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Taylor Polynomials: The Lagrange Error Bound
Louis A. Talman
Department of Mathematical & Computer Sciences
Metropolitan State College of Denver
February 21, 2005
Revisions April 14, 2008
In order to understand the rˆole played by the Lagrange remainder and the Lagrange error
bound in the study of power series, let’s carry the standard examination of the geometric
series a little farther than is usually done. That standard examination usually ends with
the observation that
∑^ n
k=
x
k
1 − x
n+
1 − x
and so the sums on the left side of (1) converge to f (x) = 1/(1 − x) iff |x| < 1. Of course,
this conclusion arises from the fact that the limit on the right-hand side exists iff |x| < 1.
But (1) contains other information from which we can learn a great deal more about the
convergence of the geometric series.
Let us begin by rewriting (1) as
∑^ n
k=
x
k
1 − x
x
n+
1 − x
This latter equation can be rewritten in two different ways:
1 − x
∑^ n
k=
x
k
x
n+
1 − x
, and (2)
1 − x
∑^ n
k=
x
k
x
n+
1 − x
Notice that equation (2) has the form
f (x) =
∑^ n
k=
akx
k
This is exactly the form that we see in Taylor’s formula with Lagrange remainder, which
we will state very soon. But let us defer thoughts connected with this observation for a
while.
It is equation (3) that we want to deal with now. Let h be any number with 0 < h < 1. If
x ∈ [−h, h], then x ≤ h < 1, so | 1 − x| = 1 − x ≥ 1 − h. This guarantees that
| 1 − x|
1 − x
1 − h
Consequently, we may write, from (3) and (5),
1 − x
∑^ n
k=
x
k
x
n+
1 − x
h
n+
1 − h
no matter what the number x ∈ [−h, h] may be. Thus, on any interval of the form [−h, h],
where 0 < h < 1, the values of the polynomial
n k=0 x
k approximate the values of the
function f : x 7 → 1 /(1 − x) uniformly well.
This turns out to be an extremely important piece of information. It has among its conse-
quences the fact that if 0 < h < 1 the definite integrals of the polynomials 1 + x + · · · + x
n
on intervals [x 1 , x 2 ] ⊆ [−h, h] approximate the definite integrals of f well.
Now let’s note that
d
dx
1 − x
(1 − x)^2
An analysis very similar to the one above, but of the polynomials
d
dx
∑^ n
k=
x
k
∑^ n
k=
kx
k− 1 , (9)
yields—when 0 < h < 1—the inequality
∣ ∣ ∣ ∣ ∣
(1 − x)^2
∑^ n
k=
kx
k− 1
h
n [(1 + h)n + 1]
(1 − h)^2
valid for every x ∈ [−h, h]. The fraction on the right side of (10) goes to zero as n grows
without bound (use L’Hˆopital’s rule together with the fact that |h| < 1 to see this), and
so the derivatives of the polynomials
∑n
k=0 x
k approximate the derivative of the function
f uniformly well on any interval of the form [−h, h], where 0 < h < 1. In other words,
