In Exercises 7 – 12, 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 8.8.1; show
work verifying these formula.
\(f(x) = \frac{x}{(1+x)} ;\) at c=1
Answer
\(f^(c) = \frac{c}{1+c} ; f(1) = \frac{1}{2}\)
\(f'(c) = \frac{-1}{(1 + c)^{2}}; f'(1) = \frac{-1}{4}\)
\(f''(c) = \frac{2}{(1 + c)^{3}} ;f''(1) = \frac{2}{8}\)
\(f'''(c) = \frac{-6}{(1 + c)^{4}} ;f'''(1) = \frac{-6}{16}\)
\(f''''(c) = \frac{24}{(1 + c)^{5}};f''''(1) = \frac{24}{32}\)
The Taylor Series of f(x) centered at c=1 is ,
\(\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ (n) }(1 ) }{ n! } } { (x-1 ) }^{ n }\)
substitute the derivatives at c=1:
\[\begin{split} f(x) &= f(1) + f'(1)\frac{(x-1)^1}{1!} + f''(1)\frac{(x-1)^2}{2!}+... \end{split}\]
\[\begin{split} f(x) &= \frac{1}{2} - \frac{1}{4}(x-1)^1 + \frac{2}{8}(x-1)^2 - \frac{6}{16}(x-1)^3 + \frac{24}{32}(x-1)^4 +.. \end{split}\]
We could generalize this formula in terms of n ,
\(= \sum_{n=0}^{\infty}\frac{-1^n}{2^{n+1}}(x-1)^n\)