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Re: Through[(a+b+b)[x]]

• To: mathgroup at smc.vnet.net
• Subject: [mg109176] Re: Through[(a+b+b)[x]]
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Fri, 16 Apr 2010 05:49:17 -0400 (EDT)

On 15 Apr 2010, at 12:13, Derek Yates wrote:

> Through[(a+b)[x]] yields a[x]+b[x] as expected, but Through[(a+b+b)
> [x]] yields a[x]+(2b)[x]. Through[(2b)[x]] yields 2[x]b[x]. Now, I can
> obviously get around this in this specific case, but generically is
> there a way to solve this so that Through[(a+b+b)[x]] yields a[x]
> +2b[x]? The case where I envisage this happening is when a sum of
> functions is supplied (say, for a given value of y, Through[(f[y]+g[y]
> +h[y]+j[y])[x]] and for some values of y, g == h. Then one will end up
> with the problem above. Other than some post processing using pattern
> matching, which feels a bit clunky, I can't think of a way around this.
>

The only way I can see of doing this without some post-processing is by using Unevaluated:

Through[Unevaluated[(a + b + b)[x]]]

a(x)+2 b(x)

Andrzej Kozlowski