diff --git a/documents/math-minimal-distance-to-cubic-function/constant-functions.tex b/documents/math-minimal-distance-to-cubic-function/constant-functions.tex index 9e26861..298403c 100644 --- a/documents/math-minimal-distance-to-cubic-function/constant-functions.tex +++ b/documents/math-minimal-distance-to-cubic-function/constant-functions.tex @@ -43,6 +43,7 @@ The situation can be seen in Figure~\ref{fig:constant-min-distance}. \label{fig:constant-min-distance} \end{figure} +The point $(x, f(x))$ with minimal distance can be calculated directly: \begin{align} d_{P,f}(x) &= \sqrt{(x_P - x)^2 + (y_P - f(x))^2}\\ &= \sqrt{(x_P^2 - 2x_P x + x^2) + (y_P^2 - 2 y_P c + c^2)} \\ diff --git a/documents/math-minimal-distance-to-cubic-function/cubic-functions.tex b/documents/math-minimal-distance-to-cubic-function/cubic-functions.tex index f23a781..f1c3f2c 100644 --- a/documents/math-minimal-distance-to-cubic-function/cubic-functions.tex +++ b/documents/math-minimal-distance-to-cubic-function/cubic-functions.tex @@ -62,7 +62,7 @@ $b, c, d \in \mdr$ be a function. Then you could solve the following problem for $x$: \begin{align} - 0 &\stackrel{!}{=} \left ((d_{P,f}(x))^2 \right )' + 0 &\stackrel{!}{=} \left ((d_{P,f}(x))^2 \right )'\\ &=-2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\ &= 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\ &= f(x) \cdot f'(x) - y_p f'(x) + x - x_p\\ diff --git a/documents/math-minimal-distance-to-cubic-function/linear-functions.tex b/documents/math-minimal-distance-to-cubic-function/linear-functions.tex index 0f14f36..2637c67 100644 --- a/documents/math-minimal-distance-to-cubic-function/linear-functions.tex +++ b/documents/math-minimal-distance-to-cubic-function/linear-functions.tex @@ -27,7 +27,7 @@ $t \in \mdr$ be a linear function. enlargelimits=true, tension=0.08] \addplot[domain=-5:5, thick,samples=50, red] {0.5*x}; - \addplot[domain=-5:5, thick,samples=50, blue] {-2*x+6}; + \addplot[domain=-5:5, thick,samples=50, blue, dashed] {-2*x+6}; \addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)}; \addplot[blue, nodes near coords=$f_\bot$,every node near coord/.style={anchor=225}] coordinates {(1.5, 3)}; \addplot[red, nodes near coords=$f$,every node near coord/.style={anchor=225}] coordinates {(0.9, 0.5)}; diff --git a/documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.pdf b/documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.pdf index f06e85e..dea2df5 100644 Binary files a/documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.pdf and b/documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.pdf differ diff --git a/documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.tex b/documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.tex index 38effd2..4962d39 100644 --- a/documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.tex +++ b/documents/math-minimal-distance-to-cubic-function/math-minimal-distance-to-cubic-function.tex @@ -40,13 +40,13 @@ \newtheorem{proof}{Proof:} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\title{Minimal distance to a cubic function} +\title{Minimal distance to polynomial functions of degree 3 or less} \author{Martin Thoma} \hypersetup{ pdfauthor = {Martin Thoma}, - pdfkeywords = {}, - pdftitle = {Minimal Distance} + pdfkeywords = {minimal distance, polynomial, function, degree 3, cubic, spline}, + pdftitle = {Minimal distance to polynomial functions of degree 3 or less} } \def\mdr{\ensuremath{\mathbb{R}}} diff --git a/documents/math-minimal-distance-to-cubic-function/quadratic-case-2.2.tex b/documents/math-minimal-distance-to-cubic-function/quadratic-case-2.2.tex new file mode 100644 index 0000000..37b795d --- /dev/null +++ b/documents/math-minimal-distance-to-cubic-function/quadratic-case-2.2.tex @@ -0,0 +1,50 @@ +\todo[inline]{calculate...