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March 19th, 2018, 12:35 PM   #1
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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: $\displaystyle den = \sqrt(B^2-4Ay)$
$\displaystyle G =\frac{B + den}{2}$
$\displaystyle g =\frac{B - den}{2}$
$\displaystyle H = \frac{C - y}{2} + \frac{B * (C - y) - 2 * A * D}{2 * den}$
$\displaystyle H = \frac{C - y}{2} - \frac{B * (C - y) - 2 * A * D}{2 * den}$
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
Is know this behavior? Is possible two or even three cubic root bad?
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March 19th, 2018, 01:13 PM   #2
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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.
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March 19th, 2018, 03:32 PM   #3
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I found solutions, more important than roots $\displaystyle y_i$ is sum $\displaystyle B^2-4Ay_i$. If is near zero - it is bad. I must search root from cubic equation, which $\displaystyle B^2-4Ay_i$ 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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