Most Maxima functions expect simplified expressions, and aren't
guaranteed to work on unsimplified expressions; mquotient never
appears in simplified expressions. The Maxima evaluator (meval)
simplifies everything for you, but if you're working in Lisp, you need
to be careful about calling simplify whenever necessary.
-s
On 4/29/08, Rupert Swarbrick <rswarbrick at googlemail.com> wrote:
> "Stavros Macrakis" <macrakis at alum.mit.edu> writes:
>
> > On Tue, Apr 29, 2008 at 7:27 AM, Rupert Swarbrick
> <rswarbrick at googlemail.com>
> > wrote:
> >
> > You get some function, f(w) and want to know the power of the leading
> > term in the taylor series of f about 0.
> > ...hipow( rat( first( taylor( f, w, 0, 0 ) ) ) )
> > (well actually it's in lisp, but it's basically this)
> >
> > But if the power is negative then hipow returns 0.
> >
> >
> > I don't have this problem (Maxima 5.14 GCL).
> >
> > taylor(atan(x)/x^3,x,0,4) => 1/x^2-1/3+x^2/5-x^4/7
> > ff: first(%) => 1/x^2
> > hipow(rat(ff),x) => -2
> > hipow(ff,x) => -2
> >
> > Could you please show us your exact code so we can reproduce and debug the
> > problem? I wonder if it has something to do with rat variable ordering?
> (Why
> > do you have rat in there?)
> >
> > -s
>
> Ahah! I've hit it again, but now I know why:
>
> MAXIMA> ($hipow '((MQUOTIENT) 2 $x) '$x)
> 1
> MAXIMA> ($hipow '((mtimes) 2 ((mexpt) $x -1)) '$x)
> -1
> MAXIMA>
>
> The problem is that $hipow doesn't know about mquotient - is there a
> nice way to do the transformation above? (which I did by hand!)
>
> Feeling (very marginally) less stupid,
>
> Rupert
>
>
>
> P.S. No, I have no idea why I ended up with rat in there.
>
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