Waldek,

Do you mean that 'sqrt(x^2)' is generated internally in transformation
Kurt's (1), resulting in incorrect sign?

I suppose that you must have meant to write the transformation using
'sqrt(x^2)/x', i.e.

  'f(sqrt(x^2))' into (1 + sqrt(x^2)/x)/2*f(x) + (1 - sqrt(x^2)/x)/2*f(-x)

It looks interesting but I guess this is for real-valued x only?  For complex z

  sqrt(z*conjugate(z))/z = exp(argument(z))

is the "complex sign".  Transforming

  'f(sqrt(z*conjugate(z)))' into (1 + sqrt(z*conjugate(z))/z)/2*f(z) +
(1 - sqrt(z*conjugate(z))/z)/2*f(-z)

(assuming conjugate existed), looks funny to me.

Bill.

On 26 September 2014 14:06, Waldek Hebisch <[email protected]> wrote:
>
> Bill Page wrote:
> >
> > It seems to me that algebraically any transformation that eliminates or
> > adds an extra sqrt is likely incorrect.  So I would say that (1) in your
> > example is very suspicious.
> >
> > On 25 September 2014 20:04, kfp <[email protected]> wrote:
> >
> > >
> > > Indeed. Since we get
> > >
> > > (1) -> normalize sqrt(1-sin(z)^2)
> > >
> > >              z 2
> > >          tan(-)  - 1
> > >              2
> > >    (1)  -----------
> > >              z 2
> > >          tan(-)  + 1
> > >              2
>
> We have problem when roots exist in base field.  That is
> lone 'sqrt(x)' does not cause correctness problems.
> So, as long as there are no roots in base field we can
> add them.  OTOH 'sqrt(x^2)' is problematic.  In principle
> we could transform 'f(sqrt(x^2))' into
>
> (1 + sqrt(x^2))/2*f(x) + (1 - sqrt(x^2))/2*f(-x)
>
> and then handle 'f(x)' and 'f(-x)' separately.  However,
> there are many places where we would need to implement
> such transformation.  Also this leads to ugly results.
>
>
> --
>                               Waldek Hebisch
> [email protected]
>
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