9.10 Taylor and Maclaurin Series Colin Maclaurin 1698-1746.

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9.10 Taylor and Maclaurin Series Colin Maclaurin 1698-1746

Transcript of 9.10 Taylor and Maclaurin Series Colin Maclaurin 1698-1746.

9.10 Taylor and Maclaurin Series

Colin Maclaurin1698-1746

In the previous lesson we were able to take certain functions that could be

written “in the form” of 1

a

r and write them as a geometric power series.

Today we are going to make this more general so that we can write any function as a power series!

Suppose we have a function represented by the polynomial:

2 31 2 3( ) ( ) ( ) ( ) ...of x a a x c a x c a x c

How could we determine what the coefficients are in terms of f ?na

If we let x = c, all the terms after the first become 0 and we have:

( ) of c a

If we differentiate term by term, we get:

2 31 2 3 4'( ) 2 ( ) 3 ( ) 4 ( ) ...f x a a x c a x c a x c

Again, if we let x = c, all the terms after the first become 0 and we have:

1'( )f c a

2 31 2 3 4'( ) 2 ( ) 3 ( ) 4 ( ) ...f x a a x c a x c a x c

If we differentiate again we get:

22 3 4"( ) 2 2 3 ( ) 3 4 ( ) ...f x a a x c a x c

Again, if we let x = c, all the terms after the first become 0 and we have:

2"( ) 2f c a

Let’s differentiate one more time to get:

23 4 5"'( ) 2 3 2 3 4 ( ) 3 4 5 ( ) ...f x a a x c a x c

Finally when x = c:

3 3"'( ) 2 3 3!f c a a

So what is the pattern??

In each case when x = c, we got:

( ) ( ) !nnf c n a

Solving this equation for we get:na

( ) ( )

!

n

n

f ca

n

Where have we seen this before?

These are the coefficients of the Taylor polynomial that represents f(x).

( ) ( )2

0

( ) '( ) "( ) ( )( ) ( ) ( ) ( ) ... ( ) ...

! 1! 2! !

n nn n

n

f c f c f c f cx c f c x c x c x c

n n

is the Taylor series for f(x) at c.

If c = 0, then what would we call this??

The Maclaurin series for f(x)!

This is useful because now we can find a series representationfor any function for which we can find its derivatives at c!

Definition of Taylor Series

Ex. 1 (Together): Form the Maclaurin series for and find its interval of convergence.

( ) sinf x x

( ) sinf x x (0) sin 0 0f

'( ) cosf x x '(0) cos0 1f

"( ) sinf x x "(0) sin 0 0f

(3) ( ) cosf x x(3) (0) cos0 1f

(4) ( ) sinf x x (4) (0) sin 0 0f (5) ( ) cosf x x (5) (0) cos0 1f

( )2 3 4 5

0

(0) 0 ( 1) 0 10 (1) ...

! 2! 3! 4! 5!

nn

n

fx x x x x x

n

( )2 3 4 5

0

(0) 0 ( 1) 0 10 (1) ...

! 2! 3! 4! 5!

nn

n

fx x x x x x

n

2 1

0

( 1)

(2 1)!

n n

n

x

n

2 3

2 1

(2 3)!lim

(2 1)!

n

nn

x n

x n

To find the interval of convergence, use the Ratio Test:

2

lim(2 3)(2 2)n

x

n n

0 1

Converges for all x, so the interval of convergence is:

( , )

3 5 7

...3! 5! 7!

x x xx

We know this series converges for all x, but what does it converge to??

Does it necessarily converge to ?sin x

Recall that a Taylor polynomial of n terms has a remainder:

( )2'( ) "( ) ( )

( ) ( ) ( ) ( ) ... ( ) ( )1! 2! !

nn

n

f c f c f cf x f c x c x c x c R x

n

What is the remainder equal to??

( 1)1( )

( ) ( )( 1)!

nn

n

f zR x x c

n

In otherwords, the series could agree with sinx at its derivatives but maybe not at points in between the derivatives.

How could we tell if the series really does converge to sinx everywhere(and in the long run), and not just at its derivatives?

