summaryrefslogtreecommitdiff
path: root/Master/texmf-dist/source/latex/l3kernel/l3fp-extended.dtx
diff options
context:
space:
mode:
Diffstat (limited to 'Master/texmf-dist/source/latex/l3kernel/l3fp-extended.dtx')
-rw-r--r--Master/texmf-dist/source/latex/l3kernel/l3fp-extended.dtx33
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