I fiddled around with this equation while on the airplane today with Maxima. If you want to do this, you should be aware of Maxima's "allroots" function, which is capable of quickly solving numerically quintic polynomials. Maxima also has a "realroots" function, which works, but doesn't tell you what is happening to the complex roots. Also, Maxima has the capability to plot the various roots, after you have stripped them into real & imaginary parts. If you're just trying to understand what's going on with this equation, the restriction to x>1 and c>(1+a)(1+b) isn't necessary. If c=(1+a)(1+b), then x=1 is always a solution, so you can then focus on what's happening with the complex roots. For some equations, this simplification might cause trouble, but for this equation, nothing particularly different happens when c gets somewhat larger. As Gene has already pointed out, there should be only one real root to this equation, with the complex roots in conjugate pairs. The only possibility for more real roots is if the derivative of the equation has real roots. But this would require that sqrt(9a^2-2ab+9b^2)-3(a+b) > 0 which never happens when a,b>0. So the derivative equation has all non-real roots, which then implies that the quintic has only one real root. (We could probably also have used Sturm sequences to figure this out, but I haven't figured out how to get Maxima to do them yet.) When c=(1+a)(1+b), and b=a, and a=1, we have x^5-1, which has the 5 roots in the unit circle. Allowing a to get bigger (even _much_ bigger), the "circle" of roots gets bigger and more ovoid, but the behavior doesn't change very much. When b starts differing from a, we get a slightly different ovoid shape, but the behavior still doesn't change very much. So, in answer to your original question about initial approximations to the real solution, we should keep the following things in mind: 1. The sum of all the roots is zero, since the coefficient of x^4 is zero. So the roots surround the origin. 2. The product of all the roots is c, which gives some idea of the size of the mushed/ovoid "circle". However, when c=(1+a)(1+b), x=1 is always a solution, so the position of the real solution may not get large even when c is very large. So, my recommendation is to start any approximation for a real root at x=1 ! Newton's convergence is not so hot around multiple roots. Cyclotomic polynomials can look like multiple roots from far away, but we seem to be starting close enough that Newton has its usual quadratic vigor, so the convergence is very fast. There are more complex iterations that attempt to approximate _all_ of the roots "simultaneously", starting from an initial approximation to each root. I believe Henrici studied some of these methods. However, I almost always found these methods to be too much trouble, and not really "simultaneous" at all. The reason is that they only start working well when at least one root is approximated very well, and then the others come "into view" essentially one by one (or as conjugate pairs). So you do almost as well by finding the roots one by one (or by conjugate pairs), and then "deflating" the polynomial by dividing out the known factor. There is also a classical Newton method for approximating a conjugate pair of roots simultaneously; it guesses a quadratic factor for the polynomial & iterates. I did a quick Google search to find the name of this algorithm, but wasn't able to find it. If anyone cares, I'll do a deeper search. I think that it appeared in one of the early ACM publications. P.S. I also tried experiments with a general expression for a quintic with 1 real and 2 pairs of conjugate complex roots and equated coefficients with this particular quintic. I was able to algebraically eliminate the many of the variables, and got a relatively compact algebraic relationship among the rest. This is fortuituous, as a very large expression factored as the eighth (where did eight come from ??) power of a much smaller expression. However, it didn't provide much insight. At 03:23 AM 2/2/2007, Christian Boyer wrote:
Sorry, some mistakes in my previous hypothesis.
With the 5th-degree equation:
x(x² + a)(x² + b) = c
x, a, b, c real numbers a, b > 0 c > (1+a)(1+b) x > 1
what is the fastest method to get an approximated value of x? (very few steps)
Christian.