In Exercise 8, find a formula for the \(n^{th}\) 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. The formulas for several of these are found in Key Idea 32; show work verifying these formula.
\[f(x) = \frac{1}{x}; c = 1\]
The function’s derivatives: \[ \begin{align} f'(x) &= \frac{1}{x} \\ f''(x) &= \frac{-1}{x^2} \\ f'''(x) &= \frac{2}{x^3} \\ f^{4}(x) &= \frac{-6}{x^4} \\ f^{5}(x) &= \frac{24}{x^5} \\ \end{align} \]
Evaluating at \(c = 1\) (\(c = 0\) is undefined): \[ \begin{align} f'(1) &= \frac{1}{1} = 1 \\ f''(1) &= \frac{-1}{1^2} = -1 \\ f'''(1) &= \frac{2}{1^3} = 2 \\ f^{4}(1) &= \frac{-6}{1^4} = -6 \\ f^{5}(1) &= \frac{24}{1^5} = 24 \\ \end{align} \]
Plugging into the Taylor series centered at \(c=1\): \[ \begin{align} f(x) &= \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^{n} = f(c) + \frac{f'(c)}{1!}(x-c)^{1} + \frac{f''(c)}{2!}(x-c)^{2} + \frac{f'''(c)}{3!}(x-c)^{3} + \frac{f''''(c)}{4!}(x-c)^{4} + \frac{f^{5}(c)}{5!}(x-c)^{5}... \\ &= \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (x-1)^{n} = f(1) + \frac{f'(1)}{1!}(x-1)^{1} + \frac{f''(1)}{2!}(x-1)^{2} + \frac{f'''(1)}{3!}(x-1)^{3} + \frac{f''''(1)}{4!}(x-1)^{4} + \frac{f^{5}(1)}{5!}(x-1)^{5}... \\ &= \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (x-1)^{n} = 1 + \frac{1}{1!}(x-1)^{1} + \frac{(-1)}{2!}(x-1)^{2} + \frac{2}{3!}(x-1)^{3} + \frac{(-6)}{4!}(x-1)^{4} + \frac{24}{5!}(x-1)^{5}... \\ &= \sum_{n=0}^{\infty} \frac{f^{(n)}(1)}{n!} (x-1)^{n} = 1 + (x-1) - \frac{(x-1)^{2}}{2} + \frac{(x-1)^{3}}{3} + \frac{(x-1)^{4}}{4} + \frac{(x-1)^{5}}{5}... \\ \frac{1}{x} &= \sum_{n=0}^{\infty} (-1)^{n+1}\frac{(x-c)^{n}}{n} \end{align} \]