( 1)1( )

We need lim ( ) ( ) 0( 1)!

nn

nn

f zR x x c

n

0

( )To say that ( ) ( )

!

nn

n

f cf x x c

n

Convergence of a Taylor Series

If for all x in the interval I centered at c, then

the Taylor series for f converges and equals f(x).

lim ( ) 0nnR x

Ex. 2 We already showed that the Maclaurin series for converges

for all x. Now show that it in fact converges to for all x.

sin x

sin x

1( 1)1( )

0 ( ) ( )( 1)! ( 1)!

nnn

n

xf zR x x c

n n

Since or ( 1) ( ) sinnf x x ( 1) ( ) cosnf x x

Taylor’s inequality tells us that our remainder is boundedby the maximum value for the (n+1)th derivative. What is that?

1

0 ( )( 1)!

n

n

xR x

n

1

lim 0( 1)!

n

n

x

n

lim ( ) 0nnR x

Therefore the Maclaurin series for converges to the

function for all x.

sin x

sin x

2 1

0

( 1)sin

(2 1)!

n n

n

xx

n

For certain functions (categorized as transcendental functions) if the Taylor Polynomial is infinite, becoming a Taylor Series, then the polynomial is actually

equivalent to the original function.

WOW!!!

http://calculusapplets.com/taylor.html

The diagram visually illustrates the convergence of the Maclaurin series for sin x by comparing the graphs of the Maclaurin polynomials P1(x), P3(x), P5(x), and P7(x) with the graph of the sine function. Notice that as the degree of the polynomial increases, its graph more closely resembles that of the sine function.

Taylor Series and Maclaurin Series

Guidelines for Finding a Taylor Series

1) Differentiate f(x) several times and evaluate each derivative at c.Try to recognize a pattern in these numbers.

2) Use the sequence developed in the first step to form the Taylor coefficients and determine the interval of

convergence for the resulting power series.

( )( ), '( ), "( ), "'( ),..., ( ),...nf c f c f c f c f c

( ) ( )

!

n

n

f ca

n

3) Within this interval of convergence, determine whether or not the series converges to f(x).

Ex. 3 (You try) Find the Taylor series for centered at c = 2.( ) xf x e

Since e is always its own derivative, ( ) 2(2)nf e( )

0

(2)( 2)

!

nn

n

fx

n

2

0

( 2)!

n

n

ex

n

Find the interval of convergence using the Ratio Test:

2 1

2

( 2) !lim

( 1)! ( 2)

n

nn

e x n

n e x

( 2)

lim1n

x

n

0 1

( , ) Interval of convergence:

1 lim ( ) lim ( 2) 0?( 1)!

xn

nn n

eR x x

n

10 ( ) ( 2)( 1)!

xn

n

eR x x

n

Ratio Test?

lim ( ) 0nn

R x

Therefore the Taylor series for   ( )  at x 2 

converges to the function  ( )   for all x.       

x

x

f x e

f x e

1( 2) lim

( 1)!

nx

n

xe

n

To show convergence for all x, we need numbers between 2 and x such that:

2

0

( 2) ?!

x n

n

ee x

n

Deriving Taylor Series from a Basic List

Ex. 4 Find the Maclaurin series for 2( ) sinf x x

Since we already know that Maclaurin series for we can usethat to build our series:

sin x

3 5 7

sin ...3! 5! 7!

x x xx x

6 10 142 2sin( ) ...

3! 5! 7!

x x xx x

( )

0

(0)

!

nn

n

fx

n

2(2 1)

0

( 1)(2 1)!

nn

n

x

n

Ex. 5 Given that find the power series for

( ) cosf x x

2 4 6

cos 1 ...2! 4! 6!

x x xx

cos x

No need to start from scratch…

2 4 6

1 ...2! 4! 6!

x x x

2 3

1 ...2! 4! 6!

x x x

Group-work

Series FRQ Practice Problem

Homework

Pg. 685 1-11 odd, 21-25 odd, 31, 43-46

Day 2 pg. 685 15-19 odd, MMM 220