diff options
Diffstat (limited to 'Master/texmf-dist/source/latex/l3kernel/l3fp-extended.dtx')
| -rw-r--r-- | Master/texmf-dist/source/latex/l3kernel/l3fp-extended.dtx | 33 |
1 files changed, 16 insertions, 17 deletions
diff --git a/Master/texmf-dist/source/latex/l3kernel/l3fp-extended.dtx b/Master/texmf-dist/source/latex/l3kernel/l3fp-extended.dtx index 286d320973c..1144d72e0ac 100644 --- a/Master/texmf-dist/source/latex/l3kernel/l3fp-extended.dtx +++ b/Master/texmf-dist/source/latex/l3kernel/l3fp-extended.dtx @@ -38,7 +38,7 @@ % {latex-team@latex-project.org}^^A % }^^A % } -% \date{Released 2017/05/29} +% \date{Released 2017/07/15} % % \maketitle % @@ -180,8 +180,8 @@ % $10^{-24}$, and leaves \meta{continuation} \Arg{c_1} \ldots{} % \Arg{c_6} |;| in the input stream, where each of the \meta{c_i} are % blocks of $4$~digits, except \meta{c_1}, which is any \TeX{} -% integer. Note that indices for \meta{b} start at~$0$: a second -% operand of |{0001}{0000}{0000}| will leave the first operand +% integer. Note that indices for \meta{b} start at~$0$: for instance +% a second operand of |{0001}{0000}{0000}| leaves the first operand % unchanged (rather than dividing it by $10^{4}$, as % \cs{@@_fixed_mul:wwn} would). % \begin{macrocode} @@ -235,7 +235,7 @@ % The \texttt{ii} auxiliary receives $Q_{i}$, $n$, and $a_{i}$ as % arguments. It adds $Q_{i}$ to a surrounding integer expression, and % starts a new one with the initial value $9999$, which ensures that -% the result of this expression will have $5$ digits. The auxiliary +% the result of this expression has $5$ digits. The auxiliary % also computes $a_{i}-n\cdot Q_{i}$, placing the result in front of % the $4$ digits of $a_{i+1}$. The resulting $a'_{i+1} = 10^{4} % (a_{i} - n \cdot Q_{i}) + a_{i+1}$ serves as the first argument for @@ -450,7 +450,7 @@ % common parts of the three functions. % % For definiteness, consider the task of computing $a\times b + c$. -% We will perform carries in +% We perform carries in % \begin{align*} % a \times b + c = % & (a_{1} \cdot b_{1} + c_{1} c_{2})\cdot 10^{-8} \\ @@ -479,7 +479,7 @@ % first level, calls the \texttt{i} auxiliary with arguments described % later, and adds a trailing ${} + c_{5}c_{6}$ |;| % \Arg{continuation}~|;|. The ${} + c_{5}c_{6}$ piece, which is -% omitted for \cs{@@_fixed_one_minus_mul:wwn}, will be taken in the +% omitted for \cs{@@_fixed_one_minus_mul:wwn}, is taken in the % integer expression for the $10^{-24}$ level. % \begin{macrocode} \cs_new:Npn \@@_fixed_mul_add:wwwn #1; #2; #3#4#5#6#7#8; @@ -564,7 +564,7 @@ % & b_{1} + a_{4} \cdot b_{2} + a_{3} \cdot b_{3} + a_{2} \cdot b_{4} + a_{1} \\ % & b_{2} + a_{4} \cdot b_{3} + a_{3} \cdot b_{4} + a_{2} . % \end{align*} -% Obviously, those expressions make no mathematical sense: we will +% Obviously, those expressions make no mathematical sense: we % complete them with $a_{5} \cdot {}$ and ${} \cdot b_{5}$, and with % $a_{6} \cdot b_{1} + a_{5} \cdot {}$ and ${} \cdot b_{5} + a_{1} % \cdot b_{6}$, and of course with the trailing ${} + c_{5} c_{6}$. @@ -626,8 +626,7 @@ % {\@@_ep_to_fixed_auxi:www, \@@_ep_to_fixed_auxii:nnnnnnnwn} % Converts an extended-precision number with an exponent at most~$4$ % and a first block