From 4027d3af0f1a3d447da4eb97f16908538420ffe4 Mon Sep 17 00:00:00 2001
From: Martin Thoma <info@martin-thoma.de>
Date: Sun, 6 Mar 2016 15:10:39 +0100
Subject: [PATCH] Remove document as it is out of scope for this project

---
 documents/c2-bezier-spline/Makefile           |  8 ---
 .../c2-bezier-spline/minimal-document.tex     | 50 -------------------
 2 files changed, 58 deletions(-)
 delete mode 100644 documents/c2-bezier-spline/Makefile
 delete mode 100644 documents/c2-bezier-spline/minimal-document.tex

diff --git a/documents/c2-bezier-spline/Makefile b/documents/c2-bezier-spline/Makefile
deleted file mode 100644
index 4bc4573..0000000
--- a/documents/c2-bezier-spline/Makefile
+++ /dev/null
@@ -1,8 +0,0 @@
-SOURCE=minimal-document
-
-make:
-	pdflatex $(SOURCE).tex -output-format=pdf
-	make clean
-
-clean:
-	rm -rf  $(TARGET) *.class *.html *.log *.aux *.out
diff --git a/documents/c2-bezier-spline/minimal-document.tex b/documents/c2-bezier-spline/minimal-document.tex
deleted file mode 100644
index 91d1d4c..0000000
--- a/documents/c2-bezier-spline/minimal-document.tex
+++ /dev/null
@@ -1,50 +0,0 @@
-\documentclass[a4paper]{scrartcl}
-\usepackage[ngerman]{babel}
-\usepackage[utf8]{inputenc}
-\usepackage{amssymb,amsmath}
-
-\begin{document}
-    Wann ist ein kubischer Bezier-Splines $S(u) = [F(2 \cdot u), G(2(u-1))]$,
-    welche Kontrollpunkte in $\mathbb{R}^2$ haben, $C^2$-stetig?
-
-    Sei im folgenden $a = P_0$, $b = P_1$, $c = P_2$, $d= P_3$. Dann gilt:
-
-    \begin{align}
-        F(x) &= \sum_{i=0}^3 \mathbf{b}_i^n P_i\\
-             &= \sum_{i=0}^3 \begin{pmatrix}n\\i\end{pmatrix} x^{i} (1-x)^{n-i}  P_i\\
-             &= (1-x)^3 P_0 + 3 x (1-x)^2 P_1 + 3 x^2 (1-x) P_2 + x^3 P_3\\
-             &= -a x^3+3 a x^2-3 a x+a+3 b x^3-6 b x^2+3 b x-3 c x^3+3 c x^2+d x^3\\
-             &= (-a + 3 b-3 c+d) x^3  + (3 a-6 b+3 c) x^2 + (-3 a+3 b) x + a\\
-        F'(x)&= 3 (-a + 3 b-3 c+d) x^2 + 2 (3 a-6 b+3 c) x + (-3 a+3 b)\\
-        F''(x) &= 6 (-a + 3 b-3 c+d) x + 2 (3 a-6 b+3 c)
-    \end{align}
-
-    Damit $S$ nun $C^0$-stetig ist, muss $F(1) = G(0)$ gelten. Also:
-
-    \begin{align}
-        G(0) &= F(1)\\
-    \Leftrightarrow a_G
-        &= (-a_F + 3 b_F-3 c_F+d_F)  + (3 a_F-6 b_F+3 c_F) + (-3 a_F+3 b_F) + (2+a_F)\\
-    \Leftrightarrow a_G &= d_F\\
-    \end{align}
-
-    Damit $S$ nun $C^1$-stetig ist, muss zusätzlich $F'(1) = G'(0)$ gelten. Also:
-
-    \begin{align}
-        G'(0) &= F'(1)\\
-    \Leftrightarrow -3 a_G+3 b_G
-        &= 3 (-a_F + 3 b_F-3 c_F+d_F) + 2 (3 a_F-6 b_F+3 c_F) + (-3 a_F+3 b_F)\\
-    \Leftrightarrow -3 (a_G-b_G) &= 3 (- c_F+d_F)\\
-    \Leftrightarrow -a_G + b_G &= - c_F + d_F
-    \end{align}
-
-    Damit $S$ nun $C^2$-stetig ist, muss zusätzlich $F''(1) = G''(0)$ gelten. Also:
-
-    \begin{align}
-        G''(0) &= F''(1)\\
-        \Leftrightarrow 2 (3 a_G-6 b_G+3 c_G)
-                     &= 6 (-a_F + 3 b_F-3 c_F+d_F) + 2 (3 a_F-6 b_F+3 c_F)\\
-        \Leftrightarrow 6 (a_G - 2 b_G + c_G) &= 6 (b_F-2 c_F+d_F)\\
-        \Leftrightarrow a_G - 2 b_G + c_G &= b_F-2 c_F+d_F
-    \end{align}
-\end{document}