# approximating $(1-e^{-x})^2$ near $x=0$ with $x^2$ via Taylor expansion

I would like to show that $(1-e^{-x} )^2$ is approximated well near $x=0$ with $x^2$ via Taylor expansion but can't quite seem to complete the job.

I know that by expanding the exponential into its Taylor series and tidying some terms I end up with $(1-e^{-x})^2=(x-\frac{x^2}{2}+\frac{x^3}{3!}-\frac{x^4}{4!}...)^2$.

By raw calculation I can see that this ends up approximating $x^2$ nicely but I do not know how to show this explicitly.

• What is your definition of "approximated well"? If you expand the square, you can see that $(1-e^{-x})^{2} = x^{2} + \mathcal{O}(x^{3})$ which already seems like a sufficient argument to me. Commented Oct 26, 2015 at 4:54
• As long as $x^2$ gets closer to $(1-e^{-x})^2$ as we get closer to x=0 then I'm happy. Marty's answer below does it nicely. Commented Oct 26, 2015 at 8:10

\begin{align*} (1-e^{-x})^2 &=1-2e^{-x}+e^{-2x}\\ &=1-2\sum_{n=0}^{\infty} \frac{(-1)^nx^n}{n!}+\sum_{n=0}^{\infty} \frac{(-2)^nx^{n}}{n!}\\ &=1-\sum_{n=0}^{\infty} \frac{(2(-1)^n-(-2)^n)x^n}{n!}\\ &=1-\sum_{n=0}^{\infty} \frac{(-1)^n(2-2^n)x^n}{n!}\\ &=1-(1-0x-x^2+\sum_{n=3}^{\infty} \frac{(-1)^n(2-2^n)x^n}{n!})\\ &=x^2-\sum_{n=3}^{\infty} \frac{(-1)^n(2-2^n)x^n}{n!}\\ \end{align*}