The Integral Form of the Remainder
in Taylor’s Theorem
MATH 141H
Jonathan Rosenberg
April 24, 2006
Let f be a smooth function near x = 0. For x close to 0, we can write f (x) in terms of
f (0) by using the Fundamental Theorem of Calculus:
Z x
f (x) = f (0) +
f 0 (t) dt.
0
Now integrate by parts, setting u = f 0 (t), du = f 00 (t) dt, v = t − x, dv = dt. (Remember,
the variable of integration is t, and we’re thinking of x as a constant.) We get
Z x
f (x) = f (0) +
f 0 (t) dt
0
Z x
t=x
0
= f (0) + (t − x)f (t) t=0 −
(t − x)f 00 (t) dt
0
Z x
= f (0) + xf 0 (0) −
(t − x)f 00 (t) dt.
0
Now repeat the process. Again, integrate by parts, this time with u = f 00 (t), du = f 000 (t) dt,
v = (t − x)2 /2, dv = (t − x) dt. We get
Z x
0
f (x) = f (0) + xf (0) −
(t − x)f 00 (t) dt
0
it=x Z x (t − x)2
h (t − x)2
00
f (t)
f 000 (t) dt
+
= f (0) −
2
2
t=0
Z x0
2
(t − x)2 000
x
f (t) dt.
= f (0) + xf 0 (0) + f 00 (0) +
2
2
0
1
Continuing this process over and over, we see eventually that
xn (n)
f (x) = f (0) + xf (0) + · · · + f (0) + Rn (x)
n!
0
where the remainder Rn (x) is given by the formula
Z x
Z x
(x − t)n (n+1)
(t − x)n (n+1)
n
f
(t) dt =
f
(t) dt.
Rn (x) = (−1)
n!
n!
0
0
In principle this is an exact formula, but in practice it’s usually impossible to compute.
However, let’s assume for simplicity that x > 0 (the case x < 0 is similar) and assume that
a ≤ f (n+1) (t) ≤ b,
0 ≤ t ≤ x.
In other words, a is a lower bound for f (n+1) (t) on the interval [0, x], and b is an upper bound
for f (n+1) (t) on the same interval. Then we get
Z x
Z x
Z x
(x − t)n
(x − t)n (n+1)
(x − t)n
a dt ≤ Rn (x) =
f
(t) dt ≤
b dt.
(∗∗)
n!
n!
n!
0
0
0
But
Z
x
0
x
h
n+1 it=x
(t − x)n
n (t − x)
dt = (−1)
n!
(n + 1)! t=0
0
h
n+1 i
0
xn+1
(−x)
= (−1)n
−
= −(−1)n (−1)n+1
(n + 1)!
(n + 1)!
(n + 1)!
n+1
x
=
.
(n + 1)!
(x − t)n
dt = (−1)n
n!
Z
Plugging this into (∗∗), we see that
a
xn+1
xn+1
≤ Rn (x) ≤ b
,
(n + 1)!
(n + 1)!
which is Lagrange’s estimate for the remainder.
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