\section{Approximation}

\generic{Observation:}
For small values of $a$, $\log_2(1 + a)$ is approximately equal to $a$. \par
Note that this equality is exact for $a = 0$ and $a = 1$, since $\log_2(1) = 0$ and $\log_2(2) = 1$.

\vspace{2mm}

We'll add a \say{correction term} $\varepsilon$ to this approximation, so that $\log_2(1 + a) \approx a + \varepsilon$.

TODO: why? Graphs.

\problem{}
Use the fact that $\log_2(1 + a) \approx a + \varepsilon$ to approximate $\log_2(x_f)$ in terms of $x_i$. \par

\vspace{5mm}

Namely, show that
\begin{equation*}
	\log_2(x_f) ~=~ \frac{1}{2^{23}}(x_i) - 127 + \varepsilon
\end{equation*}
for some error term $\varepsilon$


\begin{solution}
	Let $E$ and $F$ be the exponent and float bits of $x_f$. \par
	We then have:
	\begin{align*}
		\log_2(x_f)
		&=~ \log_2 \left( 2^{E-127} \times \left(1 + \frac{F}{2^{23}}\right) \right) \\
		&=~ E-127 + \log_2\left(1 + \frac{F}{2^{23}}\right) \\
		&\approx~ E-127 + \frac{F}{2^{23}} + \varepsilon \\
		&=~ \frac{1}{2^{23}}(2^{23}E + F) - 127 + \varepsilon \\
		&=~ \frac{1}{2^{23}}(x_i) - 127 + \varepsilon
	\end{align*}
\end{solution}

\vfill
\pagebreak