516
8.5. TAYLOR POLYNOMIALS AND TAYLOR SERIES
polynomial for f centered at x = a and if M is an upper bound for
f (n+1) (x)
on the
interval [a, c], then
|P n (c) − f (c)| ≤ M
|c − a| n+1
(n + 1)!
.
Exercises
1. In this exercise we investigation the Taylor series of polynomial functions.
(a) Find the 3rd order Taylor polynomial centered at a = 0 for f (x) = x 3 − 2x 2 +
3x − 1. Does your answer surprise you? Explain.
(b) Without doing any additional computation, find the 4th, 12th, and 100th order
Taylor polynomials (centered at a = 0) for f (x) = x 3 − 2x 2 + 3x − 1. Why
should you expect this?
(c) Now suppose f (x) is a degree m polynomial. Completely describe the nth
order Taylor polynomial (centered at a = 0) for each n.
2. The examples we have considered in this section have all been for Taylor polynomials
and series centered at 0, but Taylor polynomials and series can be centered at any
value of a. We look at examples of such Taylor polynomials in this exercise.
(a) Let f (x) = sin(x). Find the Taylor polynomials up through order four of f
centered at x =
π
2 . Then find the Taylor series for f (x) centered at x =
π
2 . Why
should you have expected the result?
(b) Let f (x) = ln(x). Find the Taylor polynomials up through order four of f
centered at x = 1. Then find the Taylor series for f (x) centered at x = 1.
(c) Use your result from (b) to determine which Taylor polynomial will approximate
ln(2) to two decimal places. Explain in detail how you know you have the
desired accuracy.
3. We can use known Taylor series to obtain other Taylor series, and we explore that idea
in this exercise, as a preview of work in the following section.
(a) Calculate the first four derivatives of sin(x 2 ) and hence find the fourth order
Taylor polynomial for sin(x 2 ) centered at a = 0.
(b) Part (a) demonstrates the brute force approach to computing Taylor polynomials
and series. Now we find an easier method that utilizes a known Taylor series.
Recall that the Taylor series centered at 0 for f (x) = sin(x) is
∞
k=0
(−1)
k x 2k+1
(2k + 1)!
.
(8.24)
8.5. TAYLOR POLYNOMIALS AND TAYLOR SERIES
polynomial for f centered at x = a and if M is an upper bound for
f (n+1) (x)
on the
interval [a, c], then
|P n (c) − f (c)| ≤ M
|c − a| n+1
(n + 1)!
.
Exercises
1. In this exercise we investigation the Taylor series of polynomial functions.
(a) Find the 3rd order Taylor polynomial centered at a = 0 for f (x) = x 3 − 2x 2 +
3x − 1. Does your answer surprise you? Explain.
(b) Without doing any additional computation, find the 4th, 12th, and 100th order
Taylor polynomials (centered at a = 0) for f (x) = x 3 − 2x 2 + 3x − 1. Why
should you expect this?
(c) Now suppose f (x) is a degree m polynomial. Completely describe the nth
order Taylor polynomial (centered at a = 0) for each n.
2. The examples we have considered in this section have all been for Taylor polynomials
and series centered at 0, but Taylor polynomials and series can be centered at any
value of a. We look at examples of such Taylor polynomials in this exercise.
(a) Let f (x) = sin(x). Find the Taylor polynomials up through order four of f
centered at x =
π
2 . Then find the Taylor series for f (x) centered at x =
π
2 . Why
should you have expected the result?
(b) Let f (x) = ln(x). Find the Taylor polynomials up through order four of f
centered at x = 1. Then find the Taylor series for f (x) centered at x = 1.
(c) Use your result from (b) to determine which Taylor polynomial will approximate
ln(2) to two decimal places. Explain in detail how you know you have the
desired accuracy.
3. We can use known Taylor series to obtain other Taylor series, and we explore that idea
in this exercise, as a preview of work in the following section.
(a) Calculate the first four derivatives of sin(x 2 ) and hence find the fourth order
Taylor polynomial for sin(x 2 ) centered at a = 0.
(b) Part (a) demonstrates the brute force approach to computing Taylor polynomials
and series. Now we find an easier method that utilizes a known Taylor series.
Recall that the Taylor series centered at 0 for f (x) = sin(x) is
∞
k=0
(−1)
k x 2k+1
(2k + 1)!
.
(8.24)