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Remove trailing spaces
The commands find . -type f -name '*.md' -exec sed --in-place 's/[[:space:]]\+$//' {} \+ and find . -type f -name '*.tex' -exec sed --in-place 's/[[:space:]]\+$//' {} \+ were used to do so.
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@ -6,12 +6,12 @@ on the graph of $f$:
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\[d_{P,f} (x) := \sqrt{(x - x_P)^2 + (f(x) - y_P)^2}\]
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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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\[M = \Set{x \in D | d_{P,f}(x) = \min_{\overline{x} \in D} d_{P,f}(\overline{x})}\]
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But minimizing $d_{P,f}$ is the same as minimizing
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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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In order to solve the minimal distance problem, Fermat's theorem
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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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@ -22,12 +22,12 @@ about stationary points will be tremendously usefull:
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So in fact you can calculate the roots of $(d_{P,f}(x))'$ or $(d_{P,f}(x)^2)'$ to get
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candidates for minimal distance.
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$(d_{P,f}(x)^2)'$ is a polynomial if $f$ is a polynomial. So if $f$
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is a polynomial, we can always get a finite number of candidates by
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$(d_{P,f}(x)^2)'$ is a polynomial if $f$ is a polynomial. So if $f$
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is a polynomial, we can always get a finite number of candidates by
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finding roots of $(d_{P,f}(x)^2)'$. But this gets difficult when $f$
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has degree 3 or higher as explained in Theorem~\ref{thm:no-finite-solution}.
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Another problem one has to bear in mind is that these candidates
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include all points with minimal distance, but might also contain
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Another problem one has to bear in mind is that these candidates
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include all points with minimal distance, but might also contain
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more. Example~\ref{ex:false-positive} shows such a situation.
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Let $S_n$ be the function that returns the set of solutions for a
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