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Square root in FunctionField

asked 2012-11-29 01:39:45 -0500

Noud gravatar image

Hi,

I'm working with $q$-functions and I would like to define a free algebra over the algebraic field with an extra parameter $q$ for which I can also take roots i.e. I would like to do something like this

sage: F.<q> = FunctionAlgebra(AA)
sage: A.<a,b,c,d> = FreeAlgebra(F, 4)
sage: (1+q)**(1/2)*a*b
sqrt(1+q)*a*b

The function FunctionAlgebra is not the correct function, since it only includes (integer) powers of $q$. Is it possible to extend this FunctionAlgebra to something where sums of powers of $q$ also have roots? Or is there an other function for which I can do this?

Best, Noud

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answered 2012-12-01 08:59:51 -0500

Noud gravatar image

updated 2012-12-01 09:00:20 -0500

I found the answer myself. Instead of using AA you should use the SR (Symbolic Ring). An example:

sage: q = var('q')
sage: F.<a,b,c,d> = FreeAlgebra(SR, 4)
sage: (1+q)**(1/2)*b*a
(sqrt(q+1))*b*a

This SR ring is still a bit of a mystery to me, but it seems to work.

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Asked: 2012-11-29 01:39:45 -0500

Seen: 115 times

Last updated: Dec 01 '12