less than $10^{8}$ to a fixed point number whose -% first block will have $12$~digits, hopefully starting with many -% zeros. +% first block has $12$~digits, hopefully starting with many zeros. % \begin{macrocode} \cs_new:Npn \@@_ep_to_fixed:wwn #1,#2 { @@ -779,7 +778,7 @@ % $\meta{d_1}\cdots\meta{d_6}$, the condition translates to % $\meta{n_1},\meta{d_1}\in[1000,9999]$. % -% We will first find an integer estimate $a \simeq 10^{8} / \meta{d}$ by +% We first find an integer estimate $a \simeq 10^{8} / \meta{d}$ by % computing % \begin{align*} % \alpha &= \eTeXfrac{10^{9}}{\meta{d_1}+1} \\ @@ -795,7 +794,7 @@ % \meta{d}$, while when $\meta{d_2}=9999$ one gets $a = % 10^{3}\alpha-1250 \simeq 10^{12} / (\meta{d_1} + 1) \simeq 10^{8} / % \meta{d}$. The shift by $1250$ helps to ensure that $a$ is an -% underestimate of the correct value. We will prove that +% underestimate of the correct value. We shall prove that % \[ % 1 - 1.755\cdot 10^{-5} < \frac{\meta{d}a}{10^{8}} < 1 . % \] @@ -812,8 +811,8 @@ % % Let us prove the upper bound first (multiplied by $10^{15}$). Note % that $10^{7} \meta{d} < 10^{3} \meta{d_1} + 10^{-1} (\meta{d_2} + 1)$, -% and that \eTeX{}'s division $\eTeXfrac{\meta{d_2}}{10}$ will at most -% underestimate $10^{-1}(\meta{d_2} + 1)$ by $0.5$, as can be checked +% and that \eTeX{}'s division $\eTeXfrac{\meta{d_2}}{10}$ underestimates +% $10^{-1}(\meta{d_2} + 1)$ by $0.5$ at most, as can be checked % for each possible last digit of \meta{d_2}. Then, % \begin{align} % 10^{7} \meta{d}a @@ -877,7 +876,7 @@ % reached for one of the extreme values $[y/10]=0$ or $[y/10]=100$, and % we easily check the bound for those values. % -% We have proven that the algorithm will give us a precise enough +% We have proven that the algorithm gives us a precise enough % answer. Incidentally, the upper bound that we derived tells us that % $a < 10^{8}/\meta{d} \leq 10^{9}$, hence we can compute $a$ safely as % a \TeX{} integer, and even add $10^{9}$ to it to ease grabbing of all @@ -916,7 +915,7 @@ % } % The \texttt{esti} function evaluates $\alpha=10^{9} / (\meta{d_1} + % 1)$, which is used twice in the expression for $a$, and combines the -% exponents |#1| and~|#4| (with a shift by~$1$ because we will compute +% exponents |#1| and~|#4| (with a shift by~$1$ because we later compute % $\meta{n}/(10\meta{d})$. Then the \texttt{estii} function evaluates % $10^{9} + a$, and puts the exponent~|#2| after the % continuation~|#7|: from there on we can forget exponents and focus @@ -1008,8 +1007,8 @@ % method: essentially $r \mapsto (r + 10^{8} / (x_{1} r)) / 2$, starting % from a guess that optimizes the number of steps before convergence. % In fact, just as there is a slight shift when computing divisions to -% ensure that some inequalities hold, we will replace $10^{8}$ by a -% slightly larger number which will ensure that $r^2 x \geq 10^{4}$. +% ensure that some inequalities hold, we replace $10^{8}$ by a +% slightly larger number which ensures that $r^2 x \geq 10^{4}$. % This also causes $r \in [101, 1003]$. Another correction to the above % is that the input is actually normalized to $[0.1,1)$, and we use % either $10^{8}$ or $10^{9}$ in the Newton method, depending on the |