Re: Re: Confusing results with N[expr]?
- To: mathgroup at smc.vnet.net
- Subject: [mg62384] Re: [mg62353] Re: Confusing results with N[expr]?
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Wed, 23 Nov 2005 01:12:02 -0500 (EST)
- References: <dlp320$1bs$1@smc.vnet.net> <200511220941.EAA23606@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
On 22 Nov 2005, at 18:41, Steven HANCOCK wrote: > There may be a clue in the following, equally distressing nonsense... I don't think there is any relation (nor is it, for that matter, very distressing). > > In[2]:= > Map[f, a^b] > > Out[2]= > f[b] > f[a] > > In[3]:= > f /@ a^b > > Out[3]= > b > a > This is because f/@ a^b is simply Map[f,a]^b. But f /@ (a^b) f[a]^f[b] > (* MapAll[f, expr] or f //@ expr applies f to every subexpression in > expr. *) > > In[4]:= > MapAll[f, a^b] > > Out[4]= > f[b] > f[f[a] ] > > In[5]:= > f //@ a^b > > Out[5]= > b > f[a] > Again: In[5]:= f //@ (a^b) Out[5]= f[f[a]^f[b]] Andrzej Kozlowski > >> >> >
- References:
- Re: Confusing results with N[expr]?
- From: Steven HANCOCK <Steven.Hancock@cern.ch>
- Re: Confusing results with N[expr]?