THE TAYLOR REMAINDER THEOREM
JAMES KEESLING
In this post we give a proof of the Taylor Remainder Theorem. It is a very simple proof
and only assumes Rolle’s Theorem.
Rolle’s Theorem. Let f (x) be differentiable on [a, b] and suppose that f (a) = f (b). Then
there is a point a < ξ < b such that f 0 (ξ) = 0.
Taylor Remainder Theorem. Suppose that f (x) is (N + 1) times differentiable on the
interval [a, b] with a < x0 < b. Let a < x0 < b. Then there is a point ξ between x0 and x
such that the following holds.
f (x) = f (x0 ) + f 0 (x0 )(x − x0 ) +
=
N
X
f (n)
n=0
n!
(x − x0 )n +
f 00
f (N ) (x0 )
f (N +1) (ξ)
(x − x0 )2 + · · · +
(x − x0 )N +
(x − x0 )N +1
2
N!
(N + 1)!
f (N +1) (ξ)
(x − x0 )N +1
(N + 1)!
Proof. Let f, a, b, x and x0 be as in the statement of the theorem. Let R be defined by
the following equation.
f (x) = f (x0 )+f 0 (x0 )(x−x0 )+
f 00
f (N ) (x0 )
R
(x−x0 )2 +· · ·+
(x−x0 )N +
(x−x0 )N +1
2
N!
(N + 1)!
Define F (ξ) by the following formula.
F (ξ) =
N
X
f (n) (ξ)
n=0
n!
(x − ξ)n +
R
(x − ξ)N +1
(N + 1)!
Then we have the following.
F 0 (ξ) = f 0 (ξ) +
=
N
X
n=0
(N
+1)
f
(ξ)
f (n+1) (ξ)
f (n) (ξ)
(x − ξ)n −
(x − ξ)n−1
n!
(n − 1)!
(x − ξ)N −
R
(x − ξ)N
N!
N!
(x − ξ)N (N +1)
=
(f
(ξ) − R)
N!
1
!
−
R
(x − ξ)N
N!
2
JAMES KEESLING
Clearly F (x0 ) = F (x) = f (x). So, by Rolle’s Theorem, there is a point ξ between x0 and
x such that F 0 (ξ) = 0. At that point ξ we have R = f (N +1) (ξ) and thus we have
f (x) =
N
X
f (n)
n=0
This is what was to be proved.
n!
(x − x0 )n +
f (N +1) (ξ)
(x − x0 )N +1 .
(N + 1)!