If the Taylor series for a function is truncated at the term, what is the difference between and the value given by the Taylor polynomial? That is, what is the error involved in using the Taylor polynomial to approximate the function?

Suppose you expand f around c, and that f is -times continuously differentiable on an open interval containing c. Taylor's Theorem with the Remainder Term says that if x is another point in this interval, then

where z is a number in the open interval between x and c.

is the degree Taylor polynomial at
c. The other term on the right is called the *
Lagrange remainder term*:

The appearance of z, a point between x and c, and the fact that it's being plugged into a derivative suggest that there is a connection between this result and the Mean Value Theorem. In fact, for the result says

where z is between x and c. This is the Mean Value Theorem.

On the one hand, this reflects the fact that Taylor's theorem is
proved using a generalization of the Mean Value Theorem. On the other
hand, this shows that you can regard a Taylor expansion as an
*extension* of the Mean Value Theorem.

There is also an expression for the error which involves an integral. I won't discuss it here.

* Example.* Compute the Remainder Term for .

For the *third* remainder term, I need the *fourth*
derivative:

The Remainder Term is

where z is a number between x and 1.

* Example.* Compute the Remainder Term for .

Since I want the Remainder Term, I need to find an expression for the derivative. I'll compute derivative until I see a pattern:

Notice that it's easier to see the pattern if you don't multiply out the power of 4.

Thus,

The Remainder Term is

where z is a number between x and 3.

There are several things you might do with the Remainder Term:

1. Estimate the error in using to estimate on a given interval . (The interval and the degree n are fixed; you want to find the error.)

2. Find the smallest value of n for which approximates to within a given error ("tolerance") on a given interval . (The interval and the error are fixed; you want to find the degree.)

3. Find the largest interval on which approximates to within a given error ("tolerance"). (The degree and the error are fixed; you want to find the interval.)

* Example.* The Maclaurin series for is

What is the largest error which might result from using the first three terms of the series to approximate , if ?

The remainder term is

where . I want to estimate the maximum size of . I take absolute values, because I don't care whether the error is positive or negative, only how large it is.

, and you can check by taking derivatives that . Thus, . So

Since I want the largest possible error, I want to see how large the terms and could be.

Remember that z is between 0 and x, and . So

First, means that

How large can be, given that ? As z goes from 0 to 1, *decreases*, so it is largest if
. This means that

In general, to estimate the z-term you'd have to find the absolute max on the interval for z. If you know that the z-term is either increasing or decreasing, you can check its value at the interval endpoints, and take the largest.

Using the estimates for and , I have

The error is no greater than .

I can check this by plotting the difference between the degree Taylor polynomial and .

From the picture, it looks as though the maximum error is around 0.15 (in absolute value). The estimated error was pretty conservative.

* Example.* (a) Compute for , and express
using and the remainder term.

Since I want , I need the fourth derivative:

Thus,

Now

Therefore,

where z is between 0 and x.

(b) Use to approximate the largest error that occurs in using to approximate for .

I have

I'll estimate the z and x-terms one at a time.

Since , I have

Since and z is between 0 and x, it follows that . On this interval, decreases, so it attains its largest value at . Therefore,

Thus,

The error is no greater than .

* Example.* Find the smallest value of n for
which the degree Taylor series for at approximates on the interval with an
error no greater than .

Notice that

So

First, I'll estimate how large the z and x-terms can be. Since and since is an increasing function, I have

Since and is an increasing function, I have

Thus,

Therefore, I want the smallest n for which

I can't solve this inequality algebraically, so I'll have to use trial-and-error:

The smallest value of n is .

You can also use the Remainder Term to estimate the error in using a Taylor polynomial to approximate an integral.

* Example.* Calvin wants to impress Phoebe
Small by using the MacLaurin series for to approximate to within 0.0001. How many terms of the series should
he use?

The Maclaurin series for is

(Substitute in the standard series for .) I want to know how many terms of the series to use to approximate the integral.

Since , , , and in general, . Therefore,

In the integral, x goes from 0 to 0.5, and z is a number between 0 (the expansion point) and x. Therefore, I know that z is a number between 0 and 0.5. Taking the worst possible case, the largest could be is . Replace with e to obtain

Insert this into the integral (remembering to multiply by x):

I want the smallest value of n for which this ugly mess is less than 0.0001. The easiest way to do this is by trial: Plug in successive values of n until you discover that is the smallest value that works.

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Copyright 2013 by Bruce Ikenaga