(07/29/2009, 05:03 PM)Tetratophile Wrote: ... the similarity between 
and 
strikes me.
I had similar feeling about these two pictures. Various holomorphic fastly growing functions shouild give similar pictures.
Quote:just a question... why do you show the , not the 
, the square root of gamma function?
Beacuse
is LOGO of the Physics Department of the Moscow State University, and there was an opinion that such a function cannot have any sense.
I do things which are declared to be "not doable". If one publishes an article with statement, that the "half-iteration of Gamma cannot be evaluated", then I hope to evaluate it as well.
Quote:... !(z) and Gamma(z) are bassically the same: !(z-1) = gamma(z). but is it easier to extract the root of 
?
The procedure for sqrt of Gamma should be similar to that for the factorial.
So, I do not think that
is much easier; basically, the same algorithm should work for sqrt(Gamma), and I expect it to give the similar picture.
We can evaluate the half-iteration of various holomorphic functions, not only exponential or factorial; and not only the half-iteration, but any iteration (even the complex one). I suggest the examples:
In nonlinear acoustics, in may have sense to characterize the nonlinearities in the
attenuation of shock waves in a homogeneous tube. (This could find an application in some advanced muffler that uses nonlinear acoustic effects to withdraw the energy of the sound waves without to disturb the flux of the gas). The analysis of the nonlinear response, id est, the Transfer Function H, may be boosted with the superfunction; in particular, the TransferFunction of a tube, which is half-shorter, is sqrt(H).
If the snowball rolls down from the hill, getting heavier, and the mass at the bottom can be characterzed with the TransferFunction H of the initial mass,
then the TransferFunction of the half-hill is sqrt(H).
If a small drop of liquid diffuses through the oversaturated wapor in a vertical tube of length h, and its mass at the output is TransferFunction H of the imput mass,
then the TransferFunction of the similar tube of length h/2 is sqrt(H).
If the logarithm of power of light at the output of the homogeneous nonlinear optical fiber is TransferFunction H of the logarithm of its input power,
then the TransferFunction of a half of this fiber is sqrt(H).
Can you suggest more such examples?
)
and )
are shown with lines=const)
and=const)
. 
are shown with thick black lines.
const
are shown with thick red lines.
is shown with dark pink.
are shown with thick blue lines.
const are shown with green. \Big)=z!)
and \Big)=\exp(z))
