# Taylor Series

A little examination using derivatives brings the following conclusion:

If f has a power series representation at a, that is,

then

.

In other words,

.

This series is called the Taylor series of f at a. In the special case where a=0, this is called the Maclaurin series. Of course, the statement "if f has a power series representation" is an important one. For some functions, you can create the above series, but it will not converge to the function value. Functions with a power series representation at a point a are called analytic at a.

In order for f to be analytic, we want f to equal its power series, or, in other words, to be the limit of the sequence of partial sums. The partial sums are

.

The term T[n](x) is called the nth-degree Taylor polynomial of f at a. We want f(x)=lim T[n](x), or we want R[n](x) to go to zero as n gets large, where R[n](x)=f(x)-T[n](x).

For instance, since we know that

,

the 5th degree Taylor polynomial of 1/(1-x) is

.

To help us determine analyticity(?), we have the following theorem:

If f(x)=T[n](x)+R[n](x), where T[n](x) is the nth-degree Taylor polynomial of f at a, and the limit of R[n](x) as n goes to infinity is 0 for |x-a|<R, then f is equal to the sum of its Taylor series on the interval |x-a|<R, that is, f is analytic at a.

A helpful result for all this is the following:

If f has n+1 derivatives in an inteval I that contains the number a, then for x in I there is a number z strictly between x and a such that the remainder term in the Taylor series can be expressed as

.

This makes finding the limit of R[n](x) much easier. Remember, that z in this formula depends on x; namely, it must be between a and x.