diff --git a/hw4/hw4.pdf b/hw4/hw4.pdf index 581436e..5023d69 100644 Binary files a/hw4/hw4.pdf and b/hw4/hw4.pdf differ diff --git a/hw4/hw4.tex b/hw4/hw4.tex index d82a35c..71619c9 100644 --- a/hw4/hw4.tex +++ b/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)