c2(x  a)2 ... cn(x  a)n ...
n0
Given:
is a Power Series centered at x = a.
Find a Power Series for:
If x 1,
1 x  x2  x3  x4 ... xn ...  1
1 x
1 on (1,1)
1 x
x
1 x
1
1 2x
Again,
Given:
If x 1,
1 x  x2  x3  x4 ... xn ...  1
1 x
Find a Power Series for:
1
(1 x)2
Now, write down the series for:
and use it to write one for :

Answer:
 nxn1
n1
1
1 x
ln(1 x)

(1)n xn1
Answer:  n 1
n0
Copy down Theorem 1 p. 478 and Theorem 2 p. 479
9.2 Taylor Series
P( x)  a0  a1x  a2 x2  a3x3  a4 x4
Given:
P(0) 1
P '(0)  2
P ''(0)  3
P '''(0)  4
P IV (0)  5
Find the Taylor Polynomial…..
P( x) 1 2x  3 x2  2 x3  5 x4
2
3
24
Find the 4th order Taylor Polynomial of
ln(1 x)
Construct a Power Series for:
Taylor Series generated by f
at x=0:
Answer:
sin x
and
cos x
at x = 0
nth term answers:

2n1
sin x   (1)n x
(2n 1)!
n0

2n
cos x   (1)n x
(2n)!
n0
 f k (0)
n (0)
f
''(0)
f
'''(0)
f
2
3
n
f (0)  f '(0) x 
x 
x ...
x ...  
xk
2!
3!
n!
k 0 k !
9.2 cont’d.
p. 489 Taylor Series centered at x = a.
Copy it down!!!!
p. 491 5 most important series: Copy them down, know them!!!
Occasionally you need 6, 7…….
9.3 Taylor’s Theorem
Adequate substitution: a Taylor series that
is off from the actual by less than 0.0001
Truncation Error:
NEXT TERM!!!!
If f has derivatives of all orders in an open interval, I, containing a,
then for each positive integer, n, and for each x in the interval:
n (a)
f
''(
a
)
f
2
f ( x)  f (a)  f '(a)( x  a) 
( x  a) ...
( x  a)n  Rn( x)
2!
n!
n1(c)
f
Where: Rn ( x)  (n 1)! ( x  a)n1 Largest value of derivative..f part
If Rn ( x)  0 as n  for all x in I, we say that the Taylor Series
generated by f at x = a converges to f on I.
Theorem 4 – Remainder Estimation Theorem. (do examples)
9.4 Radius of Convergence
A convergent series is a number and may be treated as such….
Theorem 5 - Convergence Theorem for Power Series -There are 3 possibilities for

 cn( x  a)n
n0
1. There is a  R such that the series diverges for x  a  R but
converges for x  a  R. The series may or may not converge at the endpoints.
2. The series converges for every x. (R=)
3. The series converges at x = a, but diverges elsewhere. (R=0)
R = radius of convergence and the set of x-values for which the series
converges is called the Interval of Convergence.
Theorem 6 – nth term test
for divergence

a  0.
 an diverges if nlim
 n
n1
9.4 cont’d.
Direct Comparison Test (DCT) (non-negative terms)
Greatest Power Rules!!!!!
Let  an be a series with no negative terms...
(a)  an converges if there is a convergent series  cn with an  cn
(b)  an diverges if there is a divergent series  dn with an  dn
Absolute Convergence - If the series  an of absolute values converges,
then  an converges absolutely.
Theorem 8…..
Ratio Test – (Powers and Factorials)
an1
Let  an be a series with  terms and with nlim
L
 an
Then: (a) the series converges if L < 1.
(b) the series diverges if L > 1.
Telescoping series:
(c) the test is inconclusive if L = 1.
p. 510……
9.5 Testing Convergence at Endpoints
Theorem 10 - Integral Test - Let {an} be a sequence of  terms.
Suppose that an  f (n), where f is a cts., , decreasing function of x.
Then the series


 an and the Integral N f ( x)dx either
n N
both converge or both diverge.

1
P-series test:  n p converges if p > 1, diverges if p  1.
n1
Limit Comparison Test (LCT) Suppose that an  0, bn  0:
an  c, 0  c , then a & b both converge or diverge
(1) If nlim
n
n


 b
n
an  0, and b converges then a converges.
(2) If nlim
n
 n
 b
n
an , and b diverges then a diverges.
(3) If nlim
n
n


 b
n
9.5 cont’d.
Alternating series Test (Liebniz’s Theorem)
The series

 (1)n1un  u1  u2  u3  u4 ...
n1
converges if all three of the following conditions are met:
(1) Each un is positive.
(2) un  un1 for all n...
Error is next term sign
(3) nlim
u  0.
 n
included…………
Look at examples 4 -6 pp. 518 - 520
Absolute and Conditional Convergence……