misc

documents/math-minimal-distance-to-cubic-function/constant-functions.tex

@@ -1,6 +1,7 @@
 \chapter{Constant functions}
 \section{Defined on $\mdr$}
 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}.
 
 \begin{figure}[htp]
     \centering
@@ -42,11 +43,21 @@ Let $f:\mdr \rightarrow \mdr$, $f(x) := c$ with $c \in \mdr$ be a constant funct
     \label{fig:constant-min-distance}
 \end{figure}
 
+\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}\\
+ \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}.
+See Figure~\ref{fig:constant-min-distance} to see that intuition
+yields to the same results.
 
-This means:
+This result means:
 
 \[S_0(f, P) = \Set{x_P} \text{ with } P = (x_P, y_P)\]
 \clearpage

documents/math-minimal-distance-to-cubic-function/linear-functions.tex

@@ -109,8 +109,16 @@ $a \leq b$, $m \neq 0$  be a linear function.
 \end{figure}
 
 The point with minimum distance can be found by:
-\[\underset{x\in\mdr}{\arg \min d_{P,f}(x)} = \begin{cases}
+\[\underset{x\in[a,b]}{\arg \min d_{P,f}(x)} = \begin{cases}
  S_1(f, P) &\text{if } S_1(f, P) \cap [a,b] \neq \emptyset\\
    \Set{a} &\text{if } S_1(f, P) \ni x < a\\
    \Set{b} &\text{if } S_1(f, P) \ni x > b
     \end{cases}\]
+
+Because:
+\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} x^2 - 2x_P x + (x_P^2 + y_P^2 - 2 y_P c + c^2)\\
+   &=\underset{x\in[a,b]}{\arg \min} x^2 - 2 x_P x + x_P^2\\
+   &=\underset{x\in[a,b]}{\arg \min} (x-x_P)^2\\
+\end{align}

documents/math-minimal-distance-to-cubic-function/problem-description.tex

@@ -11,16 +11,19 @@ Now there is finite set $M = \Set{x_1, \dots, x_n} \subseteq D$ of minima for gi
 But minimizing $d_{P,f}$ is the same as minimizing 
 $d_{P,f}^2 = x_p^2 - 2x_p x + x^2 + y_p^2 - 2y_p f(x) + f(x)^2$.
 
-\begin{theorem}[Fermat's theorem about stationary points]\label{thm:required-extremum-property}
+In order to solve the minimal distance problem, Fermat's theorem
+about stationary points will be tremendously usefull:
+
+\begin{theorem}[Fermat's theorem about stationary points]\label{thm:fermats-theorem}
     Let $x_0$ be a local extremum of a differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$.
 
     Then: $f'(x_0) = 0$.
 \end{theorem}
 
 Let $S_n$ be the function that returns the set of solutions for a
-polynomial of degree $n$ and a point:
+polynomial $f$ of degree $n$ and a point $P$:
 
 \[S_n: \Set{\text{Polynomials of degree } n \text{ defined on } \mdr} \times \mdr^2 \rightarrow \mathcal{P}({\mdr})\]
-\[S_n(f, P) := \underset{x\in\mdr}{\arg \min d_{P,f}(x)}\]
+\[S_n(f, P) := \underset{x\in\mdr}{\arg \min d_{P,f}(x)} = M\]
 
 If possible, I will explicitly give this function.