firstname.lastname@example.org wrote: > Hmmm! Tried following this approach, i.e. solving f(b)=0 with Netwon's and then plugging b in a=g(b), but it doesn't work. The solution doesn't satisfy the initial system. Try for example > > L=-1 > U=1 > p=0.0013498980316301 > > I get > > a=-1 > b=1 > > which clearly is wrong, i.e., a and b don't solve the initial system, even though f(b)=f(1)=0 and a=g(b)=g(1)=-1. I rewrite the initial system here: > > ((a-L)^2/a)*(2/(a+b))-p=0 > ((b-U)^2/b)*(2/(a+b))-p=0 >
In first message you wrote:
> with a>L>0, b>U>0.
Now you put L < 0, which violates this condition.
> Am I doing something wrong, or is FriCAS/Derive solution wrong?
FriCAS solution is "generic" in the sense that:
1) All solutions of your system are contained in FriCAS solution 2) It solves your system if there is no division by zero when plugging FriCAS solution into your system 3) Unless there are some special relation between L and U there will be no division by zero when plugging FriCAS solution into your system.
In case of your system special relations are: L = 0, U = 0, L + U = 0. If no of the above relations hold then you need not worry about division by zero (in particular no worry when U>0 and L>0). When one of relations holds, you need to decide if you want to you your original system (that is throw out (a, b) which cause division by zero), or you want to simplify it and solve new, simpler system.
BTW: Since a = -1, b = 1 is the only real solution that FriCAS found, this means that your orignal system even after simplification has no real solutions.