Polynomial approximation for a function using Maclaurin's Theorem:

\(f(x) = f(0) + x \:.\: f'(0) + \dfrac{x^2}{2!} \:.\: f''(0) + \dfrac{x^3}{3!} \:.\: f'''(0) + \dotso + \dfrac{x^n}{n!} \:.\: f^n(0)\)

Example 01

Expand the function \(f(x)=\sin x\) in ascending powers up to and including the term in \(x^3\)

\begin{equation*} \begin{split} & f'(x)=\cos x, \quad f''(x)=-\sin x, \quad f'''(x)=-\cos x \\\\ & \sin x=\sin 0+x \:.\: \cos 0+\frac{x^2}{2!} \:.\: -\sin 0+\frac{x^3}{3!} \:.\: -\cos 0 \\\\ & \sin x=0+x+0-\frac{1}{6}x^3+ \dotso \\\\ & \sin x=x-\frac{1}{6}x^3+ \dotso \end{split} \end{equation*}

Example 02

Expand the function \(f(x)=e^x\) in ascending powers up to and including the term in \(x^3\)

\begin{equation*} \begin{split} & f'(x)=e^x, \quad f''(x)=e^x, \quad f'''(x)=e^x \\\\ & e^x=e^0 +x \:.\: e^0+\frac{x^2}{2!} \:.\: e^0+\frac{x^3}{3!} \:.\: e^0+ \dotso \\\\ & e^x=1+x+\frac{1}{2}x^2+\frac{1}{6}x^3+ \dotso \end{split} \end{equation*}