Re: Confusing results with N[expr]?

*To*: mathgroup at smc.vnet.net*Subject*: [mg62391] Re: Confusing results with N[expr]?*From*: tt at tt.com*Date*: Wed, 23 Nov 2005 01:12:07 -0500 (EST)*Organization*: National Research Council, Ottawa, Canada*References*: <dlp320$1bs$1@smc.vnet.net> <dlup9p$n95$1@smc.vnet.net>*Reply-to*: tt at tt.com*Sender*: owner-wri-mathgroup at wolfram.com

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[1]:= >$Version > >Out[1]= >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[2]:= >Map[f, a^b] > >Out[2]= > f[b] >f[a] > >In[3]:= >f /@ a^b > >Out[3]= > b >a >