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

**References**:**Re: integrate***From:*"David W. Cantrell" <DWCantrell@sigmaxi.net>