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-rw-r--r--Master/texmf-dist/source/latex/l3kernel/l3fp-expo.dtx70
1 files changed, 35 insertions, 35 deletions
diff --git a/Master/texmf-dist/source/latex/l3kernel/l3fp-expo.dtx b/Master/texmf-dist/source/latex/l3kernel/l3fp-expo.dtx
index b0731d99868..edbfdcc57d6 100644
--- a/Master/texmf-dist/source/latex/l3kernel/l3fp-expo.dtx
+++ b/Master/texmf-dist/source/latex/l3kernel/l3fp-expo.dtx
@@ -38,7 +38,7 @@
% {latex-team@latex-project.org}^^A
% }^^A
% }
-% \date{Released 2017/07/15}
+% \date{Released 2017/07/19}
%
% \maketitle
%
@@ -191,15 +191,13 @@
% \end{macro}
%
% \begin{macro}[aux, EXP]{\@@_ln_significand:NNNNnnnN}
-% \begin{quote}
-% \cs{@@_ln_significand:NNNNnnnN} \meta{X_1} \Arg{X_2} \Arg{X_3}
-% \Arg{X_4} \meta{continuation}
-% \end{quote}
+% \begin{syntax}
+% \cs{@@_ln_significand:NNNNnnnN} \meta{X_1} \Arg{X_2} \Arg{X_3} \Arg{X_4} \meta{continuation}
+% \end{syntax}
% This function expands to
-% \begin{quote}
-% \meta{continuation} \Arg{Y_1} \Arg{Y_2} \Arg{Y_3} \Arg{Y_4}
-% \Arg{Y_5} \Arg{Y_6} |;|
-% \end{quote}
+% \begin{syntax}
+% \meta{continuation} \Arg{Y_1} \Arg{Y_2} \Arg{Y_3} \Arg{Y_4} \Arg{Y_5} \Arg{Y_6} |;|
+% \end{syntax}
% where $Y = - \ln(X)$ as an extended fixed point.
% \begin{macrocode}
\cs_new:Npn \@@_ln_significand:NNNNnnnN #1#2#3#4
@@ -316,10 +314,9 @@
% a faithful rounding).
% ^^A todo: doc
%
-% \begin{quote}
-% \cs{@@_ln_x_iv:wnnnnnnnn}
-% \meta{1 or 2} \meta{8d} |;| \Arg{4d} \Arg{4d} \meta{fixed-tl}
-% \end{quote}
+% \begin{syntax}
+% \cs{@@_ln_x_iv:wnnnnnnnn} \meta{1 or 2} \meta{8d} |;| \Arg{4d} \Arg{4d} \meta{fixed-tl}
+% \end{syntax}
% The number is $x$. Compute $y$ by adding 1 to the five first digits.
% \begin{macrocode}
\cs_new:Npn \@@_ln_x_iv:wnnnnnnnn #1; #2#3#4#5 #6#7#8#9
@@ -357,7 +354,7 @@
% We now have essentially
% ^^A todo: determine error on $Q_{6}$ (probably $6.7$),
% ^^A todo: conclude the final result is off by $<10^{-23}$
-% \begin{quote}
+% \begin{syntax}
% \cs{@@_ln_div_after:Nw} \meta{fixed tl}
% \cs{@@_div_significand_pack:NNN} $10^6 + Q_{1}$
% \cs{@@_div_significand_pack:NNN} $10^6 + Q_{2}$
@@ -366,16 +363,17 @@
% \cs{@@_div_significand_pack:NNN} $10^6 + Q_{5}$
% \cs{@@_div_significand_pack:NNN} $10^6 + Q_{6}$ |;|
% \meta{exponent} |;| \meta{continuation}
-% \end{quote}
+% \end{syntax}
% where \meta{fixed tl} holds the logarithm of a number
% in $[1,10]$, and \meta{exponent} is
% the exponent. Also, the expansion is done backwards. Then
% \cs{@@_div_significand_pack:NNN} puts things in the
% correct order to add the $Q_{i}$ together and put semicolons
% between each piece. Once those have been expanded, we get
-% \begin{quote}
-% \cs{@@_ln_div_after:Nw} \meta{fixed-tl} \meta{1d} |;| \meta{4d} |;| \meta{4d} |;| \meta{4d} |;| \meta{4d} |;| \meta{4d} |;| \meta{4d} |;| \meta{exponent} |;|
-% \end{quote}
+% \begin{syntax}
+% \cs{@@_ln_div_after:Nw} \meta{fixed-tl} \meta{1d} |;| \meta{4d} |;| \meta{4d} |;|
+% ~~\meta{4d} |;| \meta{4d} |;| \meta{4d} |;| \meta{4d} |;| \meta{exponent} |;|
+% \end{syntax}
% ^^A todo: redoc.
