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next (failed?) try to calculate the quadratic
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Martin Thoma
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8 changed files with 153 additions and 64 deletions
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@ -43,6 +43,7 @@ The situation can be seen in Figure~\ref{fig:constant-min-distance}.
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\label{fig:constant-min-distance}
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\end{figure}
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The point $(x, f(x))$ with minimal distance can be calculated directly:
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\begin{align}
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d_{P,f}(x) &= \sqrt{(x_P - x)^2 + (y_P - f(x))^2}\\
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&= \sqrt{(x_P^2 - 2x_P x + x^2) + (y_P^2 - 2 y_P c + c^2)} \\
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@ -62,7 +62,7 @@ $b, c, d \in \mdr$ be a function.
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Then you could solve the following problem for $x$:
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\begin{align}
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0 &\stackrel{!}{=} \left ((d_{P,f}(x))^2 \right )'
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0 &\stackrel{!}{=} \left ((d_{P,f}(x))^2 \right )'\\
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&=-2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
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&= 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\
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&= f(x) \cdot f'(x) - y_p f'(x) + x - x_p\\
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@ -27,7 +27,7 @@ $t \in \mdr$ be a linear function.
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enlargelimits=true,
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tension=0.08]
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\addplot[domain=-5:5, thick,samples=50, red] {0.5*x};
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\addplot[domain=-5:5, thick,samples=50, blue] {-2*x+6};
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\addplot[domain=-5:5, thick,samples=50, blue, dashed] {-2*x+6};
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\addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)};
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\addplot[blue, nodes near coords=$f_\bot$,every node near coord/.style={anchor=225}] coordinates {(1.5, 3)};
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\addplot[red, nodes near coords=$f$,every node near coord/.style={anchor=225}] coordinates {(0.9, 0.5)};
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Binary file not shown.
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@ -40,13 +40,13 @@
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\newtheorem{proof}{Proof:}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\title{Minimal distance to a cubic function}
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\title{Minimal distance to polynomial functions of degree 3 or less}
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\author{Martin Thoma}
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\hypersetup{
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pdfauthor = {Martin Thoma},
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pdfkeywords = {},
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pdftitle = {Minimal Distance}
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pdfkeywords = {minimal distance, polynomial, function, degree 3, cubic, spline},
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pdftitle = {Minimal distance to polynomial functions of degree 3 or less}
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}
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\def\mdr{\ensuremath{\mathbb{R}}}
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@ -0,0 +1,50 @@
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\todo[inline]{calculate...}
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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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\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:
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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{\alpha^3(1-3i\sqrt{3} - 3 \cdot 3 + \sqrt{27} i)}{12 t^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(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}}\\
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&= \frac{-3\alpha^2((1+2\sqrt{3}i - 3)+(- i\sqrt{3}+2\cdot 3 + i\sqrt{3}))}{12 t \sqrt[3]{12}}\\
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&= \frac{-\alpha^2(4+2\sqrt{3}i)}{4t\sqrt[3]{12}}\\
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&= \frac{-\alpha^2(2+\sqrt{3}i)}{2t\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(1+i\sqrt{3})\alpha (1-2i\sqrt{3} - 3)t}{4 \sqrt[3]{12 \cdot 18^2}}\\
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&= \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}}\\
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&= \frac{3 \alpha t(4-4\sqrt{3}i)}{4 \cdot 2 \cdot 3 \sqrt[3]{2 \cdot 3^2}}\\
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&= \frac{\alpha t(1-\sqrt{3}i)}{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{(1-3i\sqrt{3} - 3 \cdot 3 + i \sqrt{27}) t^3}{8 \cdot 18}\\
