This week, we’ll work out some Taylor Series expansions of popular functions.
\(f(x)=\frac{1}{(1-x)}\)
\(f(x)=e^{x}\)
\(f(x)=ln(1+x)\)
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 a R-Markdown document.
\(f(x)=\frac{1}{(1-x)}\)
\(f^{'}(x)=\frac{1}{(1-x)^2}\) \(f^{''}(x)=\frac{2}{(1-x)^3}\) \(f^{'''}(x)=\frac{6}{(1-x)^4}\) \(f^{''''}(x)=\frac{24}{(1-x)^4}\)
\(f(0)=1\), \(f^{'}(0)=1\) ,\(f^{''}(0)=2\), \(f^{'''}(0)=6\), \(f^{''''}(0)=24\)
Taylor Series Expansion
\(f(0)+f^{'}(0)(x-0)+\frac{f^{''}}{2!}(x-0)+\frac{f^{'''}}{3!}(x-0)+\frac{f^{''''}}{4!}(x-0)\)
\(=1 + x^2 + x^3 + x^4\)
\(f(x)=e^{x}\)
\(f^{'}(x)=e^{a}\) \(f^{''}(x)=e^{a}\) \(f^{'''}(x)=e^{a}\) \(f^{''''}(x)=e^{a}\)
\(f(0)=1\), \(f^{'}(0)=1\) ,\(f^{''}(0)=1\), \(f^{'''}(0)=1\), \(f^{''''}(0)=1\)
\(f(0)+f^{'}(0)(x-0)+\frac{f^{''}}{2!}(x-0)+\frac{f^{'''}}{3!}(x-0)+\frac{f^{''''}}{4!}(x-0)\)
\(=1 + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}\)
\(f(x)=ln(1+x)\)
\(f^{'}(x)=\frac{1}{(1+a)}\) \(f^{''}(x)=\frac{-1}{(1+a)^2}\) \(f^{'''}(x)=\frac{2}{(1+a)^3}\) \(f^{''''}(x)=\frac{-6}{(1+a)^4}\)
\(f(0)=0\), \(f^{'}(0)=1\) ,\(f^{''}(0)=-1\), \(f^{'''}(0)=2\), \(f^{''''}(0)=-6\)
\(f(0)+f^{'}(0)(x-0)+\frac{f^{''}}{2!}(x-0)+\frac{f^{'''}}{3!}(x-0)+\frac{f^{''''}}{4!}(x-0)\)
\(=x- \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}\)