the gamma distribution must be equivalent to an exponential for alpha=1 so the pdf for x=0 must be equal to 1/beta not 0.
To the definition of the GAMMADIST function, GAMMADIST(0;alpha,beta,0) is always zero.
Please let core developers close bug. This is not invalid as being 0 will break several assumptions.
references in support of defining 0^0=1 rather than 'undefined' (note that no references suggest to _define_ 0^0=0, every time lim 0^x x->0 is used is to argue for undefined, not to argue for 0^0=0
and of course, in these days and age google is the ultimate reference :-)
$ echo "0^0" | bc
Furthermore the claim that
"To the definition of the GAMMADIST function, GAMMADIST(0;alpha,beta,0) is
need a bit more substantiation. based on the behavior of the function one could argue for undefined in the name of purity or 1 in the name of prolongation by continuity.
Note: ODFF says: If Cumulative is FALSE(), GAMMADIST returns 0 if x < 0 and the value
so if the above function is to be defined for x=0, extending by continuation for x -> 0 we have to have GAMMADIST(0,1,1,0) = 1
Sorry, i must accuse me, you are right. I'm wrong, In 3.5 I'll correct this
in interpre6.cxx we must pay attention to the case x=0.
For alpha < 1 -> #div/0
for alpha = 1 -> 1/beta
for alpha > 1 -> 0
I'll implement this for 3.5.0
@Luca, is this correct.
(In reply to comment #5)
> in interpre6.cxx we must pay attention to the case x=0.
> For alpha < 1 -> #div/0
> for alpha = 1 -> 1/beta
> for alpha > 1 -> 0
> I'll implement this for 3.5.0
> @Luca, is this correct.
That should be fine, thanks.
fixed in master with commit 9f397368cce0b94f76d4f1d74150ccc75c498e8b