Use the definition to find the Taylor series centered at $c = 1 \text { for the function }$ $f ( x ) = \frac { 1 } { x }$
A)
$$\frac { 1 } { x } = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } x ^ { n }$$
B)
$$\frac { 1 } { x } = \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } ( x + 1 ) ^ { n }$$
C)
$$\frac { 1 } { x } = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n + 1 } ( x - 1 ) ^ { n }$$
D)
$$\frac { 1 } { x } = \sum _ { n = 1 } ^ { \infty } ( - 1 ) ^ { n } ( x ) ^ { n }$$
E)
$$\frac { 1 } { x } = \sum _ { n = 0 } ^ { \infty } ( - 1 ) ^ { n } ( x - 1 ) ^ { n }$$

Question

Quizplus · Accepted Answer

The Taylor series for a function $f(x)$ centered at $c=1$ is given by
$$\sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!}(x-1)^n$$
where $f^{(n)}(x)$ represents the $n$th derivative of $f(x)$ evaluated at $x$. 
For $f(x) = \frac{1}{x}$, we have
$$f'(x) = -\frac{1}{x^2}, \quad f''(x) = \frac{2}{x^3}, \quad f^{(3)}(x) = -\frac{6}{x^4}, \quad f^{(4)}(x) = \frac{24}{x^5}, \quad \ldots$$
Evaluating these derivatives at $x=1$, we have $f^{(n)}(1) = (-1)^{n+1}(n-1)!$, and therefore the Taylor series for $f(x) = \frac{1}{x}$ centered at $c=1$ is given by 
$$\frac{1}{x} = \sum_{n=0}^{\infty}(-1)^n(x-1)^n,$$
which matches choice E.

Use the Definition to Find the Taylor Series Centered At c