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Topic: Through[Divide[f1, f2][x]]
Replies: 3   Last Post: May 28, 2013 3:48 AM

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Ray Koopman

Posts: 3,383
Registered: 12/7/04
Re: Through[Divide[f1, f2][x]]
Posted: May 28, 2013 3:48 AM
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On Sun, May 26, 2013 @ 11:23 AM, Andrzej Kozlowski <> wrote:
> On 26 May 2013, at 11:06, Ray Koopman <> 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

Through does do that. When I saw Bob's response to the original post
I asked myself "How did I miss that?" I thought I had tried it and
found that it didn't work. The only explanation I can think of is
that I had the "reverse map" metaphor fixed too firmly in my mind,
to the point that I used Map-like syntax [{f1,f2},x] instead of

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