hw4: done 1,2,3,4,5

hw4/hw4.tex

@@ -37,16 +37,16 @@ $$$$
 0.5 \cdot \left(-\frac{4}{3}x^3 - \frac{2}{3}x^2 + \frac{4}{3}x + \frac{2}{3}\right) + 
 0.4 \cdot \left(\frac{2}{3}x^3 + \frac{2}{3}x^2 -\frac{1}{6}x - \frac{1}{6}\right) = -\frac{2}{5}x^3 + \frac{3}{5}x^2 - \frac{2}{5}x + \frac{3}{5}$$
 
-$$\frac{\max_{x \in [-1,1]} |f^{(n+1)}|}{5!} = 
-\frac{\max_{x \in [-1,1]}|\frac{7680}{|2x+3|^6}|}{120} =
-\max_{x \in [-1,1]}\frac{64}{|2x+3|^6} = 64$$
+$$\frac{\max_{x \in [-1,1]} |f^{(n+1)}|}{4!} = 
+\frac{\max_{x \in [-1,1]}|\frac{768}{|2x+3|^6}|}{24} =
+\max_{x \in [-1,1]}\frac{32}{|2x+3|^6} = 32$$
 
 $$\max_{x \in [-1,1]} \left|(x - 1)\left(x - \frac{1}{2}\right)
 \left(x + \frac{1}{2}\right)(x+1)\right| = 
 \max_{x \in [-1,1]} \left| x^4 - \frac{5}{4}x^2 + \frac{1}{4}\right| = \frac{1}{4}$$
 
-$$\max_{x \in [-1,1]} |f(x) - p(x)| \leq \frac{\max_{x \in [-1,1]} |f^{(n+1)}|}{5!} \max_{x \in [-1,1]} \left|(x - 1)\left(x - \frac{1}{2}\right)
-\left(x + \frac{1}{2}\right)(x+1)\right| =$$$$= 64 \cdot \frac{1}{4} = 8 \leq 8$$
+$$\max_{x \in [-1,1]} |f(x) - p(x)| \leq \frac{\max_{x \in [-1,1]} |f^{(n+1)}|}{4!} \max_{x \in [-1,1]} \left|(x - 1)\left(x - \frac{1}{2}\right)
+\left(x + \frac{1}{2}\right)(x+1)\right| =$$$$= 32 \cdot \frac{1}{4} = 8 \leq 8$$
 
 The statement above is true so p satisfies the error estimate:
 
@@ -76,6 +76,16 @@ $$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.
 
+Now we use the Horner method to compute $p(0.5)$:
+
+$$y=a_n = a_2 = \frac{5}{3}$$
+
+$$i = 1$$
+$$y = y (x - x_1) + a_1 = \frac{5}{3} (0.5 - 1) + a_1 - 3 = -\frac{5}{6} - 3 = -\frac{23}{6}$$
+
+$$i = 0$$
+$$y = y (x - x_0) + a_0 = -\frac{23}{6} \cdot 0.5 + 0 = -\frac{23}{12}$$
+
 \section*{Question 4}
 \subsection*{Point a)}
 
@@ -174,4 +184,126 @@ a_0\\a_1\\a_2\\a_{-1}\\
 y_0\\y_1\\y_2\\y_{3}\\
 \end{bmatrix}\]
 
