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Re: Controlled evaluation of functions


The reason Sin doesn't further evaluate is that you have exact expressions
and Mathematica knows no further rules for it. But make i in your Table an
approximate number and Sin will evaluate and behave just like your k.

So if you don't want k to evaluate, don't give it a definition! Or rather,
make it a rule and then use the rule when you want evaluation. And in
general, if you want controlled evaluation use rules instead of definitions.

krule = k -> (#^2 &);
f[i_, x_] := k[i x]

g[x_] = Table[f[i, x], {i, 3}]
{k[x], k[2 x], k[3 x]}

{3, 0, 1}.g[y]
% /. krule
3 k[y] + k[3 y]

David Park
djmp at

From: Brett Patterson [mailto:muckle.moose at]
To: mathgroup at

Consider the following behaviour:

  In[1]:= f[i_, x_] := Sin[i x]

  In[2]:= g[x_] = Table[f[i, x], {i, 3}]

  Out[2]= {Sin[x], Sin[2 x], Sin[3 x]}

  In[3]:= {3, 0, 1} . g[y]

  Out[3]= 3 Sin[y] + Sin[3 y]

This is what I want to do, but using my own function instead of Sin.
However, this is the result:

  In[4]:= k[x_] := x^2           (* This is my alternative to Sin *)

  In[5]:= f[i_, x_] := k[i x]

  In[6]:= g[x_] = Table[f[i, x], {i, 3}]

  Out[6]= {x^2, 4 x^2, 9 x^2}    (* I want {k[x], k[2 x], k[3 x]} *)

  In[7]:= {3, 0, 1} . g[y]

  Out[7]= 12 y^2                 (* I want 3 k[y] + k[3 y] *)

How can I get the function k to behave like Sin, so that it is not

Note that in my real application, k is a lot more complex and has
conditions on its arguments, etc.


Brett Patterson

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