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March 19th, 2018, 12:35 PM  #1 
Newbie Joined: Oct 2013 Posts: 20 Thanks: 0  Quartic equations  interesting problem
I can solve all square and cubic equations with complex coefficients. For quartic I use method from book "Solution of cubic and quartic equations"  S.Neumark. For $\displaystyle Ax^4+Bx^3+Cx^2+Dx+E$ In this method we first solving auxiliary cubic equation: $\displaystyle 1 2Cx^2+ (C^2 + BD  4AE)x (BCD  BBE  ADD)$ next we choose y = any(?) complex root from three roots, Next compute: $\displaystyle den = \sqrt(B^24Ay)$ $\displaystyle G =\frac{B + den}{2}$ $\displaystyle g =\frac{B  den}{2}$ $\displaystyle H = \frac{C  y}{2} + \frac{B * (C  y)  2 * A * D}{2 * den}$ $\displaystyle H = \frac{C  y}{2}  \frac{B * (C  y)  2 * A * D}{2 * den}$ and 4 roots will be just two + two roots of two square equation: For $\displaystyle Ax^2+Gx+H$ and $\displaystyle Ax^2+gx+h$ only double square equations (B==D==0) need other simpler method  compute roots of square roots. Is OK, but 100/million random equations (random coefficient from 2 to 2 step 0.25) is bad. Bad if chosen cubic root is special: if real or imaginary part is zero, or first or second is 0.5 or 0.25. My test: Code: 0.50, 0.50 0.50, 0.00 0.50, 0.25 0.50, 0.50 0.25, 0.25 0.25, 0.50 0.25, 0.00 0.25, 0.25 0.00, 0.25 0.00, 0.50 0.00, 1.00 0.00, 0.25 0.00, 0.50 0.25, 0.25 0.25, 0.50 0.25, 0.00 0.25, 0.25 0.25, 0.50 0.50, 0.25 0.50, 0.00 0.50, 0.50 1.00, 0.00 
March 19th, 2018, 01:13 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,710 Thanks: 675 
It is not clear what parameters you are varying. In any case, from past experience, computer calculations can at times give weird results due to rounding effects. For 4th degree or higher polynomial equations I found that Sturm sequence approach works much better for real solutions.

March 19th, 2018, 03:32 PM  #3 
Newbie Joined: Oct 2013 Posts: 20 Thanks: 0 
I found solutions, more important than roots $\displaystyle y_i$ is sum $\displaystyle B^24Ay_i$. If is near zero  it is bad. I must search root from cubic equation, which $\displaystyle B^24Ay_i$ has maximal modulus. If all are near zero  quartic equation has one quadruple solution. Problem is solved, but error propagation makes big errors in quadratic equations.


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