The * Taylor series* for at is

(By convention, .) When , the series is called a * Maclaurin series*.

You can construct the series on the right provided that f is infinitely differentiable on an interval containing c. You already know how to determine the interval of convergence of the series. However, the fact that the series converges at x does not imply that the series converges to .

* Example.* The function

is infinitely differentiable everywhere. In particular, all the derivatives of f at 0 vanish, and the Maclaurin series for f is identically 0.

Hence, the Maclaurin series for f converges for all x, but only converges to at .

The following result ([1], page 418) gives a sufficient condition for the Taylor series of a function to converge to the function:

* Theorem.* Let be infinitely
differentiable on , and let . Suppose
there is a constant M such that for all
, and for all x in , where N is a
neighborhood of c. Then for all ,

In other words, under reasonable conditions:

- You can construct a Taylor series by computing the derivatives of f.

- The series will converge to f on an interval around the expansion point. (You can find the interval of convergence as usual.)

* Example.* Find the first four nonzero terms
and the general term of the Taylor series for at
and at by computing the derivatives of f.

For , for all n. The Taylor series at is

For , for all n. The Taylor series at is

* Terminology.* If you truncate the series
after the -degree term, what's left is the -* degree Taylor polynomial*. For
example, the third degree polynomial of at
is

Note that the "n" here refers to the largest *power of
x*, not the number of terms. For example, the Taylor series for at is

The degree Taylor polynomial and the degree Taylor polynomial are equal:

* Example.* Find the degree Taylor
polynomial for at .

Thus,

The degree Taylor polynomial is

It's tedious to have to compute lots of derivatives, and in many cases you can derive a series from another, known series. Here are the series expansions for several important functions:

* Example.* Find the Taylor series for
at . What is its interval of convergence?

Use

I'm expanding at , so I want the result to come out in powers of . This is easy --- just set :

The u-series converges for , so the x-series converges for , or .

* Example.* The quantity occurs in special relativity. (v
is the velocity of an object, and c is the speed of light.)
Approximate
using the first two nonzero terms of the binomial series.

so for ,

Take :

The approximation is good as long as v is small compared to c.

* Example.* Find the Taylor series for at . What is its interval of convergence?

I want things to come out in powers of , so I'll write the function in terms of :

I'll use the series for . To do this, I need on the bottom:

Let in the series for . Then

Hence,

The u-series converges for , so the x-series converges for , or .

* Example.* Find the Taylor series for at .

Since I'm expanding at , the answer should have the form

where the b's are numbers. That is, the answer must come out in terms of powers of .

*Start with the function you're trying to expand.* To get
's in the answer, write the given function in terms of :

(Notice that the work has to be legal algebra.)

I'll break up the fraction and do the pieces separately.

I want to "match" each piece against the standard series . Here's the first piece:

Expand by setting in :

Here's the second piece:

That is,

* Example.* What is the Maclaurin series for
? What is the Taylor series for at ?

The Maclaurin series for a polynomial is the polynomial: .

To obtain the Taylor expansion at , write the function in terms of :

It's also possible to construct power series by integrating or
differentiating other power series. *A power series may be
integrated or differentiated term-by-term in the interior of its
interval of convergence.* (You will need to check convergence at
the endpoints separately.)

* Example.* The Maclaurin series for is

(Put in the series for .) It converges for .

Integrate the series from 0 to u:

This series will converge for . The left side blows up at . On the other hand, if ,

The right side *does* converges (by the Alternating Series
Test), so the series converges for .

* Example.* Find the Taylor series for at .

I'll use the fact that a Taylor series can be integrated term-by-term on the interval where it converges absolutely.

(I integrated from 2 to x because I want the expansion at .) Now find the series at for :

Plug this series back into the integral and integrate term-by-term:

* Example.* Find for .

The degree term is . On the other hand, Taylor's formula says that the degree term is . Equating the coefficients, I get

[1] Tom M. Apostol, *Mathematical Analysis*. Reading,
Massachusetts: Addision-Wesley Publishing Company, Inc., 1957.

Copyright 2005 by Bruce Ikenaga