Am Dienstag, 24. Juli 2018 22:37:35 UTC+2 schrieb Erik Krause:
>
> Am 23.07.2018 um 17:26 schrieb [email protected] <javascript:>: 
>
> > Non-zero parameters a and c introduce singularities at r=0, something a 
> real lens does not have. 
>
> Since the correction function itself doesn't have a singularity at r=0 
>
I insist to disagree.

You are talking about a polynomial that covers the radial part in polar 
coordinates description. Polynomials do not have singularities, yes, in 
that you are right.

But we have to consider a two-dimensional mapping function. Using polar 
coordinates, it factors, this is nice, the phi mapping is identity, even 
nicer.
But polar corrdinates are slightly peculiar at vector-r=0, and to have them 
differentiable there the function in r must be odd.
(More general, odd f(r) for f(r)*sin(2n*phi+phi0) and even f(r) for 
f(r)*sin((2n+1)*phi+phi0) 

As the two-dimensional lens function is differentiable in vector-r=0, 
introducing non-differentiable functions is not a good idea.

and all that changes at r=0 is the slope of the curve there is a change 
> in magnification only in the center. Since a single mathematical point 
> can't be magnified there is no singularity either. The point in the 
> center stays in the center. 
>
Please compute, in cartesian coordinates, the partial derivative d^2/dxdx 
of the 2-dim mapping function with c=1 as parameter.
Hint: you'll encounter some x/abs(x) - like terms.
 
 

> In the wiki discussion you write "sqrt() has a singularity at 0". This 
> is not true. sqrt(0) is 0 
>
In simple language: singularities can be present even if the function is 
continuous.

Best regards
Klaus

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