5
$\begingroup$

I am trying to determine some numerical difficulties that arise from a couple problems, and a good way to re-write them to avoid those errors.

For instance, I have:

1) $\sqrt{x+\dfrac{1}{x}} - \sqrt{x-\dfrac{1}{x}}$ where $x\gg 1$

I think that since these two terms approximately equal each other, there will be cancellation error. So I multiplied the numerator and denominator by the conjugate yielding:

$\dfrac{\dfrac{2}{x}}{\sqrt{x+\dfrac{1}{x}}+\sqrt{x-\dfrac{1}{x}}}$

I think that this should get rid of the cancellation error, does anyone see anything wrong with this attempt?

If this looks right, then I will show my attempt on the second problem, but I hope to verify my method first.

2) $\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}}$ where $a\approx 0$ and $b\approx 1$

Thanks!

  • 0
    Under which values of x do you experience problems in either expressions? I am not aware of the exact value of re-writing the expression in your case. Could you provide an example?2012-08-26

1 Answers 1

3

For 1, you have successfully avoided the cancellation. If you want, you could go to $\frac {\frac 2x}{\sqrt x (\sqrt{1+\frac 1{x^2} }+\sqrt{1-\frac 1{x^2} })}=\frac 2{x^{\frac 32} (\sqrt{1+\frac 1{x^2} }+\sqrt{1-\frac 1{x^2} })}\approx x^{-\frac 32}$ but I am not sure that is an improvement.

For 2, you could have $\frac 1{a^2}$ overflow where $\frac 1a$ does not. To avoid this, you could rewrite it as $\frac 1a \sqrt {1+\frac {a^2}{b^2}}$. That still squares $a$, but if it underflows maybe it gets set to zero and you are OK.