512
8.5. TAYLOR POLYNOMIALS AND TAYLOR SERIES
for 0 ≤ k ≤ n. Furthermore, since P n (x) is a polynomial of degree less than or equal to n,
we know that
P
(n+1)
n
(x) = 0.
Thus, since E
(n+1)
n
(x) = f (n+1) (x) − P
(n+1)
n
(x), it follows that
E
(n+1)
n
(x) = f
(n+1) (x)
for all x.
Suppose that we want to approximate f (x) at a number c close to 0 using P n (c). If we
assume | f (n+1) (t)| is bounded by some number M on [0, c], so that
f
(n+1) (t) ≤ M
for all 0 ≤ t ≤ c, then we can say that
E
(n+1)
n
(t) = f
(n+1) (t) ≤ M
for all t between 0 and c. Equivalently,
− M ≤ E
(n+1)
n
(t) ≤ M
(8.22)
on [0, c]. Next, we integrate the three terms in the inequality (8.22) from t = 0 to t = x,
and thus find that
x
0
−M dt ≤
x
0
E
(n+1)
n
(t) dt ≤
x
0
M dt
for every value of x in [0, c]. Since E
(n)
n (0) = 0, the First FTC tells us that
−M x ≤ E
(n)
n (x) ≤ M x
for every x in [0, c].
Integrating the most recent inequality, we obtain
x
0
−Mt dt ≤
x
0
E
(n)
n (t) dt ≤
x
0
Mt dt
and thus
−M
x 2
2
≤ E
(n−1)
n
(x) ≤ M
x 2
2
for all x in [0, c].
Integrating n times, we arrive at
−M
x n+1
(n + 1)!
≤ E n (x) ≤ M
x n+1
(n + 1)!