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&=- \frac{t^3 (-8)}{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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+ \frac{-\alpha^2(2+\sqrt{3}i)}{2t\sqrt[3]{12}}
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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{{}=} + \color{blue}\alpha\color{black} \left (\frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t}
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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{\alpha^2(2(1+i\sqrt{3})-(2+\sqrt{3}i))}{2t\sqrt[3]{12}}
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+ \frac{t^3}{18}
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+ \beta\\
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&= \frac{-24 \alpha^3 + (3\sqrt[3]{18}t^2)(\alpha^2\sqrt{3}i) + 2t^3+36 t^3 \beta}{36t^3}
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\end{align}
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@ -0,0 +1,90 @@
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\todo[inline]{calculate...}
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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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\begin{align}
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x^3 &= \left (\frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} - \frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}} \right)^3\\
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&= \left (\frac{(2\sqrt[3]{18})(1-i \sqrt{3}) \alpha - (\sqrt[3]{12} \cdot t)(1+i\sqrt{3}) t}{\sqrt[3]{12} \cdot t \cdot 2 \cdot \sqrt[3]{18}} \right)^3\\
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&= \left (\frac{2\sqrt[3]{18}a \alpha (1-i \sqrt{3}) - \sqrt[3]{12} \cdot t^2(1+i\sqrt{3})}{2t \sqrt[3]{2^3 \cdot 3^3}} \right )^3\\
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&= 12 \cdot \bigg (\frac{\overbrace{\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2(1+i\sqrt{3})}^{\text{numerator}}}{12t} \bigg )^3
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\end{align}
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Now calculate the numerator$^3$. Remember, that $(1-i \sqrt{3})^2 = -2 (1+i \sqrt{3})$
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and $(1 \pm i \sqrt{3})^3 = -8$.
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\begin{align}
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\left (\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2(1+i\sqrt{3}) \right )^3 &=
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12 \alpha^3 (1-i\sqrt{3})^3 \\
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&\hphantom{{}=}- 3 \sqrt[3]{12^2} \alpha^2(1-i\sqrt{3})^2 (t^2(1+i \sqrt{3}))\\
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&\hphantom{{}=}+ 3 \sqrt[3]{12\hphantom{^2}} \alpha\hphantom{^2} (1-i\sqrt{3}) t^4 (1+i\sqrt{3})^2 - t^6 (1+i\sqrt{3})^3\\
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&= 12 \alpha^3 \cdot (-8) \\
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&\hphantom{{}=}- 3 \cdot 2 \sqrt[3]{18} \alpha^2(-2(1+i\sqrt{3}))(t^2(1+i \sqrt{3}))\\
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&\hphantom{{}=}+ 3 \sqrt[3]{12} \alpha (1-i\sqrt{3}) t^4 (-2(1-i\sqrt{3})) - t^6 (-8)\\
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&= -96 \alpha^3 + 12 \sqrt[3]{18} \alpha^2 t^2 (1+i \sqrt{3})^2\\
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&\hphantom{{}=}- 6 \sqrt[3]{12} \alpha (1-i\sqrt{3})^2 t^4 +8 t^6\\
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&= -96 \alpha^3 - 24 \sqrt[3]{18} \alpha^2 t^2 (1-i \sqrt{3})\\
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&\hphantom{{}=}+ 12 \sqrt[3]{12} \alpha t^4 (1+i \sqrt{3}) +8 t^6
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\end{align}
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\goodbreak
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Now 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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&= \frac{-96 \alpha^3 - 24 \sqrt[3]{18} \alpha^2 t^2 (1-i \sqrt{3}) + 12 \sqrt[3]{12} \alpha t^4 (1+i \sqrt{3}) +8 t^6}{12^2 t^3}\\
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&\hphantom{{}=}+\alpha \left (\sqrt[3]{12} \cdot \frac{\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2(1+i\sqrt{3})}{12t} \right ) + \beta
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\end{align}
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\todo[inline]{the calculation above seems to be wrong / too long. Next try}
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When you insert this in Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance}
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you get:\footnote{Remember, that $(1+i\sqrt{3})^2 = -2 (1-i \sqrt{3})$ and $(1-i \sqrt{3})^2 = -2 (1+i \sqrt{3})$
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and $(1 \pm i \sqrt{3})^3 = -8$}
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\begin{align}
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0 &\stackrel{!}{=} \left( \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}} \right)^3
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+ \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right )
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+ \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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&= \frac{(1-i \sqrt{3})^3 \alpha^3}{12 \cdot t^3}
