SECTION 5.4 Polynomial Approximations 299
Having been reminded of theMean Value Theorem, perhaps now
Taylor’s Theorem with Remainderwon’t seem so strange. Here it
is.
Taylor’s Theorem with Remainder.Letfbe an infinitely-differentiable
function on an open intervalI. Ifa andxare inI, and ifn is a non-
negative integer, then there exists a real number c betweena and x
such that
f(x) =f(a) +f′(a)(x−a) +
f′′(a)
2!(x−a)^2 +······+
f(n)(a)
n!(x−a)n+
f(n+1)(c)
(n+ 1)!(x−a)n+1
︸ ︷︷ ︸
this is the remainder.
Proof.^22 We start by proving that, for alln≥ 0
f(x) =f(a)+f′(a)(x−a)+
f′′(a)
2!+···+
f(n)(a)
n!(x−1)n+∫x
af(n+1)(t)
n!(x−t)ndt.Note that since ∫x
a f′(t)dt=f(x)−f(a),then a simple rearrangement gives
f(x) =f(a) +∫x
a f′(t)dt,which is the above statement whenn= 0. We take now as our induction
hypothesis, that
f(x) =f(a)+f′(a)(x−a)+
f′′(a)
2!(x−a)^2 +···+f(n−1)(a)
(n−1)!(x−a)n+∫x
af(n)(t)
(n−1)!(x−t)n−^1 dtis true.
We evaluate the integral using integration by parts with the substi-
tution
(^22) Very few textbooks at this level provide a proof; however, since the two main ingredients are
induction and integration by parts, I felt that giving the proof would be instructive reading for the
serious student.