Show that $\exp(x)-\frac{1+\frac{x}{2}}{1-\frac{x}{2}} = O(x^3)$

Question

I'm trying to show the following statement:

$\exp(x)-\frac{1+\frac{x}{2}}{1-\frac{x}{2}} = O(x^3)$

I know that this is an example of Padé Approximation of the exponential. But I am not allowed to just take the Padé Table from Wiki and take it for $m=n=1$.

So, I would be glad if someone could help me and give me a hint how to do this. Thanks in Advance!

Answer 1

Consider, using Taylor series around $x=0$ $$\frac 1{1-\frac x 2}=1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8}+\frac{x^4}{16}+O\left(x^5\right)$$ Multiply both sides by $({1+\frac x 2})$ to get $$\frac {1+\frac x 2}{1-\frac x 2}=1+x+\frac{x^2}{2}+\frac{x^3}{4}+\frac{x^4}{8}+O\left(x^5\right)$$ Using now Taylor expansion of $e^x$, then $$e^x-\frac {1+\frac x 2}{1-\frac x 2}=\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+O\left(x^5\right)\right)-\left(1+x+\frac{x^2}{2}+\frac{x^3}{4}+\frac{x^4}{8}+O\left(x^5\right) \right)$$ Simplify and get $$e^x-\frac {1+\frac x 2}{1-\frac x 2}=-\frac{x^3}{12}-\frac{x^4}{12}+O\left(x^5\right)$$

Answer 2

This is equivalent to showing $\frac{1+x/2}{1-x/2}=1+x+\frac{x^2}{2}+cx^3+o(x^3)$ with $c\ne \frac{1}{6}$. To prove this use $\frac{1}{1-x/2}=1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{8}+o(x^3)$. You should find $c=\frac{1}{4}$.