Chapter 9
AP Calculus BC
9.1 Power Series1 1 1 1 ... 12 4 8 16
1 1 1 11 ...52 3 4
1 1 1 1 1 1 ... ???
Infinite Series:
1 2 31
1 2
... ...
, are terms is the n term
nkk
thn
a a a a a
a a a
Partial Sums:1 1
2 1 2
3 1 2 3
1
nn k
k
S a
S a a
S a a a
S a
If the sequence of partial sums has a limit S, as ninfinity, then we say the series Converges to S. Otherwise it Diverges.
Geometric Series:2 3 1 1
1... ...
converges if 1 and
diverges if 1
n n
na ar ar ar ar ar
r
r
Interval of convergence : -1 < r < 1Is the IOC for Geom. Series.
converges to sum 1ar
Representing functions by series:
2 3 4
If 1, then the Geometric Series formula assures us that:
11 ... ...1
n
x
x x x x xx
0 is a Power Series centered at x=0 and it converges on the interval (-1,1)n
nx
Definition of a Power Series: An expression of the form:1 2
0 1 20
... ...n nn n
nc x c c x c x c x
is a Power Series centered at x=0.
1 20 1 2
0( ) ( ) ( ) ... ( ) ...n nn n
nc x a c c x a c x a c x a
is a Power Series centered at x = a.
Given:
2 3 4
If 1,
11 ... ...1
n
x
x x x x xx
Find a Power Series for: 1 on ( 1,1)1
11
1 2
xxx
x
2 3 4
If 1,
11 ... ...1
n
x
x x x x xx
Again, Given:
Find a Power Series for: 21
(1 )x Answer: 1
1
n
nnx
Now, write down the series for:1
1 x
and use it to write one for : ln(1 )x Answer:1
0
( 1)1n n
n
xn
Copy down Theorem 1 p. 478 and Theorem 2 p. 479
9.2 Taylor Series2 3 4
0 1 2 3 4( )P x a a x a x a x a x
Given: (0) 1'(0) 2''(0) 3'''(0) 4
(0) 5IV
PPPPP
Find the Taylor Polynomial…..2 3 43 2 5( ) 1 2
2 3 24P x x x x x Answer:
Find the 4th order Taylor Polynomial ofln(1 )x
Construct a Power Series for: sin
cosat x = 0
xandx
nth term answers:2 1
0
2
0
sin ( 1)(2 1)!
cos ( 1)(2 )!
nn
n
nn
n
xxn
xxn
Taylor Series generated by f at x=0:
2 3
0
''(0) '''(0) (0) (0)(0) '(0) ... ...2! 3! ! !
n kn k
k
f f f ff f x x x x xn 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 TheoremAdequate 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:
2''( ) ( )( ) ( ) '( )( ) ( ) ... ( ) ( )2! !
n nn
f a f af x f a f a x a x a x a R xn
Where:1 1( )( ) ( )
( 1)!n n
nf cR x x an Largest value of derivative..f part
If ( ) 0 as for all x in I, we say that the Taylor Series generated by f at x = a converges to f on I.
nR x n
Theorem 4 – Remainder Estimation Theorem. (do examples)
9.4 Radius of ConvergenceA convergent series is a number and may be treated as such….
0
Theorem 5 - Convergence Theorem for Power Series --
There are 3 possibilities for ( )
1. There is a R such that the series diverges for but
converges for . The series may or may n
nn
nc x a
x a R
x a R
ot 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 1
diverges if lim 0.n nnna a
9.4 cont’d.Direct Comparison Test (DCT) (non-negative terms) Greatest Power Rules!!!!!
Let be a series with no negative terms...(a) converges if there is a convergent series with (b) diverges if there is a divergent series with
n
n n n n
n n n n
aa c a ca d a d
Absolute Convergence - If the series of absolute values converges,
then converges absolutely.n
n
a
a
Theorem 8…..
Ratio Test – (Powers and Factorials)1Let be a series with terms and with lim
Then: (a) the series converges if L < 1.(b) the series diverges if L > 1.(c) the test is inconclusive if L = 1.
nn n n
aa La
Telescoping series:p. 510……
9.5 Testing Convergence at EndpointsTheorem 10 - Integral Test - Let { } be a sequence of terms. Suppose that ( ), where f is a cts., , decreasing function of x.
Then the series and the Integral ( ) either
both conv
n
n
n Nn N
aa f n
a f x dx
erge or both diverge.
P-series test:1
1 converges if p > 1, diverges if p 1.pn n
Limit Comparison Test (LCT) Suppose that 0, 0:
(1) If lim , 0 , then & both converge or diverge
(2) If lim 0, and converges then converges.
(3) If lim , an
n nn
n nnnn
n nnnn
nn
a ba c c a bba b abab
d diverges then diverges.n nb a
9.5 cont’d.Alternating series Test (Liebniz’s Theorem)
11 2 3 4
1
1
The series ( 1) ...
converges if all three of the following conditions are met:(1) Each is positive.(2) for all n...
(3) lim 0.
nn
n
n
n n
nn
u u u u u
uu u
u
Error is next term sign included…………
Look at examples 4 -6 pp. 518 - 520
Absolute and Conditional Convergence……
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