MathGroup Archive 2013

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

Search the Archive

Re: Through[Divide[f1, f2][x]]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg130938] Re: Through[Divide[f1, f2][x]]
  • From: Andrzej Kozlowski <akozlowski at gmail.com>
  • Date: Mon, 27 May 2013 04:24:04 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • Delivered-to: l-mathgroup@wolfram.com
  • Delivered-to: mathgroup-outx@smc.vnet.net
  • Delivered-to: mathgroup-newsendx@smc.vnet.net
  • References: <20130526090600.D7C18633A@smc.vnet.net>

On 26 May 2013, at 11:06, Ray Koopman <koopman at sfu.ca> wrote:

> Either of these will give you f1[x]/f2[x]:
> 
>  Divide@@{f1@#,f2@#}&@x
> 
>  Divide@@(#@x&)/@{f1,f2}
> 
> Sometimes it would be nice to have a "reverse map", that 
> applied each of a list of functions to a single argument, 
> instead of a single function to each of a list of arguments.

I am not sure I understand what you mean. Why doesn't Through do that?

Through[{f, g, h}[x]]

{f[x], g[x], h[x]}

Andrzej Kozlowski



  • Prev by Date: Re: How to extract from list
  • Next by Date: Re: How to use Mathematica find the solution of an
  • Previous by thread: Re: Through[Divide[f1, f2][x]]
  • Next by thread: Re: Through[Divide[f1, f2][x]]