hw1 almost done
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@ -45,7 +45,7 @@ A = [1 -2 0 ; -2 1 -2; 0 -2 1]
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Z = zeros(9,9)
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B = ones(9,9) * 3
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C = (eye(9) - 1) * -1
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D = diag([[1:5],[4:-1:1]])
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D = diag([1:5,4:-1:1])
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E = repmat(transpose(1:9), 1, 5)
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%% Problem 3
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@ -87,7 +87,8 @@ function xmax = getxmax
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x = y;
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y = y * 2;
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end
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xmax = typecast(bitor(typecast(x, 'uint64'), 0x000FFFFFFFFFFFFF), 'double');
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xmax = typecast(bitor(typecast(x, 'uint64'), 0x000FFFFFFFFFFFFF), ...
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'double');
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end
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% geteps does not differ from eps, as getxmax does not differ from realmax.
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% However, getxmin returns the nearest positive floating point value to 0
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\relax
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This is pdfTeX, Version 3.14159265-2.6-1.40.19 (TeX Live 2018) (preloaded format=pdflatex 2019.1.23) 3 MAR 2020 21:53
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entering extended mode
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restricted \write18 enabled.
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**hw1.tex
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) (./hw1.aux)
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fb>
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Output written on hw1.pdf (2 pages, 79760 bytes).
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@ -0,0 +1,59 @@
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\documentclass[12pt]{article}
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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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\newcolumntype{L}{>{$}l<{$}}
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\title{Howework 1 -- 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 5}
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\subsection*{Point a)}
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$1.1110_2 * 2^4$, so $d_0=1$, $d_1=1$, $d_2=1$, $d_3=1$, $d_4=0$, and $e=4$.
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\subsection*{Point b)}
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$1.1110_2 * 2^1$, so $d_0=1$, $d_1=1$, $d_2=1$, $d_3=1$, $d_4=0$, and $e=1$.
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\subsection*{Point c)}
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$0.00048828125$, or $2^{-4} * 2^{-7}$
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\subsection*{Point d)}
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$248$, or $(2 - 2^{-4}) * 2^7$
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\subsection*{Point e)}
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The best approximation for 30.1 is 30, or $1.1110_2 * 2^4$, so $d_0=1$, $d_1=1$, $d_2=1$, $d_3=1$, $d_4=0$, and $e=4$.
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This representation is not exact. The number 30.1 cannot be represented exactly in this format or any format with $e=2$.
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\section*{Question 6}
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\subsection*{Point a)}
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$$f(x) = \frac{1}{x} - \frac{1}{x + 1} = \frac{x + 1}{x (x + 1)} - \frac{x}{x (x + 1)}
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= \frac{x + 1 - x}{x (x + 1)} = \frac{1}{x (x + 1)} = g(x) $$
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\subsection*{Point b)}
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$$ f(10_f) = \frac{1}{1.00 * 10^1} - \frac{1}{(1.00 * 10^1) + 1} =
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(1.00 * 10^{-1}) - \frac{1}{1.10 * 10^1} \approx$$
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$$\approx (1.00 * 10^{-1}) - (0.91 * 10^{-1}) = 9.00 * 10^{-3} = 0.009 $$
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$$ g(10_f) = \frac{1}{(1.00 * 10^1) ((1.00 * 10^1) + 1)} = \frac{1}{1.00 * 10^1 * 1.10 * 10^1} \approx
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(9.09 * 10^{-3}) = 0.00909$$
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$$ f(100_f) = \frac{1}{1.00 * 10^2} - \frac{1}{(1.00 * 10^2) + 1} =
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(1.00 * 10^{-2}) - \frac{1}{1.01 * 10^2} \approx$$
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$$\approx (1.00 * 10^{-2}) - (0.99 * 10^{-2}) = 1.00 * 10^{-4} = 0.0001 $$
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$$ g(100_f) = \frac{1}{(1.00 * 10^2) ((1.00 * 10^2) + 1)} = \frac{1}{1.00 * 10^2 * 1.01 * 10^2} \approx
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(0.99 * 10^{-4}) = 0.000099$$
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$$ f(1000_f) = \frac{1}{1.00 * 10^3} - \frac{1}{(1.00 * 10^3) + 1} \approx
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(1.00 * 10^{-3}) - \frac{1}{1.00 * 10^3} =$$
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$$= (1.00 * 10^{-3}) - (1.00 * 10^{-3}) = 0 $$
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$$ g(1000_f) = \frac{1}{(1.00 * 10^3) ((1.00 * 10^3) + 1)} \approx \frac{1}{1.00 * 10^3 * 1.00 * 10^3} =
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(1.00 * 10^{-6}) = 0.000001$$
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\subsection*{Point c)}
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\begin{tabular}{L | L | L | L}
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\text{Value} & \text{Correct} & \text{Abs. Error} & \text{Rel. error} \\
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f(10_f) & 0.\overline{90} & 9.\overline{09} * 10^{-5} & 8.2645 * 10^{-7} \\
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g(10_f) & 0.\overline{90} & 9.\overline{09} * 10^{-7} & 8.26 * 10^{-9} \\
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f(100_f) & 9.\overline{9009} * 10^{-5} & 9.901 * 10^{-7} & 10^{-2} \\
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g(100_f) & 9.\overline{9009} * 10^{-5} & 9.9 * 10^{-9} & 10^{-4} \\
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f(1000_f) & 9.99 * 10^{-7} & 9.99 * 10^{-7} & 1 \\
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f(1000_f) & 9.99 * 10^{-7} & 10^{-9} & 10^{-3} \\
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\end{tabular}
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\end{document}
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