Calculus- Early Transcendentals, 2021a

9.11. Taylor’s Theorem 373
Note that we can now approximate the value of sin(x) to within 0.005 by using simple trigonometric
identities to translate x into the interval [−π/2,π/2].
We can extract a bit more information from this example. If we do not limit the value of x, we still
have
∣ ∣ ∣∣∣∣
f (N+1) (z) ∣∣∣∣ (N + 1)! xN+1 ≤
x N+1
∣(N + 1)! ∣
so that sinx is represented by
If we can show that
for each x then
sinx =
N
∑
n=0
∞
∑
n=0
f (n) (0)
n!
x n ±
x N+1
∣(N + 1)! ∣ .
lim
x N+1
∣(N + 1)! ∣ = 0
N→∞
f (n) (0)
x n =
n!
∞
∑
n=0
(−1) n x 2n+1
(2n + 1)! ,
that is, the sine function is actually equal to its Maclaurin series for all x. How can we prove that the
limit is zero? Suppose that N is larger than |x|, andletM be the largest integer less than |x| (if M = 0the
following is even easier). Then
|x| |x|
N N − 1 ··· |x| |x| |x| ···|x|
M + 1 M M − 1 2
|x| |x|
≤ · 1 · 1···1 · ···|x| |x|
N + 1 M M − 1 2 1
= |x| |x| M
N + 1 M! .
|x N+1 |
(N + 1)! = |x|
N + 1
|x|
The quantity |x| M /M! is a constant, so
lim
N→∞
|x| |x| M
N + 1 M! = 0
and by the Squeeze Theorem (9.6)
lim
x N+1
N→∞∣(N + 1)! ∣ = 0
as desired. Essentially the same argument works for cosx and e x ; unfortunately, it is more difficult to show
that most functions are equal to their Maclaurin series.
Example 9.58
Find a polynomial approximation for e x near x = 2 accurate to ±0.005.
|x|
1
Solution. From Taylor’s theorem:
e x =
N
e
∑
2
n=0
n! (x − 2)n + ez
(N + 1)! (x − 2)N+1 ,