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64 lines
3.6 KiB
TeX
64 lines
3.6 KiB
TeX
The second solution of $x^3+\alpha x + \beta=0$ is
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\[x = \frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t}
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-\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}}\]
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We will verify it in multiple steps. First, calculate $x^3$:
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\begin{align}
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x^3 &= \underbrace{\left (\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^3}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}}
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\underbrace{- 3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^2 \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}} \right)}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}}}\\
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&\hphantom{{}=}+ \underbrace{3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right) \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^2}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}}}
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\underbrace{- \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^3}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {4}}}}
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\end{align}
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Now simplify the summands of $x^3$:
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\begin{align}
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\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} &=
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\left (\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^3\\
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&= \frac{-8\alpha^3}{12 t^3}\\
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&= \frac{-2 \alpha^3}{3 t^3}\\
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\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} &=- 3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^2 \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}} \right)\\
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&= \frac{-3\alpha^2(-2(1-i\sqrt{3}))(1-i\sqrt{3})t}{t^2 \sqrt[3]{2^4 \cdot 3^2} \cdot 2 \sqrt[3]{2 \cdot 3^2}}\\
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&= \frac{6\alpha^2 t (-2 (1+i \sqrt{3}))}{12 t^2 \sqrt[3]{12}}\\
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&= \frac{- \alpha^2 (1+i\sqrt{3})}{t\sqrt[3]{12}}\\
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\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}} &= 3 \left(\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right) \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^2\\
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&= \frac{3\alpha t (1+i\sqrt{3})(-2(1+i\sqrt{3}))}{4 \cdot \sqrt[3]{12 \cdot 18^2}}\\
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&= \frac{-\alpha t (-2 (1 - i \sqrt{3}))}{2 \sqrt[3]{12 \cdot 4 \cdot 3}}\\
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&= \frac{\alpha t (1-i\sqrt{3})}{\sqrt[3]{2^4 \cdot 3^2}}\\
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&= \frac{\alpha t (1-i\sqrt{3}}{2 \sqrt[3]{18}}\\
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\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {4}}} &= - \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^3\\
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&=- \frac{(-8) t^3}{8 \cdot 18}\\
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&= \frac{t^3}{18}
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\end{align}
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Now get back to the original equation:
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\begin{align}
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0 &\stackrel{!}{=} x^3 + \alpha x + \beta \\
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&= \left (\frac{-2 \alpha^3}{3 t^3}
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+ \color{red}\frac{-\alpha^2(1+\sqrt{3}i)}{t\sqrt[3]{12}}\color{black}
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+ \color{blue}\frac{\alpha t(1-\sqrt{3}i)}{2\sqrt[3]{18}}\color{black}
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+ \frac{t^3}{18} \right )\\
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&\hphantom{{}=} + \alpha \left (\color{red}\frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \color{black}
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\color{blue}-\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}} \color{black} \right ) + \beta\\
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&= \frac{-2 \alpha^3}{3 t^3}
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+ \frac{t^3}{18}
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+ \beta\\
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&= \frac{-12 \alpha^3 + t^6+18 t^3 \beta}{18t^3}
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\end{align}
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Now continue with only the numerator
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\begin{align}
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0 &\stackrel{!}{=}
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- 12 \alpha^3
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+ (\sqrt{3(4 \alpha^3 + 27 \beta^2)}-9\beta)^2
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+ 18 (\sqrt{3(4 \alpha^3 + 27 \beta^2)} - 9 \beta) \beta\\
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&=
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\color{red}- 12 \alpha^3 \color{black}+
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\left (
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3(\color{red}4 \alpha^3\color{black} + \color{blue}27 \beta^2 \color{black})
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\color{orange}- 2 \cdot \sqrt{3(4 \alpha^3 + 27 \beta^2)} \cdot 9\beta\color{black}
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+ \color{blue}81 \beta^2\color{black}
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\right )\\
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&\hphantom{{}=}+ 18 \beta (\color{orange}\sqrt{3(4 \alpha^3 + 27 \beta^2)}\color{black} \color{blue}- 9 \beta\color{black}) \\
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&= 81 \beta^2 + 81 \beta^2 - 2 \cdot 81 \beta^2\\
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&= 0
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\end{align}
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