+\subsection*{Question 5}
+
+$$1 = y_0 = s(x_0) = 0 = a_{-1}B_3(1) +a_0B_3(0)
++a_1B_3(-1)
++a_2B_3(-2)
++a_3B_3(-3) = \frac{1}{6}a_{-1} +  \frac{2}{3}a_0 + \frac{1}{6}a_1$$
+
+$$5 = y_1 = s(x_1) = 1 = a_{-1}B_3(2) +a_0B_3(1)
++a_1B_3(0)
++a_2B_3(-1)
++a_3B_3(-2) = \frac{1}{6}a_{0} + \frac{2}{3}a_1 + \frac{1}{6}a_2$$
+
+$$1 = y_2 = s(x_2) = 2 = a_{-1}B_3(3) +a_0B_3(2)
++a_1B_3(1)
++a_2B_3(0)
++a_3B_3(-1) = \frac{1}{6}a_{1} + \frac{2}{3}a_2 + \frac{1}{6}a_3$$
+
+$$s''(0) =  a_{-1}B''_3(1) +a_0B''_3(0)
++a_1B''_3(-1)
++a_2B''_3(-2)
++a_3B''_3(-3) = a_{-1} - 2 a_0 + a_{1}$$
+
+$$s''(2) =  a_{-1}B''_3(3) +a_0B''_3(2)
++a_1B''_3(1)
++a_2B''_3(-0)
++a_3B''_3(-1) = a_{1} - 2 a_2 + a_{3}$$
+
+\[\frac{1}{6} \cdot
+\begin{bmatrix}
+1&-2&1&0&0\\
+1&4&1&0&0\\
+0&1&4&1&0\\
+0&0&1&4&1\\
+0&0&1&-2&1\\
+\end{bmatrix}
+\begin{bmatrix}
+a_{-1}\\a_0\\a_1\\a_2\\a_3\\
+\end{bmatrix}
+=
+\begin{bmatrix}
+0\\y_0\\y_1\\y_2\\0\\
+\end{bmatrix}\]
+
+We then use Gaussian \textit{ellimination} to solve the system.
+
+\[
+\begin{array}{@{}ccccc|c@{}}
+1&-2&1&0&0&0\\
+1&4&1&0&0&6\\
+0&1&4&1&0&30\\
+0&0&1&4&1&6\\
+0&0&1&-2&1&0\\
+\end{array}
+\qquad
+\begin{array}{@{}ccccc|c@{}}
+1&-2&1&0&0&0\\
+0&6&0&0&0&6\\
+0&1&4&1&0&30\\
+0&0&1&4&1&6\\
+0&0&1&-2&1&0\\
+\end{array}
+\qquad
+\begin{array}{@{}ccccc|c@{}}
+1&-2&1&0&0&0\\
+0&1&4&1&0&30\\
+0&6&0&0&0&6\\
+0&0&1&4&1&6\\
+0&0&1&-2&1&0\\
+\end{array}
+\]
+\[
+\begin{array}{@{}ccccc|c@{}}
+1&0&9&2&0&60\\
+0&1&4&1&0&30\\
+0&0&-24&-6&0&-174\\
+0&0&1&4&1&6\\
+0&0&1&-2&1&0\\
+\end{array}
+\qquad
+\begin{array}{@{}ccccc|c@{}}
+1&0&9&2&0&60\\
+0&1&4&1&0&30\\
+0&0&1&4&1&6\\
+0&0&-24&-6&0&-174\\
+0&0&1&-2&1&0\\
+\end{array}
+\qquad
+\begin{array}{@{}ccccc|c@{}}
+1&0&0&-34&-9&6\\
+0&1&0&-15&-4&6\\
+0&0&1&4&1&6\\
+0&0&0&90&24&-30\\
+0&0&0&6&0&-6\\
+\end{array}
+\]
+\[
+\begin{array}{@{}ccccc|c@{}}
+1&0&0&-34&-9&6\\
+0&1&0&-15&-4&6\\
+0&0&1&4&1&6\\
+0&0&0&1&4/15&-1/3\\
+0&0&0&6&0&-6\\
+\end{array}
+\qquad
+\begin{array}{@{}ccccc|c@{}}
+  1 & 0 & 0 & 0 & 1/15  & -16/3  \\
+  0 & 1 & 0 & 0 & 0     & 1      \\
+  0 & 0 & 1 & 0 & -1/15 & 22/3   \\
+  0 & 0 & 0 & 1 & 4/15  & -1/3   \\
+  0 & 0 & 0 & 0 & 8/5   & -8     \\
+\end{array}
+\qquad
+\begin{array}{@{}ccccc|c@{}}
+  1 & 0 & 0 & 0 & 0 & -5  \\
+  0 & 1 & 0 & 0 & 0 & 1      \\
+  0 & 0 & 1 & 0 & 0 & 7   \\
+  0 & 0 & 0 & 1 & 0 & 1   \\
+  0 & 0 & 0 & 0 & 1 & -5     \\
+\end{array}
+\]
+
+Therefore, the coefficients are $a_{-1} = a_3 = -5$, $a_0 = a_2 = 1$, and $a_1 = 7$.
 \end{document}