383 lines
9.9 KiB
TeX
383 lines
9.9 KiB
TeX
% vim: set ts=2 sw=2 et tw=80:
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\documentclass[12pt,a4paper]{article}
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\usepackage[utf8]{inputenc} \usepackage[margin=2cm]{geometry}
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\usepackage{amstext}
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\usepackage{amsmath}
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\usepackage{array}
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\usepackage[utf8]{inputenc}
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\usepackage[margin=2cm]{geometry}
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\usepackage{amstext}
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\usepackage{array}
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\newcommand{\lra}{\Leftrightarrow}
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\newcolumntype{L}{>{$}l<{$}}
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\title{Midterm -- Introduction to Computational Science}
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\author{Claudio Maggioni}
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\begin{document}
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\maketitle
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\section*{Question 1}
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\subsection*{Point a)}
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$$7.125_{10} = (1 + 2^{-1} + 2^{-2} + 2^{-5}) * {2^2}_{10} = 0 | 1100 1000 0000 | 110_F$$
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$$0.8_{10} = (1 + 2^{-1} + 2^{-4} + 2^{-5}+ 2^{-8} + 2^{-9} + 2^{-12}) * {2^{-1}}_{10} \approx 0 | 100110011001|011_F$$
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$$0.046875_{10} = (2^{-2} + 2^{-3}) * {2^{-3}}_{10} = 0 | 0011 0000 0000|000_F$$
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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.
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\subsection*{Point b)}
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$$1|011010111000|110_F = -(1+2^{-2} +2^{-3} +2^{-5} +2^{-7} +2^{-8} + 2^{-9}) * {2^{2}}_{10} \approx
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-5.6796875$$
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$$1|101010101010|010_F = -(1+2^{-1} +2^{-3} +2^{-5} +2^{-7} +2^{-9} + 2^{-11}) * {2^{-2}}_{10} \approx
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-0.4166259766$$
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\subsection*{Point c)}
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$$1|0000 0000 0000|001_F = 2^{-3}_{10} = 0.125_{10}$$
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\subsection*{Point d)}
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$$1|1111 1111 1111|111_F = $$
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$$=(1 + 2^{-1}+ 2^{-2} + 2^{-3} +2^{-4} + 2^{-5} +2^{-6} + 2^{-7} + 2^{-8} + 2^{-9}+ 2^{-10}+ 2^{-11}+ 2^{-12}) * {2^3}_{10} = $$
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$$ = 15.998046875_{10}$$
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\subsection*{Point e)}
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With 12 independent binary choices (bits to flip), there are $2^{12}$ different denormalized numbers in this encoding.
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\subsection*{Point f)}
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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
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as $1|1000 0000 0000|000_F$
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\section*{Question 2}
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\subsection*{Point a)}
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$$ \sqrt[3]{1 + x} - 1 = (\sqrt[3]{1 + x} - 1) \cdot
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\frac{\sqrt[3]{(1 + x)^2} + \sqrt[3]{1 + x} + 1}{ \sqrt[3]{(1 + x)^2} + \sqrt[3]{1 + x} + 1} = \frac{(1 + x) - 1}{\sqrt[3]{(1 + x)^2} + \sqrt[3]{1 + x} + 1} =$$
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$$ \frac{x}{\sqrt[3]{(1 + x)^2} + \sqrt[3]{1 + x} + 1} $$
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\subsection*{Point b)}
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$$ \frac{1 - cos(x)}{sin(x)} = \frac{sin^2(x)cos^2(x) - cos(x)}{sin(x)} \cdot \frac{sin(x)}{cos(x)} \cdot \frac{cos(x)}{sin(x)} = (sin^2(x)cos(x) - 1)\cdot\frac{cos(x)}{sin(x)}$$
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\subsection*{Point c)}
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$$ \frac{1}{1-\sqrt{x^2-1}} = \frac{1+\sqrt{x^2-1}}{(1-\sqrt{x^2-1})(1+\sqrt{x^2-1})} =
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\frac{1+\sqrt{x^2-1}}{1 - (x^2-1)} = -\frac{1+\sqrt{x^2-1}}{x^2} $$
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\subsection*{Point d)}
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$$ x^3\cdot\left(\frac{x}{x^2-1}-\frac{1}{x}\right) = x^3\cdot\left(\frac{x^2-x^2+1}{x^3-x}\right) =
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\frac{x^2}{x^2-1}$$
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\subsection*{Point e)}
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$$ \frac{1}{x} - \frac{1}{x+1} = \frac{x + 1 - x}{x^2 + x} = \frac{1}{x^2 + x}$$
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\section*{Question 3}
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\subsection*{Point a)}
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Consider the Taylor expansion with $a=x$ of $f(x+h)$ and $f(x-h)$:
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$$f(x+h) \geq f(x) + \frac{f'(x)}{1}\cdot h + \frac{f''(x)}{2}\cdot h^2 +
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\frac{f'''(x)}{6}\cdot h^3$$
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$$f(x-h) \geq f(x) - \frac{f'(x)}{1}\cdot h + \frac{f''(x)}{2}\cdot h^2 -
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\frac{f'''(x)}{6}\cdot h^3$$
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Then, we can derive that:
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$$\frac{f(x + h) - f(x - h)}{2h} \geq
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\frac{1}{2h} \cdot
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\left(2hf'(x) + \frac{2h^3f'''(x)}{6} \right) =
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f'(x) + \frac{h^2f'''(x)}{6}
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$$
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So:
