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\begin{document}

\begin{center}
\begin{tabular}{cccc}
    Initialization&amp;$D$&amp;$0001.1010\ 000$&amp;\\
    &amp;$-D=\overline{D}+1$&amp;$1110.0101\ 111$&amp;(+ 1 ulp)\\
    &amp;&amp;&amp;\\
    &amp;$WS_{-1}=X$&amp;$0001.0000\ 010$&amp;\\
    &amp;$WC_{-1}$&amp;$0000.0000\ 000$&amp;\\
    \hdashline\\
    Step 0:&amp;$WS_{-1}$&amp;$0001.0000\ 010$&amp;\\
    &amp;$WC_{-1}$&amp;$0000.0000\ 00\mathbf{1}$&amp;($W_{msbs}=0001\ \text{so}\ q_0=1$)\\
    &amp;$-q_0D$&amp;$1110.0101\ 111$&amp;\\
    \cline{2-3}
    &amp;$sum$&amp;$1111.0101\ 100$&amp;$\ll1$\\
    &amp;$carry$&amp;$0000.0000\ 110$&amp;$\ll1$\\
    \hdashline\\
    Step 1:&amp;$WS_0$&amp;$1110.1011\ 000$&amp;\\
    &amp;$WC_0$&amp;$0000.0001\ 10\mathbf{0}$&amp;($W_{msbs}=1110\ \text{so}\ q_1=-1$)\\
    &amp;$-q_1D$&amp;$0001.1010\ 000$&amp;\\
    \cline{2-3}
    &amp;$sum$&amp;$1111.0000\ 100$&amp;$\ll1$\\
    &amp;$carry$&amp;$0001.0110\ 000$&amp;$\ll1$\\
    \hdashline\\
    Step 2:&amp;$WS_1$&amp;$1110.0001\ 000$&amp;\\
    &amp;$WC_1$&amp;$0010.1100\ 00\mathbf{1}$&amp;($W_{msbs}=0000\ \text{so}\ q_2=1$)\\
    &amp;$-q_2D$&amp;$1110.0101\ 111$&amp;\\
    \cline{2-3}
    &amp;$sum$&amp;$0010.1000\ 110$&amp;$\ll1$\\
    &amp;$carry$&amp;$1100.1010\ 010$&amp;$\ll1$\\
    \hdashline\\
    Step 3:&amp;$WS_2$&amp;$0101.0001\ 100$&amp;\\
    &amp;$WC_2$&amp;$1001.0100\ 10\mathbf{0}$&amp;($W_{msbs}=1110\ \text{so}\ q_3=-1$)\\
    &amp;$-q_3D$&amp;$0001.1010\ 000$&amp;\\
    \cline{2-3}
    &amp;$sum$&amp;$1101.1111\ 000$&amp;\\
    &amp;$carry$&amp;$0010.0001\ 000$&amp;$sum+carry=0$, terminate.\\
    \hdashline\\
    Terminate&amp;Quotient&amp;0.101
\end{tabular}
\end{center}

\vfill
\eject

X = 1.0110\ 011 (179/128)

D = 1.0011\ 000 (152/128)

Q = 1.0010\ 1101\ 0

D[1.3] = 1.001, so we use the ``"1.001" column of chart 13.X. This means we select a quotient bit of 2 if the partial remainder is greater than or equal to 3.5, a quotient bit of 1 if the partial is greater or equal to than 1.0, a zero if the partial is greater than or equal to -1.5, -1 if the partial is greater than or equal to -3.75, and a -2 otherwise.
\begin{center}
\begin{tabular}{cccc}
    Initialization&amp;$D$&amp;$0001.0011\ 000$&amp;\\
    &amp;$2D$&amp;$0010.0110\ 000$&amp;\\
    &amp;$-D=\overline{D}+1$&amp;$1110.1100\ 111$&amp;(+ 1 ulp)\\
    &amp;$-2D=\overline{2D}+1$&amp;$1101.1001\ 111$&amp;(+ 1 ulp)\\
    &amp;&amp;&amp;\\
    &amp;$X=WS$&amp;$0001.0110\ 011$&amp;\\
    &amp;$WC$&amp;$0000.0000\ 000$&amp;\\
    \hdashline\\
    Step 4:&amp;$WS$&amp;$0001.0110\ 011$&amp;\\
    &amp;$WC$&amp;$0000.0000\ 00\mathbf{1}$&amp;($RW_{msbs}=0001.010\ \text{so}\ q_4=1$)\\
    &amp;$-q_7D$&amp;$1110.1100\ 111$&amp;\\
    \cline{2-3}
    &amp;$WS$&amp;$1111.1010\ 101$&amp;$\ll2$\\
    &amp;$WC$&amp;$0000.1000\ 110$&amp;$\ll2$\\
    \hdashline\\
    Step 3:&amp;$WS$&amp;$1110.1010\ 100$&amp;\\
    &amp;$WC$&amp;$0010.0011\ 000$&amp;($RW_{msbs}=0000.110\ \text{so}\ q_3=1$)\\
    &amp;$-q_6D$&amp;$0000.0000\ 000$&amp;\\
    \cline{2-3}
    &amp;$WS$&amp;$1100.1001\ 100$&amp;$\ll2$\\
    &amp;$WC$&amp;$0100.0100\ 000$&amp;$\ll2$\\
    \hdashline\\
    Step 2:&amp;$WS$&amp;$0010.0110\ 000$&amp;\\
    &amp;$WC$&amp;$0001.0000\ 00\mathbf{1}$&amp;($RW_{msbs}=0011.010\ \text{so}\ q_2=-1$)\\
    &amp;$-q_5D$&amp;$1110.0101\ 111$&amp;\\
    \cline{2-3}
\end{tabular}
\end{center}

page 269 306

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\large{\bf{
Math for the recurrence relation}}

**going to have to change notation for sure, change the subscripts for steps and might have to get rid of some exponents**
\begin{align*}
    w[j+1] &amp;= r^{j+1}\big(x-S[j+1]^2\big)\\
    &amp;= r^{j+1}\big(x-(S[j]+s_{j+1}r^{-(j+1)})^2\big)\\
    &amp;= r^{j+1}x-r^{j+1}\big(S[j]^2+2S[j]s_{j+1}r^{-(j+1)}+s^2_{j+1}r^{-2(j+1)}\big)\\
    &amp;= r^{j+1}\big(x-S[j]^2\big)-\big(2S[j]s_{j+1}+s_{j+1}^2r^{-(j+1)}\big)\\
    &amp;= rw[j]-\big(2S[j]s_{j+1}+s_{j+1}^2r^{-(j+1)}\big)\\
    &amp;= rw[j]+F[j]
\end{align*}
where
\begin{align*}
    F[j]=-\big(2S[j]s_{j+1}+s_{j+1}^2r^{-(j+1)}\big)
\end{align*}

Since there is a term of $S$ in the expression of $F$, we must come up with a way to represent $S$ using only zeros and ones, rather than using the bit set $\{-a,\ldots,a\}$. This is done using on-the-fly conversion just as we did to compute the quotient for the divider. We keep a running copy of $S$, but we also keep the value $SM=S-1$. The logic is still the same for computing $S$ and $SM$ on the next step; see figure 13.15. 

Now that $S$ is in a form such that we can use it in a CSA, we need to compute $F$. To do so, 

\end{document}
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