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RE: Arbitrary selection of a particular cube root of a negative number
*To*: mathgroup at smc.vnet.net
*Subject*: [mg32589] RE: [mg32540] Arbitrary selection of a particular cube root of a negative number
*From*: "Florian Jaccard" <jaccardf at eicn.ch>
*Date*: Wed, 30 Jan 2002 03:19:11 -0500 (EST)
*Sender*: owner-wri-mathgroup at wolfram.com
Hello !
If you have to work only on the real numbers, wich seems to be the case if I
understand correctly your question, you can load the
<<Miscellaneous`RealOnly`
package.
When the pyckage is on, (-x)^(1/3) with x<0, always gives -(x^(1/3)).
This package can be replaced by a better one you find on the Mathgroup site
:
http://library.wolfram.com/mathgroup/search/?words=0211-396&restrict=http%3A
%2F%2Flibrary.wolfram.com%2Fmathgroup%2Farchive%2F
Meilleures salutations
Florian Jaccard
EICN-HES
e-mail : jaccardf at eicn.ch
-----Message d'origine-----
De : Michael Peter Bakich [mailto:mike.bakich at cox-internet.com]
Envoyé : sam., 26. janvier 2002 10:08
À : mathgroup at smc.vnet.net
Objet : [mg32540] Arbitrary selection of a particular cube root of a
negative number
For a cube root of a negative value there are three roots possible. I
have found that the root mathematica chooses can be influenced by the
bounds chosen for a Plot3D graphic. I have noted that I get a different
graph (not just a case of more or less detail-or detail from a different
area of the graph) depending on which bounds I choose. The different
graphs correspond to different choices of the cube root (the expressions
I am using are complicated and graph interpretation is not an easy
"given", but I am fairly certain of this). I fixed this in one case by
writing my own cube root finder (if the values is negative, multiply by
-1 or apply absolute value, then take cube root, then return root
multiplied by cube root of minus one of your choice), but is there a
setting that can accomplish this for me? Since I am solving for a
variable and so getting a very large expression containing cube roots, I
then have to edit in my own cube root function in place of the one
supplied by Mathematica in the solve process. This has to be done for
every "run" and this is clumsy and time consuming.
Mike Bakich
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