1 2 + 1 6 23 3 3 + 1 2 1 6 23 3 3 {\displaystyle {\sqrt[{3}]{{\frac {1}{2}}+{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}+{\sqrt[{3}]{{\frac {1}{2}}-{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}}
二进制约为1.010100110010000010110111010011101100101
八进制约为1.2462026723545104533260274211370405060463
十进制约为1.324717957244746025960908854478097340734
十六进制约为1.5320B74ECA44ADAC178897C41461334737F8172F

塑胶数银数是一元三次方程 x 3 = x + 1 {\displaystyle x^{3}=x+1\,} 的唯一一个实数根,其值为

1 2 + 1 6 23 3 3 + 1 2 1 6 23 3 3 {\displaystyle {\sqrt[{3}]{{\frac {1}{2}}+{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}+{\sqrt[{3}]{{\frac {1}{2}}-{\frac {1}{6}}{\sqrt {\frac {23}{3}}}}}}

约等于 1.3247179572447460259609 {\displaystyle 1.3247179572447460259609}

塑胶数对于佩兰数列和巴都万数列,就如黄金分割对于斐波那契数列——是两项的比的极限。它亦是最小的皮索数。

塑胶数的来源

塑胶数是方程 x 3 = x + 1 {\displaystyle x^{3}=x+1\,} 的唯一实数根。

对于方程 x 3 = x + 1 {\displaystyle x^{3}=x+1\,} ,现将等式右边变为0,即

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

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

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

得到

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

等式两边同时乘 y 3 {\displaystyle y^{3}}

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

λ = 1 3 {\displaystyle \lambda ={\frac {1}{3}}\,} ,将其带入上面方程,并设 z = y 3 {\displaystyle z=y^{3}\,} ,得到一个 z {\displaystyle z} 的二次方程

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

解得

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

根据 z = y 3 {\displaystyle z=y^{3}\,} ,得

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

y {\displaystyle y} 有实数解

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

根据 y {\displaystyle y} λ {\displaystyle \lambda } 的关系,得 y = x 2 + 1 2 x 2 4 3 {\displaystyle y={\tfrac {x}{2}}+{\tfrac {1}{2}}{\sqrt {x^{2}-{\tfrac {4}{3}}}}\,} ,得 x {\displaystyle x} 的实数解

x = 1 2 + 1 6 23 3 3 + 1 2 1 6 23 3 3 {\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}}}}}\,}

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