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$$ \left|f'(x) - \frac{f(x + h) - f(x - h)}{2h}\right| \leq \left|f'(x) - \left(f'(x) + \frac{h^2f'''(x)}{6}\right)
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\right| = \frac{h^2|f'''(x)|}{6}
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\Rightarrow
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C = \frac{f'''(x)}{6}$$
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\subsection*{Point b)}
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In order to find a valid constant for the entire domain $[-10,10]$ we find the constant for the value of $x$ that maximizes $f'''(x)$, hence the
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$\sup$s.
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\paragraph{Function 1}
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$$f(x) = e^{-x^2} \hspace{1cm} f'''(x) = -4e^{-x^2} (2x^3 - 3x) \hspace{1cm} C = \sup_{x \in [-10,10]} \frac{f'''(x)}{6} = 2 \sqrt{1-\sqrt{\frac{2}{3}}}e^{\sqrt{\frac{3}{2}} - \frac{3}{2}} $$
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\paragraph{Function 2}
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$$f(x) = x^2 \hspace{1cm} f'''(x) = 0 \hspace{1cm} C = \sup_{x \in [-10,10]} \frac{f'''(x)}{6} = 0 $$
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\paragraph{Function 3}
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$$f(x) = sin(x) \hspace{1cm} f'''(x) = -cos(x) \hspace{1cm} C = \sup_{x \in [-10,10]} \frac{f'''(x)}{6} = \frac{1}{6} $$
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\section*{Question 4}
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\subsection*{Point a)}
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\[
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A_1 =
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\begin{bmatrix}
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1 & 1 & 1 & 1 & 1 \\
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2 & 4 & 4 & 4 & 4 \\
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3 & 7 & 10 & 10 & 10 \\
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4 & 10 & 16 & 20 & 20 \\
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5 & 13 & 22 & 30 & 35 \\
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\end{bmatrix},
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b =
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\begin{bmatrix}
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1 \\ 2 \\ 3 \\ 4 \\ 5 \\
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\end{bmatrix}
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\]
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\[
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l_1 =
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\begin{bmatrix}
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1 \\ 2 \\ 3 \\ 4 \\ 5 \\
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\end{bmatrix},
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u_1 =
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\begin{bmatrix}
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1 & 1 & 1 & 1 & 1 \\
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\end{bmatrix},
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A_2
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\begin{bmatrix}
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0 & 0 & 0 & 0 & 0 \\
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0 & 2 & 2 & 2 & 2 \\
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0 & 4 & 7 & 7 & 7 \\
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0 & 6 & 12 & 16 & 16 \\
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0 & 8 & 17 & 25 & 30 \\
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\end{bmatrix}
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\]
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\[
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l_2 =
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\begin{bmatrix}
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0 \\ 1 \\ 2 \\ 3 \\ 4 \\
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\end{bmatrix},
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u_2 =
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\begin{bmatrix}
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0 & 2 & 2 & 2 & 2 \\
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\end{bmatrix},
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A_3
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\begin{bmatrix}
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 3 & 3 & 3 \\
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0 & 0 & 6 & 10 & 10 \\
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0 & 0 & 9 & 17 & 22 \\
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\end{bmatrix}
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\]
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\[
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l_3 =
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\begin{bmatrix}
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0 \\ 0 \\ 1 \\ 2 \\ 3 \\
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\end{bmatrix},
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u_3 =
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\begin{bmatrix}
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0 & 0 & 3 & 3 & 3 \\
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\end{bmatrix},
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A_4
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\begin{bmatrix}
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 4 & 4 \\
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0 & 0 & 0 & 8 & 13 \\
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\end{bmatrix}
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\]
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\[
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l_4 =
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\begin{bmatrix}
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0 \\ 0 \\ 0 \\ 1 \\ 2 \\
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\end{bmatrix},
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u_4 =
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\begin{bmatrix}
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0 & 0 & 0 & 4 & 4 \\
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\end{bmatrix},
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A_5
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\begin{bmatrix}
