summaryrefslogtreecommitdiff
path: root/doc
diff options
context:
space:
mode:
authorenge <enge@211d60ee-9f03-0410-a15a-8952a2c7a4e4>2012-09-17 12:17:16 +0000
committerenge <enge@211d60ee-9f03-0410-a15a-8952a2c7a4e4>2012-09-17 12:17:16 +0000
commit27620e76c573a1899d69ee534a0d2fb91c062c71 (patch)
tree231ac0134da8dc405e91b07e61f9b5142ff51176 /doc
parentdb579a51a75f18aa21b688c076b1e47557833728 (diff)
downloadmpc-27620e76c573a1899d69ee534a0d2fb91c062c71.tar.gz
algorithms.tex: layout
git-svn-id: svn://scm.gforge.inria.fr/svn/mpc/trunk@1265 211d60ee-9f03-0410-a15a-8952a2c7a4e4
Diffstat (limited to 'doc')
-rw-r--r--doc/algorithms.tex15
1 files changed, 10 insertions, 5 deletions
diff --git a/doc/algorithms.tex b/doc/algorithms.tex
index f3707fa..b7002bd 100644
--- a/doc/algorithms.tex
+++ b/doc/algorithms.tex
@@ -1782,19 +1782,24 @@ in the place of $n - 1$.
We assume we compute the univariate AGM defined by $\AGM(1,z)$.
It is easy to see
-that after one AGM iteration the values $\circ((1+z)/2)$ and
-$\circ(\sqrt{z})$ are in the same quadrant (in the first quadrant if the
+that after one AGM iteration the values
+$\round \left( \frac {1+z}{2} \right)$ and
+$\round (\sqrt{z})$ are in the same quadrant (in the first quadrant if the
imaginary part of $z$ is nonnegative, in the fourth quadrant otherwise).
Taking into account symmetries, we thus assume $z$ is in the first quadrant.
Let $\corr {a_n}, \corr {b_n}$ be the values of the AGM iteration performed
-with infinite precision, and $\appro {a_n}, \appro {b_n}$ those performed with $p$-bit precision
+with infinite precision, and
+$\appro {a_n}, \appro {b_n}$ those performed with $p$-bit precision
and rounding towards $+\infty$. We have $\corr {a_0} = a_0 = 1$ and
$\corr {b_0} = b_0 = z$ (assuming $z$ is exactly representable in precision
$p$).
-Each iteration computes $\appro {a_n} = \circ((\appro {a_{n-1}} + \appro {b_{n-1}})/2)$ and
-$\appro {b_n} = \circ(\sqrt{\appro {a_{n-1}} \appro {b_{n-1}}})$.
+Each iteration computes
+$\appro {a_n} = \round \left(
+\frac {\appro {a_{n-1}} + \appro {b_{n-1}}}{2} \right)$ and
+$\appro {b_n} = \round \left(
+\sqrt {\appro {a_{n-1}} \appro {b_{n-1}}} \right)$.
Assume by induction we can write
$\appro {a_{n-1}} = \corr {a_{n-1}} (1 + \mu_{n-1})^{e_{n-1}}$
and $\appro {b_{n-1}} = \corr {b_{n-1}} (1 + \nu_{n-1})^{e_{n-1}}$