Question

The third term in Maclaurin series of $x{e}^{-x}$ is?

• A. $\frac{x^3}{2}$
• B. $\frac{x^2}{2}$
• C. $\frac{x^3}{3}$
• D. $\frac{x}{2}$

Answer 1

The correct option is A $\frac{x^3}{2}$

The Maclaurin series is given by $f\left(x\right)={\sum }_{k=0}^{\infty }\frac{f^{\left(k\right)}\left(a\right)}{k!}{x}^k$ where $a=0$
We have $f\left(x\right)=x{e}^{-x}$
Since we have to find the third term, let us take $n=5$.
$\therefore f\left(x\right)\approx {\sum }_{k=0}^5\frac{f^{\left(k\right)}\left(0\right)}{k!}{x}^k$
$f^{\left(0\right)}\left(x\right)=x{e}^{-x},\Rightarrow f^{\left(0\right)}\left(0\right)=0$
$f^{\left(1\right)}\left(x\right)=(-x+1)e^{-x},\Rightarrow f^{\left(1\right)}\left(0\right)=1$
$f^{\left(2\right)}\left(x\right)=(x-2)e^{-x},\Rightarrow f^{\left(2\right)}\left(0\right)=-2$
$f^{\left(3\right)}\left(x\right)=(-x+3)e^{-x},\Rightarrow f^{\left(3\right)}\left(0\right)=3$
$f^{\left(4\right)}\left(x\right)=(x-4)e^{-x},\Rightarrow f^{\left(4\right)}\left(0\right)=-4$
$f^{\left(5\right)}\left(x\right)=(-x+5)e^{-x},\Rightarrow f^{\left(5\right)}\left(0\right)=5$
$\therefore f\left(x\right)\approx \frac{0}{0!}{x}^0+\frac{1}{1!}{x}^1+\frac{-2}{2!}{x}^2+\frac{3}{3!}{x}^3+\frac{-4}{4!}{x}^4+\frac{5}{5!}{x}^5$
$\Rightarrow f\left(x\right)\approx x-x^2+\frac{1}{2}{x}^3-\frac{1}{6}{x}^4+\frac{1}{24}{x}^5$
Thus the third term is $\frac{1}{2}{x}^3$.