[Axiom-developer] [AxiomProblems] (t>0)$POLY INT

[Bill Page]

Changes http://page.axiom-developer.org/zope/mathaction/AxiomProblems/diff
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But in 'D(N(t,0,3),t)' you are not calling the function N
with numeric parameters. In 'N(t,0,3)' the type of t is
'Variable t'. Ultimately 'N(t,0,3)=0' because of your funtion
definition 'N0(t|(t<0) or (t>1))==0'. This is '0' because
't>1' is 'true' when 't' is of type 'Variable t'. You can
see why if you use the option ')set message bottomup on' to
see the mode map selection
\begin{axiom}
)set message bottomup on
t>1
\end{axiom}
So Axiom interprets both 't' and '1' as being of type 'POLY INT'
and the function '>' is defined by the lexical ordering of the
polynomials.

I will grant that this result is counter-intuitive, but I think
that once you understand why Axiom gives this result then you
will be in a good position to understand the rest of Axiom's
type system!
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forwarded from http://page.axiom-developer.org/zope/mathaction/address@hidden