diff options
Diffstat (limited to 'Master/texmf-dist/source/latex/l3kernel/l3fp-expo.dtx')
-rw-r--r-- | Master/texmf-dist/source/latex/l3kernel/l3fp-expo.dtx | 70 |
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 |