9.3 Taylor’s Theorem: Error Analysis for Series

Tacoma Narrows Bridge: November 7, 1940

Last time in BC…

...4

)1(3

)1(2

)1()1(ln)(432

xxxxxxf

1

1 )1()1(n

nn

nx

So the Taylor Series for ln x centered at x = 1 is given by…

Use the first two terms of the Taylor Series for ln x centered at x = 1 to approximate:

21ln

23ln

2)15.1()15.1(

2

2)15.0()15.0(

2

375.0225.05.0

625.0225.05.0

068147.0|)625.0()5.0ln(| erroractual

0417.03

)15.0( 3

error

Recall that the Taylor Series for ln x centered at x = 1 is given by…

Find the maximum error bound for each approximation.

21ln

23ln 0417.0

3)15.1( 3

error

03047.0|625.0)5.1ln(| erroractual

Wait! How is the actual error bigger than the error bound for ln 0.5?

Because the series is alternating, we can start with…

f (x) ln x (x 1) (x 1)2

2

(x 1)3

3

(x 1)4

4... ( 1)n1 (x 1)n

nn1

And now, the exciting conclusion of Chapter 9…

Since each term of a convergent alternating series moves the partial sum a little closer to the limit:

This is also a good tool to remember because it is easier than the Lagrange Error Bound…which you’ll find out about soon enough…

Muhahahahahahaaa!

Alternating Series Estimation TheoremFor a convergent alternating series, the truncation error is less than the first missing term, and is the same sign as that term.

Taylor’s Theorem with Remainder

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 I:

2

2! !

nn

nf fa af f f Rx a a x a x a x a x

n

Lagrange Error Bound

11

1 !

nn

n

f cR x x a

n

In this case, c is the number between x and a that will give us the largest result for )(xRn

11

1 !

nn

n

f cR x x a

n

This remainder term is just like the Alternating Series error (note

that it uses the n + 1 term) except for the cf n 1

If our Taylor Series had alternating terms:

Does any part of this look familiar?

If our Taylor Series did not have alternating terms:

nn

n axn

afxR )()!1()()(

1

11

1 !

nn

n

f cR x x a

n

This is just the next term of the series which is all we need if it is an Alternating Series

is the part that makes the Lagrange Error Bound more complicated.

Note that working with cf n 1

Taylor’s Theorem with Remainder

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 I:

2

2! !

nn

nf fa af f f Rx a a x a x a x a x

n

Lagrange Error Bound

11

1 !

nn

n

f cR x x a

n

Now let’s go back to our last problem…

Why this is the case involves a mind-bending proof so we just won’t do it here.

068147.0|)625.0()5.0ln(| erroractual

0417.03

)15.0( 3

error

Recall that the Taylor Series for ln x centered at x = 1 is given by…

Find the maximum error bound for each approximation.

21ln

23ln 0417.0

3)15.1( 3

error

03047.0|625.0)5.1ln(| erroractual

Wait! How is the actual error bigger than the error bound for ln 0.5?

Because the series is alternating, we can start with…

1

1432 )1()1(...

4)1(

3)1(

2)1()1(ln

n

nn

nxxxxxx

First of all, when plugging in ½ for x, what happens to your series?

1

132 )5.0()1(...

3)15.0(

2)15.0()15.0(

21ln

n

nn

n

Note that when x = ½, the series is no longer alternating.

So now what do we do?

Since the Remainder Term will work for any Taylor Series, we’ll have to use it to find our error bound in this case

1

1432 )1()1(...

4)1(

3)1(

2)1()1(ln

n

nn

nxxxxxx

Recall that the Taylor Series for ln x centered at x = 1 is given by…

1

)5.0(...3

0625.0225.05.0

21ln

n

n

n

xxf 1)( 1

11)1( f

2

1)(x

xf 1

11)1( 2 f

3

2)(x

xf 33 22)( c

ccf

!3cf

2c 3

3!

xxf ln)( 01ln)1( f

xxf ln)( The Taylor Series for centered at x = 1

Since we used terms up through n = 2, we will need to go to n = 3 to find our Remainder Term(error bound):

This is the part of the error bound formula that we need

The third derivative gives us this coefficient:

We saw that plugging in ½ for x makes each term of the series positive and therefore it is no longer an alternating series. So we need to use the Remainder Term which is also called…

21ln 0417.0

3)15.0( 3

error

068147.0|)625.0()5.0ln(| erroractual

3)15.0(!3

)(

cferror )125.0(!3

2 3

c The third derivative

of ln x at x = c

What value of c will give us the maximum error?

Normally, we wouldn’t care about the actual value of c but in this case, we need to find out what value of c will give us the maximum value for 2c–3.

The Lagrange Error Bound 11

)()!1()(

nn

axn

cf

3)15.0(!3

)(

cferror )125.0(!3

2 3

c The third derivative

of ln x at x = c

The question is what value of c between x and a will give us the maximum error?

So we are looking for a number for c between 0.5 and 1.

Let’s rewrite it as

c = 0.5

2c 3 which has its largest value when c is smallest.

And therefore…

3)15.0(!3

)(

cferror )125.0(!3

2 3

c

2(0.5) 3

3!(0.125)

81

616

31

068147.0|)625.0()5.0ln(| erroractual

Which is larger than the actual error!

And we always want the error bound to be larger than the actual error

Let’s try using Lagrange on an alternating series

ln(1 x) x x 2

2We know that since this is an alternating series, the error bound would be

x 3

3But let’s apply Lagrange (which works on all Taylor Series)…

3

!3)( xcferror

3)1(

2)(x

xf

The third derivative of ln(1+ x) is

error 2(1c) 3

3!x 3

x 3

3(1c)3The value of c that will maximize the error is 0 so…

x 3

3(1c)3 x 3

3Which is the same as the Alternating Series error bound

Lagrange Form of the Remainder:

11

1 !

nn

n

f cR x x a

n

Remainder Estimation Theorem:

If M is the maximum value of on the interval between a and x, then:

1nf x

1

1 !n

nMR x x a

n

Most text books will describe the error bound two ways:

and

Note from the way that it is described above that M is just another way of saying that you have to maximize cf n 1

Remember that the only difference you need to worry about between Alternating Series error and La Grange is finding cf n 1