[cfp-interest 3507] Re: hypot(x, y) - C2Y F.10.5.4

[Paul Zimmermann]

       Hi Damian,

> > 	hypot (x, y), hypot (y, x), and hypot (x, -y) are equivalent.
> >
> > Is that complete?

yes it is. Let us name hypot(x, y) = hypot(y, x) rule (A),
and hypot(x, y) = hypot(x, -y) rule (B).

Applying (A) on hypot(x, -y) you get hypot(-y,x).
Applying (B) you get hypot(-y,-x).
Applying (B) on hypot(y, x) you get hypot(y, -x).

Applying (A) on hypot(-y,-x) and hypot(y, -x)
you get hypot(-x,-y) and hypot(-x,y).

Thus you get all 8 possible forms.

Paul