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added some ideas
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Martin Thoma
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@ -15,6 +15,7 @@
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\usepackage[framed,amsmath,thmmarks,hyperref]{ntheorem}
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\usepackage{framed}
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\usepackage{nicefrac}
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\usepackage{siunitx}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% Define theorems %
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@ -334,7 +335,7 @@ Because:
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Then move $f_1$ and $P_1$ by $\frac{b^2}{4a}-c$ in $y$ direction. You get:
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\[f_2(x) = ax^2\;\;\;\text{ and }\;\;\; P_2 = \left (x_p+\frac{b}{2a},\;\; y_p+\frac{b^2}{4a}-c \right )\]
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As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points
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\textbf{Case 1:} As $f_2(x) = ax^2$ is symmetric to the $y$ axis, only points
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$P = (0, w)$ could possilby have three minima.
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Then compute:
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@ -353,8 +354,43 @@ should get as close to $0$ as possilbe when we want to minimize
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$d_{P,{f_2}}$. For $w \leq \nicefrac{1}{2a}$ you only have $x = 0$ as a minimum.
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For all other points $P = (0, w)$, there are exactly two minima $x_{1,2} = \pm \sqrt{aw - \nicefrac{1}{2}}$.
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So the solution is given by
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\textbf{Case 2:} $P = (z, w)$ is not on the symmetry axis, so $z \neq 0$. Then you compute:
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\begin{align}
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d_{P,{f_2}}(x) &= \sqrt{(x-z)^2 + (f(x)-w)^2}\\
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&= \sqrt{(x^2 - 2zx + z^2) + ((ax^2)^2 - 2 awx^2 - w^2)}\\
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&= \sqrt{a^2x^4 + (1- 2 aw)x^2 +(- 2z)x + z^2 - w^2}\\
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0 &\stackrel{!}{=} d_{P, {f_2}}'(x) \\
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&= 4a^2x^3 + 2(1- 2 aw)x +(- 2z)\\
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&= 2 \left (2a^2x^2 + (1- 2 aw) \right )x - 2z\\
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\Leftrightarrow z &\stackrel{!}{=} (2a^2x^2 + (1- 2 aw)) x\\
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\Leftrightarrow 0 &\stackrel{!}{=} 2 a^2 x^3 + (1- 2 aw)x - z\label{line:solution-equation}
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\end{align}
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The solution for this equation was computated with Wolfram|Alpha.
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I will only verify that the solution is correct. As there is only
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one solution in this case, we only have to check this one.
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\begin{align}
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t &:= \sqrt[3]{108a^4z + \sqrt{\num{11664}a^8 z^2 + 864 a^6 (1-2aw)^3}}\\
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x &= \frac{t}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2}(1-2aw)}{t}\\
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\xRightarrow{x \text{ in line }\ref{line:solution-equation}} 0 &\stackrel{!}{=} 2a^2 \left ( \frac{t}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2}(1-2aw)}{t} \right )^3 + (1-2aw) (\frac{t}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2}(1-2aw)}{t}) - z\\
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&= 2a^2 \underbrace{\left ( \frac{t}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2}(1-2aw)}{t} \right )^3}_{=: \alpha} + \frac{t \cdot (1-2aw)}{6 \sqrt[3]{2} a^2} - \frac{\sqrt[3]{2} (1-2aw)^2}{t} - z
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\end{align}
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Now compute $\alpha$. We know that $(a-b)^3 = a^3 - 3a^2 b +3ab^2 - b^3$:
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\begin{align}
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\alpha &= \left (\frac{t}{6 \sqrt[3]{2} a^2} \right )^3 - 3 \left (\frac{t}{6 \sqrt[3]{2} a^2} \right )^2 \left (\frac{\sqrt[3]{2}(1-2aw)}{t} \right ) + 3 \left (\frac{t}{6 \sqrt[3]{2} a^2} \right ) \left (\frac{\sqrt[3]{2}(1-2aw)}{t} \right )^2 - \left (\frac{\sqrt[3]{2}(1-2aw)}{t} \right )^3\\
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&= \frac{t^3}{216 \cdot 2 \cdot a^6} - \frac{3t^2}{36 \sqrt[3]{4} a^4} \cdot \frac{\sqrt[3]{2}(1-2aw)}{t}
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+ \frac{3t}{6 \sqrt[3]{2} a^2} \cdot \frac{\sqrt[3]{4} (1-2aw)^2}{t^2} - \frac{2 (1-2aw)^3}{t^3}\\
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&= \frac{t^3}{432a^6} - \frac{t (1-2aw)}{12\sqrt[3]{2} a^4} + \frac{\sqrt[3]{2} (1-2aw)^2}{2ta^2} - \frac{2 (1-2aw)^3}{t^3}\\
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&= \frac{t^3}{432a^6} + \frac{t^2 (2aw-1) + 6 \sqrt[3]{4} a^2 (1-2aw)^2}{12\sqrt[3]{2} a^4} - \frac{2 (1-2aw)^3}{t^3}\\
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&= \frac{t^3- 2 (1-2aw)^3 \cdot 432a^6}{432a^6 t^3} + \frac{t^2 (2aw-1) + 6 \sqrt[3]{4} a^2 (1-2aw)^2}{12\sqrt[3]{2} a^4}\\
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&= \frac{\sqrt[3]{2}(t^3- 2 (1-2aw)^3 \cdot 432a^6) + 36 a^2 t^3 (t^2 (2aw-1) + 6 \sqrt[3]{4} a^2 (1-2aw)^2)}{432 \sqrt[3]{2} a^6 t^3}\\
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&= \frac{\sqrt[3]{2} t^3 - \sqrt[3]{2} (1-2aw)^3 \cdot 864a^6 + 36 a^2 t^3 (2awt^1-t^2 + 6 \sqrt[3]{4} a^2 (1-2aw)^2)}{6 \sqrt[3]{2} a^2 t \cdot 2a^2 \cdot 36 a^2 t^2}
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\end{align}
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\goodbreak
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So the solution is given by
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\begin{align*}
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x_S &:= - \frac{b}{2a} \;\;\;\;\; \text{(the symmetry axis)}\\
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\underset{x\in\mdr}{\arg \min d_{P,f}(x)} &= \begin{cases}
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@ -459,6 +495,68 @@ $\tilde{e}$.
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This means, that there is no solution formula for the problem of
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finding the closest points on a cubic function to a given point.
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\subsection{Another approach}
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Just like we moved the function $f$ and the point to get in a
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nicer situation, we can apply this approach for cubic functions.
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\begin{figure}[htp]
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\centering
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\begin{tikzpicture}
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\begin{axis}[
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legend pos=south east,
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axis x line=middle,
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axis y line=middle,
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grid = major,
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width=0.8\linewidth,
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height=8cm,
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grid style={dashed, gray!30},
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xmin=-3, % start the diagram at this x-coordinate
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xmax= 3, % end the diagram at this x-coordinate
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ymin=-3, % start the diagram at this y-coordinate
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ymax= 3, % end the diagram at this y-coordinate
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axis background/.style={fill=white},
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xlabel=$x$,
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ylabel=$y$,
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%xticklabels={-2,-1.6,...,7},
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%yticklabels={-8,-7,...,8},
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tick align=outside,
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minor tick num=-3,
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enlargelimits=true,
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tension=0.08]
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\addplot[domain=-3:3, thick,samples=50, red] {x*x*x};
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\addplot[domain=-3:3, thick,samples=50, green] {x*x*x+x};
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\addplot[domain=-3:3, thick,samples=50, orange] {x*x*x-x};
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\addplot[domain=-3:3, thick,samples=50, blue, dotted] {x*x*x+2*x};
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\addplot[domain=-3:3, thick,samples=50, lime, dashed] {x*x*x+3*x};
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\addlegendentry{$f_1(x)=x^3$}
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\addlegendentry{$f_2(x)=x^3 + x$}
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\addlegendentry{$f_1(x)=x^3 - x$}
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\addlegendentry{$f_2(x)=x^3 + 2 \cdot x$}
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\addlegendentry{$f_2(x)=x^3 + 3 \cdot x$}
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\end{axis}
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\end{tikzpicture}
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\caption{Cubic functions with $b = d = 0$}
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\end{figure}
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First, we move $f_0$ by $\frac{b}{3a}$ to the right, so
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\[f_1(x) = ax^3 + \frac{b^2 (c-1)}{3a} x + \frac{2b^3}{27 a^2} - \frac{bc}{3a} + d \;\;\;\text{ and }\;\;\;P_1 = (x_P + \frac{b}{3a}, y_P)\]
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because
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\begin{align}
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f_1(x) &= a \left (x - \frac{b}{3a} \right )^3 + b \left (x-\frac{b}{3a} \right )^2 + c \left (x-\frac{b}{3a} \right ) + d\\
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&= a \left (x^3 - 3 \frac{b}{3a}x^2 + 3 (\frac{b}{3a})^2 x - \frac{b^3}{27a^3} \right )
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+b \left (x^2 - \frac{2b}{3a} x + \frac{b^2}{9a^2} \right )
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+c x - \frac{bc}{3a} + d\\
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&= ax^3 - bx^2 + \frac{b^2}{3a}x - \frac{b^3}{27 a^2}\\
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& \;\;\;\;\;\;+ bx^2 - \frac{2b^2}{3a}x + \frac{b^3}{9a^2}\\
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& \;\;\;\;\;\;\;\;\;\;\;\; + c x - \frac{bc}{3a} + d\\
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&= ax^3 + \frac{b^2}{3a}\left (1-2+c \right )x + \frac{b^3}{9a^2} \left (1-\frac{1}{3} \right )- \frac{bc}{3a} + d
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\end{align}
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\todo[inline]{Which way to move might be clever?}
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\subsection{Number of points with minimal distance}
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As there is an algebraic equation of degree 5, there cannot be more
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than 5 solutions.
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