2025-01-22 21:50:31 -08:00
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\section{Approximation}
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\generic{Observation:}
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For small values of $a$, $\log_2(1 + a)$ is approximately equal to $a$. \par
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Note that this equality is exact for $a = 0$ and $a = 1$, since $\log_2(1) = 0$ and $\log_2(2) = 1$.
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\vspace{2mm}
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We'll add a \say{correction term} $\varepsilon$ to this approximation, so that $\log_2(1 + a) \approx a + \varepsilon$.
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TODO: why? Graphs.
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\problem{}
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2025-02-08 14:54:40 -08:00
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Use the fact that $\log_2(1 + a) \approx a + \varepsilon$ to approximate $\log_2(x_f)$ in terms of $x_i$. \par
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\vspace{5mm}
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Namely, show that
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\begin{equation*}
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\log_2(x_f) ~=~ \frac{1}{2^{23}}(x_i) - 127 + \varepsilon
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\end{equation*}
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for some error term $\varepsilon$
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2025-01-22 21:50:31 -08:00
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\begin{solution}
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Let $E$ and $F$ be the exponent and float bits of $x_f$. \par
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We then have:
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\begin{align*}
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\log_2(x_f)
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&=~ \log_2 \left( 2^{E-127} \times \left(1 + \frac{F}{2^{23}}\right) \right) \\
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&=~ E-127 + \log_2\left(1 + \frac{F}{2^{23}}\right) \\
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&\approx~ E-127 + \frac{F}{2^{23}} + \varepsilon \\
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&=~ \frac{1}{2^{23}}(2^{23}E + F) - 127 + \varepsilon \\
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&=~ \frac{1}{2^{23}}(x_i) - 127 + \varepsilon
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\end{align*}
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\end{solution}
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\vfill
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\pagebreak
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