塑膠數

塑膠數是方程${\displaystyle x^{3}=x+1\,}$的唯一實數根。

對於方程${\displaystyle x^{3}=x+1\,}$，現將等式右邊變為0，即

${\displaystyle x^{3}-x-1=0\,}$

由勘根定理可判斷出該實根大小介於1與2之間，設

${\displaystyle x={\frac {\lambda }{y}}+y\,}$，

則

${\displaystyle y={\frac {x}{2}}+{\frac {1}{2}}{\sqrt {x^{2}-4\lambda }}\,}$

得到

${\displaystyle -1-y-{\frac {\lambda }{y}}+\left(y+{\frac {\lambda }{y}}\right)^{3}=0\,}$

等式兩邊同時乘 ${\displaystyle y^{3}}$ 得

${\displaystyle y^{6}+y^{4}\left(3\lambda -1\right)-y^{3}+y^{2}\left(3\lambda ^{2}-\lambda \right)+\lambda ^{3}=0\,}$

令${\displaystyle \lambda ={\frac {1}{3}}\,}$，將其帶入上面方程，并設${\displaystyle z=y^{3}\,}$，得到一個${\displaystyle z}$的二次方程

${\displaystyle z^{2}-z+{\frac {1}{27}}=0\,}$

解得

${\displaystyle z={\frac {1}{18}}\left(9+{\sqrt {69}}\right)\,}$

根據${\displaystyle z=y^{3}\,}$，得

${\displaystyle y^{3}={\frac {1}{18}}\left(9+{\sqrt {69}}\right)\,}$

則${\displaystyle y}$有實數解

${\displaystyle y={\sqrt[{3}]{{\frac {1}{2}}+{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}\,}$

根據${\displaystyle y}$与${\displaystyle \lambda }$的關係，得${\displaystyle y={\tfrac {x}{2}}+{\tfrac {1}{2}}{\sqrt {x^{2}-{\tfrac {4}{3}}}}\,}$，得${\displaystyle x}$的實數解

${\displaystyle x={\sqrt[{3}]{{\frac {1}{2}}+{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}+{\sqrt[{3}]{{\frac {1}{2}}-{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}\,}$