This preview shows page 1. Sign up to view the full content.
Unformatted text preview: imit:
Theorem.
Let x0 ∈ R and let M > 0 be any constant. Then
M  x0 k
=0
k→∞
k!
lim Example.
Let f (x) = cos(x) and a = 0. Let x0 be any point in R. Taylor’s Theorem shows that for
each k there exists a point ck between 0 and x0 such that
 Rk,a (x0 ) = f (k+1) (ck ) k+1
x

(k + 1)! 0 We have seen that if f (x) = cos(x), then f (x) = − sin(x), f (x) = − cos(x), f (x) =
sin(x) and f (4) (x) = cos(x). Since the fourth derivative is again cos(x), the 5th, 6th, 7th and 8th derivative will be, respectively, f (5) (x) = − sin(x), f (6) (x) = − cos(x),
f (7) (x) = sin(x) and f (8) (x) = cos(x). The pattern will then be repeated for the 9th,
10th, 11th and 12th derivatives, and then for every group of four derivatives thereafter.
In fact, what we have just shown is that if f (x) = cos(x), then for any k f (k) (x) = if k = 4j cos(x) − sin(x) if k = 4j + 1 . − cos(x) if k = 4j + 2 sin(x)
if k = 4j + 3
where j = 0, 1, 2, · · · . However, this means that no matter what k is and no matter where
ck...
View Full
Document