tehno said:
Well,I don't have that software.
That's why I asked for help in a cross check in first place
))
Ah, but you see, I don't have the software either, otherwise I would certainly have been able to help you out.
Maybe someone else will step in here ?
But I have an interesting question of even higher importance:
Suppose we have to charaterize set of all REAL solutions for system of two equations:
f(x,y)=0
g(x,y)=0
__________________
If f(x,y),g(x,y) are both polynomials of order n>4 such
that f(x,y)=g(y,x) is satisfied,and the polynomials have rational coeficients,
what theorem(s) can answer the question concerning the real solutions set characterisation and number of real solutions of the system judging just on the base of the coeficients and their signs?
Edit:{Of course,n is also known }
I'm afraid that's a little out of my depth. But if the degree of the polynomial is greater than 4 (meaning 5 and above), then there is no guarantee an analytical solution exists. I believe there are classes of quintics (and higher orders) where such solutions can be found, but I am not well up on that theory. I assume the same considerations should apply to simultaneous equations of such high order (except that when one "merges" the equations so that they get a single polynomial in one variable, the degreee is likely to be higher than either of the two separate equations).
I'm afraid I cannot really help with the question of the simultaneous equations in arbitrary order - maybe one of the mathematicians here can assist you better.
Considering only single variable polynomial equations of degree 4 and below (quartics and simpler), a guaranteed analytical solution exists, even though it's tedious to find. I just put up a tutorial linking a good site giving the general method of solution, check out this reference :
https://www.physicsforums.com/showthread.php?t=119284
If you don't want to solve, but only want to ascertain how many roots are real or complex, that's fairly simple. It becomes a simple problem of reduction and deduction.
Let's say we have a quartic in x called f(x). For simplicity, always take the coefficient of x^4 to be positive (in fact, let it be positive one).
f(x) = 0 can have no real solutions (2 sets of complex conjugate roots), 2 real and 2 complex solutions or 4 real solutions. (Note that repeated roots are possible). The problem is to determine which case holds.
The easiest way is to sketch the curve and consider the turning points of the curve. The quartic is guaranteed to either one or three turning points.
If the quartic has only one turning point, then the turning point can occur above (that is, f(x) is positive at that point), at (tangential to, meaning f(x) is zero at that point) or below the x-axis (f(x) negative at that point). If it occurs above, there are no real roots (hence 4 complex roots). If it occurs below, then there are exactly 2 distinct real roots (and 2 complex roots). If it occurs tangential to the x-axis, then there are repeated real roots.
If the quartic has three turning points and they all occur above the x-axis, there are no real solutions. If all 3 turning points occur below the x-axis, there are 2 distinct real roots. And if two turning points occur below the x-axis and one occurs at or above the x-axis, there are 4 real roots (in the case where the turning point forms a tangent to the axis, there is a repeated root). Simple, no ?
But how to find the turning point(s) ? For this, examine the derivative of the quartic.
At the turning points, f'(x) = 0. The problem is reduced to solving a cubic. The cubic is guaranteed to have one real root. In addition it may have another 2 real roots or a pair of complex conjugate roots.
You can either simply solve the cubic, or repeat the process to deduce how many real roots the cubic has. A cubic function has either no turning points or two turning points. It may also have a single point of inflexion. You can determine which case holds by examining the derivative of the cubic. f''(x) is a quadratic function. If the discriminant of the quadratic is negative, then there are no stationary points (turning or inflexion points) for the cubic f'(x). If the discriminant is zero, then there is a single inflexion point. If the disc. is positive, there are 2 real turning points on the cubic. For a cubic of the form x^3 + bx^2 + cx + d = 0, the problem reduces to determining the sign of b^2 - 3c.
You can go through the same logical process to determine, based on the position of the turning points, whether the cubic has a single real root or 3 real roots. Then use those deductions to work out the nature of the roots of the quartic.
This process (of determing merely the nature of the roots) should be generalisable to any degree of polynomial equation. The difficulty lies in actually finding those roots.
