Thanks for reminding me about the testing. I fixed the Confluent
Hypergeometric U() errors for my tests; but neglected to revise
test_hyperg.c by killing the U() fixme's. I have now done that and the
tests all succed for U(). I will send you the modified test_hyperg.c;
the other file hyperg_U.c is unchanged. It should be reviewed and
double checked on another system. If it passes then somebody can take
out the "fixme" for U() permanently.
I have done nothing for M() except try to find out where the egregious
errors are and understand them. I haven't even figured out how to fix
the thing I mentioned below; I will probably try killing the Kummer
Transform and see how bad the regular summation is with x<0 ; otherwise ...
Ray
On 10/03/2014 11:12 AM, Patrick Alken wrote:
I confess I'm not an expert on these functions, and have never used
them in my work, and it doesn't look like other experts are involved
in this discussion.
The two files you sent earlier - I added them to the bug tracker - do
they contain a fix for all these bugs you've found, or is it an
incomplete fix? I imagine implementing a highly robust code for these
functions which handle all possible argument ranges would be a large
task.
Patrick
On 10/02/2014 04:30 PM, Raymond Rogers wrote:
On 10/02/2014 02:16 PM, Patrick Alken wrote:
On 10/02/2014 12:11 PM, Raymond Rogers wrote:
I believe I have found some problems and perhaps a solution in the
1F1 code.
Could somebody (s) evaluate the following in Maple and Mathematica and
post the results?
1F1( -1, -2, -4)
The online Mathematica answer is: -1.000000000000000000000
Whereas GSL gives 0.0549469166662
If anybody is willing to discuss the calculation details; I would be
grateful. I have reached a conclusion that should be double checked.
Ray
My mathematica says -1.
Let me give a short form of why GSL (and some other algorithms) go
wrong.
Let b<a<0 be negative integers. Then we have a finite sum (polynomial)
if x>0 or x<0 . But in order to avoid certain calculation probems ; if
x<0 Kummers transform is applied M(a,b,x)=e^x M(b-a,b,-x) . Notice
that b<(b-a)<0 afterwards so the same summation is applied to M(b-a,b,x)
yielding another polynomial.
But if this were valid we could say.
e^x = ratio of two polynomials; which isn't true.
--
The primary use of conversation is to satisfy the impulse to talk
George Santanyana
