[EMAIL PROTECTED] wrote:

>Hi,
>
>I have recieved the following note from a user:
>
>"""
>In SciPy 0.3.x the ufuncs were overloaded by more "intelligent" versions.
>A very attractive feature was that sqrt(-1) would yield 1j as in Matlab.
>Then you can program formulas directly (e.g., roots of a 2nd order
>polynomial) and the right answer is always achieved. In the Matlab-Python
>battle in mathematics education, this feature is important.
>
>Now in SciPy 0.5.x sqrt(-1) yields nan. A lot of code we have, especially
>for introductory numerics and physics courses, is now broken.
>This has already made my colleagues at the University skeptical to
>Python as "this lack of backward compatibility would never happen in Matlab".
>  
>
This was a consequence of moving scipy_base into NumPy but not exposing 
the scimath library in NumPy.   It would be a very easy thing to put 
from numpy.lib.scimath import *
into the scipy name-space.

I'm supportive of that as a backward-compatibility measure.

>Another problem related to Numeric and numpy is that in these courses we
>use ScientificPython several places, which applies Numeric and will
>continue to do so. You then easily get a mix of numpy and Numeric
>in scripts, which may cause problems and at least extra overhead.
>Just converting to numpy in your own scripts isn't enough if you call
>up libraries using and returning Numeric.
>  
>
>"""
>
>I wonder, what are the reasons that numpy.sqrt(-1) returns nan?
>  
>
Because that is backwards compatible.  You have to construct a 
function-wrapper in order to handle the negative case correctly.  The 
function wrapper is going to be slower.  Thus, it is placed in a 
separate library.

>Could sqrt(-1) made to return 1j again? 
>
Not in NumPy.  But, in scipy it could.

>If not, shouldn't
>  
>
>numpy.sqrt(-1) raise a ValueError instead of returning silently nan?
>  
>
This is user adjustable.  You change the error mode to raise on 
'invalid' instead of pass silently which is now the default.

-Travis



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