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- 3 \left (\frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} \right )^2 \frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}
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+ 3 \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} \left (\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}} \right)^2\\
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&\hphantom{{}=}
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+ \frac{(1+i\sqrt{3})^3 t^3}{2^3 \cdot 18}
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+ \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right )
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+ \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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&= \frac{-8 \alpha^3}{12t^3}
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- 3 \frac{-2(1+i \sqrt{3}) \alpha^2}{\sqrt[3]{12^2} t^2} \frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}
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+ 3 \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} \frac{-2(1-i\sqrt{3}) t^2}{4\sqrt[3]{18^2}}\\
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&\hphantom{{}=}
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+ \frac{-8 t^3}{2^3 \cdot 18}
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+ \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right )
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+ \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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&= \frac{-2 \alpha^3}{3t^3}
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+ \frac{6 \alpha^2 t (-2)(1-i \sqrt{3})}{(\sqrt[3]{12^2} t^2)(2\sqrt[3]{18})}
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+ \frac{12 \alpha t^2 (1+i \sqrt{3})}{(\sqrt[3]{12} \cdot t)(4\sqrt[3]{18^2)}}\\
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&\hphantom{{}=}
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+ \frac{- t^3}{18}
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+ \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right )
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+ \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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&= \frac{-2 \alpha^3}{3t^3}
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+ \frac{-6 \alpha^2 (1-i \sqrt{3})}{6 \sqrt[3]{12} t}
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+ \frac{3 \alpha t (1+i \sqrt{3})}{6\sqrt[3]{18}}
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+ \frac{- t^3}{18}\\
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&\hphantom{{}=}+ \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right )
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+ \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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&= \frac{-2 \alpha^3}{3t^3}
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+ \frac{-\alpha^2 (1-i \sqrt{3})}{\sqrt[3]{12} t}
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+ \frac{\alpha t (1+i \sqrt{3})}{2\sqrt[3]{18}}
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+ \frac{- t^3}{18}\\
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&\hphantom{{}=}+ \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right )
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+ \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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&= \frac{12 \cdot (-2 \alpha^3) +(6 \sqrt[3]{18}t^2)(-\alpha^2 (1-i \sqrt{3}))+ (3 \sqrt[3]{12})(\alpha t (1+i \sqrt{3})) + (2t^3)(- t^3)}{36t^3}\\
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&\hphantom{{}=}+ \frac{(6 \sqrt[3]{18})((1-i \sqrt{3}) \alpha) - (3 \sqrt[3]{12})((1+i\sqrt{3}) t) + 36t^3 \beta}{36t^3}\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\end{align}
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\goodbreak
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Now calculate only the numerator:
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\begin{align}
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0 &\stackrel{!}{=} -12 \alpha^3 - 6 \sqrt[3]{18} t^2 \alpha^2 (1 - i \sqrt{3})
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+ 3 \sqrt[3]{12} \alpha t (1+i\sqrt{3}) - 2t^6\\
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&\hphantom{{}=} + 6\sqrt[3]{18} \alpha (1- i \sqrt{3})
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- 3 \sqrt[3]{12} t (1+i \sqrt{3}) + 36 t^3 \beta
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\end{align}
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@ -42,7 +42,7 @@ $b, c \in \mdr$ be a quadratic function.
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\subsection{Calculate points with minimal distance}
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In this case, $d_{P,f}^2$ is polynomial of degree 4.
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We use Theorem~\ref{thm:required-extremum-property}:\nobreak
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We use Theorem~\ref{thm:fermats-theorem}:\nobreak
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\begin{align}
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0 &\overset{!}{=} (d_{P,f}^2)'\\
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&= -2 x_p + 2x -2y_p f'(x) + \left (f(x)^2 \right )'\\
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@ -57,7 +57,7 @@ We use Theorem~\ref{thm:required-extremum-property}:\nobreak
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This is an algebraic equation of degree 3.