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 5 \\
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\end{bmatrix}
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\]
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\[
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l_5 =
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\begin{bmatrix}
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0 \\ 0 \\ 0 \\ 0 \\ 1 \\
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\end{bmatrix},
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u_5 =
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\begin{bmatrix}
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0 & 0 & 0 & 0 & 5 \\
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\end{bmatrix},
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L =
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\begin{bmatrix}
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1 & 0 & 0 & 0 & 0 \\
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2 & 1 & 0 & 0 & 0 \\
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3 & 2 & 1 & 0 & 0 \\
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4 & 3 & 2 & 1 & 0 \\
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5 & 4 & 3 & 2 & 1 \\
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\end{bmatrix},
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U =
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\begin{bmatrix}
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1 & 1 & 1 & 1 & 1 \\
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0 & 2 & 2 & 2 & 2 \\
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0 & 0 & 3 & 3 & 3 \\
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0 & 0 & 0 & 4 & 4 \\
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0 & 0 & 0 & 0 & 5 \\
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\end{bmatrix}
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\]
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\subsection*{Point b)}
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\[
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l_1 =
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\begin{bmatrix}
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1 \\ 2 \\ 3 \\ 4 \\ 5 \\
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\end{bmatrix},
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e^{T}_1 =
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\begin{bmatrix}
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1 & 0 & 0 & 0 & 0 \\
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\end{bmatrix},
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L_1 =
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\begin{bmatrix}
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0 & 0 & 0 & 0 & 0 \\
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-2 & 1 & 0 & 0 & 0 \\
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-3 & 0 & 1 & 0 & 0 \\
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-4 & 0 & 0 & 1 & 0 \\
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-5 & 0 & 0 & 0 & 1 \\
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\end{bmatrix}
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\]
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\[
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l_2 =
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\begin{bmatrix}
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0 \\ 1 \\ 2 \\ 3 \\ 4 \\
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\end{bmatrix},
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e^{T}_2 =
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\begin{bmatrix}
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0 & 1 & 0 & 0 & 0 \\
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\end{bmatrix},
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L_2 =
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\begin{bmatrix}
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1 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & -2 & 1 & 0 & 0 \\
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0 & -3 & 0 & 1 & 0 \\
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0 & -4 & 0 & 0 & 1 \\
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\end{bmatrix}
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\]
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\[
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l_3 =
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\begin{bmatrix}
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0 \\ 0 \\ 1 \\ 2 \\ 3 \\
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\end{bmatrix},
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e^{T}_3 =
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\begin{bmatrix}
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0 & 0 & 1 & 0 & 0 \\
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\end{bmatrix},
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L_3 =
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\begin{bmatrix}
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1 & 0 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & -2 & 1 & 0 \\
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0 & 0 & -3 & 0 & 1 \\
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\end{bmatrix}
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\]
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\[
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l_4 =
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\begin{bmatrix}
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0 \\ 0 \\ 0 \\ 1 \\ 2 \\
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\end{bmatrix},
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e^{T}_4 =
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\begin{bmatrix}
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0 & 0 & 0 & 1 & 0 \\
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\end{bmatrix},
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L_4 =
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\begin{bmatrix}
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1 & 0 & 0 & 0 & 0 \\
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0 & 1 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 & 0 \\
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0 & 0 & 0 & 0 & 0 \\
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0 & 0 & 0 & -2 & 1 \\
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\end{bmatrix}
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\]
