Tsallis Constraints Sum(i) ei p(ei) power q = Eave and Sum (i) p(ei)=1 Mapping to One Constraint Sum(i) p(ei)=1?

It is possible to solve for the Tsallis distribution by maximizing Shannon’s entropy - Sum over i F(p(ei)) p(ei) power q (with F(p(ei)) = -ei/T) subject to the two constraints Sum over i ei p(ei) power q = Eave and Sum over i p(ei)=1. One may argue that p(ei) power q has physical meaning and the result is a power law (ei/T) power 1/(1-q) for ei/T>>1 (and (1+(1-q)(-ei/T)) power 1/(1-q) in general. Alternatively, one may seek a power law distribution which behaves as (ei/T) power 1/(1-q) (we write the power as 1/(1-q)) and then ask: What g(p) in the constraint Sum over i g(p(ei)) ei together with Sum over i p(ei) =1 allows for - Sum over i g(p(ei)) F(p(ei)) to be maximized. In this way one may find g=p power q. We suggest here that there exists a second way to consider the maximization of entropy problem. We argue that instead of seeking two independent constraints Sum over i g(p(ei)) ei =Eave (where g(p) = p power q) and Sum over i p(ei)=1, one may maximize entropy subject to a single constraint, i.e. Sum over i g(p) ei becomes proportional to Sum over i p(ei)=1. In other words, one solves a differently constrained system with the a priori knowledge that p(ei) behaves as (ei/T) power 1/(1-q) for ei/T>>1.