This sort of problem can be solved by identifying the generator with the smallest power. In this case `g = x**1/2` so `sqrt(x) + x -> y + y**2` which factors as `y*(1 + y) -> sqrt(x)*(1 + sqrt(x))`.
/c On Friday, February 24, 2023 at 3:58:37 AM UTC-6 atharv....@gmail.com wrote: > Lack of ability to factor out square roots seems to be a particular cause > for simplification issues like #23641 > <https://github.com/sympy/sympy/issues/23641>. For example, one might > expect factor(sqrt(x)+x) to return sqrt(x)*(1+sqrt(x)), but it does not do > anything. In this specific case, we may be able to avoid simplification > issues by using radsimp. However, in the more general scenario, it would be > very beneficial to be able to factor out radicals "naturally". > > One potential issue I see with this is that there are many expressions > that would factor 'indefinitely' if we were able to factor using radicals. > For instance, factor_rad(x+1) might return (sqrt(x)+I)*(sqrt(x)-I). > However, there are many other > <https://www.youtube.com/watch?v=pf5wc6E4Dxc> factorizations of this > expression into different radicals. > > The simplest fix for this would be to implement a way to specify a > "irreducible degree" with default setting 1. For instance, in the above > examples, we would choose 1/2. > > I would attempt to implement this right away, but I am new to both > open-source and SymPy, so I need some guidance. In particular, I don't > quite understand how _symbolic_factor works, and how exactly formal mode is > implemented for factor. > -- You received this message because you are subscribed to the Google Groups "sympy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sympy+unsubscr...@googlegroups.com. To view this discussion on the web visit https://groups.google.com/d/msgid/sympy/2a8d8514-40ee-42c5-80a9-b817742b8589n%40googlegroups.com.