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\subsection*{Point c)}
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$$Ly = b$$
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\[
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L =
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\begin{bmatrix}
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1 & 0 & 0 & 0 & 0 \\
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2 & 1 & 0 & 0 & 0 \\
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3 & 2 & 1 & 0 & 0 \\
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4 & 3 & 2 & 1 & 0 \\
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5 & 4 & 3 & 2 & 1 \\
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\end{bmatrix},
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b =
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\begin{bmatrix}
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1 \\ 2 \\ 3 \\ 4 \\ 5 \\
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\end{bmatrix}
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\]
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\[
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L =
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\begin{bmatrix}
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1 & 0 & 0 & 0 & 0 \\
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2 & 1 & 0 & 0 & 0 \\
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3 & 2 & 1 & 0 & 0 \\
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4 & 3 & 2 & 1 & 0 \\
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5 & 4 & 3 & 2 & 1 \\
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\end{bmatrix},
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b =
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\begin{bmatrix}
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1 \\ 2 \\ 3 \\ 4 \\ 5 \\
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\end{bmatrix}
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\]
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$$y_1 = \frac{1}{1} = 1$$
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$$y_2 = \frac{2 - 2 \cdot 1}{1} = 0$$
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$$y_3 = \frac{3 - 3 \cdot 1 - 2 \cdot 0}{1} = 0$$
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$$y_4 = \frac{4 - 4 \cdot 1 - 3 \cdot 0 - 2 \cdot 0}{1} = 0$$
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$$y_5 = \frac{5 - 5 \cdot 1 - 4 \cdot 0 - 3 \cdot 0 - 2 \cdot 0}{1} = 0$$
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\[
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y =
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\begin{bmatrix}
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1 \\ 0 \\ 0 \\ 0 \\ 0 \\
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\end{bmatrix}
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\]
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\subsection*{Point d)}
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$$Ux = y$$
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\[
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U =
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\begin{bmatrix}
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1 & 1 & 1 & 1 & 1 \\
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0 & 2 & 2 & 2 & 2 \\
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0 & 0 & 3 & 3 & 3 \\
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0 & 0 & 0 & 4 & 4 \\
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0 & 0 & 0 & 0 & 5 \\
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\end{bmatrix}
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y =
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\begin{bmatrix}
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1 \\ 0 \\ 0 \\ 0 \\ 0 \\
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\end{bmatrix}
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\]
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$$x_5 = \frac{0}{5} = 0$$
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$$x_4 = \frac{0 - 4 \cdot 0}{4} = 0$$
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$$x_3 = \frac{0 - 3 \cdot 0 - 3 \cdot 0}{3} = 0$$
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$$x_2 = \frac{0 - 2 \cdot 0 - 2 \cdot 0 - 2 \cdot 0}{2} = 0$$
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$$x_1 = \frac{0 - 1 \cdot 0 - 1 \cdot 0 - 1 \cdot 0 - 1 \cdot }{1} = 0$$
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\[
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x =
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\begin{bmatrix}
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0 \\ 0 \\ 0 \\ 0 \\ 0 \\
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\end{bmatrix}
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\]
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\section*{Question 5}
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\subsection*{Point a)}
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$$f(x) = x \hspace{2cm} K_{abs} = |f'(x)| = 1 \hspace{2cm} K_{rel} = \left|\frac{1 \cdot x}{x}\right| = 1$$
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\subsection*{Point b)}
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$$f(x) = \sqrt[3]{x} \hspace{2cm} K_{abs} = |f'(x)| = \frac{1}{3\sqrt[3]{x^2}} \hspace{2cm}
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K_{rel} = \left|\frac{1}{3\sqrt[3]{x^2}} \cdot \frac{x}{\sqrt[3]{x}}\right| = \frac{1}{3}$$
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\subsection*{Point c)}
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$$f(x) = \frac{1}{x} \hspace{2cm} K_{abs} = |f'(x)| = \frac{1}{x^2} \hspace{2cm}
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K_{rel} = \left|\frac{-x}{x^2} \cdot \frac{1}{\frac{1}{x}}\right| = 1$$
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\subsection*{Point d)}
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$$f(x) = e^x \hspace{2cm} K_{abs} = |f'(x)| = e^x \hspace{2cm}
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K_{rel} = \left|\frac{xe^x}{e^x}\right| = |x|$$
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\subsection*{Point e)}
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Cases \textit{a)},\textit{b)} and \textit{c)} are well-conditioned for any $x$ since their $K_rel$
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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 $|x|$.
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\end{document} |