Hello,
Please consider the following code:
k=var('k')
f=function('f')(x)
solve(-1/3*log(f(x) + 1) + 1/3*log(f(x) - 2) == -k+x, f,to_poly_solve=True)
even if I call the solve function as
solve(-1/3*log(f(x) + 1) + 1/3*log(f(x) - 2) == -k+x, f(x),to_poly_solve=True)
or
solve(-1/3*log(f + 1) + 1/3*log(f - 2) == -k+x, f,to_poly_solve=True)
or
solve(-1/3*log(f+ 1) + 1/3*log(f - 2) == -k+x, f(x),to_poly_solve=True)
This always throws back [] at me.
However, if I substitue the function f with the variable z as shown below
k=var('k')
f=function('f')(x)
z=var('z')
solve((-1/3*log(f(x) + 1) + 1/3*log(f(x) - 2) == -k+x).subs(f(x)==z), z,to_poly_solve=True)
I get an answer
[z == (2*e^(3*k) + e^(3*x))/(e^(3*k) - e^(3*x))]
Under normal circumstances (where I don't need to use to_poly_solve=True) solve solves for the function.
Is there anyway to solve for a function (without the need to substitute it with a variable) when to_poly_solve=True is enabled?
Thanks in advance
