Taylor Series and Taylor’s Theorem

When is a function given by its Taylor Series?

So where were we?

21

e if 0( )0 if 0

x xf xx

Facts:• f is continuous and has derivatives of all orders at x = 0.• f (n)(0)=0 for all n.

This tells us that the Maclaurin Series for f is zero everywhere!

The Maclaurin Series for f converges everywhere, but is equal to f only at x = 0!

This tells us that….

Our ability write down a Taylor series for a function is not in itself a guarantee that the series will any anything to do with the function, even on its interval of convergence!

However, Ostebee and Zorn assures that….

“Taylor’s theorem guarantees that this unfortunate event seldom occurs.”

In other words, the functions that are not given by their Taylor series are pretty weird. Most of our “everyday” functions ARE given by their Taylor Series.

Recall Taylor’s TheoremSuppose that f is repeatedly differentiable on an interval I containing x0 and that

is the nth order Taylor polynomial based at x0. Suppose that Kn+1 is a number such that for all z in I,

Then for x in I,

(Page 504 in OZ)

2 30 1 0 2 0 3 0 0( ) ( ) ( ) ( ) ( )n

n nP x a a x x a x x a x x a x x

( 1)1| ( ) | .n

nf z K

11

0| ( ) ( ) |1 !

nnn

Kf x P x x xn

Pinning this down

Recall that Pn is the nth partial sum of theTaylor Series of f based at x0.

And thus

Measures the error made by Pn(x) in approximating f (x). Taylor’s theorem gives us an upper bound on this error!

| ( ) ( ) |nf x P x

The Taylor series for f will converge to f if and only if for all x| f (x) - Pn(x) |

goes to zero as n →∞. Taylor’s theorem can help us establish this.

Using Taylor’s Theorem

( ) sin( )f x x

1. Find the Taylor series for f that is based at x = /4.2. Show that this Taylor series converges to f for all values of x.

1. Taylor Series for f (x) = sin(x)n f (n)(x) f (n)( ) an= f (n)( )/n!

0

1

2

3

4

sin( )x

cos( )x

sin( )x

cos( )x

sin( )x

4

11 1 1

1!2 2a

1sin 4 2

1sin 4 2

1cos 4 2

1cos 4 2

1sin 4 2

4

01 1 1

0!2 2a

21 1

2!2a

31 1

3!2a

41 1

4!2a

2 3 4 51 1 1 1 1 1 1 1 1 14 4 4 4 42 3! 4! 5!2 2 2 2 2 2

x x x x x

We start with the general set-up for Taylor’s Theorem.

What is Kn+1?

It follows that for all x

Show that this converges to sin(x)

11sin( ) ( ) 41 !

n

nx P x xn

What happensto this quantity

As n→∞?

1

1 sin( ) 1 for all and all .n

n

d x n xdx

1

41 !

nx

n

We start with the general set-up for Taylor’s Theorem.

What is Kn+1?

It follows that for all x

Show that this converges to sin(x)

11sin( ) ( ) 41 !

n

nx P x xn

1

1 sin( ) 1 for all and all .n

n

d x n xdx

1

41 !

nx

n

Notice that I didn’t have to know what Pn was in order to gather this information. (In

other words, our second question is independent of our

first.)

Now it’s your turn

Repeat this exercise with the Maclaurin series for f (x) = cos(2x) .

1. Find the Maclaurin series for f (x) = cos(2x).2. Show that this series converges to f for all values of x.

1. Taylor Series for f (x) = cos(2x)

n f (n)(x) f (n)(0) an= f (n)(0)/n!

0

1

2

3

4

2sin(2 )x

22 cos(2 )x

32 sin(2 )x

42 cos(2 )x

1 0a

2 22 cos 0 2

4 42 cos(0) 2

2sin 0 0

32 sin(0) 0

cos 0 1 0 1a

2

222!

a

3 0a 4

424!

a

4 62 4 62 2 21

2! 4! 6!x x x

cos(2 )x

2

2

0

( 1) 22 !

n nn

n

xn

We start with the general set-up for Taylor’s Theorem.

What is Kn+1?

It follows that for all x,

Show that this converges to cos(2x)

11cos(2 ) ( )

1 !nn

nKx P x xn

1

12cos(2 ) ( )1 !

nn

nx P x xn

This quantity goes

to 0 as n→∞!

11

1 cos(2 ) 2 for all and all .n

nn

d x n xdx

12=

1 !

nxn

Epilogue---Two points of view

Power series as functions

00

( )nn

n

a x x

00

( ) ( )nn

n

f x a x x

First a series . . .

. . . Then a function

Taylor Series

( )f x

( )0

00

( ) ( )!

nn

n

f x x xn

First a function . . .

. . . Then a series

Guarantees that f is equal to the power series where the power series converges.

No a priori guarantee that f is equal to its Taylor series.

Why the Taylor series, then?

Power series as functions

00

( )nn

n

a x x

00

( ) ( )nn

n

f x a x x

First a series . . .

. . . Then a function

Taylor Series

( )f x

( )0

00

( ) ( )!

nn

n

f x x xn

First a function . . .

. . . Then a series

Guarantees that the power series we started with is, in fact, the TAYLOR SERIES FOR f .

If f is equal to any power series at all, that power series must be the Taylor series for f. That’s why that’s were we look!