2013-12-12 16:22:13 +01:00
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\chapter{Description of the Problem}
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Let $f: D \rightarrow \mdr$ with $D \subseteq \mdr$ be a polynomial function and $P \in \mdr^2$
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be a point. Let $d_{P,f}: \mdr \rightarrow \mdr_0^+$
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2013-12-14 00:20:46 +01:00
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be the Euklidean distance of a point $P$ to a point $\left (x, f(x) \right )$
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2013-12-12 16:22:13 +01:00
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on the graph of $f$:
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2013-12-21 19:10:35 +01:00
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\[d_{P,f} (x) := \sqrt{(x - x_P)^2 + (f(x) - y_P)^2}\]
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2013-12-12 16:22:13 +01:00
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Now there is finite set $M = \Set{x_1, \dots, x_n} \subseteq D$ of minima for given $f$ and $P$:
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\[M = \Set{x \in D | d_{P,f}(x) = \min_{\overline{x} \in D} d_{P,f}(\overline{x})}\]
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2013-12-14 00:20:46 +01:00
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But minimizing $d_{P,f}$ is the same as minimizing
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$d_{P,f}^2 = x_p^2 - 2x_p x + x^2 + y_p^2 - 2y_p f(x) + f(x)^2$.
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2013-12-12 16:22:13 +01:00
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2013-12-16 10:51:15 +01:00
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In order to solve the minimal distance problem, Fermat's theorem
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about stationary points will be tremendously usefull:
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\begin{theorem}[Fermat's theorem about stationary points]\label{thm:fermats-theorem}
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Let $x_0$ be a local extremum of a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$.
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Then: $f'(x_0) = 0$.
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\end{theorem}
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2013-12-12 23:05:04 +01:00
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2013-12-21 19:10:35 +01:00
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So in fact you can calculate the roots of $(d_{P,f}(x))'$ to get
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candidates for minimal distance. These candidates include all points
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with minimal distance, but might also contain more. Example~\ref{ex:false-positive}
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shows such a situation.
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2013-12-12 23:05:04 +01:00
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Let $S_n$ be the function that returns the set of solutions for a
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polynomial $f$ of degree $n$ and a point $P$:
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\[S_n: \Set{\text{Polynomials of degree } n \text{ defined on } \mdr} \times \mdr^2 \rightarrow \mathcal{P}({\mdr})\]
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\[S_n(f, P) := \underset{x\in\mdr}{\arg \min d_{P,f}(x)} = M\]
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2013-12-12 23:05:04 +01:00
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If possible, I will explicitly give this function.
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