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Re: Confusing results with N[expr]?
In the following code, just put parenthesis around a^b and you get the good result. f/@ a^b is the same as Map[f,a]^b (and Map[f,a]=a) f/@(a^b) is the same as Map[f,a^b] /@ has precedence over ^. GL On Tue, 22 Nov 2005 09:44:25 +0000 (UTC), Steven HANCOCK <Steven.Hancock at cern.ch> wrote: >There may be a clue in the following, equally distressing nonsense... > >In:= >$Version > >Out= >5.1 for Microsoft Windows (October 25, 2004) > >(* Map[f, expr] or f /@ expr applies f to each element on the first >level in expr. *) > >In:= >Map[f, a^b] > >Out= > f[b] >f[a] > >In:= >f /@ a^b > >Out= > b >a >