Taylor's Remainder Theorem

This is the most important topic of MATH 117/119.

Error of a Taylor Polynomial

What is the error when approximating using a Taylor polynomial?

How do we estimate this error? With repeated Integration by Parts 😳.

Here's what Cauchy did:

Alternatively, using FTC 2:

Note that we assume . If , then the bounds of the integral would be switched.

Either way, we end up with as an and something for the remainder.

IBP:

Let and . Then:

All together:

This puts in terms of and some remainder, which is an exact value for .

Keep integrating by parts:

Let and

All together:

notice that the bracketed terms are nothing but a Taylor polynomial (wtf). We now have in terms of . But the remainder ensures we have an exact value for .

Observation

Each time we run an IBP, we create a new Taylor polynomial term, plus the integral which makes it an exact value.

One more IBP:

Let and
All together (skipping the actual IBP):

If we keep going:

which means that the integral is our error.

Definition

The error of a Taylor polynomial is:

Note that this is heavily related to Connection with FTC.

Taylor's Remainder Theorem

Theorem

The Remainder of a Taylor Polynomial is defined by:

or sometimes:

where is some value between and which we cannot determine.

And we can re-write as:

Taylor's Inequality

Notice, if we can bound the derivative, we can bound the error as well.

If for all and for some constant , then:

Definition

The error in using an -order Taylor polynomial as an approximation to satisfies the inequality:

For all and for all , if , then:

Bounding the Error

Example

Let . Find , the second degree Taylor Polynomial centred at . Then, find the upper bound of the error using to approximate .

solution

As such,

To bound our error, must look for the remainder, first by finding the third derivative.

Using the Taylor Inequality:

Now, we must find , which is defined by .

we seek to find a value for which will yield the largest number on the left side (since we seek the upper bound). We chose :

Plugging this back into our Taylor inequality:

This means the error when using to approximate will be no more than in magnitude.

Example

What is our error when approximating

solution
We know:

If we stop at terms, the error estimate is:

Now, we must find , which is defined by . We know , so we have . Now, we want to pick a value of that will yield the greatest value for the left side of the equation. In this case, it would be .

Therefore, we can say , so . But we can't bound with , otherwise it's pointless, so we can pick some number greater than : .

As such, we expect our Taylor polynomial to be approximately with error:

at . This means using our Taylor polynomial, we fill find with an error of magnitude no more than .

More generally, for any :

also written as

where

Example

Lets flip it around. How many terms are needed to estimate to ?

solution
We know:

and setting means:

which is

Listing out our options for :

so the number that satisfies the inequality would be (since .

Example

Estimate to decimal places ()

solution
We have

In this example, we have centred at (remember, is the centre of our approximation, is where we wish to use it).

Listing out our derivatives:

we find that

We wish to bound it for .

Biggest at .

Since is our final term, we have (recall from the definition of the remainder), and obtain:

Trying values for :

we find satisfies the desired inequality, thus should use .

Obtaining our Taylor polynomial:

Actual value:

The error is , which is less than the maximum-allowed error of .

With Big-O

In Conjunction with Taylor's Inequality

This is where things get crazy

Recall the definition of a Taylor polynomial:

Recall Taylor's inequality:

for some for all .

Using Big O notation we can easily write the Taylor remainder theorem:

Definition

We can rearrange:

Why can we do this?

Example

Remember, we seek the highest allowable exponent

Example