Should numpy.sqrt(-1) return 1j rather than nan?

[James Graham]

Greg Willden wrote:
> On 10/11/06, *Bill Baxter* <wbaxter at gmail.com 
> <mailto:wbaxter at gmail.com>> wrote:
> 
>     On 10/12/06, Greg Willden <gregwillden at gmail.com
>     <mailto:gregwillden at gmail.com>> wrote:
>      > Speed should not take precedence over correctness.
> 
>     Unless your goal is speed.  Then speed should take precedence over
>     correctness. 
> 
> 
> Huh?
> Person 1:  "Hey you should use function X."
> Person 2:  "No, it doesn't give me the correct answer."
> Person 1: "Who cares?  It's fast!"
> 
> What kind of logic is that?
> 
> I'm having serious doubts about my conversion to Numpy.

It's more like
Person 1: Function x is designed to be fast, have predictable memory usage and 
so be suitable for a wide variety of applications. Alternatively there is 
function y which is slower but handles a larger numerical domain. y Might be 
more sutiable for your needs; you should use it.
Person 2: Numpy is broken! Function x should just do what I want!

Really, given that both options are available, I don't see a big issue here. 
Claims that returning j for sqrt(-1) is "more correct" seem highly dubious as a) 
NumPy is a numerical library, not an algebraic one, hence the fact that it 
behaves in a type-dependent way is not so surprising, and b) no one complains 
that functions like sqrt return only positive floats, despite the fact that, 
mathematically, this function is multivalued.

-- 
"Eternity's a terrible thought. I mean, where's it all going to end?"
  -- Tom Stoppard, Rosencrantz and Guildenstern are Dead

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