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fixed some errors in calculation

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Martin Thoma 2013-12-19 17:36:50 +01:00
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commit 4440bd4e62
2 changed files with 35 additions and 22 deletions
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documents/math-minimal-distance-to-cubic-function/quadratic-functions.tex
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@ -207,42 +207,55 @@ $t$:
\textbf{Case 2.2:}
\todo[inline]{calculate...}
\[x = \frac{(1+i \sqrt{3})a}{\sqrt[3]{12} \cdot t}
\[x = \frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t}
-\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}}\]
\begin{align}
x^3 &= \underbrace{\left (\frac{(1+i\sqrt{3})a}{\sqrt[3]{12} \cdot t} \right)^3}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}}
\underbrace{- 3 \left(\frac{(1+i\sqrt{3})a}{\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}}}}\\
&\hphantom{{}=}+ \underbrace{3 \left(\frac{(1+i\sqrt{3})a}{\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}}}}
x^3 &= \underbrace{\left (\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^3}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}}}
\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}}}}\\
&\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}}}}
\underbrace{- \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^3}_{=: \raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {4}}}}
\end{align}
Now simplify the summands:
\begin{align}
\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {1}}} &=
\frac{a^3(1+3i\sqrt{3} - 3 \cdot 3 - \sqrt{27} i)}{12 t^3}\\
&= \frac{a^3((3\sqrt{3}- \sqrt{27})i - 8)}{12 t^3}\\
&= \frac{-8a^3}{12 t^3}\\
&= \frac{-2a^3}{3 t^3}\\
\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {2}}} &=- 3 \left(\frac{(1+i\sqrt{3})a}{\sqrt[3]{12} \cdot t} \right)^2 \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}} \right)\\
&= \frac{3a(1+2\sqrt{3}i-3)(1-i\sqrt{3})}{t \cdot 2 \cdot 2 \sqrt[3]{3 \cdot 3 \cdot 2 \cdot 18}}\\
&= \frac{3a(1+2\sqrt{3}i - 3- i\sqrt{3}+2\cdot 3 + i\sqrt[3]{3})}{4t \cdot 3 \sqrt{6}}\\
&= \frac{a(1-3+4\sqrt{3}i + 6)}{4t\sqrt[3]{6}}\\
&= \frac{a(4+4\sqrt{3}i)}{4t \sqrt[3]{6}}\\
&= \frac{a(1+\sqrt{3}i)}{t \sqrt[3]{6}}\\
\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {3}}} &= 3 \left(\frac{(1+i\sqrt{3})a}{\sqrt[3]{12} \cdot t} \right) \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^2\\
&= \frac{3(1+i\sqrt{3})a (1-2i\sqrt{3} - 3)t}{\sqrt[3]{12 \cdot 18^2}}\\
&= \frac{3at((1-2i\sqrt{3}-3)+(i\sqrt{3} + 2\cdot 3 - 3i\sqrt{3}))}{\sqrt[3]{2^2 \cdot 3 \cdot (2 \cdot 3^2)^2}}\\
&= \frac{3at(4+i(-4\sqrt{3})}{4\cdot3 \sqrt[3]{9}}\\
&= \frac{at(1-\sqrt{3}i}{\sqrt[3]{9}}\\
\left (\frac{(1+i\sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} \right)^3\\
&= \frac{\alpha^3(1-3i\sqrt{3} - 3 \cdot 3 + \sqrt{27} i)}{12 t^3}\\
&= \frac{-8\alpha^3}{12 t^3}\\
&= \frac{-2 \alpha^3}{3 t^3}\\
\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)\\
&= \frac{-3\alpha^2(1+2\sqrt{3}i-3)(1-i\sqrt{3})t}{t^2 \sqrt[3]{2^4 \cdot 3^2} \cdot 2 \sqrt[3]{2 \cdot 3^2}}\\
&= \frac{-3\alpha^2((1+2\sqrt{3}i - 3)+(- i\sqrt{3}+2\cdot 3 + i\sqrt{3}))}{12 t \sqrt[3]{12}}\\
&= \frac{-\alpha^2(4+2\sqrt{3}i)}{4t\sqrt[3]{12}}\\
&= \frac{-\alpha^2(2+\sqrt{3}i)}{2t\sqrt[3]{12}}\\
\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\\
&= \frac{3(1+i\sqrt{3})\alpha (1-2i\sqrt{3} - 3)t}{4 \sqrt[3]{12 \cdot 18^2}}\\
&= \frac{3 \alpha t((1-2i\sqrt{3}-3)+(i\sqrt{3} + 2\cdot 3 - 3i\sqrt{3}))}{4 \sqrt[3]{2^2 \cdot 3 \cdot (2 \cdot 3^2)^2}}\\
&= \frac{3 \alpha t(4-4\sqrt{3}i)}{4 \cdot 2 \cdot 3 \sqrt[3]{2 \cdot 3^2}}\\
&= \frac{\alpha t(1-\sqrt{3}i)}{2\sqrt[3]{18}}\\
\raisebox{.5pt}{\textcircled{\raisebox{-.9pt} {4}}} &= - \left(\frac{(1-i\sqrt{3})t}{2 \sqrt[3]{18}}\right)^3\\
&= -\frac{(1-3i\sqrt{3} - 3 \cdot 3 + i \sqrt{27}) t^3}{8 \cdot 18}\\
&=- \frac{t^3 (-8)}{2^4 \cdot 3^2}\\
&= \frac{8t^3}{2^4 \cdot 3^2}\\
&=- \frac{t^3 (-8)}{8 \cdot 18}\\
&= \frac{t^3}{18}
\end{align}
Now get back to the original equation:
\begin{align}
0 &\stackrel{!}{=} x^3 + \alpha x + \beta \\
&= \left (\frac{-2 \alpha^3}{3 t^3}
+ \frac{-\alpha^2(2+\sqrt{3}i)}{2t\sqrt[3]{12}}
+ \color{blue}\frac{\alpha t(1-\sqrt{3}i)}{2\sqrt[3]{18}}\color{black}
+ \frac{t^3}{18} \right )\\
&\hphantom{{}=} + \color{blue}\alpha\color{black} \left (\frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t}
\color{blue}-\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}} \color{black} \right ) + \beta\\
&= \frac{-2 \alpha^3}{3 t^3}
+ \frac{\alpha^2(2(1+i\sqrt{3})-(2+\sqrt{3}i))}{2t\sqrt[3]{12}}
+ \frac{t^3}{18}
+ \beta\\
&= \frac{-24 \alpha^3 + (3\sqrt[3]{18}t^2)(\alpha^2\sqrt{3}i) + 2t^3+36 t^3 \beta}{36t^3}
\end{align}
\textbf{Case 2.3:}
\todo[inline]{calculate...}