8.5. TAYLOR POLYNOMIALS AND TAYLOR SERIES
505
So having
P
(k)
n (a) = f
(k) (a)
means that
k!c k = f
(k) (a)
and therefore
c k =
f (k) (a)
k!
for each value of k. In this expression for c k , we have found the formula for the degree n
polynomial approximation of f that we seek.
The nth order Taylor polynomial of f centered at x = a is given by
P n (x) = f (a) + f
′ (a)(x − a) +
f ′′ (a)
2!
(x − a)
2 + · · · +
f (n) (a)
n!
(x − a)
n
=
n
k=0
f (k) (a)
k!
(x − a)
k .
This degree n polynomial approximates f (x) near x = a and has the property that
P
(k)
n (a) = f (k) (a) for k = 0 . . . n.
Example 8.2. Determine the third order Taylor polynomial for f (x) = e x , as well as the
general nth order Taylor polynomial for f centered at x = 0.
Solution. We know that f ′ (x) = e x and so f ′′ (x) = e x and f ′′′ (x) = e x . Thus,
f (0) = f
′ (0) = f
′′ (0) = f
′′′ (0) = 1.
So the third order Taylor polynomial of f (x) = e x centered at x = 0 is
P 3 (x) = f (0) + f
′ (0)(x − 0) +
f ′′ (0)
2!
(x − 0)
2 +
f ′′′ (0)
3!
(x − 0)
3 = 1 + x +
x 2
2
+
x 3
6
.
In general, for the exponential function f we have f (k) (x) = e x for every positive integer
k. Thus, the kth term in the nth order Taylor polynomial for f (x) centered at x = 0 is
f (k) (0)
k!
(x − 0)
k =
1
k!
x
k .
Therefore, the nth order Taylor polynomial for f (x) = e x centered at x = 0 is
P n (x) = 1 + x +
x 2
2!
+ · · · +
1
n!
x
n =
n
k=0
x k
k!
.