NORMALIZED CUBIC EQUATION ROOTS

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navy    Normalized Cubic Equation

The plot shows the roots of the cubic equation in one diagram.
It serves as the general case,

$z_0$,

by using a linear transformation
for both the independent and dependent variable.  It should be
useful for rough calculations or starting points for Newton's
method.  Explicit formulas for the real and imaginary component
of each root are given so the ambiguity of root choice is not left
to the reader.


$\mathrm{ }z\bullet \sqrt[3]{\left(\frac{2}{n}\right)}\mathrm{ }\mathrm{versus}\mathrm{ }x=\frac{4\bullet m^3}{n^2}$

$\mathrm{ }z=z_0+\frac{b_0}{3\bullet a_0}$





$\mathrm{ }a\bullet z^3\left(x\right)+b\bullet z^2+c\bullet z+d=0$



$\mathrm{ }x=\frac{4\bullet m^3}{n^2}$

 

$\mathrm{ }a=1$

$\mathrm{ }b=1$


$\mathrm{ }m=\frac{1}{9}-\frac{c}{3}$

$\mathrm{ }n=\frac{c}{3}-d-\frac{2}{\mathrm{27}}$

$\mathrm{ }c=-3\bullet m+\frac{1}{3}$


$\mathrm{ }d=-m-n-\frac{1}{\mathrm{27}}$





 Normalized Cubic Equation   The single real root is blue
Analytical expressions for the curves are developed at
 cubic_exp.xml  from
the expressions reported at
 Weisstein                             
 wikipedia  and by   
 Knaust                        
The real and imaginary components of the derived expressions were found
to be exact fits.






$\mathrm{For} \frac{4\bullet m^3}{n^2}<0, z_{\mathrm{}}=\sqrt[3]{1+\sqrt{1-\frac{4\bullet m^3}{n^2}}}-\sqrt[3]{-1+\sqrt{1-\frac{4\bullet m^3}{n^2}}}-\frac{1}{3}$








$\mathrm{For} 0<\frac{4\bullet m^3}{n^2}<1, z_{\mathrm{}}=\sqrt[3]{1+\sqrt{1-\frac{4\bullet m^3}{n^2}}}+\sqrt[3]{+1+\sqrt{1-\frac{4\bullet m^3}{n^2}}}-\frac{1}{3}$








Exact agreement for the first real root when all roots are real.



$z=\sqrt[3]{\frac{n}{2}}\bullet \sqrt[3]{\frac{4\bullet m^3}{n^2}}\bullet 2\bullet \cos \left(\frac{0\bullet \pi }{6}+\frac{\arctan \left(\sqrt{\frac{4\bullet m^3}{n^2}-1}\right)}{3}\right)-\frac{1}{3}$





The imaginary components of the complex roots are red color coded curves.



$\mathrm{For} \frac{4\bullet m^3}{n^2}<0, z_{\mathrm{i}}=\pm \frac{\sqrt{3}}{2}\bullet \left(\sqrt[3]{1+\sqrt{1-\frac{4\bullet m^3}{n^2}}}-\sqrt[3]{-1+\sqrt{1-\frac{4\bullet m^3}{n^2}}}\right)$








$\mathrm{For} 0<\frac{4\bullet m^3}{n^2}<1, z_{\mathrm{i}}=\pm \frac{\sqrt{3}}{2}\bullet \left(\sqrt[3]{1+\sqrt{1-\frac{4\bullet m^3}{n^2}}}+\sqrt[3]{1-\sqrt{1-\frac{4\bullet m^3}{n^2}}}\right)$





 The real component of the complex roots is color coded orange.
$\mathrm{For} \frac{4\bullet m^3}{n^2}<0, z_{\mathrm{r}}=\frac{1}{2}\bullet \left(\sqrt[3]{1+\sqrt{1-\frac{4\bullet m^3}{n^2}}}-\sqrt[3]{-1+\sqrt{1-\frac{4\bullet m^3}{n^2}}}\right)$



 The real component of the complex roots is color coded orange.
$\mathrm{For} 0<\frac{4\bullet m^3}{n^2}<1, z_{\mathrm{r}}=\frac{1}{2}\bullet \left(\sqrt[3]{1+\sqrt{1-\frac{4\bullet m^3}{n^2}}}+\sqrt[3]{+1+\sqrt{1-\frac{4\bullet m^3}{n^2}}}\right)$









 fuchsia

 cyan

 magenta

 pink

 green

$z=\sqrt[3]{\frac{n}{2}}\bullet \sqrt[3]{\frac{4\bullet m^3}{n^2}}\bullet \cos \left(\frac{0\bullet \pi }{6}+\frac{\arctan \left(\sqrt{\frac{4\bullet m^3}{n^2}-1}\right)}{3}\right)-\frac{1}{3}\pm \sqrt[3]{\frac{n}{2}}\bullet \sqrt[3]{\frac{4\bullet m^3}{n^2}}\bullet \sqrt{3}\bullet \sin \left(\frac{0\bullet \pi }{6}+\frac{\arctan \left(\sqrt{\frac{4\bullet m^3}{n^2}-1}\right)}{3}\right)$



 maroon


For an angle of
 pi/7 
the
 solution 
is transcendental
but for pi/60 the solution is only irrational   


$2\bullet \cos \left(2\bullet \pi /7\right)={\left(\frac{7}{\mathrm{27}}\bullet \frac{\sqrt{1+3^3}}{2}\right)}^{\left(1/3\right)}2\bullet \sin \left(\frac{\pi -\frac{2}{3}\mathrm{atan}\left(\sqrt{\mathrm{27}}\right)}{2}\right)-\frac{1}{3}$



 darkorange

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overzealous autocrats. 




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Copyright © 2009

Wm. C.

Corwin