LECTURE 18: TAYLOR POLYNOMIALS 2
Definition. We say functions f, g : I → R are equal to order n at a ∈ I if
f (x) − g(x)
= 0,
lim
x→a (x − a)n
In this language, the main theorem of the previous lecture can be stated thus:
Theorem. If f : I → R is n-times differentiable at a ∈ I, then f and Pn,a are
equal to order n.
In fact, Pn,a is the unique polynomial of degree ≤ n with this property. To see
this we need the following theorem.
Theorem. Let P and Q be polynomials in (x − a), both of degree ≤ n and suppose
P and Q are equal to order n. Then P = Q.
Note that P is a polynomial in (x − a) of degree ≤ n if for some choice of
a0 , a1 , . . . , an ∈ R one has
n
X
P (x) =
aj (x − a)j .
j=0
Corollary. If f : I → R is n-times differentiable at a ∈ I and P is a polynomial of
degree ≤ n which equals f to order n, then P = Pn,a,f is a Taylor polynomial.
The corollary is in fact useful for computing the Taylor polynomial of certain
functions.
Example. Consider the Taylor polynomial of
Z x
dt
arctan x :=
2
0 1+t
of degree n about 0. By direct computation we have
1
arctan0 (x) =
arctan0 (0) = 1,
1 + x2
−2x
arctan00 (x) =
arctan00 (0) = 0,
(1 + x2 )2
−2(1 + x2 )2 + 8x2 (1 + x2 )
arctan000 (0) = −2.
arctan000 (x) =
(1 + x2 )4
It is difficult to see a pattern emerging from these derivatives and so ostensibly
the computation of the coefficients of the Taylor polynomials appears to be a hard
problem. However, we can get around this problem using the corollary. First note
that one may express 1/(1 + t2 ) as a polynomial of degree n plus a remainder term:
1
t2n+2
= 1 − t2 + t4 − t6 + · · · + (−1)n t2n + (−1)n+1
.
2
1+t
1 + t2
Thus,
x
x
t2n+2
dt
2
0 1+t
0
Z x 2n+2
x3
x5
x7
x2n+1
t
=x−
+
−
+ · · · + (−1)n
+ (−1)n+1
dt.
2
3
5
7
2n + 1
0 1+t
Z
arctan(x) =
1 − t2 + t4 − t6 + · · · + (−1)n t2n dt + (−1)n+1
1
Z
2
LECTURE 18: TAYLOR POLYNOMIALS 2
Let
x3
x5
x7
x2n+1
+
−
+ · · · + (−1)n
3
5
7
2n + 1
so that, by the Corollary, if P and arctan are equal to order 2n + 1 at 0, then
P = P2n+1,0 . To show this is the case we are required to prove that
Z x 2n+2
1
t
lim
dt = 0,
x→0 |x|2n+1 0 1 + t2
but since
Z x t2n+2 Z x
|x|2n+3
2n+2
t
dt
=
≤
dt
,
2
2n + 3
0
0 1+t
this clearly holds.
As an application of this result, note that one can work backwards by comparing
the coefficients of P and Pn,a (which, of course, agree) to compute the values of the
derivatives of arctan at 0.
P (x) := x −
Jonathan Hickman, Department of mathematics, University of Chicago, 5734 S. University Avenue, Eckhart hall Room 414, Chicago, Illinois, 60637.
E-mail address: [email protected]