% Just as with division, we know that the first two digits
% are |1| and |0| because of bounds on the final result of
@@ -407,9 +405,11 @@
}
% \end{macrocode}
%
-% \begin{quote}
-% \cs{@@_ln_t_large:NNw} \meta{sign}\meta{fixed tl} \meta{t_1}|;| \meta{t_2} |;| \meta{t_3}|;| \meta{t_4}|;| \meta{t_5} |;| \meta{t_6}|;| \meta{exponent} |;| \meta{continuation}
-% \end{quote}
+% \begin{syntax}
+% \cs{@@_ln_t_large:NNw} \meta{sign} \meta{fixed tl}
+% ~~\meta{t_1}|;| \meta{t_2} |;| \meta{t_3}|;| \meta{t_4}|;| \meta{t_5} |;| \meta{t_6}|;|
+% ~~\meta{exponent} |;| \meta{continuation}
+% \end{syntax}
% Compute the square $|t|^2$, and keep $|t|$ at the end with its
% sign. We know that $|t|<0.1765$, so every piece has at most $4$
% digits. However, since we were not careful in \cs{@@_ln_t_small:w},
@@ -455,13 +455,13 @@
%
% \begin{macro}{\@@_ln_Taylor:wwNw}
% Denoting $T=t^2$, we get
-% \begin{quote}
+% \begin{syntax}
% \cs{@@_ln_Taylor:wwNw}
-% \Arg{T_1} \Arg{T_2} \Arg{T_3} \Arg{T_4} \Arg{T_5} \Arg{T_6} |;| |;|
-% \Arg{(2t)_1} \Arg{(2t)_2} \Arg{(2t)_3} \Arg{(2t)_4} \Arg{(2t)_5} \Arg{(2t)_6} |;|
-% |{| \cs{@@_ln_c:NwNw} \meta{sign} |}|
-% \meta{fixed tl} \meta{exponent} |;| \meta{continuation}
-% \end{quote}
+% ~~\Arg{T_1} \Arg{T_2} \Arg{T_3} \Arg{T_4} \Arg{T_5} \Arg{T_6} |;| |;|
+% ~~\Arg{(2t)_1} \Arg{(2t)_2} \Arg{(2t)_3} \Arg{(2t)_4} \Arg{(2t)_5} \Arg{(2t)_6} |;|
+% ~~|{| \cs{@@_ln_c:NwNw} \meta{sign} |}|
+% ~~\meta{fixed tl} \meta{exponent} |;| \meta{continuation}
+% \end{syntax}
% And we want to compute
% \[
% \ln\left(\frac{1+t}{1-t}\right)
@@ -508,11 +508,11 @@
% \end{macro}
%
% \begin{macro}{\@@_ln_c:NwNw}
-% \begin{quote}
+% \begin{syntax}
% \cs{@@_ln_c:NwNw} \meta{sign}
-% \Arg{r_1} \Arg{r_2} \Arg{r_3} \Arg{r_4} \Arg{r_5} \Arg{r_6} |;|
-% \meta{fixed tl} \meta{exponent} |;| \meta{continuation}
-% \end{quote}
+% ~~\Arg{r_1} \Arg{r_2} \Arg{r_3} \Arg{r_4} \Arg{r_5} \Arg{r_6} |;|
+% ~~\meta{fixed tl} \meta{exponent} |;| \meta{continuation}
+% \end{syntax}
% We are now reduced to finding $\ln(c)$ and $\meta{exponent}\ln(10)$
% in a table, and adding it to the mixture. The first step is to
% get $\ln(c) - \ln(x) = - \ln(a)$, then we get $|b|\ln(10)$ and add
@@ -535,11 +535,11 @@
% \end{macro}
%
% \begin{macro}{\@@_ln_exponent:wn}
-% \begin{quote}\raggedright
+% \begin{syntax}
% \cs{@@_ln_exponent:wn}
-% \Arg{s_1} \Arg{s_2} \Arg{s_3} \Arg{s_4} \Arg{s_5} \Arg{s_6} |;|
-% \Arg{exponent}
-% \end{quote}
+% ~~\Arg{s_1} \Arg{s_2} \Arg{s_3} \Arg{s_4} \Arg{s_5} \Arg{s_6} |;|
+% ~~\Arg{exponent}
+% \end{syntax}
% Compute \meta{exponent} times $\ln(10)$. Apart from the cases where
% \meta{exponent} is $0$ or $1$, the result is necessarily at
% least $\ln(10) \simeq 2.3$ in magnitude. We can thus drop the least