Section 9.7 – Taylor Theorem

Taylor’s Theorem

If is a function with continuous derivatives on (or ), then there is a number in (or ) such that

2 1( ) ( 1)

...2! ! 1 !

n nn ng a x a g a x a g c x ag x g a g a x a

n n

Like all of the “Value Theorems,” this is an existence theorem.

The approximate value of using the Taylor Polynomial centered at

The actual value of

An unknown function of because a derivative is evaluated at some point

in the interval.

Taylor’s Theorem Re-Worded

If is the nth degree Taylor Polynomial for at , then

where is the “remainder.”

Technical Note:

The amount needed for the approximate to be equal to the actual.

The approximate value.

The actual value.

Example 1Let , express in terms of a third-degree Taylor polynomial centered at and the remainder.

𝑇 3 (𝑥 )=1+𝑥+𝑥2

2 !+𝑥

3

3 !

𝑔 (2 )=𝑇 3 (2 )+𝑅3 (2 )

𝑒2=1+2+ 22

2 !+ 2

3

3 !+𝑅3 (2 )

Find the third degree Taylor polynomial:

Use the Taylor Theorem Equation:

We can calculate this value since we know the function. But what if we did not know the function?

7.389=6.333+𝑅3 (2 )

The Remainder in Taylor’s Theorem

Also known as the Lagrange Form of the Remainder, for some in the interval .

1( 1)

1 !

nn

n

g c x aR x

n

According to Taylor’s Theorem, the remainder (or error) is one degree higher than the Taylor Polynomial used to approximate the actual value.

Remember, this is the same result from power zooming!

Example 1 ContinuedLet , express in terms of a third-degree Taylor polynomial centered at and the remainder.

7.389=6.333+𝑅3 (2 )

𝑔(𝑛+1) (𝑐 ) (𝑥−𝑎)𝑛+1

(𝑛+1 ) !

7.389=6.333+𝑔 (3+1 ) (𝑐 ) (2−0 )3+1

(3+1 )!

7.389=6.333+𝑒𝑐24

4 !

This equation is true for some

value on 1.0557=𝑒𝑐24

4 !𝑐=0 .45969 Notice

What would happen if we could find the

Taylor polynomial but didn’t know the

function and could not

calculate this value?

Example 1 ContinuedLet , express in terms of a third-degree Taylor polynomial centered at and the remainder.

𝑔 (2 )=6.333+𝑅3 (2 )

𝑔(𝑛+1) (𝑐 ) (𝑥−𝑎)𝑛+1

(𝑛+1 ) !

𝑔 (2 )=6.333+𝑔(3+1 ) (𝑐 ) (2−0 )3+1

(3+1 ) !

𝑔 (2 )=6.333+𝑒𝑐 24

4 !

Seek Comfort in the Familiar: Taylor’s Theorem, like the Intermediate Value, Extreme Value, and Mean Value Theorem is an existence theorem.

All Taylor’s Theorem says is this equation

MUST be true for SOME value on

We can usually NOT

find the value for .

Example 1 ContinuedLet , express in terms of a third-degree Taylor polynomial centered at and the remainder.

𝑔 (2 )=6.333+𝑒𝑐 24

4 !

Instead of actually calculating the remainder (or error). The best we can usually do is find a bound for the error. In other words, something the error

is always less than. The worst case scenario for your approximation.

Try to find the maximum value of the error on

is strictly increasing on .

Thus reaches a maximum when .

Therefore OR

The Lagrange Error Bound

One useful consequence of Taylor’s Theorem is that:

1

1 !

n

n

M x aR x

n

The Error

The ABSOLUTE VALUE of the maximum value of on

Use the absolute value to

keep things positive when

calculating error.

Example 2Determine the degree of the Maclaurin polynomial in order to approximate the square root of to within 0.0001.

We want to use to approximate .Investigate the error function on the interval :

𝑅𝑛 (𝑥 )=𝑔 (𝑛+1 ) (𝑐 ) (𝑥−𝑎 )𝑛+1

(𝑛+1 ) !

𝑅𝑛 (0.5 )=𝑒𝑐 (0.5−0 )𝑛+1

(𝑛+1 ) !

0.0001≥𝑒0.5 (0.5 )𝑛+1

(𝑛+1 )!

The derivative is always

We are trying to

approximate , so .

The error has to be less than

Use a calculator to find a value for that

satisfies the inequality.

n Max Error

0 0.82436

1 0.20609

2 0.03435

3 0.00429

4 0.00043

5 0.00004

5th Degree Polynomial

The series is centered at 0 so

Worst case scenario:

Example 3If the third degree Maclaurin polynomial for is used to approximate , determine the accuracy of the approximation.

We want to use to approximate .

Investigate the error function on the interval :

𝑅𝑛 (𝑥 )=𝑔 (𝑛+1 ) (𝑐 ) (𝑥−𝑎 )𝑛+1

(𝑛+1 ) !

𝑅3 (0.1 )=𝑔 (3+1) (𝑐 ) (0.1−0 )3+ 1

(3+1 ) !

|𝑅3 (0.1 )|< 1∙0.14

4 !Or

We are trying to approximate , so . Using the 3rd degree

polynomial, so .

Worst case scenario is that the derivative equals

Extension: Taylor Theorem and The Mean Value Theorem

Let in Taylor’s Theorem for some in :

'f b f a f c b a

'f b f a

f cb a

Taylor Series is 1 term long

Remainder

Mean Value Theorem is a special case of Taylors Theorem.

Solve for the derivative