Waldek Hebisch <[EMAIL PROTECTED]> writes:
> I guess I should explicitly write parenthesis here. I say
>
> normalize(acos (cos(z))
>
> evaluated numerically _have to_ give different values than
>
> (acos$Float)(cos (z))
>
> because normalize acos cos z must be meromorphic, so the
> only possibility is z + 2*k*%pi (note that k has to be constant).
> The simplest version is with k = 0. Numeric version must be different
> (in numeric version k can not be a constant).
You also wrote:
> From analitical point of view Expression consists of meromorphic
> functions in some unspecified domain
OK, so possibly the problem arose because some people wanted that EXPR INT
would also contain non-meromorphic functions.
> To put things in different words: normalize is supposed to find
> kernels k_1, k_2,..,k_n,...k_m such that k_1,...,k_n are
> algebraically independent, k_{n+1},...,k_m are algebraic and
> the differential field generated by subexpressions of our
> expression is Q(k_1,...,k_m). The important part is that
> we need _a field_.
I suppose that this is a requirement of the Risch algorithm? And is
equivalent to saying that the result of normalize should be meromorphic?
If so, we really want (and you said this previously)
* one domain with "general functions" - this could contain definitions
with conditionals, where
* each conditional expression is a meromorphic functions.
Correct?
> Now take f = (sqrt(z^2) - z)*(sqrt(z^2) + z).
> On real line using branch cut interpretation f is 0. So
> the corresponding ring contains zero divisors, so it is not
> isomorphic to field. In other words, when you start forming
> compositions numeric version will get "removable" branch cuts.
> If you try to add the same cuts in normalize you will get
> zero divisors and your ring will no loger be a field.
I'll have to digest that properly. Dinner first.
Martin
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