Albo

Chapter Nine

The Taylor Polynomial
9.1 Introduction
Let f be a function and let F be a collection of "nice" functions. The approximation
problem is simply to find a function g ∈ F that is "close" to the given function f . There are
two issues immediately. How is the collection F selected, and what do we mean by
"close"? The answers depend on the problem at hand. Presumably we want to dosomething to f that is difficult or impossible (This might be something as simple as finding
f ( x ) for some x . ). The collection F would thus consist of functions to which it is easy to

do that which we wish to do to f . Our measure of how close one function is to another
would try to reflect the closeness of the results of our operations. Now, what are we
talking about here. Suppose, forexample, we wish to find

f ( x ) . Our collection F of

functions should include functions that are easy to evaluate at x , and two function would
be "close" simply if there values are close. We might, for instance, want to evaluate sin x
for all x i s some interval I. The collection F could be a collection of second degree
polynomials. The approximation problem is then to find elementsof F that make the
"distance" max{|sin x − p (x )|: x ∈ I } as small as possible. Similarly, we might want to find
the integral of some function f over an interval I . Here we would want F to consist of
functions easily integrated and measure the distance between functions by the difference of
their integrals over I . In the previous chapter, we found the "best" straight line
approximation toa set of data points. In that case, the collection F consisted of all
nonvertical straight lines, and we measured the distance between functions by the sum of
the squares of their differences on a specified set of points { x1 , x2 ,K , x n } .

You can

imagine many other examples.
9.2 The Taylor Polynomial
We look first at a simple but useful problem: Given a nice function f : D ⊂ R → R ,a
point a in the interior of the domain D , and an integer n , find a polynomial p of degree
≤ n such that

9.1

p (a ) = f (a )
p '( a ) = f '( a )
p ''( a ) = f ' ' ( a )
M
p ( n ) (a ) = f

( n)

(a )

We solve the problem by the Behold Method. Simply verify that

p ( x ) = f ( a ) + f '( a )( x − a ) +

f ''( a )
f '''( a )
f ( n ) (a )
( x − a) 2 +
( x − a ) 3 +K+(x − a ) n
2!
3!
n!

does the job! It is also fairly easy to see that this polynomial is the only polynomial of
degree ≤ n that does the job. Suppose q is also a polynomial with degree g ≤ n such that
p (a ) = f (a )
p '( a ) = f '( a )
p ''( a ) = f ' ' ( a )
M
p ( n ) (a ) = f

( n)

(a )

and consider the function r = p − q . Note that r is also a polynomial of degree ≤ n . Butr ( a ) = r '( a ) = r ''( a ) =K = r ( n ) ( a ) = 0 .

Or, in other words, r has a zero of order n + 1 , and the only way this can happen is if
r ( x ) ≡ 0 for all x . That is, p ( x ) ≡ q ( x ) identically.

Example
Let f (x ) = sin x and let a = 0 . Let's find the Taylor polynomial for a few different
values of n. For n = 1, we have simply p1 ( x) = f (a ) + f '( a )( x − a ) = sin 0 +cos 0( x) = x .
Note that for n = 2, we have p 2 ( x) = sin 0 + cos0 (x ) − sin 0( x 2 ) = x , also. Let's take a look
at the next Taylor polynomial. Here p 3 (x ) = x −
at the graph of p3 and f . We shall use Maple.

9.2

x3
. Let's draw some pictures; we'll look
6

What we see is that the Taylor polynomial looks like a pretty good approximation as long
as we don't get too far awayfrom a = 0. Let us continue. Convince yourself that p 4 = p 3 ,
and p 5 ( x) = x −

x3
x5
+
. Another picture:
6 120

9.3

Exercises
1 . Find the Taylor polynomial of degree n for f (x ) = e x at a = 0.
2 . Find the Taylor polynomial of degree n for f (x ) = x 3 at a = 1.
3 . Find the Taylor polynomial of degree 3 for f (x ) = log x at a = 1.
4 . Find the Taylor polynomial of...