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authorvlefevre <vlefevre@280ebfd0-de03-0410-8827-d642c229c3f4>2017-12-18 12:22:26 +0000
committervlefevre <vlefevre@280ebfd0-de03-0410-8827-d642c229c3f4>2017-12-18 12:22:26 +0000
commitd9ad9c695dcc0197dcec82930f840caa8d063d31 (patch)
treee0a0361674ce396f2041b1309774ad8d155c6e01
parentd5677d5bd42a35b5060396d9d38da00f2cf1650c (diff)
downloadmpfr-d9ad9c695dcc0197dcec82930f840caa8d063d31.tar.gz
[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
-rw-r--r--doc/algorithms.tex4
1 files changed, 2 insertions, 2 deletions
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 $x=1/2$,