For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion. Please submit your assignment as an R- Markdown document.
\[f(x) = \frac{1}{1-x} == (1-x)^{-1}\]
Solve for first, second, third derivative and evaluate at 0
\[f'(x) = -1*(1-x)^{-2}*-1 == (1-x)^{-2}\] \[f'(0) = (1-0)^{-2} == 1\]
\[f''(x) = 2*(1-x)^{-3}\] \[f''(0) = 2*(1-0)^{-3} = 2\]
\[f^{3'}(x) = 6*(1-x)^{-4}\] \[f^{3'}(0) = 6*(1-0)^{-4} = 6\]
Putting it all together
\[f(x) = \frac{1}{1!}x + \frac{2}{2!}x^{2} + \frac{6}{3!}x^{3} + ... == x + x^{2} + x^{3} + ... + x^{n}\] Therefore,
\[f(x) = \sum_{n=0}^{\infty}x^{n}\]
\[f(x) = e^{x}\] All derivatives are e^x, therefore all derivatives evaulated at 0 are 1.
So we can skip right to the creation of the Taylor Series:
\[f(x) = \frac{1}{1!}x + \frac{1}{2!}x^{2} + ... + \frac{1}{n!}x^{n}\] Which can also be represented as
\[f(x) = \sum_{n=0}^{\infty}\frac{x^{n}}{n!}\]
\[ln(1+x)\] First derivative
\[f'(x) = \frac{1}{1+x}\] \[f'(0) = \frac{1}{1+0} == 1\] Second derivative
\[f''(x) = -1*(1+x)^{-2}\] \[f''(0) = -1*(1+0)^{-2} == -1\] Third derivative
\[f^{3'}(x) = 2*(1+x)^{-3}\] \[f^{3'}(0) = 2*(1+0)^{-3} == 2\] Fourth derivative
\[f^{4'}(x) = -6*(1+x)^{-4}\]
\[f^{4'}(0) = -6*(1+0)^{-4} == -6\] Putting it together:
\[f(x) = \frac{1}{1!}x + \frac{-1}{2!}x^{2} + \frac{2}{3!}x^{3} + \frac{-6}{4!}x^{4}\] \[f(x) = x + \frac{-1}{2}x^{2} + \frac{1}{3}x^{3} + \frac{-1}{4}x^{4}\] \[f(x) = \sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^{n}}{n}\]
\[f(x) = x^{1/2}\] \[f'(x) = \frac{1}{2}*x^{-1/2}\] \[f'(0) = \frac{1}{2}*0^{-1/2} == undefined\]