Cubic Theory

When programming to find the roots of a cubic, the statements executed will be different depending on the discriminate and the sign of a term. The real and complex components will have to be explicitly evaluated. For $a\bullet z^3+b\bullet z^2+c\bullet z^{\mathrm{}}+d=0$ the discriminate is $D=\frac{4\bullet m^3}{n^2}$ , where m = 1/9 - c/3 and n = c/3 - d - 2/27 . The above discriminate is defined so as to be the independent variable for a normalized plot of the real and imaginary components of the roots. The relationship of the roots can be visualized and a starting approximation for obtaining the correct root with Newton's method, can be obtained. Also, the plot could be used as a nomograph for rough answers. The correct log-log plots of subsections of the graph could be referenced also.

If D<0, then $z={\left(\mathrm{n}/2\right)}^{\left(1/3\right)}\bullet \left({\left(1\pm \sqrt{1-\frac{4\bullet m^3}{n^2}}\right)}^{\left(1/3\right)}+{\left(-1\pm \sqrt{1-\frac{4\bullet m^3}{n^2}}\right)}^{\left(1/3\right)}\right)-\frac{1}{3}$

The real component of the two complex roots is $\mathrm{z2r}=-\frac{{\left(\mathrm{n}/2\right)}^{\left(1/3\right)}}{2}\bullet \left({\left(1\pm \sqrt{1-\frac{4\bullet m^3}{n^2}}\right)}^{\left(1/3\right)}+{\left(-1\pm \sqrt{1-\frac{4\bullet m^3}{n^2}}\right)}^{\left(1/3\right)}\right)-\frac{1}{3}$ The imaginary component of the complex roots is $\mathrm{z2i}=\pm {\left(\mathrm{n}/2\right)}^{\left(1/3\right)}\bullet \frac{\sqrt{3}}{2}\bullet \left({\left(1\pm \sqrt{1-\frac{4\bullet m^3}{n^2}}\right)}^{\left(1/3\right)}-{\left(-1\pm \sqrt{1-\frac{4\bullet m^3}{n^2}}\right)}^{\left(1/3\right)}\right)$

If 0<D<1, then $z={\left(\mathrm{n}/2\right)}^{\left(1/3\right)}\bullet \left({\left(1\pm \sqrt{1-\frac{4\bullet m^3}{n^2}}\right)}^{\left(1/3\right)}-{\left(-1\pm \sqrt{1-\frac{4\bullet m^3}{n^2}}\right)}^{\left(1/3\right)}\right)-\frac{1}{3}$

The real component of the two complex roots is $\mathrm{z2r}=-\frac{{\left(\mathrm{n}/2\right)}^{\left(1/3\right)}}{2}\bullet \left({\left(1\pm \sqrt{1-\frac{4\bullet m^3}{n^2}}\right)}^{\left(1/3\right)}-{\left(-1\pm \sqrt{1-\frac{4\bullet m^3}{n^2}}\right)}^{\left(1/3\right)}\right)-\frac{1}{3}$ The imaginary component of the complex roots is $\mathrm{z2i}=\pm {\left(\mathrm{n}/2\right)}^{\left(1/3\right)}\bullet \frac{\sqrt{3}}{2}\bullet \left({\left(1\pm \sqrt{1-\frac{4\bullet m^3}{n^2}}\right)}^{\left(1/3\right)}+{\left(-1\pm \sqrt{1-\frac{4\bullet m^3}{n^2}}\right)}^{\left(1/3\right)}\right)$

If D > 1, Then the three roots are real but all manipulations go through complex phases involving the projections of a vector r having magnitude $r=\sqrt{1+\left(\frac{4\bullet m^3}{n^2}-1\right)}$ . The terms $S=\sqrt[3]{1\pm \; \sqrt{1-\frac{4\bullet m^3}{n^2}}}$ and $T=\sqrt[3]{-1\pm \; \sqrt{1-\frac{4\bullet m^3}{n^2}}}$ are complex with magnitude r. They can be written in exponential form. $S=\sqrt[3]{r}\bullet {ℯ}^{\frac{\mathrm{\pm \; }\mathrm{atan}\sqrt{\frac{4\bullet m^3}{n^2}-1}\bullet i}{3}}$ and $T=\sqrt[3]{r}\bullet {ℯ}^{\left(\frac{\pi }{3}\bullet i\mp \frac{\mathrm{atan}\sqrt{\frac{4\bullet m^3}{n^2}-1}}{3}\bullet i\right)}$ .   S+T then can be factored? $S+T=\sqrt[3]{r}\bullet {ℯ}^{\frac{\mathrm{atan}\sqrt{\frac{4\bullet m^3}{n^2}-1}\bullet i}{3}}\bullet \left(1+{ℯ}^{\frac{\pi }{3}\bullet i}\right)$ . Now $1+{ℯ}^{\frac{\pi }{3}\bullet i}=1+\frac{1}{2}+\frac{\sqrt{3}}{2}\bullet i=\frac{3}{2}+\frac{\sqrt{3}}{2}\bullet i$ . So, since $x\bullet \sin \left(a+\frac{1}{3}\mathrm{atan}\left(\sqrt{\frac{4\bullet m^3}{n^2}-1}\right)\right)=x\bullet \sin \left(a\right)\bullet \sin \left(\frac{1}{3}\mathrm{atan}\left(\sqrt{\frac{4\bullet m^3}{n^2}-1}\right)\right)+x\bullet \cos \left(a\right)\bullet \cos \left(\frac{1}{3}\mathrm{atan}\left(\sqrt{\frac{4\bullet m^3}{n^2}-1}\right)\right)$ . and using $x\bullet \sin \left(a\right)=\frac{3}{2}\mathrm{ \; and\; }x\bullet \cos \left(a\right)=-\frac{\sqrt{3}}{2}$ . Thus the first real root is

$z={\left(\frac{n\bullet r}{2}\right)}^{\left(1/3\right)}\sqrt{3}\bullet \sin \left(-\frac{\pi }{6}+\frac{1}{3}\mathrm{atan}\left(\sqrt{\frac{4\bullet m^3}{n^2}-1}\right)\right)-\frac{1}{3}$ .

Likewise $S-T=\sqrt[3]{r}\bullet$

$\mathrm{zi}?=\pm \; \sqrt[3]{\frac{n\bullet r}{2}}\bullet \frac{1}{3}\bullet \cos \left(\frac{\pi }{2}-\frac{1}{3}\mathrm{atan}\left(\sqrt{\frac{4\bullet m^3}{n^2}-1}\right)\right)$

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

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

Divide the circle into 120 divisions wthout transendental formulas $\mathrm{16}\bullet \sin \left(\pi /\mathrm{60}\right)=$ $2\bullet \left(1-\sqrt{3}\right)\bullet \sqrt{5+\sqrt{5}}+\sqrt{2}\bullet \left(\sqrt{5}-1\right)\bullet \left(\sqrt{3}+1\right)$