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On 2010-08-26 15:39, Tobias Burnus wrote:
For norm2 (L2 norm), one essentially does: tmp = max(array) ! = L_infinity norm norm2 = tmp * sqrt( (array/tmp)**2) to avoid an overflow.
Just a random idea: the maximum precision that such a single pass algorithm can achieve is probably obtained by using and modifying the binary exponent instead of multiplying with a scale factor as this renders the scale transformations exact.
-- Toon Moene - e-mail: toon@moene.org - phone: +31 346 214290 Saturnushof 14, 3738 XG Maartensdijk, The Netherlands At home: http://moene.org/~toon/; weather: http://moene.org/~hirlam/ Progress of GNU Fortran: http://gcc.gnu.org/gcc-4.5/changes.html#Fortran
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