SECTION 5.4 Polynomial Approximations 301
The above remainder (i.e., error term) is called theLagrange form
of the error.
We’ll conclude this subsection with some examples.
Example 1. As a warm-up, let’s prove that the Maclaurin series for
cosxactually converges tof(x) = cosxfor allx. We have, byTaylor’s
Theorem with Remainder, that
cosx= 1−
x^2
2!
+
x^4
4!
−···±
x^2 n
(2n!)
+
f(2n+1)(c)
(2n+ 1)!
x^2 n+1,
for some real numbercbetween 0 andx. Since all derivatives of cosx
are±sinxor±cosx, we see that
∣∣
∣∣
∣f
(2n+1)(c)
∣∣
∣∣
∣≤1. This means that for
fixedx, if we letn→∞, then the remainder
f(2n+1)(c)
(2n+ 1)!
x^2 n+1→ 0 ,
proving that
cosx = 1−
x^2
2!
+
x^4
4!
−···=
∑∞
n=0
(−1)n
x^2 n
(2n)!
.
Example 2. Here’s a similar example. ByTaylor’s Theorem with
Remainder, we have forf(x) = ln(1 +x), that
ln(1 +x) =x−
x^2
2
+
x^3
3
−···±
xn
n
+
f(n+1)(c)
(n+ 1)!
xn+1,
for some real numbercbetween 0 andx. It is easy to verify that the
Maclaurin series for ln(1+x) has interval of convergence− 1 < x≤1, so
we need to insist thatxis in this interval. Sincef(n+1)(c) =
n!
(1 +c)n+1
,
we see that the error term satisfies
∣∣
∣∣
∣
f(n+1)(c)
(n+ 1)!
xn+1
∣∣
∣∣
∣=
∣∣
∣∣
∣
n!xn+1
(1 +c)n+1(n+ 1)!
∣∣
∣∣
∣=
∣∣
∣∣
∣
xn+1
(1 +c)n+1(n+ 1)
∣∣
∣∣
∣