Re: [Maxima-discuss] Another solve toughie

[Michel T. <ta...@lp...>]

Le 04/03/2026 à 22:52, Richard Fateman a écrit :
> solve( y =  (sqrt(4*x+1)-sqrt(x+1))/(sqrt(9*x+1)-sqrt(x+1)) , x);
>
> a solution is x= -((4*(y-1)*y*(8*y-3))/((2*y-1)*(2*y+3)*(4*y-3)*(4*y+1)))
>
> a challenge might also be:  prove that
> f(x) := if x=0 then 3/8 else 
> (sqrt(4*x+1)-sqrt(x+1))/(sqrt(9*x+1)-sqrt(x+1))
> is continuous everywhere.
>
>
> RJF


This equation is very interesting. First thing obvious, how is it that 
one can find such a simple solution for such  a complicated equation? A 
way to get this result is the following: assume s=sqrt(x+1), 
t=sqrt(9x+1), u=sqrt(4x+1), then our equation reads y*(t-s)-u+s=0 so we 
get a system in which we eliminate s,t,u.

(%i2) eliminate([y*(t-s)-u+s,u^2-4*x-1,t^2-9*x-1,s^2-x-1],[s,t,u]);

(%o2) [x^4*(x*(64*y^4+32*y^3-92*y^2+12*y+9)+32*y^3-44*y^2+12*y)^4]

from which it is obvious that the solution (note that the expression is 
simple in x, complicated in y, i don't know why) is:

(%i3) solve(%,x);

(%o3) [x = -((32*y^3-44*y^2+12*y)/(64*y^4+32*y^3-92*y^2+12*y+9)),x = 0]

(%i4) factor(%);

(%o4) [x = -((4*(y-1)*y*(8*y-3))/((2*y-1)*(2*y+3)*(4*y-3)*(4*y+1))),x = 0]

as claimed above. Since eliminate may produce spurious solutions, it is 
necessary to look at the graphs of y=y(x)and x=x(y) to see that they are 
indeed inverses to one another- assuming we are interested in real 
solutions. For y=y(x) one must have x>=-1/9, and x can go to + infinity, 
going through x=0 at which one has an indeterminate form that Taylor 
reduces to 3/8.

plot2d((sqrt(4*x+1)-sqrt(x+1))/(sqrt(9*x+1)-sqrt(x+1)),[x,-1/9,10]);

shows that y(x) is increasing monotonically from (4-sqrt(10))/4 = 0.2094 
for x=-1/9 (as Taylor shows) to 0.5 at x= + infinity.So when examining 
x=x(y) one has to restrict to 0.2094 < y < 0.5. Plotting shows x=x(y)  
is also monotonically increasing from -1/9 to infinity obtained when 
y=1/2  as obvious from %o4. It is then clear that we have a correct 
solution to the equation, and that this is the *only* real solution to 
the equation.

However i have seen a small problem.

radcan(subst(x= -((32*y^3-44*y^2+12*y)/(64*y^4+32*y^3-92*y^2+12*y+9)),
(sqrt(4*x+1)-sqrt(x+1))/(sqrt(9*x+1)-sqrt(x+1)));

yields not y as one may expect, but (3*y-3)/(8*y-3).  It happens that y 
-> (3*y-3)/(8*y-3) is an involution (its square is identity). I suppose 
this is because radcan makes some choices that are inappropriate here. 
This points to the dangers floating around the use of (sqrt(z))^2=z and 
such stuff.

This is the occasion to remark that augmenting maxima solve program by 
stuff such as taking the log of both sides, or taking the square of both 
sides (as i have shown in the same thread), is not necessarily well 
advised, and will certainly be ineffective for examples as the present 
one. There are things that must be done with conscient thought of the 
end user.


-- 
Michel Talon