From d9ad9c695dcc0197dcec82930f840caa8d063d31 Mon Sep 17 00:00:00 2001
From: vlefevre <vlefevre@280ebfd0-de03-0410-8827-d642c229c3f4>
Date: Mon, 18 Dec 2017 12:22:26 +0000
Subject: [doc/algorithms.tex] mpfr_tanh: missing absolute value; added a
 \cdot.

git-svn-id: svn://scm.gforge.inria.fr/svn/mpfr/trunk@11999 280ebfd0-de03-0410-8827-d642c229c3f4
---
 doc/algorithms.tex | 4 ++--
 1 file changed, 2 insertions(+), 2 deletions(-)

(limited to 'doc')

diff --git a/doc/algorithms.tex b/doc/algorithms.tex
index 4cedf794a..4ab148f40 100644
--- a/doc/algorithms.tex
+++ b/doc/algorithms.tex
@@ -1865,10 +1865,10 @@ and then $s = \tanh(x) \cdot (1+\theta_4)^{2^k+4}$.
 
 \begin{lemma}
 For $|x| \leq 1/2$, and $|y| \leq |x|^{-1/2}$, we have:
-\[ |(1+x)^y-1| \leq 2.5 \cdot |y| \cdot x. \]
+\[ |(1+x)^y-1| \leq 2.5 \cdot |y| \cdot |x|. \]
 \end{lemma}
 \begin{proof}
-We have $(1+x)^y = e^{y \log (1+x)}$,
+We have $(1+x)^y = e^{y \cdot \log (1+x)}$,
 with $|y \cdot \log (1+x)| \leq |x|^{-1/2} \cdot \left|\log (1+x)\right|$.
 The function $|x|^{-1/2} \cdot \log (1+x)$ is increasing on $[-1/2,1/2]$, and
 takes as values $\approx -0.980$ in $x=-1/2$ and $\approx 0.573$ in $x=1/2$,
-- 
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