} + +\[x = \frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} + -\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}}\] + +\begin{align} + x^3 &= \underbrace{\left (\frac{(1+i\sqrt{3})\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}}} &= + \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)}{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} diff --git a/documents/math-minimal-distance-to-cubic-function/quadratic-case-2.3.tex b/documents/math-minimal-distance-to-cubic-function/quadratic-case-2.3.tex new file mode 100644 index 0000000..8b9c8ef --- /dev/null +++ b/documents/math-minimal-distance-to-cubic-function/quadratic-case-2.3.tex @@ -0,0 +1,90 @@ +\todo[inline]{calculate...} + +\[x = \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} + -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\] + +\begin{align} + x^3 &= \left (\frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} - \frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}} \right)^3\\ + &= \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\\ + &= \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\\ + &= 12 \cdot \bigg (\frac{\overbrace{\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2(1+i\sqrt{3})}^{\text{numerator}}}{12t} \bigg )^3 +\end{align} + +Now calculate the numerator$^3$. Remember, that $(1-i \sqrt{3})^2 = -2 (1+i \sqrt{3})$ +and $(1 \pm i \sqrt{3})^3 = -8$. +\begin{align} + \left (\sqrt[3]{12} \alpha (1-i \sqrt{3}) - t^2(1+i\sqrt{3}) \right )^3 &= + 12 \alpha^3 (1-i\sqrt{3})^3 \\ + &\hphantom{{}=}- 3 \sqrt[3]{12^2} \alpha^2(1-i\sqrt{3})^2 (t^2(1+i \sqrt{3}))\\ + &\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\\ + &= 12 \alpha^3 \cdot (-8) \\ + &\hphantom{{}=}- 3 \cdot 2 \sqrt[3]{18} \alpha^2(-2(1+i\sqrt{3}))(t^2(1+i \sqrt{3}))\\ + &\hphantom{{}=}+ 3 \sqrt[3]{12} \alpha (1-i\sqrt{3}) t^4 (-2(1-i\sqrt{3})) - t^6 (-8)\\ + &= -96 \alpha^3 + 12 \sqrt[3]{18} \alpha^2 t^2 (1+i \sqrt{3})^2\\ + &\hphantom{{}=}- 6 \sqrt[3]{12} \alpha (1-i\sqrt{3})^2 t^4 +8 t^6\\ + &= -96 \alpha^3 - 24 \sqrt[3]{18} \alpha^2 t^2 (1-i \sqrt{3})\\ + &\hphantom{{}=}+ 12 \sqrt[3]{12} \alpha t^4 (1+i \sqrt{3}) +8 t^6 +\end{align} +\goodbreak +Now back to the original equation: +\begin{align} +0 &\stackrel{!}{=} x^3 + \alpha x + \beta\\ + &= \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}\\ + &\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 +\end{align} + +\todo[inline]{the calculation above seems to be wrong / too long. Next try} + +When you insert this in Equation~\ref{eq:simple-cubic-equation-for-quadratic-distance} +you get:\footnote{Remember, that $(1+i\sqrt{3})^2 = -2 (1-i \sqrt{3})$ and $(1-i \sqrt{3})^2 = -2 (1+i \sqrt{3})$ +and $(1 \pm i \sqrt{3})^3 = -8$} +\begin{align} + 0 &\stackrel{!}{=} \left( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} + -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}} \right)^3 + + \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right ) + + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + &= \frac{(1-i \sqrt{3})^3 \alpha^3}{12 \cdot t^3} + - 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}} + + 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\\ + &\hphantom{{}=} + + \frac{(1+i\sqrt{3})^3 t^3}{2^3 \cdot 18} + + \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right ) + + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + &= \frac{-8 \alpha^3}{12t^3} + - 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}} + + 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}}\\ + &\hphantom{{}=} + + \frac{-8 t^3}{2^3 \cdot 18} + + \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right ) + + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + &= \frac{-2 \alpha^3}{3t^3} + + \frac{6 \alpha^2 t (-2)(1-i \sqrt{3})}{(\sqrt[3]{12^2} t^2)(2\sqrt[3]{18})} + + \frac{12 \alpha t^2 (1+i \sqrt{3})}{(\sqrt[3]{12} \cdot t)(4\sqrt[3]{18^2)}}\\ + &\hphantom{{}=} + + \frac{- t^3}{18} + + \alpha \left ( \frac{(1-i \sqrt{3}) \alpha}{\sqrt[3]{12} \cdot t} -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\right ) + + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + &= \frac{-2 \alpha^3}{3t^3} + + \frac{-6 \alpha^2 (1-i \sqrt{3})}{6 \sqrt[3]{12} t} + + \frac{3 \alpha t (1+i \sqrt{3})}{6\sqrt[3]{18}} + + \frac{- t^3}{18}\\ + &\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 ) + + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + &= \frac{-2 \alpha^3}{3t^3} + + \frac{-\alpha^2 (1-i \sqrt{3})}{\sqrt[3]{12} t} + + \frac{\alpha t (1+i \sqrt{3})}{2\sqrt[3]{18}} + + \frac{- t^3}{18}\\ + &\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 ) + + \beta\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + &= \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}\\ + &\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}\\%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\end{align} + +\goodbreak +Now calculate only the numerator: +\begin{align} + 0 &\stackrel{!}{=} -12 \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^6\\ + &\hphantom{{}=} + 6\sqrt[3]{18} \alpha (1- i \sqrt{3}) + - 3 \sqrt[3]{12} t (1+i \sqrt{3}) + 36 t^3 \beta +\end{align} diff --git a/documents/math-minimal-distance-to-cubic-function/quadratic-functions.tex b/documents/math-minimal-distance-to-cubic-function/quadratic-functions.tex index ceae46c..af5ef00 100644 --- a/documents/math-minimal-distance-to-cubic-function/quadratic-functions.tex +++ b/documents/math-minimal-distance-to-cubic-function/quadratic-functions.tex @@ -42,7 +42,7 @@ $b, c \in \mdr$ be a quadratic function. \subsection{Calculate points with minimal distance} In this case, $d_{P,f}^2$ is polynomial of degree 4. -We use Theorem~\ref{thm:required-extremum-property}:\nobreak +We use Theorem~\ref{thm:fermats-theorem}:\nobreak \begin{align} 0 &\overset{!}{=} (d_{P,f}^2)'\\ &= -2 x_p + 2x -2y_p f'(x) + \left (f(x)^2 \right )'\\ @@ -57,7 +57,7 @@ We use Theorem~\ref{thm:required-extremum-property}:\nobreak This is an algebraic equation of degree 3. There can be up to 3 solutions in such an equation. Those solutions can be found with a closed formula. But not every solution of the -equation given by Theorem~\ref{thm:required-extremum-property} +equation given by Theorem~\ref{thm:fermats-theorem} has to be a solution to the given problem. \begin{example} @@ -134,7 +134,7 @@ We start with the graph that was moved so that $f_2 = ax^2$. \textbf{Case 1:} $P$ is on the symmetry axis, hence $x_P = - \frac{b}{2a}$. -In this case, we have already found the solution. If $y_P + \frac{b^2}{4a} - c > \frac{1}{2a}$, +In this case, we have already found the solution. If $w = y_P + \frac{b^2}{4a} - c > \frac{1}{2a}$, then there are two solutions: \[x_{1,2} = \pm \sqrt{aw - \nicefrac{1}{2}}\] Otherwise, there is only one solution $x_1 = 0$. @@ -206,62 +206,10 @@ $t$: \end{align} \textbf{Case 2.2:} -\todo[inline]{calculate...} -\[x = \frac{(1+i \sqrt{3})\alpha}{\sqrt[3]{12} \cdot t} - -\frac{(1-i\sqrt{3}) t}{2\sqrt[3]{18}}\] +\input{quadratic-case-2.2} -\begin{align} - 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}}} &= - \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)}{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...} - -\[x = \frac{(1-i \sqrt{3})a}{\sqrt[3]{12} \cdot t} - -\frac{(1+i\sqrt{3}) t}{2\sqrt[3]{18}}\] +\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\\