[Axiom-developer] [ExampleSolution1] Still a problem.

[wyscc]

Changes http://page.axiom-developer.org/zope/mathaction/ExampleSolution1/diff
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<pre>From wyscc, March 16, 2005 05:06:00</pre>



The above does not really solve the problem on differentiation of a piecewise 
function, which, in my opinion, is an unreasonable expectation in general 
because of the multitude of ways to define the conditions; it may help if one 
uses the Heaviside function. The numerical definition of <code>N0(t)</code> is 
actually totally ignored in the definition of <code>N(t,i,p)</code> and thus 
also in the differentiation process, by the use of 
<code>operator('No)(t)</code>. Indeed, in the expression for <code>N</code> or 
its derivative with respect to <code>t</code>, the  "function" <code>N0</code> 
is still an <code>operator</code>, and as such, <code>N0(2)</code> is 
undefined! In other words, there is a distinction between the numerically 
defined function <code>N0</code> and the operator <code>N0</code>. This is 
illustrated below.

Aside: I am getting into something I don't quite understand: the first group of 
code is not meant to be there, but it somehow returns something wrong. If the 
commented line <code>--dNdt(t)</code> is removed, the result for the 
<code>subst</code> line is what I expect. If the order of the commands is as 
for the second group, the result is ok too.
Finally if I copied the block to the end and run it a second time, everything 
is also ok. But his may go away after I save. Image is in: 
http://page.axiom-developer.org/zope/mathaction/images/1141703130-18px.png. 

\begin{axiom}
N(2,0,3)
dNdt(t)==D(N(t,0,3),t)
--dNdt(t)
subst(dNdt(t), t=2)
dNdt(2)
\end{axiom}

Compared with

\begin{axiom}
dNdt(t)==D(N(t,0,3),t)
subst(dNdt(t), t=2)
N(2,0,3)
dNdt(2)
\end{axiom}

Notice that the evaluation for <code>N0(2)</code> is not really done. One way 
to avoid this error is to use substitution instead of a function call, as done 
in the second line above. 

The last function call <code>dNdt(2)</code> is deliberate, to illustrate a 
common error when mixing numeric and symbolic computation: first define the 
derivative as a function of <code>t</code> and then evaluate the derivative at 
some value of <code>t</code>. This does not work because when the derivative is 
called the system (Axiom, or other systems) will substitute the value of 
<code>t</code> before differentiating. 

\begin{axiom}
N(2,0,3)
dNdt(t)==D(N(t,0,3),t)
--dNdt(t)
subst(dNdt(t), t=2)
dNdt(2)
\end{axiom}


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