1

I found somebody using the following approximation for $\sqrt{x^2 + y^2}$:

$\sqrt{x^2 + y^2} \approx (x+y) \sqrt{2/\pi},$

where $x$ and $y$ are positive numbers smaller than 1. It appears in this context:

http://www.cepii.fr/CEPII/en/publications/wp/abstract.asp?NoDoc=2726; p.18

So $x^2$ and $y^2$ are variances of a random value which describes the reporting error in trade data (assumed to be log-normally distributed).

What's the rationale for this approximation?

I tried several values for $x$ and $y$ and the approximation isn't too bad. E.g. $x=0.5$ and $y=0.3$ yields 0.64 instead of 0.58, and even for $x=0.9$ and $y=0.1$, the error is still acceptable (0.80 instead of 0.91). In most practical cases, we expect $x$ and $y$ to be of somewhat similar size (let's say a maximum ratio $x/y$ of 2).

Thanks!

Christian
  • 111
  • 3
    What do you mean? In what range is this meant to be useful? At $(1,1)$ you get $2\approx 1.59577$, not very good. If $x,y$ are large it gets much worse. If $x,y<0$ then it is terrible. – lulu Jun 15 '17 at 10:57
  • Can you clarify your question? As it stands, it really doesn't seem to make any sense. – lulu Jun 15 '17 at 11:02
  • In which context did you see this approximation ? – Claude Leibovici Jun 15 '17 at 11:02
  • I believe the author meant rationale. Seems like a simple error :) – user2662833 Jun 15 '17 at 12:58
  • This approximation certainly doesn't work very well as an approximation. For example, suppose you chose some x and some y that makes $x^2 + y^2 = a$, this illustrates how good the approximation is. I'd say it's pretty bad except when a is almost 1. – user2662833 Jun 15 '17 at 13:05
  • Working backwards a bit, $x+y=\sqrt{(x+y)^2}=\sqrt{x^2+2xy+y^2}$. So your question amounts to justifying approximating $\frac{x^2+y^2}{x^2+2xy+y^2}$ by $\frac{2}{\pi}$ over some region. Yet this function is clearly quite far from being constant even on a region like $[0,1]^2$. – Ian Jun 15 '17 at 14:36

0 Answers0