塑胶数塑胶数或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 塑胶数塑胶数数表—无理数 2 {\displaystyle \color {blue}{\sqrt {2}}} φ {\displaystyle \color {blue}\varphi } 3 {\displaystyle \color {blue}{\sqrt {3}}} 5 {\displaystyle \color {blue}{\sqrt {5}}} δ S {\displaystyle \color {blue}\delta _{S}} e {\displaystyle \color {blue}e} π {\displaystyle \color {blue}\pi } 命名数字ρ名称塑胶数识别种类无理数符号 ρ {\displaystyle \rho } 位数数列编号 A060006性质以此为根的多项式或函数 x 3 − x − 1 = 0 {\displaystyle x^{3}-x-1=0\,} 表示方式值 ρ ≈ {\displaystyle \rho \approx } 1.3247179572...代数形式 9 + 69 18 3 + 9 − 69 18 3 {\displaystyle {\sqrt[{3}]{\frac {9+{\sqrt {69}}}{18}}}+{\sqrt[{3}]{\frac {9-{\sqrt {69}}}{18}}}} 二进制1.010100110010000010110111…八进制1.246202672354510453326027…十进制1.324717957244746025960908…十六进制1.5320B74ECA44ADAC178897C4… 查论编 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} （OEIS数列A060006）。 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\,} 由勘根定理可判断出该实根大小介于1与2之间，设 x = λ y + y {\displaystyle x={\frac {\lambda }{y}}+y\,} ， 则 y = x 2 + 1 2 x 2 − 4 λ {\displaystyle y={\frac {x}{2}}+{\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}} 得 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\,} 令 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)\,} 则 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} 与 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}}}}}\,} 参考文献 ^ Sequence A072117 in the OEIS
A060006
![{\displaystyle {\sqrt[{3}]{\frac {9+{\sqrt {69}}}{18}}}+{\sqrt[{3}]{\frac {9-{\sqrt {69}}}{18}}}}](/media/math_img/4330/a1816877ef2e344fbf8c9255d18f8a409012741a.svg)
![{\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}}}}}}](/media/math_img/4330/632f36cfd2cc1e85c932d625257359ebf7ed3330.svg)
