Yes, the treatment of radicals should be explained clearly somewhere. But
it is not quite so simple as you might think.
High-school algebra tells us that sqrt(x)*sqrt(y) = sqrt(x*y). In
particular, %i*sqrt(x)=sqrt(-1)*sqrt(x)=sqrt(-x). *But* sqrt(-1)*sqrt(-2) =
%i*%i*sqrt(2)= - sqrt(2) and sqrt((-1)*(-2))=sqrt(2)
So the basic premise of rootscontract is problematic. I suspect that Dieter
Kaiser is thinking through cases like this to make them more correct. The
problem is that the fully correct answer may be more complicated than you
think....
-s
On Mon, Aug 2, 2010 at 15:22, mok-kong shen <mok-kong.shen at t-online.de>wrote:
> Stavros Macrakis wrote:
>
>> Why do you have radcan in there? Radcan does not guarantee any
>> particular root. Why not just rootscontract(sqrt(...)/sqrt(...))?
>>
>
> Thank you once again. I am beginner. The manual says only that
> "radscan(expr) simplifies expr" and hence I didn't expect that
> the result could under circumstances involve a change of sign.
> Would it be favourable that the manual gives a warning about this?
>
> M. K. Shen
>
>
>