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)nx^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)∣∣
∣∣
∣