\chapter{Constant functions} \section{Defined on $\mdr$} \begin{lemma} Let $f:\mdr \rightarrow \mdr$, $f(x) := c$ with $c \in \mdr$ be a constant function. Then $(x_P, f(x_P))$ is the only point on the graph of $f$ with minimal distance to $P$. \end{lemma} The situation can be seen in Figure~\ref{fig:constant-min-distance}. \begin{figure}[htp] \centering \begin{tikzpicture} \begin{axis}[ legend pos=north west, legend cell align=left, axis x line=middle, axis y line=middle, grid = major, width=0.8\linewidth, height=8cm, grid style={dashed, gray!30}, xmin=-5, % start the diagram at this x-coordinate xmax= 5, % end the diagram at this x-coordinate ymin= 0, % start the diagram at this y-coordinate ymax= 3, % end the diagram at this y-coordinate axis background/.style={fill=white}, xlabel=$x$, ylabel=$y$, tick align=outside, minor tick num=-3, enlargelimits=true, tension=0.08] \addplot[domain=-5:5, thick,samples=50, red] {1}; \addplot[domain=-5:5, thick,samples=50, green] {2}; \addplot[domain=-5:5, thick,samples=50, blue, densely dotted] {3}; \addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)}; \addplot[blue, mark = *, nodes near coords=$P_{h,\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(2, 3)}; \addplot[green, mark = x, nodes near coords=$P_{g,\text{min}}$,every node near coord/.style={anchor=120}] coordinates {(2, 2)}; \addplot[red, mark = *, nodes near coords=$P_{f,\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(2, 1)}; \draw[thick, dashed] (axis cs:2,0) -- (axis cs:2,3); \addlegendentry{$f(x)=1$} \addlegendentry{$g(x)=2$} \addlegendentry{$h(x)=3$} \end{axis} \end{tikzpicture} \caption{Three constant functions and their points with minimal distance} \label{fig:constant-min-distance} \end{figure} \begin{proof} The point $(x, f(x))$ with minimal distance can be calculated directly: \begin{align} d_{P,f}(x) &= \sqrt{(x - x_P)^2 + (f(x) - y_P)^2}\\ &= \sqrt{(x^2 - 2x_P x + x_P^2) + (c^2 - 2 c y_P + y_P^2)} \\ &= \sqrt{x^2 - 2 x_P x + (x_P^2 + c^2 - 2 c y_P + y_P^2)}\label{eq:constant-function-distance}\\ \xRightarrow{\text{Theorem}~\ref{thm:fermats-theorem}} 0 &\stackrel{!}{=} (d_{P,f}(x)^2)'\\ &= 2x - 2x_P\\ \Leftrightarrow x &\stackrel{!}{=} x_P \end{align} So $(x_P,f(x_P))$ is the only point with minimal distance to $P$. $\qed$ \end{proof} This result means: \[S_0(f, P) = \Set{x_P} \text{ with } P = (x_P, y_P)\] \clearpage \section{Defined on a closed interval $[a,b] \subseteq \mdr$} \begin{theorem}[Solution formula for constant functions] Let $f:[a,b] \rightarrow \mdr$, $f(x) := c$ with $a,b,c \in \mdr$ and $a \leq b$ be a constant function. Then the point $(x, f(x))$ of $f$ with minimal distance to $P$ is given by: \[\underset{x\in [a,b]}{\arg \min d_{P,f}(x)} = \begin{cases} S_0(f,P) &\text{if } S_0(f,P) \cap [a,b] \neq \emptyset \\ \Set{a} &\text{if } S_0(f,P) \ni x_P < a\\ \Set{b} &\text{if } S_0(f,P) \ni x_P > b \end{cases}\] \end{theorem} \begin{figure}[htp] \centering \begin{tikzpicture} \begin{axis}[ legend pos=north west, legend cell align=left, axis x line=middle, axis y line=middle, grid = major, width=0.8\linewidth, height=8cm, grid style={dashed, gray!30}, xmin=-5, % start the diagram at this x-coordinate xmax= 5, % end the diagram at this x-coordinate ymin= 0, % start the diagram at this y-coordinate ymax= 3, % end the diagram at this y-coordinate axis background/.style={fill=white}, xlabel=$x$, ylabel=$y$, tick align=outside, minor tick num=-3, enlargelimits=true, tension=0.08] \addplot[domain=-5:-2, thick,samples=50, red] {1}; \addplot[domain=-1:3, thick,samples=50, green] {1.5}; \addplot[domain=3:5, thick,samples=50, blue, densely dotted] {3}; \addplot[black, mark = *, nodes near coords=$P$,every node near coord/.style={anchor=225}] coordinates {(2, 2)}; \addplot[blue, mark = *, nodes near coords=$P_{h,\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(3, 3)}; \addplot[green, mark = x, nodes near coords=$P_{g,\text{min}}$,every node near coord/.style={anchor=120}] coordinates {(2, 1.5)}; \addplot[red, mark = *, nodes near coords=$P_{f,\text{min}}$,every node near coord/.style={anchor=225}] coordinates {(-2, 1)}; \draw[thick, dashed] (axis cs:2,1.5) -- (axis cs:2,2); \draw[thick, dashed] (axis cs:2,2) -- (axis cs:-2,1); \draw[thick, dashed] (axis cs:2,2) -- (axis cs:3,3); \addlegendentry{$f(x)=1, D = [-5,-2]$} \addlegendentry{$g(x)=1.5, D = [-1,3]$} \addlegendentry{$h(x)=3, D = [3,5]$} \end{axis} \end{tikzpicture} \caption{Three constant functions and their points with minimal distance} \label{fig:constant-min-distance-closed-intervall} \end{figure} \begin{proof} \begin{align} \underset{x\in[a,b]}{\arg \min d_{P,f}(x)} &= \underset{x\in[a,b]}{\arg \min d_{P,f}(x)^2}\\ &=\underset{x\in[a,b]}{\arg \min} \big ((x-x_P)^2 + \overbrace{(y_P^2 - 2 y_P c + c^2)}^{\text{constant}} \big )\\ &=\underset{x\in[a,b]}{\arg \min} (x-x_P)^2 \end{align} which is optimal for $x = x_P$, but if $x_P \notin [a,b]$, you want to make this term as small as possible. It gets as small as possible when $x$ is as similar to $x_p$ as possible. This yields directly to the solution formula.$\qed$ \end{proof}