hw4: preparing for submission

hw4/hw4.tex

@@ -89,7 +89,7 @@ $$y = y (x - x_0) + a_0 = -\frac{23}{6} \cdot 0.5 + 0 = -\frac{23}{12}$$
 \section*{Question 4}
 \subsection*{Point a)}
 
-The node coordinates to which fix the quadratic spline are $(1/2, y_0), (1, y_1),(3/2, y_2),(2, y_3)$.
+The node coordinates to which fix the quadratic spline are $(1/2, y_0), (3/2, y_1),(5/2, y_2),(7/2, y_3)$.
 
 Then, we can start formulating the equations for the linear system:
 
@@ -100,40 +100,38 @@ $$y_0 = s\left(\frac{1}{2}\right) = a_{-1}B_2\left(\frac{1}{2} + 1\right) + a_{0
 + a_{1}B_2\left(\frac{1}{2} - 5\right)=$$$$
 a_{-1} \cdot 0 + a_{0} \cdot \frac{1}{2} + a_{1} \cdot \frac{1}{2} + a_{2} \cdot 0 + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = \frac{a_0 + a_1}{2}$$ 
 
-$$y_1 = s\left(1\right) = a_{-1}B_2\left(1 + 1\right) + a_{0}B_2\left(1\right) + a_{1}B_2\left(1 - 1\right) + $$$$
-+ a_{2}B_2\left(1 - 2\right)
-+ a_{-1}B_2\left(1 - 3\right)
-+ a_{0}B_2\left(1 - 4\right)
-+ a_{1}B_2\left(1 - 5\right)=$$$$
-a_{-1} \cdot 0 + a_{0} \cdot \left(\frac{1}{8}\right) + a_{1} \cdot \frac{3}{4} + a_{2} \cdot \left(\frac{1}{8}\right) + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = 
-\frac{1}{8} a_{0} + \frac{3}{4} a_{1} + \frac{1}{8} a_{2}$$ 
-
-$$y_2 = s\left(\frac{3}{2}\right) = a_{-1}B_2\left(\frac{3}{2} + 1\right) + a_{0}B_2\left(\frac{3}{2}\right) + a_{1}B_2\left(\frac{3}{2} - 1\right) + $$$$
+$$y_1 = 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) + $$$$
 + a_{2}B_2\left(\frac{3}{2} - 2\right)
 + a_{-1}B_2\left(\frac{3}{2} - 3\right)
 + a_{0}B_2\left(\frac{3}{2} - 4\right)
 + a_{1}B_2\left(\frac{3}{2} - 5\right)=$$$$
 a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot \frac{1}{2} + a_{2} \cdot \frac{1}{2} + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = \frac{a_1 + a_2}{2}$$ 
 
-$$y_3 = s\left(2\right) = a_{-1}B_2\left(2 + 1\right) + a_{0}B_2\left(2\right) + a_{1}B_2\left(2 - 1\right) + $$$$
-+ a_{2}B_2\left(2 - 2\right)
-+ a_{-1}B_2\left(2 - 3\right)
-+ a_{0}B_2\left(2 - 4\right)
-+ a_{1}B_2\left(2 - 5\right)=$$$$
-a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot \left(\frac{1}{8}\right) + a_{2} \cdot \frac{3}{4} + a_{-1} \cdot \left(\frac{1}{8}\right) + a_{0} \cdot 0 + a_{1} \cdot 0 = 
-\frac{1}{8} a_{1} + \frac{3}{4} a_{2} + \frac{1}{8} a_{-1}$$
+$$y_2 = s\left(\frac{5}{2}\right) = a_{-1}B_2\left(\frac{5}{2} + 1\right) + a_{0}B_2\left(\frac{5}{2}\right) + a_{1}B_2\left(\frac{5}{2} - 1\right) + $$$$
++ a_{2}B_2\left(\frac{5}{2} - 2\right)
++ a_{-1}B_2\left(\frac{5}{2} - 3\right)
++ a_{0}B_2\left(\frac{5}{2} - 4\right)
++ a_{1}B_2\left(\frac{5}{2} - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot 0 +  a_{1} \cdot 0 + a_{2} \cdot \frac{1}{2} + a_{-1} \cdot \frac{1}{2} + a_{0} \cdot 0 + a_{1} \cdot 0 = \frac{a_2 + a_{-1}}{2}$$ 
+
+$$y_1 = s\left(\frac{7}{2}\right) = a_{-1}B_2\left(\frac{7}{2} + 1\right) + a_{0}B_2\left(\frac{7}{2}\right) + a_{1}B_2\left(\frac{7}{2} - 1\right) + $$$$
++ a_{2}B_2\left(\frac{7}{2} - 2\right)
++ a_{-1}B_2\left(\frac{7}{2} - 3\right)
++ a_{0}B_2\left(\frac{7}{2} - 4\right)
++ a_{1}B_2\left(\frac{7}{2} - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 + a_{2} \cdot 0 + a_{-1} \cdot \frac{1}{2} + a_{0} \cdot \frac{1}{2} + a_{1} \cdot 0  = \frac{a_{-1} + a_0}{2}$$ 
 
