Pick any exercise in 8.8 of the calculus textbook. Solve and post your solution. If you have issues doing so, discuss them.
8.8.10
\(f(x)=ln(1+x); c=0\)
Taylor Polynomial:
\(f(x) = \sum\limits_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x - c)^n\)
Expand first few terms to obtain coefficients.
\[P(x) = f(c) + \frac{f'(c)}{1!}(x-c)^1 + \frac{f''(c)}{2!}(x-c)^2 + \frac{f'''(c)}{3!}(x-c)^3 + ...\]
Each \(\frac{f^{(n)}(c)}{n!}\) term is the coefficient of each polynomial.
Find coefficients:
\(f'(x) = \frac{1}{x+1} = \frac{\ln(1 + 0)}{0!}(x - 0)^0 = \ln(1) = 0\)
\(f''(x) = -\frac{1}{(x+1)^2} = \frac{\frac{1}{0+1}}{1!}(x - 0)^1 = x\)
\(f'''(x) = \frac{2}{(x+1)^3} = \frac{-\frac{1}{(0+1)^2}}{2!}(x - 0)^2 = -\frac{1}{2}x^2\)
\(f''''(x) = -\frac{6}{(x+1)^4} = \frac{\frac{1}{(0+1)^3}}{3!}(x - 0)^3 = \frac{1}{6}x^3\)
\(= \frac{-\frac{1}{(0+1)^4}}{4!}(x - 0)^4 = -\frac{1}{24}x^4\)
The formula for the Taylor Polynomial: \(0 + x - \frac{1}{2}x^2 + \frac{1}{6}x^3 - \frac{1}{24}x^4 + ...\)
Or in summation form \(\sum\limits_{n=0}^{\infty} \frac{x^n}{n!}(-1)^{n+1}\)