Page 496 Question 8

Find a formula for the nth term of the Taylor series of f(x), centered at c, by finding the coefficients of the first few powers of x and looking for a pattern. Show work verifying the formulas in Key Idea 8.8.1.

f(x) = 1/x

c = 1

\(\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x - c)^n\)

\(\sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!}(x - 1)^n\)

For n = 0:

\(f(1) = \frac{1^{-1}}{0!}(x - 1)^0 = 1\)

For n = 1:

\({f}^{'}(x) = -x^{-2}\)

\(f(1) = \frac{-1^{-2}}{1!}(x - 1)^1 = -x + 1\)

For n = 2:

\({f}^{''}(x) = 2x^{-3} . . . f^{''}(1) = 2\)

\(f(1) = \frac{2}{2!}(x - 1)^2 = (x - 1)^2\)

For n = 3:

\({f}^{'''}(x) = -6x^{-4} . . . f^{'''}(1) = -6\)

\(\frac{-6}{3!}(x - 1)^3 = -(x - 1)^3\)

A pattern emerges from these first few values. The sign alternates from + to -. The exponent for (x - 1) continues to increase by 1. The formula for this series is the following:

\((-1)^{n} (x - 1)^n\)