but
(%i391) ((-x-1)^(-bb-aa-2)*(x+2)^aa)/(((x+2)/(x+1))^aa*(1/(x+1)-(x+2)/(x
+1)^2)*(1-(x+2)/(x+1))^bb);
(%o391) ((-x-1)^(-bb-aa-2)*(x+2)^aa)/(((x+2)/(x+1))^aa*(1/(x+1)-(x+2)/(x
+1)^2)*(1-(x+2)/(x+1))^bb)
(%i392) radcan(%);
(%o392) -(x+1)^aa/(-x-1)^aa
radcan(%-radcan(%)); /* gives zero */
> 0
which is all that the documentation guarantees will work. I rely on this to tell if an expression is linear or quadratic by
subtraction of the radcan'ed expression from its simplified result. I hope that still works.
Rich
----- Original Message -----
From: "Richard Fateman" <fateman at cs.berkeley.edu>
To: "Barton Willis" <willisb at unk.edu>; "Maxima List" <maxima at math.utexas.edu>
Sent: Sunday, June 28, 2009 8:58 PM
Subject: Re: [Maxima] radcan not Idempotent
Barton Willis wrote:
> Here is an expression e for which radcan(radcan(e)) # radcan(e). I'd
> call this a bug:
>
> (%i391) ((-x-1)^(-bb-aa-2)*(x+2)^aa)/(((x+2)/(x+1))^aa*(1/(x+1)-(x+2)/(x
> +1)^2)*(1-(x+2)/(x+1))^bb);
> (%o391) ((-x-1)^(-bb-aa-2)*(x+2)^aa)/(((x+2)/(x+1))^aa*(1/(x+1)-(x+2)/(x
> +1)^2)*(1-(x+2)/(x+1))^bb)
>
> (%i392) radcan(%);
> (%o392) -(x+1)^aa/(-x-1)^aa
>
> (%i393) radcan(%);
> (%o393) -1/(-1)^aa
>
>
> Barton
>
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>
for what it is worth, commercial macsyma gets 2 different answers, but
the second one is
- (%e)^( - %i * %pi * aa)
its a bug, I think.
RJF
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