[Ifeffit] third cumulant

[grant bunker]

Matthew - as you know, it's trivial to just let b=1/a and the write  
the distribution as
(x-x)^s Exp[- (x-x0)/b]
so that the gaussian limit is obtained as b->0.  So infinite  
parameters are not a problem.

Your suggestion and other observations are quite right.  In fact, if  
convoluting distributions is ok,
you can use whatever your favorites are, since cumulants just add  
under convolution.

grant



On Jul 16, 2008, at 11:08 AM, ifeffit- 
request at millenia.cars.aps.anl.gov wrote:

> OK, here's my $0.02.  I've used the convolution of an exponential  
> tail function
> exp(-(r-r0)/w)    (r-r0)*w >= 0
> 0                        (r-r0)*r < 0
>
> with a Gaussian.  This avoids having to have parameters go to  
> infinity to approach a gaussian.  This function
> is a little unwieldy in real space but is simple in k-space.
>    mam