Question

For the function $x{e}^x$, find the Taylor approximations of order $3$ about $x=0$.

Answer 1

$f\left(x\right)=x{e}^x$

$f^1\left(x\right)=e^x+{xe}^x$

$f^{11}\left(x\right)=e^x+e^x+x{e}^x={2e}^x+x{e}^x$

$f^{111}\left(x\right)={2e}^x+e^x+{xe}^x={3e}^x+{xe}^x$

$\therefore$ ${xe}^x=f\left(0\right)+f^1\left(0\right)\left(x\right)+\frac{f^{11}\left(0\right)}{2!}.x^2+\frac{f^{111}\left(0\right)}{3!}{\left(x\right)}^3+...$

$\therefore$ ${xe}^x=0+1.x+\frac{2}{2!}{x}^2+\frac{3}{3!}{x}^3+....$

$\therefore$ ${xe}^x=1+x^2+\frac{x^3}{2!}+....$