Re: Solutions in polynomials with real powers



On Jan 30, 7:02 am, "G. A. Edgar" <e...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
In article <20070129163406.117...@xxxxxxxxxxxxxx>, David W. Cantrell

<DWCantr...@xxxxxxxxxxx> wrote:
"john" <xyz91...@xxxxxxxxx> wrote:
In Mathematica, when I did this: Solve[x^3 + x^(3/2) == -1], it
returns four solutions.
Why does it return four solutions?

Maybe because there _are_ four solutions (two conjugate pairs).

Davidthere are 6 solutions when you allow both branches of the 3/2 power

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/


If you use that interpretation of x^(3/2) then powering is no longer a
numeric function (in Mathematica-ese, Power would lose it's
NumericFunction attribute). I think the bad consequences of this would
far outweigh the benefit of regarding fractional powers as
multivalued. Among other possible drawbacks, one would also need to
regard logs as multivalued, hence also inverse trig functions, etc.

To see the parasite solutions one might do

In[6]:= InputForm[Solve[x^3 + x^(3/2) == -1, x, VerifySolutions-
False]]

Out[6]//InputForm=
{{x -> -(-1)^(1/9)}, {x -> (-1)^(2/9)}, {x -> (-1)^(4/9)},
{x -> -(-1)^(5/9)}, {x -> -(-1)^(7/9)}, {x -> (-1)^(8/9)}}


Daniel Lichtblau
Wolfram Research

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