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Re: Re: integrate
*To*: mathgroup at smc.vnet.net
*Subject*: [mg72461] Re: [mg72453] Re: integrate
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Fri, 29 Dec 2006 05:19:48 -0500 (EST)
*References*: <emgg3i$2ca$1@smc.vnet.net> <200612281036.FAA05409@smc.vnet.net>
On 28 Dec 2006, at 19:36, David W. Cantrell wrote:
>
>
> The product of the algebraic factors involving x should be just 1, but
> I was not able to get Mathematica to give that:
>
> In[8]:=
> FullSimplify[Sqrt[1 - (-1)^(1/3)*x^2]*Sqrt[1 + (-1)^(2/3)*x^2]*
> Sqrt[1/(1 - x^2 + x^4)], x > 0]
> Out[8]=
> Sqrt[1 - (-1)^(1/3)*x^2]*Sqrt[(1 + (-1)^(2/3)*x^2)/(1 - x^2 + x^4)]
>
> Maybe I didn't try hard enough. Can someone suggest how I should get
> the above to simplify to 1, perhaps assuming x > 0?
It may not be quite "simplifying" but you can certainly use
Mathematica to "prove" that this is 1.
In[1]:=
FullSimplify[(Sqrt[1 - (-1)^(1/3)*x^2]*Sqrt[(-1)^(2/3)*x^2 + 1]*
Sqrt[1/(x^4 - x^2 + 1)])^2]
Out[1]=
1
So if Mathematica is right then we know that the expression can only
be 1 or -1. Since we can show that for some chosen real value it is
equal to 1
FullSimplify[Sqrt[1 - (-1)^(1/3)*x^2]*Sqrt[((-1)^(2/3)*x^2 + 1)/(x^4
- x^2 + 1)] /.
x -> 2]
1
and since it is defined everywhere and continuous on the real line,
it has to be equal to 1 for all real x. Actually, it is clear that
the expression has to be 1 even for all complex values of x (with the
usual convention about principal values for roots and fractional
powers), (assuming of course that we agree to consider the value of
the expression to be 1 at the points where x^4 - x^2 + 1 == 0 and
were it has a removable singularity).
Andrzej Kozlowski
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