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In the output from a calculation in mathematica stands a/((R^3*c)^(1/3)), with c and a constants and R the variable. Now I want to use the outputformula for a new calculation. But Mathematica does not understand that (R^3)^(1/3) is the same as R. So I don't get (correct) answers by using this formula. But when I replace (R^3)^(1/3) manually in the outputformula by R, then it works perfecty. My question is how I can manage that mathematica understands automatically (R^3)^(1/3)? I have already tried formulas like Simplify, FullSimplify, Collect and much more formulas like that.

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closed as too localized by belisarius, Yves Klett, whuber, m_goldberg, Oleksandr R. Apr 26 '13 at 16:40

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What if R == -1 ? –  belisarius Apr 26 '13 at 13:38
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Why do you think that (R^3)^(1/3) is the same thing that R? –  Leonid Shifrin Apr 26 '13 at 13:38
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This should have a canonical version... –  Szabolcs Apr 26 '13 at 14:24
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@belisarius I've noticed that "bubbles" article gets printed here from time to time. –  Daniel Lichtblau Apr 26 '13 at 14:29
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@Ajasja, if someone writes an answer with something like "Mathematica always assumes variables to be complex unless told otherwise" explicitly noted, maybe we can make this particular question canonical. –  J. M. Apr 26 '13 at 15:59
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2 Answers 2

If you make any assumptions you have to share them with Mathematica as well. For Example:

Assuming[R > 0, FullSimplify[(R^3)^(1/3)]]
(*R*)

The default assumption is that all variables are complex. (As @J.M. noted in the comments).

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Ahh... that too. –  Mark McClure Apr 26 '13 at 13:41
    
Ah thank you for your answers! I hadn't tought about that, because R means Radius. But now it works. Thank you! –  user7122 Apr 26 '13 at 14:58
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It's just not always true that $(R^3)^{1/3} = R$. How about $R=i$, for example?

N[(I^3)^(1/3)]

(* Out: 0.866025 - 0.5 I *)

If you expect this, you might have more luck with the real-valued CubeRoot function. For example:

FullSimplify[CubeRoot[R^3]]

(* Out: R *)
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In addition to being complex, it could be quaternion or even octonion. –  J Gregory Moxness Apr 26 '13 at 15:32
    
Now, now, Gregory, if OP is still reeling from being introduced to complexes... –  J. M. Apr 26 '13 at 15:57
    
Doesn't Matmematica automatically assume that the symbol is element of the complex numbers? I think that would be a nice addition to this answer if you want to be able to say: It's just not always true that: "...". Nothing is always true if you don't specify the domain of the variables :-/. Note that Solve[-1 == x*x, x] gives I and -I, but no quaternions and that b*a gets rearranged to a*b which is not true for quaternions. –  Jacob Akkerboom Apr 26 '13 at 16:41
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@JacobAkkerboom Yes, Mathematica generally assumes that unspecified symbols represent complex quantities. However, certain mathematical functions work only with a proper subset of the complexes. Many number theoretic functions, such as EulerPhi work only with the integers, for example. The point behind my answer is that the new CubeRoot function assumes the input is real. –  Mark McClure Apr 27 '13 at 14:22
    
Thanks for the response. I reacted too strongly. I still think that it would be nice if you put in your answer whether you mean that this does not always evaluate to True, or that commonly accepted mathematical definitions make the statement not always true for complex R. I think not everybody may use the definitions Mathematica uses. For example, Mathematica evaluates Log[-1] to I*Pi... More to the point, the way Mathematica defines Power must lead to a discontinous function, if I understand correctly. I can't undo my downvote now, it's locked, but I would love to change that :). –  Jacob Akkerboom Apr 27 '13 at 21:58
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