利用者:Hrtauo/sandbox

数式

三次方程式$ax^{3}+bx^{2}+cx+d=0$の解を求める。

まず方程式を$x^{3}+px+q=0$の形で表すことを考える。

両辺をaで割って$x^{3}+{\frac {b}{a}}x^{2}+{\frac {c}{a}}x+{\frac {d}{a}}=0$

$x=y-{\frac {b}{3a}}$と置くと

${\begin{aligned}(y-{\frac {b}{3a}})^{3}+{\frac {b}{a}}(y-{\frac {b}{3a}})^{2}+{\frac {c}{a}}(y-{\frac {b}{3a}})+{\frac {d}{a}}&=y^{3}-{\frac {b}{a}}y^{2}+{\frac {b^{2}}{3a^{2}}}y-{\frac {b^{3}}{27a^{3}}}+{\frac {b}{a}}y^{2}-{\frac {2b^{2}}{3a^{2}}}y+{\frac {b^{3}}{9a^{3}}}+{\frac {c}{a}}y-{\frac {bc}{3a^{2}}}+{\frac {d}{a}}\\&=y^{3}+{\frac {b^{2}+3ac}{3a^{2}}}y+{\frac {2b^{3}-9abc+27a^{2}d}{27a^{3}}}\\&=0\end{aligned}}$

$p={\frac {b^{2}+3ac}{3a^{2}}}$,$q={\frac {2b^{3}-9abc+27a^{2}c}{27a^{3}}}$と置くと$y^{3}+py+q=0$

ここで$y=\alpha +\beta$とおくと

${\begin{alignedat}{2}y^{3}&=\alpha ^{3}+\beta ^{3}+3\alpha ^{2}\beta +3\alpha \beta ^{2}\\&=\alpha ^{3}+\beta ^{3}+3\alpha \beta (\alpha +\beta )\\&=\alpha ^{3}+\beta ^{3}+3\alpha \beta y\end{alignedat}}$

$y^{3}-3\alpha \beta y-(\alpha ^{3}+\beta ^{3})=0$

係数比較により${\begin{cases}3\alpha \beta =-p\cdots (1)\\\alpha ^{3}+\beta ^{3}=-q\cdots (2)\\\end{cases}}$

(1)の両辺を3で割って$\alpha \beta =-{\frac {p}{3}}\cdots (1)\prime$

(1)'^3により$\alpha ^{3}\beta ^{3}=-{\frac {p^{3}}{27}}$

よってα^3,β^3は二次方程式$X^{2}+qX-{\frac {p^{3}}{27}}=0$の解である。

二次方程式の解の公式を使って$X={\frac {9q\pm {\sqrt {81q^{2}+12p^{3}}}}{18}}$

よって$\alpha ^{3},\beta ^{3}={\frac {9q+{\sqrt {81q^{2}-12p^{3}}}}{18}},{\frac {9q-{\sqrt {81q^{2}-12p^{3}}}}{18}}$

(1)を満たすように3乗根をとって$\alpha ={\sqrt[{3}]{\frac {9q+{\sqrt {81q^{2}+12p^{3}}}}{18}}},{\sqrt[{3}]{\frac {9q+{\sqrt {81q^{2}+12p^{3}}}}{18}}}\omega ,{\sqrt[{3}]{\frac {9q+{\sqrt {81q^{2}+12p^{3}}}}{18}}}\omega ^{2}$,$\beta ={\sqrt[{3}]{\frac {9q-{\sqrt {81q^{2}+12p^{3}}}}{18}}},{\sqrt[{3}]{\frac {9q-{\sqrt {81q^{2}+12p^{3}}}}{18}}}\omega ^{2},{\sqrt[{3}]{\frac {9q-{\sqrt {81q^{2}+12p^{3}}}}{18}}}\omega$

よって${\begin{alignedat}{2}y&={\sqrt[{3}]{\frac {9q+{\sqrt {81q^{2}+12p^{3}}}}{18}}}+{\sqrt[{3}]{\frac {9q-{\sqrt {81q^{2}+12p^{3}}}}{18}}}\\&,{\sqrt[{3}]{\frac {9q+{\sqrt {81q^{2}+12p^{3}}}}{18}}}\omega +{\sqrt[{3}]{\frac {9q-{\sqrt {81q^{2}+12p^{3}}}}{18}}}\omega ^{2}\\&,{\sqrt[{3}]{\frac {9q+{\sqrt {81q^{2}+12p^{3}}}}{18}}}\omega ^{2}+{\sqrt[{3}]{\frac {9q-{\sqrt {81q^{2}+12p^{3}}}}{18}}}\omega \\\end{alignedat}}$

よって解は${\begin{alignedat}{2}x&={\sqrt[{3}]{\frac {9q+{\sqrt {81q^{2}+12p^{3}}}}{18}}}+{\sqrt[{3}]{\frac {9q-{\sqrt {81q^{2}+12p^{3}}}}{18}}}-{\frac {b}{3a}},\\&{\sqrt[{3}]{\frac {9q+{\sqrt {81q^{2}+12p^{3}}}}{18}}}\omega +{\sqrt[{3}]{\frac {9q-{\sqrt {81q^{2}+12p^{3}}}}{18}}}\omega ^{2}-{\frac {b}{3a}},\\&{\sqrt[{3}]{\frac {9q+{\sqrt {81q^{2}+12p^{3}}}}{18}}}\omega ^{2}+{\sqrt[{3}]{\frac {9q-{\sqrt {81q^{2}+12p^{3}}}}{18}}}\omega -{\frac {b}{3a}}\\\end{alignedat}}{\text{(た だ し }}p={\frac {b^{2}+3ac}{3a^{2}}},q={\frac {2b^{3}-9abc+27a^{2}c}{27a^{3}}}{\text{と す る }})$

夏休み用

トンネル効果 呼吸 https://www.google.com/search?q=%E5%91%BC%E5%90%B8+%E9%87%8F%E5%AD%90%E5%8A%9B%E5%AD%A6&safe=active&sca_esv=ffb312a028fe1876&sca_upv=1&ei=icyVZsqJCd-zvr0PsMqsgQI&sqi=2&start=10&sa=N&sstk=Aagrsuh6uUPfYKocbVB6yVsEwcZru8Z9KIKytetEueoGUxYttdP73u15M_JYKL-m_yhFdaqzMvVpOdWqUTSruIf6RYDSHfYR_ak-sg&ved=2ahUKEwiK8t6BtKqHAxXfma8BHTAlKyAQ8NMDegQICRAW&biw=1280&bih=673&dpr=1

コード

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