Give an example to show that computing $ \frac{x - y}{\sqrt x + \sqrt y}$ is not always more accurate for computing $fl(\sqrt x - \sqrt y)$?

Question

I assume that we are working on a machine with rounded four-digit decimal arithmetic, I know that we have this formula:

$\sqrt x - \sqrt y = \frac{x - y}{\sqrt x + \sqrt y}$

Since there is cancelation error in machine so it is better to use this formula for computing $fl(\sqrt x - \sqrt y)$ .But can you give me an example that computing simple difference for computing $fl(\sqrt x - \sqrt y)$ (I mean directly compute $\sqrt x - \sqrt y$) is more accurate than computing $\frac{x - y}{\sqrt x + \sqrt y}$ ?

I tried lots of examples but using formula is always more accurate or both of them have the same accuracy ,can any one help me?

What do you mean by $fl$? – Patrick Stevens Apr 06 '17 at 07:19 — Patrick Stevens, Apr 06 '17 at 07:19
$fl(x)$ is floating-point version of x. – haleh Apr 06 '17 at 07:25 — haleh, Apr 06 '17 at 07:25
$fl(x) = x(1+\delta ) s.t |\delta | < epsilon machine$ – haleh Apr 06 '17 at 07:26 — haleh, Apr 06 '17 at 07:26

Robert Israel · Answer 1 · 2017-04-06T07:33:17.787

The advantage of the formula is mainly when $x$ and $y$ are close together. When they are not, the direct method has a fair chance of being better. For example, with $x = 6$ and $y = 1$, the direct method gives $\sqrt{x} - \sqrt{y} = 2.449 - 1.000 = 1.449$, while the formula gives $$\frac{x - y}{\sqrt{x} + \sqrt{y}} = \frac{6 - 1}{2.449 + 1.000} = \frac{5}{3.449} = 1.450$$ and it happens that $\sqrt{6} - \sqrt{1} = 1.4494897\ldots$ is closer to $1.449$ than to $1.450$.

Give an example to show that computing $ \frac{x - y}{\sqrt x + \sqrt y}$ is not always more accurate for computing $fl(\sqrt x - \sqrt y)$?

1 Answers1