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\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.
The situation can be seen in Figure~\ref{fig:constant-min-distance}.
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}
@ -43,20 +48,19 @@ The situation can be seen in Figure~\ref{fig:constant-min-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_P - x)^2 + (y_P - f(x))^2}\\
&= \sqrt{(x_P^2 - 2x_P x + x^2) + (y_P^2 - 2 y_P c + c^2)} \\
&= \sqrt{x^2 - 2 x_P x + (x_P^2 + y_P^2 - 2 y_P c + c^2)}\label{eq:constant-function-distance}\\
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}
Then $(x_P,f(x_P))$ has
minimal distance to $P$. Every other point has higher distance.
See Figure~\ref{fig:constant-min-distance} to see that intuition
yields to the same results.
So $(x_P,f(x_P))$ is the only point with minimal distance to $P$. $\qed$
\end{proof}
This result means:
@ -124,8 +128,7 @@ given by:
\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^2 - 2x_P 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^2 - 2 x_P x + x_P^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}