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There can be up to 3 solutions in such an equation. Those solutions
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can be found with a closed formula. But not every solution of the
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equation given by Theorem~\ref{thm:required-extremum-property}
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equation given by Theorem~\ref{thm:fermats-theorem}
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has to be a solution to the given problem.
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\begin{example}
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@ -134,7 +134,7 @@ We start with the graph that was moved so that $f_2 = ax^2$.
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\textbf{Case 1:} $P$ is on the symmetry axis, hence $x_P = - \frac{b}{2a}$.
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In this case, we have already found the solution. If $y_P + \frac{b^2}{4a} - c > \frac{1}{2a}$,
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In this case, we have already found the solution. If $w = y_P + \frac{b^2}{4a} - c > \frac{1}{2a}$,
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then there are two solutions:
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\[x_{1,2} = \pm \sqrt{aw - \nicefrac{1}{2}}\]
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Otherwise, there is only one solution $x_1 = 0$.
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@ -206,62 +206,10 @@ $t$:
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\end{align}
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\textbf{Case 2.2:}
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\todo[inline]{calculate...}
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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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\input{quadratic-case-2.2}
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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:
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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{\alpha^3(1-3i\sqrt{3} - 3 \cdot 3 + \sqrt{27} i)}{12 t^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(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}}\\
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&= \frac{-3\alpha^2((1+2\sqrt{3}i - 3)+(- i\sqrt{3}+2\cdot 3 + i\sqrt{3}))}{12 t \sqrt[3]{12}}\\
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&= \frac{-\alpha^2(4+2\sqrt{3}i)}{4t\sqrt[3]{12}}\\
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&= \frac{-\alpha^2(2+\sqrt{3}i)}{2t\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(1+i\sqrt{3})\alpha (1-2i\sqrt{3} - 3)t}{4 \sqrt[3]{12 \cdot 18^2}}\\
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&= \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}}\\
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&= \frac{3 \alpha t(4-4\sqrt{3}i)}{4 \cdot 2 \cdot 3 \sqrt[3]{2 \cdot 3^2}}\\
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&= \frac{\alpha t(1-\sqrt{3}i)}{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{(1-3i\sqrt{3} - 3 \cdot 3 + i \sqrt{27}) t^3}{8 \cdot 18}\\
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&=- \frac{t^3 (-8)}{8 \cdot 18}\\
|
||||
&= \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 \\
|
||||
&= \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...}
|
||||
|
||||
\[x = \frac{(1-i \sqrt{3})a}{\sqrt[3]{12} \cdot t}
|
||||
-\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\]
|
||||
\textbf{Case 2.3:}
|
||||
\input{quadratic-case-2.3}
|
||||
|
||||
\goodbreak
|
||||
So the solution is given by
|
||||
|
|
@ -298,7 +246,7 @@ If the function (defined on $\mdr$) has only one shortest distance
|
|||
point $x$ for the given $P$, it's also easy: The point in $[a,b]$ that
|
||||
is closest to $x$ will have the sortest distance.
|
||||
|
||||
\[\underset{x\in\mdr}{\arg \min d_{P,f}(x)} = \begin{cases}
|
||||
\[\underset{x\in[a,b]}{\arg \min d_{P,f}(x)} = \begin{cases}
|
||||
S_2(f, P) \cap [a,b] &\text{if } S_2(f, P) \cap [a,b] \neq \emptyset \\
|
||||
\Set{a} &\text{if } |S_2(f, P)| = 1 \text{ and } S_2(f, P) \ni x < a\\
|
||||
\Set{b} &\text{if } |S_2(f, P)| = 1 \text{ and } S_2(f, P) \ni x > b\\
|
||||
|
|
|
|||
Loading…
Add table
Add a link
Reference in a new issue