% vim: set ts=2 sw=2 et tw=80: \documentclass[12pt,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage[margin=2cm]{geometry} \usepackage{amstext} \usepackage{amsmath} \usepackage{array} \usepackage[utf8]{inputenc} \usepackage[margin=2cm]{geometry} \usepackage{amstext} \usepackage{array} \newcommand{\lra}{\Leftrightarrow} \newcolumntype{L}{>{$}l<{$}} \title{Midterm -- Introduction to Computational Science} \author{Claudio Maggioni} \begin{document} \maketitle \section*{Question 1} \subsection*{Point a)} $$7.125_{10} = (1 + 2^{-1} + 2^{-2} + 2^{-5}) * {2^2}_{10} = 0 | 1100 1000 0000 | 110_F$$ $$0.8_{10} = (1 + 2^{-1} + 2^{-4} + 2^{-5}+ 2^{-8} + 2^{-9} + 2^{-12}) * {2^{-1}}_{10} \approx 0 | 100110011001|011_F$$ $$0.046875_{10} = (2^{-2} + 2^{-3}) * {2^{-3}}_{10} = 0 | 0011 0000 0000|000_F$$ For the last conversion, we assume that the denormalized mode of this floating point representation implicitly includes an exponent of $-3$ (this makes the first bit in the mantissa of a denormalized number weigh $2^{-3}$). This behaviour is akin to the denormalized implementation of IEEE 754 floating point numbers. \subsection*{Point b)} $$1|011010111000|110_F = -(1+2^{-2} +2^{-3} +2^{-5} +2^{-7} +2^{-8} + 2^{-9}) * {2^{2}}_{10} \approx -5.6796875$$ $$1|101010101010|010_F = -(1+2^{-1} +2^{-3} +2^{-5} +2^{-7} +2^{-9} + 2^{-11}) * {2^{-2}}_{10} \approx -0.4166259766$$ \subsection*{Point c)} $$1|0000 0000 0000|001_F = 2^{-3}_{10} = 0.125_{10}$$ \subsection*{Point d)} $$1|1111 1111 1111|111_F = $$ $$=(1 + 2^{-1}+ 2^{-2} + 2^{-3} +2^{-4} + 2^{-5} +2^{-6} + 2^{-7} + 2^{-8} + 2^{-9}+ 2^{-10}+ 2^{-11}+ 2^{-12}) * {2^3}_{10} = $$ $$ = 15.998046875_{10}$$ \subsection*{Point e)} With 12 independent binary choices (bits to flip), there are $2^{12}$ different denormalized numbers in this encoding. \subsection*{Point f)} With 12 independent binary choices (bits to flip) and 3 extra bits for the exponent, there are $2^{15}-1$ different denormalized numbers in this encoding. We subtract 1 in order to be consistent with the assumption in point \textit{a)}, since $0.125_{10}$ would be representable both as $1|0000 0000 0000|001_F$ and as $1|1000 0000 0000|000_F$ \section*{Question 2} \subsection*{Point a)} $$ \sqrt[3]{1 + x} - 1 = (\sqrt[3]{1 + x} - 1) \cdot \frac{ \sqrt[3]{1 + x} + 1}{ \sqrt[3]{1 + x} + 1} = \frac{\sqrt[3]{(1 + x)^2} - 1}{\sqrt[3]{1 + x} + 1} = \frac{\sqrt[3]{(1 + x)^2} - 1}{\sqrt[3]{1 + x} + 1} \cdot \frac{\sqrt[3]{(1 + x)^2} + 1}{\sqrt[3]{(1 + x)^2} + 1} = $$ $$\frac{(1 + x)\sqrt[3]{1 + x} - 1}{(\sqrt[3]{1 + x} + 1) \cdot (\sqrt[3]{(1 + x)^2} + 1)} $$ \subsection*{Point b)} $$ \frac{1 - cos(x)}{sin(x)} = \frac{sin^2(x)cos^2(x) - cos(x)}{sin(x)} \cdot \frac{sin(x)}{cos(x)} \cdot \frac{cos(x)}{sin(x)} = (sin^2(x)cos(x) - 1)\cdot\frac{cos(x)}{sin(x)}$$ \subsection*{Point c)} $$ \frac{1}{1-\sqrt{x^2-1}} = \frac{x}{x^2-\sqrt{x^4-x^2}}$$ \subsection*{Point d)} $$ x^3\cdot\left(\frac{x}{x^2-1}-\frac{1}{x}\right) = x^3\cdot\left(\frac{x^2-x^2+1}{x^3-x}\right) = \frac{x^2}{x^2-1}$$ \subsection*{Point e)} $$ \frac{1}{x} - \frac{1}{x+1} = \frac{x + 1 - x}{x^2 + x} = \frac{1}{x^2 + x}$$ \section*{Question 3} \subsection*{Point a)} First we point out that: $$ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \stackrel{k := -h}{=} \lim_{k \to 0} \frac{f(x - (-k)) - f(x)}{-k} = $$ $$ = \lim_{k \to 0} \frac{f(x) - f(x + k)}{k} = - \lim_{h \to 0} \frac{f(x) - f(x - h)}{h} $$ Then, we find an equivalent way to represent $f'(x)$: $$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = 2 \cdot \lim_{h \to 0} \frac{f(x + h) - f(x)}{2h} =$$ $$ = \lim_{h \to 0} \frac{f(x + h) - f(x)}{2h} + \left(- \lim_{h \to 0} \frac{f(x) - f(x - h)}{2h}\right) = \lim_{h \to 0} \frac{f(x + h) - f(x - h)}{2h} $$ Then we consider the \textit{epsilon-delta} definition of limits for the last limit: $$ \forall \epsilon > 0 \exists \delta > 0 | \forall h > 0, \text{if } 0 < h < \delta \Rightarrow$$ $$\left| \frac{f(x + h) - f(x - h)}{2h} - f'(x)\right| = \left|f'(x) - \frac{f(x + h) - f(x - h)}{2h}\right| < \epsilon$$ I GIVE UP :( \section*{Question 4} \subsection*{Point a)} \[ A_1 = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 2 & 4 & 4 & 4 & 4 \\ 3 & 7 & 10 & 10 & 10 \\ 4 & 10 & 16 & 20 & 20 \\ 5 & 13 & 22 & 30 & 35 \\ \end{bmatrix}, b = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ \end{bmatrix} \] \[ l_1 = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ \end{bmatrix}, u_1 = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ \end{bmatrix}, A_2 \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 2 & 2 & 2 & 2 \\ 0 & 4 & 7 & 7 & 7 \\ 0 & 6 & 12 & 16 & 16 \\ 0 & 8 & 17 & 25 & 30 \\ \end{bmatrix} \] \[ l_2 = \begin{bmatrix} 0 \\ 1 \\ 2 \\ 3 \\ 4 \\ \end{bmatrix}, u_2 = \begin{bmatrix} 0 & 2 & 2 & 2 & 2 \\ \end{bmatrix}, A_3 \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 3 & 3 & 3 \\ 0 & 0 & 6 & 10 & 10 \\ 0 & 0 & 9 & 17 & 22 \\ \end{bmatrix} \] \[ l_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 2 \\ 3 \\ \end{bmatrix}, u_3 = \begin{bmatrix} 0 & 0 & 3 & 3 & 3 \\ \end{bmatrix}, A_4 \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 4 & 4 \\ 0 & 0 & 0 & 8 & 13 \\ \end{bmatrix} \] \[ l_4 = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 2 \\ \end{bmatrix}, u_4 = \begin{bmatrix} 0 & 0 & 0 & 4 & 4 \\ \end{bmatrix}, A_5 \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 5 \\ \end{bmatrix} \] \[ l_5 = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \\ \end{bmatrix}, u_5 = \begin{bmatrix} 0 & 0 & 0 & 0 & 5 \\ \end{bmatrix}, L = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 & 0 \\ 3 & 2 & 1 & 0 & 0 \\ 4 & 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 & 1 \\ \end{bmatrix}, U = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 2 & 2 & 2 & 2 \\ 0 & 0 & 3 & 3 & 3 \\ 0 & 0 & 0 & 4 & 4 \\ 0 & 0 & 0 & 0 & 5 \\ \end{bmatrix} \] \subsection*{Point b)} \[ l_1 = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ \end{bmatrix}, e^{T}_1 = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ \end{bmatrix}, L_1 = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ -2 & 1 & 0 & 0 & 0 \\ -3 & 0 & 1 & 0 & 0 \\ -4 & 0 & 0 & 1 & 0 \\ -5 & 0 & 0 & 0 & 1 \\ \end{bmatrix} \] \[ l_2 = \begin{bmatrix} 0 \\ 1 \\ 2 \\ 3 \\ 4 \\ \end{bmatrix}, e^{T}_2 = \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ \end{bmatrix}, L_2 = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & -2 & 1 & 0 & 0 \\ 0 & -3 & 0 & 1 & 0 \\ 0 & -4 & 0 & 0 & 1 \\ \end{bmatrix} \] \[ l_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 2 \\ 3 \\ \end{bmatrix}, e^{T}_3 = \begin{bmatrix} 0 & 0 & 1 & 0 & 0 \\ \end{bmatrix}, L_3 = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & -2 & 1 & 0 \\ 0 & 0 & -3 & 0 & 1 \\ \end{bmatrix} \] \[ l_4 = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 2 \\ \end{bmatrix}, e^{T}_4 = \begin{bmatrix} 0 & 0 & 0 & 1 & 0 \\ \end{bmatrix}, L_4 = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2 & 1 \\ \end{bmatrix} \] \subsection*{Point c)} $$Ly = b$$ \[ L = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 & 0 \\ 3 & 2 & 1 & 0 & 0 \\ 4 & 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 & 1 \\ \end{bmatrix}, b = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ \end{bmatrix} \] \[ L = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 0 & 0 \\ 3 & 2 & 1 & 0 & 0 \\ 4 & 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 & 1 \\ \end{bmatrix}, b = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ \end{bmatrix} \] $$y_1 = \frac{1}{1} = 1$$ $$y_2 = \frac{2 - 2 \cdot 1}{1} = 0$$ $$y_3 = \frac{3 - 3 \cdot 1 - 2 \cdot 0}{1} = 0$$ $$y_4 = \frac{4 - 4 \cdot 1 - 3 \cdot 0 - 2 \cdot 0}{1} = 0$$ $$y_5 = \frac{5 - 5 \cdot 1 - 4 \cdot 0 - 3 \cdot 0 - 2 \cdot 0}{1} = 0$$ \[ y = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} \] \subsection*{Point d)} $$Ux = y$$ \[ U = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 2 & 2 & 2 & 2 \\ 0 & 0 & 3 & 3 & 3 \\ 0 & 0 & 0 & 4 & 4 \\ 0 & 0 & 0 & 0 & 5 \\ \end{bmatrix} y = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} \] $$x_5 = \frac{0}{5} = 0$$ $$x_4 = \frac{0 - 4 \cdot 0}{4} = 0$$ $$x_3 = \frac{0 - 3 \cdot 0 - 3 \cdot 0}{3} = 0$$ $$x_2 = \frac{0 - 2 \cdot 0 - 2 \cdot 0 - 2 \cdot 0}{2} = 0$$ $$x_1 = \frac{0 - 1 \cdot 0 - 1 \cdot 0 - 1 \cdot 0 - 1 \cdot }{1} = 0$$ \[ x = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix} \] \section*{Question 5} \subsection*{Point a)} $$f(x) = x \hspace{2cm} K_{abs} = |f'(x)| = 1 \hspace{2cm} K_{rel} = \left|\frac{1 \cdot x}{x}\right| = 1$$ \subsection*{Point b)} $$f(x) = \sqrt[3]{x} \hspace{2cm} K_{abs} = |f'(x)| = \frac{1}{3\sqrt[3]{x^2}} \hspace{2cm} K_{rel} = \left|\frac{1}{3\sqrt[3]{x^2}} \cdot \frac{x}{\sqrt[3]{x}}\right| = \frac{1}{3}$$ \subsection*{Point c)} $$f(x) = \frac{1}{x} \hspace{2cm} K_{abs} = |f'(x)| = \frac{1}{x^2} \hspace{2cm} K_{rel} = \left|\frac{-x}{x^2} \cdot \frac{1}{\frac{1}{x}}\right| = 1$$ \subsection*{Point d)} $$f(x) = e^x \hspace{2cm} K_{abs} = |f'(x)| = e^x \hspace{2cm} K_{rel} = \left|\frac{xe^x}{e^x}\right| = |x|$$ \subsection*{Point e)} Cases \textit{a)},\textit{b)} and \textit{c)} are well-conditioned for any $x$ since their $K_rel$ is not defined by x. Case \textit{d)} is well-conditioned only for $x$s whose absolute value is in the order of magnitude of $1$ or less, since $K_rel$ in this case is exactly $|x|$. \end{document}