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……