## Calculus 8th Edition

Use Maclaurin Series expansion to find the value of $f'''(0)$. $f(x)=\Sigma_{n=0}^{\infty}\frac{f^{n}(0)}{n!}x^{n}=f(0)+f'(0)+\frac{f''(0)}{2!}x^{2}+....+\frac{f^{n}(0)}{n!}x^{n}+...$ Given: $f(x)=2x-x^{2}+\frac{x^{3}}{3}$ $\frac{f^{'''}(0)}{1.2.3}x^{3}=\frac{x^{3}}{3}$ This implies $\frac{f^{'''}(0)}{2}=1$ $f'''(0)=2\times 1=2$ Hence, the statement is true.