66 lines
3.0 KiB
TeX
66 lines
3.0 KiB
TeX
\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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g(1000_f) & 9.99 * 10^{-7} & 10^{-9} & 10^{-3} \\
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\end{tabular}
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\subsection*{Point d)}
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$f(x)$ is clearly less accuratee than $g(x)$ when dealing with floating point numbers. This is largely due to the final
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subtraction, since that amplifies approximation errors already introduced in the division $\frac{1}{x + 1}$. Also, the multiplication
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in $\frac{1}{x (x + 1)}$ does not cause any approximation errors with the numbers tested, since the mantissa is always 1 and
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the multiplication of the exponent part is always exact.
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\end{document}
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