Use taylor polynomial approximations to avoid loss of significance error.
(x - sin(x)) / x^3
Any advice? Thanks
Note that there is loss of significance at $\displaystyle x=0$.
Thus, use the taylor series @ x=0 for sin x:
$\displaystyle \sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$.
So it turns out that
$\displaystyle \begin{aligned}\frac{x-\sin x}{x^3}&=\frac{x-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots\right)}{x^3}\\ &=\frac{1}{3!}-\frac{x^2}{5!}+\frac{x^4}{7!}-\dots\end{aligned}$
Does this make sense?