[Python-Dev] math.hypot, complex.__abs__, and documentation

[David DiCato]

Mathematically, think of nan as 'indeterminate'. When you're trying to get the magnitude of a vector, you know that it's infinite if even one of the components is infinite. So, the fact that the other component is indeterminate can be ignored. It's the same with math.pow(1.0, float('nan')); the second argument simply doesn't matter when the first is 1.0.

FWIW, these conventions also exist in the C99 standard.

Hope this helps,
- David

-----Original Message-----
From: python-dev-bounces+ddicato=microsoft.com at python.org [mailto:python-dev-bounces+ddicato=microsoft.com at python.org] On Behalf Of Steven D'Aprano
Sent: Tuesday, February 16, 2010 2:47 PM
To: python-dev at python.org
Subject: Re: [Python-Dev] math.hypot, complex.__abs__, and documentation

On Wed, 17 Feb 2010 08:19:00 am David DiCato wrote:
> I have a minor concern about certain corner cases with math.hypot and
> complex.__abs__, namely when one component is infinite and one is not
> a number. If we execute the following code:
>
> import math
> inf = float('inf')
> nan = float('nan')
> print math.hypot(inf, nan)
> print abs(complex(nan, inf))
>
> ... then we see that 'inf' is printed in both cases. The standard
> library tests (for example, test_cmath.py:test_abs()) seem to test
> for this behavior as well, and FWIW, I personally agree with this
> convention.

What's the justification for that convention? It seems wrong to me. 

If you expand out hypot and substitute a=inf and b=nan, you get:

>>> math.sqrt(inf*inf + nan*nan)
nan

which agrees with my pencil-and-paper calculation:

sqrt(inf*inf + nan*nan)
= sqrt(inf + nan)
= sqrt(nan)
= nan



-- 
Steven D'Aprano
_______________________________________________
Python-Dev mailing list
Python-Dev at python.org
http://mail.python.org/mailman/listinfo/python-dev
Unsubscribe: http://mail.python.org/mailman/options/python-dev/ddicato%40microsoft.com