hw4: done 1,2,3,4

hw4/hw4.tex

@@ -75,4 +75,103 @@ $$a_2 = f[0,1,3] = \frac{2-(-3)}{3-0} = \frac{5}{3}$$
 $$p(x) = \left(\frac{5}{3}(x-1)-3\right) x + 0 = \frac{5}{3}x^2 - \frac{14}{3}x$$
 
 The interpolating polynomials are indeed equal.
+
+\section*{Question 4}
+\subsection*{Point a)}
+
+The node coordinates to which fix the quadratic spline are $(1/2, y_0), (1, y_1),(3/2, y_2),(2, y_3)$.
+
+Then, we can start formulating the equations for the linear system:
+
+$$y_0 = s\left(\frac{1}{2}\right) = a_{-1}B_2\left(\frac{1}{2} + 1\right) + a_{0}B_2\left(\frac{1}{2}\right) + a_{1}B_2\left(\frac{1}{2} - 1\right) + $$$$
++ a_{2}B_2\left(\frac{1}{2} - 2\right)
++ a_{-1}B_2\left(\frac{1}{2} - 3\right)
++ a_{0}B_2\left(\frac{1}{2} - 4\right)
++ a_{1}B_2\left(\frac{1}{2} - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot \frac{1}{2} + a_{1} \cdot \frac{1}{2} + a_{2} \cdot 0 + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = \frac{a_0 + a_1}{2}$$ 
+
+$$y_1 = s\left(1\right) = a_{-1}B_2\left(1 + 1\right) + a_{0}B_2\left(1\right) + a_{1}B_2\left(1 - 1\right) + $$$$
++ a_{2}B_2\left(1 - 2\right)
++ a_{-1}B_2\left(1 - 3\right)
++ a_{0}B_2\left(1 - 4\right)
++ a_{1}B_2\left(1 - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot \left(\frac{1}{8}\right) + a_{1} \cdot \frac{3}{4} + a_{2} \cdot \left(\frac{1}{8}\right) + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = 
+\frac{1}{8} a_{0} + \frac{3}{4} a_{1} + \frac{1}{8} a_{2}$$ 
+
+$$y_2 = s\left(\frac{3}{2}\right) = a_{-1}B_2\left(\frac{3}{2} + 1\right) + a_{0}B_2\left(\frac{3}{2}\right) + a_{1}B_2\left(\frac{3}{2} - 1\right) + $$$$
++ a_{2}B_2\left(\frac{3}{2} - 2\right)
++ a_{-1}B_2\left(\frac{3}{2} - 3\right)
++ a_{0}B_2\left(\frac{3}{2} - 4\right)
++ a_{1}B_2\left(\frac{3}{2} - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot \frac{1}{2} + a_{2} \cdot \frac{1}{2} + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = \frac{a_1 + a_2}{2}$$ 
+
+$$y_3 = s\left(2\right) = a_{-1}B_2\left(2 + 1\right) + a_{0}B_2\left(2\right) + a_{1}B_2\left(2 - 1\right) + $$$$
++ a_{2}B_2\left(2 - 2\right)
++ a_{-1}B_2\left(2 - 3\right)
++ a_{0}B_2\left(2 - 4\right)
++ a_{1}B_2\left(2 - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot \left(\frac{1}{8}\right) + a_{2} \cdot \frac{3}{4} + a_{-1} \cdot \left(\frac{1}{8}\right) + a_{0} \cdot 0 + a_{1} \cdot 0 = 
+\frac{1}{8} a_{1} + \frac{3}{4} a_{2} + \frac{1}{8} a_{-1}$$
+
+The linear system in matrix form is:
+
+\[\frac{1}{8} \cdot
+\begin{bmatrix}
+4&4&0&0\\
+1&6&1&0\\
+0&4&4&0\\
+0&1&6&1\\
+\end{bmatrix}
+\begin{bmatrix}
+a_0\\a_1\\a_2\\a_{-1}\\
+\end{bmatrix}
+=
+\begin{bmatrix}
+y_0\\y_1\\y_2\\y_{3}\\
+\end{bmatrix}\]
+
+The determinant of that matrix is 0 so the matrix is singular.
+
+\subsection*{Point b)}
+
+The node coordinates to which fix the quadratic spline are $(0, y_0), (1, y_1),(2, y_2),(3, y_3)$.
+
+Then, we can start formulating the equations for the linear system:
+$$y_0 = s(0) = a_{-1}B_2\left(1\right) + a_{0}B_2\left(0\right) + a_{1}B_2\left(- 1\right) + $$$$
++ a_{2}B_2\left(- 2\right)
++ a_{-1}B_2\left(- 3\right)
++ a_{0}B_2\left(- 4\right)
++ a_{1}B_2\left(- 5\right)=$$$$
+a_{-1} \cdot \frac{1}{8} + a_{0} \cdot \frac{3}{4} + a_{1} \cdot \frac{1}{8} + a_{2} \cdot 0 + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = 
+\frac{1}{8} a_{-1} + \frac{3}{4} a_{0} + \frac{1}{8} a_{1}$$
+
+$$y_1 = s\left(1\right) = \frac{1}{8} a_{0} + \frac{3}{4} a_{1} + \frac{1}{8} a_{2}$$ 
+
+$$y_2 = s\left(2\right) = \frac{1}{8} a_{1} + \frac{3}{4} a_{2} + \frac{1}{8} a_{-1}$$
+
+$$y_3 = s\left(3\right) = a_{-1}B_2\left(3 + 1\right) + a_{0}B_2\left(3\right) + a_{1}B_2\left(3 - 1\right) + $$$$
++ a_{2}B_2\left(3 - 2\right)
++ a_{-1}B_2\left(3 - 3\right)
++ a_{0}B_2\left(3 - 4\right)
++ a_{1}B_2\left(3 - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 + a_{2} \cdot \frac{1}{8} + a_{-1} \cdot \frac{3}{4} + a_{0} \cdot \frac{1}{8} + a_{1} \cdot 0 = 
+\frac{1}{8} a_{2} + \frac{3}{4} a_{-1} + \frac{1}{8} a_{0}$$
+
+The linear system in matrix form is:
+
+\[\frac{1}{8} \cdot
+\begin{bmatrix}
+6&1&0&1\\
+1&6&1&0\\
+0&1&6&1\\
+1&0&1&6\\
+\end{bmatrix}
+\begin{bmatrix}
+a_0\\a_1\\a_2\\a_{-1}\\
+\end{bmatrix}
+=
+\begin{bmatrix}
+y_0\\y_1\\y_2\\y_{3}\\
+\end{bmatrix}\]
+
 \end{document}