MathGroup Archive 2010

[Date Index] [Thread Index] [Author Index]

Search the Archive

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

  • To: mathgroup at
  • Subject: [mg109176] Re: Through[(a+b+b)[x]]
  • From: Andrzej Kozlowski <akoz at>
  • 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

  • Prev by Date: BIOKMOD 5.0, a Mathematica tool for biokinetic modeling, fitting and
  • Next by Date: integrate log*sinc
  • Previous by thread: Through[(a+b+b)[x]]
  • Next by thread: Re: Through[(a+b+b)[x]]