8.8 - #11: 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 pa􏰕ern. (The formu- las for several of these are found in Key Idea 32; show work verifying these formula.)

\[f(x)=\frac{x}{(x+1)}; c=1\]

• \(f(x) = \frac{x}{(x+1)}\)
\(f^{'}(x) = \frac{1}{(x+1)^2}\), \(f^{'}(1) = 1\)
\(f^{''}(x) = -\frac{2}{(x+1)^3}\), \(f^{''}(1) = 2\)
\(f^{'''}(x) = \frac{6}{(x+1)^4}\), \(f^{'''}(1) = 6\)
\(f^4(x) = -\frac{24}{(x+1)^5}\), \(f^4(1) = 24\)
. . .
\(f^n(x) = (-1)^{(n+1)}\frac{n!}{(x+1)^{n+1}}\), \(f^n(1) = (-1)^{(n+1)}n!\)

Therefore based on Taylor series, \(\sum_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x-c)^n = \sum_{n=0}^{\infty}(-1)^{(n+1)}\frac{n!}{n!}(x-1)^n = \sum_{n=0}^{\infty}(-1)^{(n+1)}(x-1)^n\)