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

>
>>
>>
>

```

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