Multivariate Calculus - Bruce E. Shapiro

LECTURE 13. TANGENT PLANES 97
Multivariate Taylor Series
Recall the following result from Calculus I (see section 10.8 of the text):
Theorem 13.1 Taylor’s Theorem with Remainder Let f be a function whose
first n + 1 derivatives exist for all x in an interval J containing a. Then for any
x ∈ J,
f(x) = f(a) + (x − a)f ′ (a) +
where the remainder is
(x − a)2
f ′′ (a) + · · · +
2!
(x − a)n
f (n) (a) + R n (x) (13.4)
n!
R n (x) =
(x − a)(n+1)
f (n+1) (c) (13.5)
(n + 1)!
for some (unknown) number c ∈ J.
Definition 13.2 The Taylor Polynomial of order n at a (or about a) is
defined as
P n (x) = f(a) + (x − a)f ′ (a) +
(x − a)2
f ′′ (a) + · · · +
2!
Corollary 13.1 f(x) = P n (x) + R n (x) for all x ∈ J.
(x − a)n
f (n) (a) (13.6)
n!
Theorem 13.2 If f is infinitely differentiable then R n (x) → 0 as n → ∞, so that
f(x) = lim n→∞ P n (x) for all x ∈ J, or more explicitly,
f(x) =
∞∑
k=0
f (k) (a)
(x − a) k (13.7)
k!
= f(a) + (x − a)f ′ (a) + 1 2! (x − a)2 f ′′ (a) + · · · (13.8)
which is called the Taylor Series of f about the point x=a
Example 13.5 Find the Taylor Polynomial of order 3 of f(x) = √ x + 1 and the
corresponding remainder formula about the point a = 0.
Solution. Taking the first 4 derivatives,
f(x) = (x + 1) 1/2 ⇒ f(0) = 1
f ′ (x) = 1 2 (x + 1)−1/2 ⇒ f ′ (0) = 1 2
f ′′ (x) = − 1 4 (x + 1)−3/2 ⇒ f ′′ (0) = − 1 4
f ′′′ (x) = 3 8 (x + 1)−5/2 ⇒ f ′′′ (0) = 3 8
Math 250, Fall 2006 Revised December 6, 2006.