フェラーリの公式

四次方程式の解の公式

$ =(\chi-a)^4+4a(\chi-a)^3+12b(\chi-a)^2+24c(\chi-a)+3d

$ =(\chi^4-4a\chi^3+6a^2\chi^2-4a^3\chi+a^4)

$ +(4a\chi^3-12a^2\chi^2+12a^3\chi-4a^4)

$ +(12b\chi^2-24ab\chi+12a^2b)+(24c\chi-24ac)+3d

$ =\chi^4+(6a^2-12a^2+12b)\chi^2+(-4a^3+12a^3-24ab+24c)\chi+(a^4-4a^4+12a^2b-24ac+3d)

$ =\chi^4+(12b-6a^2)\chi^2+(24c-24ab+8a^3)\chi+(3d-24ac+12a^2b-3a^4)

$ =\chi^4+6(2b-a^2)\chi^2+8\left(3c-3ab+a^3\right)\chi+3(d-8ac+4a^2b-a^4)

$ =\chi^4-6\beta\chi^2+4\gamma\chi-3\delta

$ \chi={\color{red}\pm}\frac{\sqrt u}2{\color{blue}\pm}\frac12\sqrt{u-4(\Beta-\Gamma\sqrt u)}

展開すると、

$ =\chi^4+2\Beta\chi^2+\Beta^2-u\chi^2-2u\Gamma\chi-u\Gamma^2

$ =\chi^4+2(\Beta-u)\chi^2-2u\Gamma\chi+\Beta^2-u\Gamma^2

$ \left\{\begin{matrix}2(\Beta-u)=-6\beta\\-2u\Gamma=4\gamma\\\Beta^2-u\Gamma^2=-3\delta\end{matrix}\right.

$ \Leftrightarrow\left\{\begin{matrix}\Beta=u-3\beta\\\Gamma=-\frac{2\gamma}{u}\\\Beta^2-u\Gamma^2=-3\delta\end{matrix}\right.

3つめの式から

$ \left(u-3\beta\right)^2-u\left(-\frac{2\gamma}{u}\right)^2=-3\delta

$ u(u-3\beta)^2+3\delta u-4\gamma^2=0

1,2個めの式から

$ 0=\left(\chi^2+\left(u-3\beta\right)\right)^2-u\left(\chi-\frac{2\gamma}{u}\right)^2

$ \therefore

$ \chi^2+(u-3\beta)=\pm\left(\chi\sqrt u-2\gamma u^{-\frac12}\right)

$ \left\{\begin{matrix}\chi^2-\sqrt u\chi+(u-3\beta+2\gamma u^{-\frac12})=0\\\chi^2+\sqrt u\chi+(u-3\beta-2\gamma u^{-\frac12})=0\end{matrix}\right.

$ \chi=\left\{\begin{matrix}+\frac{\sqrt{u}}2\pm\frac12\sqrt{-3u+12\beta-8\gamma u^{-\frac12}}\\-\frac{\sqrt{u}}2\pm\frac12\sqrt{-3u+12\beta+8\gamma u^{-\frac12}}\end{matrix}\right.

$ \chi=\left\{\begin{matrix}+\frac{\sqrt{u}}2\pm \frac{i}2\sqrt{3u-12\beta+8\gamma u^{-\frac12}}\\-\frac{\sqrt{u}}2\pm\frac{i}2\sqrt{3u-12\beta-8\gamma u^{-\frac12}}\end{matrix}\right.

$ 0=u(u-3\beta)^2+3\delta{u}-4\gamma^2

$ =u(u^2-6\beta u+9\beta^2)+3\delta u-4\gamma^2

$ =u^3-6\beta{u}^2+(9\beta^2+3\delta)u-4\gamma^2

$ =u^3+3(-2\beta)u^2+6\left(\frac32\beta^2+\frac12\delta\right)u+2(-2\gamma^2)

$ s=(\delta-\beta^2),\ t=(2\gamma^2+\beta^3+3\beta\delta)

$ u_k=2\beta+\omega^k\,^3\sqrt{t+\sqrt{t^2+s^3}}+\omega^{3-k}\,^3\sqrt{t-\sqrt{t^2+s^3}}

$ \therefore

$ u_k=2\beta+\omega^k\,^3\sqrt{(2\gamma^2+\beta^3+3\beta\delta)+\sqrt{(2\gamma^2+\beta^3+3\beta\delta)^2+(\delta-\beta^2)^3}}

$ +\omega^{-k}\,^3\sqrt{(2\gamma^2+\beta^3+3\beta\delta)-\sqrt{(2\gamma^2+\beta^3+3\beta\delta)^2+(\delta-\beta^2)^3}}

この式変形のうち、複二次式に落とし込む所が重要