diff --git a/hw4/hw4.pdf b/hw4/hw4.pdf index 5fc10f8..376cbc4 100644 Binary files a/hw4/hw4.pdf and b/hw4/hw4.pdf differ diff --git a/hw4/hw4.tex b/hw4/hw4.tex index be039b7..a2af726 100644 --- a/hw4/hw4.tex +++ b/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}