How to deal with r^(bfloat(2/3))/r^(2.0/3)

• Subject: How to deal with r^(bfloat(2/3))/r^(2.0/3)
• From: zycentre
• Date: Thu, 13 Dec 2007 00:27:17 +0800 (CST)

(%i5) r^(bfloat(2/3))/r^(2.0/3);
                                            -.19073b-6
(%o5)                             r

The right result should be 1.
The reason is:
(%i6) (bfloat(2/3))-(2.0/3);
(%o6)                             -.19073b-6

So, I want to set a rule: when the index less than a certain value, like 1.e-6, take it as 0.

How to do this in maxima level (better not in lisp, I can't understand lisp)?

Thank you for your consideration.

Yuanchuan Zou

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