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MathGroup Archive 2004

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Re: Forcing Argument Evaluation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49873] Re: [mg49848] Forcing Argument Evaluation
  • From: DrBob <drbob at bigfoot.com>
  • Date: Thu, 5 Aug 2004 09:20:33 -0400 (EDT)
  • References: <200408041446.KAA20125@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

f is a function, in the sense that when Mathematica sees it with an argument--f[2], for instance--your definition matches that pattern, so Mathematica can substitute 2 for x in the right-hand side and get 4.

But f/2[2] doesn't match the pattern, so your definition doesn't take effect.

If you want the elements of "a" to be functions, you can do this:

f[x_] := x^2
a = {f, f[#1]/2 & , Function[y, f[y]/2]};
a[[2]][2]
2

a[[3]][3]
9/2

Here's a more complicated example:

a = NestList[Function[y, Evaluate[#1[y]^3/2]] & , f, 3]
{f, Function[y, y^6/2],
   Function[y, y^18/16],
   Function[y, y^54/8192]}

Last[a][y]
y^54/8192

Bobby

On Wed, 4 Aug 2004 10:46:25 -0400 (EDT), Scott Guthery <sguthery at mobile-mind.com> wrote:

> f[x_] := x^2;
> a = {f,f/2};
>a[[1]][2]
> 4
>a[[2]][[2]
> f/2[2]
>I know I'm missing something fundamental.
>Cheers, Scott
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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