塑膠數

Multi tool use
Multi tool use























12+162333+12−162333{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}}}}}}{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}}}}}}

二進位
約為1.010100110010000010110111010011101100101

八進位
約為1.2462026723545104533260274211370405060463

十進位
約為1.324717957244746025960908854478097340734

十六進位
約為1.5320B74ECA44ADAC178897C41461334737F8172F

塑膠數銀數是一元三次方程 x3=x+1{displaystyle x^{3}=x+1,}{displaystyle x^{3}=x+1,} 的唯一一個實數根,其值為


12+162333+12−162333{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}}}}}}{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}}}}}}

約等於1.3247179572447460259609{displaystyle 1.3247179572447460259609}{displaystyle 1.3247179572447460259609}


塑膠數對於佩蘭數列和巴都萬數列,就如黃金分割對於斐波那契數列——是兩項的比的極限。它亦是最小的皮索數。



塑膠數的來源


塑膠數是方程x3=x+1{displaystyle x^{3}=x+1,}{displaystyle x^{3}=x+1,}的唯一實數根。


對於方程x3=x+1{displaystyle x^{3}=x+1,}{displaystyle x^{3}=x+1,},現將等式右邊變為0,即


x3−x−1=0{displaystyle x^{3}-x-1=0,}{displaystyle x^{3}-x-1=0,}




x=λy+y{displaystyle x={frac {lambda }{y}}+y,}{displaystyle x={frac {lambda }{y}}+y,}




y=12x2−{displaystyle y={frac {1}{2}}{sqrt {x^{2}-4lambda }},}{displaystyle y={frac {1}{2}}{sqrt {x^{2}-4lambda }},}


得到


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


等式兩邊同時乘 y3{displaystyle y^{3}}{displaystyle y^{3}}


y6+y4(3λ1)−y3+y2(3λ2−λ)+λ3=0{displaystyle y^{6}+y^{4}left(3lambda -1right)-y^{3}+y^{2}left(3lambda ^{2}-lambda right)+lambda ^{3}=0,}{displaystyle y^{6}+y^{4}left(3lambda -1right)-y^{3}+y^{2}left(3lambda ^{2}-lambda right)+lambda ^{3}=0,}


λ=13{displaystyle lambda ={frac {1}{3}},}{displaystyle lambda ={frac {1}{3}},},將其帶入上面方程,并設z=y3{displaystyle z=y^{3},}{displaystyle z=y^{3},},得到一個z{displaystyle z}z的二次方程


z2−z+127=0{displaystyle z^{2}-z+{frac {1}{27}}=0,}{displaystyle z^{2}-z+{frac {1}{27}}=0,}


解得


z=118(9+69){displaystyle z={frac {1}{18}}left(9+{sqrt {69}}right),}{displaystyle z={frac {1}{18}}left(9+{sqrt {69}}right),}


根據z=y3{displaystyle z=y^{3},}{displaystyle z=y^{3},},得


y3=118(9+69){displaystyle y^{3}={frac {1}{18}}left(9+{sqrt {69}}right),}{displaystyle y^{3}={frac {1}{18}}left(9+{sqrt {69}}right),}


y{displaystyle y}{displaystyle y}有實數解


y=12+162333{displaystyle y={sqrt[{3}]{{frac {1}{2}}+{frac {1}{6}}{sqrt {frac {23}{3}}}}},}{displaystyle y={sqrt[{3}]{{frac {1}{2}}+{frac {1}{6}}{sqrt {frac {23}{3}}}}},}


根據y{displaystyle y}{displaystyle y}λ{displaystyle lambda }lambda 的關係,得y=x2+12x2−43{displaystyle y={tfrac {x}{2}}+{tfrac {1}{2}}{sqrt {x^{2}-{tfrac {4}{3}}}},}{displaystyle y={tfrac {x}{2}}+{tfrac {1}{2}}{sqrt {x^{2}-{tfrac {4}{3}}}},},得x{displaystyle x}x的實數解


x=12+162333+12−162333{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}}}}},}{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}}}}},}








fhJ6VqdEGifcYo B 9jxMX3YT0JBJiaO2y9,jNZo
Ev46ouLz2B,qcyh7KTM,DiipO VbX,6yOWbE3u4z,G5JPUCYiOT7BQOtSN rFpJ4E rO

Popular posts from this blog

How to pass form data using jquery Ajax to insert data in database?

Guess what letter conforming each word

Run scheduled task as local user group (not BUILTIN)