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MathGroup Archive 2002

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Re: Replacement question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg35143] Re: [mg35114] Replacement question
  • From: Andrzej Kozlowski <andrzej at lineone.net>
  • Date: Tue, 25 Jun 2002 19:55:57 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

How about something like:

In[1]:=
h[x_] := x^3; f[x_] := x^4;

In[2]:=
h[3]/f[2]

Out[2]=
27/16

In[3]:=
Block[{h, f}, In[2] /. h[3] -> h[5]]

Out[3]=
125/16

Andrzej Kozlowski
Toyama International University
JAPAN
htt


On Tuesday, June 25, 2002, at 04:41  PM, ginak wrote:

> Suppose I have something like
>
>  In [100] := h[3]/f[2.33342]
>
>  Out[100] := 24.12711
>
> and now I want to evaluate h[5]/f[2.33342], i.e. the same as in
> In[100] but replaceing h[3] with h[5].  This won't work
>
>  In [101] := In[100] /. h[3]->h[5]
>
>  Out[101] := 24.12711
>
> because In[100] is fully evaluated to 24.12711 before the rule is
> applied.  (Generally, the expressions I'm interested in are more of a
> pain to type than h[5]/f[2.33342]).  How do I tell Mathematica to 
> evaluate
> In[100] only enough to apply the given substition rule, apply the
> substitution rule, and only then proceed with the evaluation?
>
> In fact, I don't even know how to do a replacement like
>
>  In [102] := h[3]/f[2.33342] /. h[3]->h[5]
>
> for the same reason: the LHS is evaluated before the rule can be
> applied.  (Of course, in this case this replacment task is pointless,
> since it is so easy to type out the desired expression, but there are
> situations in which one can obtain a complicated expression by cutting
> and pasting, and wants to apply a substitution rule to the complicated
> expression before Mathematica evaluates it.)
>
> Thanks!
>
> G.
>
>
>
>
p://platon.c.u-tokyo.ac.jp/andrzej/



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