Barrons AP Calculus - David Bock

For x > 0, this is an alternating series with terms decreasing in magnitude and approaching 0, so

the error committed by using the first two terms is less than If then the given

approximation formula will yield accuracy to three decimal places. We therefore require that |x|^3 <

0.0015 or that |x| < 0.114.

C5. Taylor’s Formula with Remainder; Lagrange Error Bound.

When we approximate a function using a Taylor polynomial, it is important to know how large the
remainder (error) may be. If at the desired value of x the Taylor series is alternating, this issue is
easily resolved: the first omitted term serves as an upper bound on the error. However, when the
approximation involves a nonnegative Taylor series, placing an upper bound on the error is more
difficult. This issue is resolved by the Lagrange remainder.

TAYLOR’S THEOREM. If a function f and its first (n + 1) derivatives are continuous on the interval |x
− a| < r, then for each x in this interval

where

and c is some number between a and x. Rn (x) is called the Lagrange remainder.
Note that the equation above expresses f (x) as the sum of the Taylor polynomial Pn (x) and the
error that results when that polynomial is used as an approximation for f (x).
When we truncate a series after the (n + 1)st term, we can compute the error bound Rn, according
to Lagrange, if we know what to substitute for c. In practice we find, not Rn exactly, but only an upper
bound for it by assigning to c the value between a and x that determines the largest possible value of
Rn. Hence:

the Lagrange error bound.

EXAMPLE 48

Estimate the error in using the Maclaurin series generated by ex to approximate the value of e.

SOLUTION: From Example 40 we know that f (x) = ex generates the Maclaurin series

The Lagrange error bound is

To estimate e, we use x = 1. For 0 < c < 1, the maximum value of ec is e. Thus: