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removed unnecessary stuff; began developing solution formula
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Martin Thoma
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1 changed files with 28 additions and 110 deletions
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@ -306,110 +306,20 @@ As you can easily verify, only $x_1$ is a minimum of $d_{P,f}$.
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\subsection{Number of points with minimal distance}
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\subsubsection{Two points with minimal distance}
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Quadratic functions can have two points with minimal distance.
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\begin{theorem}
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A point $P$ has either one or two points on the graph of a
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quadratic function $f$ that are closest to $P$.
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\end{theorem}
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For example, let $f(x) = x^2$ and $P = (0,5)$. Then $P_{f,1} = (\sqrt{\frac{9}{2}}, \frac{9}{2})$
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has minimal distance to $P$, but also $P_{f,2} = (-\sqrt{\frac{9}{2}}, \frac{9}{2})$:
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In the following, I will do some transformations with $f = f_0$ and
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$P = P_0$ .
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\begin{proof}
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\begin{align}
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d_{P,f}(x) &= \sqrt{(x-x_P)^2 + (f(x)-y_p)^2}\\
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&= \sqrt{x^2 + (x^2-5)^2}\\
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&= \sqrt{x^2 + x^4-10x^2+25}\\
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&= \sqrt{x^4 -9x^2 + 25}\\
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&= \sqrt{x^4 -9x^2 + \frac{81}{4}+\frac{19}{4}}\\
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&= \sqrt{\left (x^2 - \frac{9}{2} \right )^2 + \frac{19}{4}}
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\end{align}
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Moving $f_0$ and $P_0$ simultaneously in $x$ or $y$ direction does
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not change the minimum distance. Furthermore, we can find the
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points with minimum distance on the moved situation and calculate
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the minimum points in the original situation.
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Obviously, $d_{P,f}$ is minimal for $x = \pm \sqrt{\frac{9}{2}} \qed$
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\end{proof}
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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=north west,
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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=0.6\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= 0, % start the diagram at this y-coordinate
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ymax= 5, % 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, orange] {x*x};
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\draw (axis cs:0,5) circle[radius=2.17];
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\draw[red, thick] (axis cs:0,5) -- (axis cs:2.121,4.5);
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\draw[red, thick] (axis cs:0,5) -- (axis cs:-2.121,4.5);
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\addlegendentry{$f(x)=x^2$}
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\end{axis}
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\end{tikzpicture}
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\caption{Two points with minimal distance}
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\end{figure}
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\subsubsection{Three points with minimal distance}
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As discussed before, there cannot be more than 3 points on the graph
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of $f$ next to $P$.
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\todo[inline]{But can there be three points? O.b.d.A: $a > 0$.
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As $c$ is moves the curve only up and down, we can o.b.d.A assume
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that $c=0$.
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$x=-\frac{b}{2a}$ is the minimum of $f$. If there are 3 points, this will
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be one of them. The other two ones are symmetric by an axis through
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$-\frac{b}{2a}$}
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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=north west,
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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=-0.7, % start the diagram at this x-coordinate
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xmax= 0.7, % end the diagram at this x-coordinate
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ymin=-0.25, % start the diagram at this y-coordinate
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ymax= 0.5, % 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=-0.7:0.7, thick,samples=50, orange] {x*x};
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\draw (axis cs:0,0.5) circle[radius=0.5];
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\draw[red, thick] (axis cs:0,0.5) -- (axis cs:0.101,0.0102);
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\draw[red, thick] (axis cs:0,0.5) -- (axis cs:-0.101,0.0102);
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\draw[red, thick] (axis cs:0,0.5) -- (axis cs:0,0);
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\addlegendentry{$f(x)=x^2$}
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\end{axis}
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\end{tikzpicture}
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\caption{3 points with minimal distance?}
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\end{figure}
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When move the $f$ and $P$ simultaneously in $x$ direction, you will not change the
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results.
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First of all, you move $f_0$ by $\frac{b}{2a}$, so
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First of all, we move $f_0$ and $P_0$ by $\frac{b}{2a}$ in $x$ direction, so
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\[f_1(x) = ax^2 - \frac{b^2}{4a} + c \;\;\;\text{ and }\;\;\; P_1 = \left (x_p+\frac{b}{2a},\;\; y_p \right )\]
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Because:
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@ -425,20 +335,28 @@ 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 $f(x) = ax^2$ is symmetric to the $y$ axis, only points
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$P = (0, y_p)$ could possilby have three minima.
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$P = (0, w)$ could possilby have three minima.
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Then compute:
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\begin{align}
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d_{P,f}(x) &= \sqrt{(x-x_P)^2 + (f(x)-y_p)^2}\\
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&= \sqrt{x^2 + (ax^2-y_p)^2}\\
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&= \sqrt{x^2 + a^2 x^4-2ay_p x^2+y_p^2}\\
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&= \sqrt{a^2 x^4 + (1-2ay_p) x^2 + y_p^2}\\
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&= \sqrt{\left (a^2 x^2 + \frac{1-2 a y_p}{2} \right )^2 + y_p^2 - (1-2 a y_p)^2}\\
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&= \sqrt{\left (a^2 x^2 + \nicefrac{1}{2}-a y_p \right )^2 + (y_p^2 - (1-2 a y_p)^2)}\\
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d_{P,f}(x) &= \sqrt{(x-x_{P})^2 + (f(x)-w)^2}\\
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&= \sqrt{x^2 + (ax^2-w)^2}\\
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&= \sqrt{x^2 + a^2 x^4-2aw x^2+w^2}\\
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&= \sqrt{a^2 x^4 + (1-2aw) x^2 + w^2}\\
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&= \sqrt{\left (a^2 x^2 + \frac{1-2 a w}{2} \right )^2 + w^2 - (1-2 a w)^2}\\
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&= \sqrt{\left (a^2 x^2 + \nicefrac{1}{2}-a w \right )^2 + (w^2 - (1-2 a w)^2)}\\
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\end{align}
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For $y_p \leq \nicefrac{1}{2a}$ you only have $x = 0$ as a minimum.
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For all other points, there are exactly two minima.
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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.
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So the solution is given by
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\[\underset{x\in\mdr}{\arg \min d_{P,f}(x)} = \begin{cases}
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x_1 = todo \text{ and } x_2 = todo &\text{if } x_P = - \frac{b}{2a} \text{ and } y_p + \frac{b^2}{4a} - c > \frac{1}{2a} \\
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x = todo &\text{if } x_P = - \frac{b}{2a} \text{ and } y_p + \frac{b^2}{4a} - c \leq \frac{1}{2a} \\
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x = todo &\text{if } x_P \neq - \frac{b}{2a}
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\end{cases}\]
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\clearpage
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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