 The linear system in matrix form is:
 
-\[\frac{1}{8} \cdot
+\[\frac{1}{2} \cdot
 \begin{bmatrix}
-4&4&0&0\\
-1&6&1&0\\
-0&4&4&0\\
-0&1&6&1\\
+0&1&1&0\\
+0&0&1&1\\
+1&0&0&1\\
+1&1&0&0\\
 \end{bmatrix}
 \begin{bmatrix}
-a_0\\a_1\\a_2\\a_{-1}\\
+a_{-1}\\a_0\\a_1\\a_2\\
 \end{bmatrix}
 =
 \begin{bmatrix}
@@ -155,9 +153,21 @@ $$y_0 = s(0) = a_{-1}B_2\left(1\right) + a_{0}B_2\left(0\right) + a_{1}B_2\left(
 a_{-1} \cdot \frac{1}{8} + a_{0} \cdot \frac{3}{4} + a_{1} \cdot \frac{1}{8} + a_{2} \cdot 0 + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = 
 \frac{1}{8} a_{-1} + \frac{3}{4} a_{0} + \frac{1}{8} a_{1}$$
 
-$$y_1 = s\left(1\right) = \frac{1}{8} a_{0} + \frac{3}{4} a_{1} + \frac{1}{8} a_{2}$$ 
+$$y_1 = s\left(1\right) = a_{-1}B_2\left(1 + 1\right) + a_{0}B_2\left(1\right) + a_{1}B_2\left(1 - 1\right) + $$$$
++ a_{2}B_2\left(1 - 2\right)
++ a_{-1}B_2\left(1 - 3\right)
++ a_{0}B_2\left(1 - 4\right)
++ a_{1}B_2\left(1 - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot \left(\frac{1}{8}\right) + a_{1} \cdot \frac{3}{4} + a_{2} \cdot \left(\frac{1}{8}\right) + a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot 0 = 
+\frac{1}{8} a_{0} + \frac{3}{4} a_{1} + \frac{1}{8} a_{2}$$ 
 
-$$y_2 = s\left(2\right) = \frac{1}{8} a_{1} + \frac{3}{4} a_{2} + \frac{1}{8} a_{-1}$$
+$$y_2 = s\left(2\right) = a_{-1}B_2\left(2 + 1\right) + a_{0}B_2\left(2\right) + a_{1}B_2\left(2 - 1\right) + $$$$
++ a_{2}B_2\left(2 - 2\right)
++ a_{-1}B_2\left(2 - 3\right)
++ a_{0}B_2\left(2 - 4\right)
++ a_{1}B_2\left(2 - 5\right)=$$$$
+a_{-1} \cdot 0 + a_{0} \cdot 0 + a_{1} \cdot \left(\frac{1}{8}\right) + a_{2} \cdot \frac{3}{4} + a_{-1} \cdot \left(\frac{1}{8}\right) + a_{0} \cdot 0 + a_{1} \cdot 0 = 
+\frac{1}{8} a_{1} + \frac{3}{4} a_{2} + \frac{1}{8} a_{-1}$$
 
 $$y_3 = s\left(3\right) = a_{-1}B_2\left(3 + 1\right) + a_{0}B_2\left(3\right) + a_{1}B_2\left(3 - 1\right) + $$$$
 + a_{2}B_2\left(3 - 2\right)