# Avoiding catastrophic cancellation with $\sqrt{1+x} - 1$ for $x$ close to $0$

I'm trying to figure out how to avoid catastrophic cancellation for the following expression $$\sqrt{1+x} - 1$$ for $$x$$ being a number very close to $$0$$.

Of course, the answer would come to $$0$$ unless the expression is changed around.

Any help is appreciated! Thanks!

You could use $$\sqrt{1+x}-1=\frac{x}{\sqrt{1+x}+1}$$
If you want accurate results without computing any square root, you could use $$[n,n]$$ Padé approximants.
These could be $$\sqrt{1+x}-1\sim \frac{2 x}{x+4}$$ $$\sqrt{1+x}-1\sim \frac{4 x (x+2)}{x (x+12)+16}$$
• I'd like to point out that these equivalences are with the limit as $x\to0$, not $x\to\infty$ as you might usually expect from $\sim$. – Jam Jan 19 at 15:53
• @Jam. I totally agree with you. In the question, it is specified "for $x$ being a number very close to $0$". – Claude Leibovici Jan 19 at 16:06
If $$x$$ is seriously small, e.g. $$\ x < 10^{-14}$$, and you don't care too much about terms of order $$O(10^{-28})$$ then why not use:
$$\sqrt{1+x}\approx1+\frac{x}{2}$$