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.
\(f'(x)=\frac{1}{(1−x)^2}\) \(f'(0)=1\)
\(f''(x)=\frac{2}{(1−x)^3}\) \(f''(0)=2\)
\(f'''(x)=\frac{6}{(1−x)^4}\) \(f'''(0)=6\) … \(f^n(x)=\frac{1}{(1−x)^{n+1}}\) \(f^n(0)=n!\) so \(\frac{1}{(1−x)}=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+...+f^n(0)\frac{x^n}{n!}\) then \(\frac{1}{(1−x)}=1+x+2\frac{x^2}{2!}+6\frac{x^3}{3!}+...=1+x+x^2+x^3+...+x^n\)
\(f(x)=e^x\) \(f(0)=1\) \(f'(x)=e^x\) \(f'(0)=1\) \(f''(x)=e^x\) \(f''(0)=1\) … \(f^n(x)=e^x\) \(f^n(0)=1\) so \(e^x=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+...+f^n(0)\frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}\)
\(f(x)=ln(1+x)\) \(f(0)=0\)
\(f'(x)=\frac{1}{(1+x)}\) \(f(0)=1\)
\(f''(x)=\frac{-1}{(1+x)^2}\) \(f'(0)=-1\)
\(f'''(x)=\frac{2}{(1+x)^3}\) \(f''(0)=2\)
\(f''''(x)=\frac{-6}{(1−x)^4}\) \(f'''(0)=-6\) … \(f^n(x)=\frac{-1^{n+1}}{(1−x)^{n+1}}\) \(f^n(0)=\) so \(ln(1+x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac{x^3}{3!}+...+f^n(0)\frac{x^n}{n!}\) then \(\frac{1}{(1−x)}=x-\frac{x^2}{2!}+2\frac{x^3}{3!}+...=x-\frac{x^2}{2}+\frac{x^3}{3}-...+\frac{(-1)^{n+1}x^n}{n}\)