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I'm having a problem concerning function addition. The following code shows an example that illustrates my problem. I had this list of functions that have the same variable. I had them by using the Total function and declared a function of that variable equal to the Total[list], like in this example:

f1[x_] := 2 x + 1;
f2[x_] := x;
f3[x_] = 4 x;
l1 = List[f1[x], f2[x], f3[x]];
o[x_] := Total[l1];
o[1]

The output of o[1] was 1+7x, instead of 8, which is what I want. Any suggestion on how to approach this problem? Thank you

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  • $\begingroup$ If you do this, you'll get 8 f1[x_] := 2 x + 1; f2[x_] := x; f3[x_] = 4 x; l1[x_] := List[f1[x], f2[x], f3[x]]; o[x_] := Total[l1[x]]; o[1] $\endgroup$
    – Nasser
    Commented Oct 7, 2017 at 19:32
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    $\begingroup$ On top of what @Nasser gave you, the issue with your code has to do with how variables are scoped. The x_ parameter is scoped differently from the x symbol you provide in l1. So you could also mostly keep your code but replace o with this: o[v_] := Total[l1] /. x -> v $\endgroup$
    – b3m2a1
    Commented Oct 7, 2017 at 19:33
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    $\begingroup$ Or include Evaluate in definition of o[x_], i.e., o[x_] := Evaluate@Total[l1]; $\endgroup$
    – Bob Hanlon
    Commented Oct 7, 2017 at 20:03
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    $\begingroup$ @BobHanlon There is a shortcut for the combination of SetDelayed and Evaluate! $\endgroup$
    – Roman
    Commented Dec 22, 2022 at 9:04

2 Answers 2

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Try this (as a side note, when defining functions using SetDelayed one doesn't have to include the semi-colon at the end of the line).

f1[x_] := 2 x + 1
f2[x_] := x
f3[x_] := 4 x

Use SetDelayed (i.e., :=) again to defrine l1

l1[x_] := List[f1[x], f2[x], f3[x]]

Inside the defintion for o use l1 with an argument

o[x_] := Total[l1[x]]

Then test

o[1]
(* 8 *)

o[2]
(* 15 *)

This is similar but not quite identical to what Nasser wrote in his comment.

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  • $\begingroup$ Is it possible to get a specific element of the l1 list, now that it is a list of functions? $\endgroup$
    – RicardoP
    Commented Oct 7, 2017 at 23:30
  • $\begingroup$ l1[x][[2]] would, for example, get element 2 $\endgroup$
    – Jack LaVigne
    Commented Oct 8, 2017 at 2:52
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My new resource function ThroughOperator can do this:

f1[x_] := 2 x + 1;
f2[x_] := x;
f3[x_] := 4 x;
o = ResourceFunction["ThroughOperator"][f1 + f2 + f3];
o /@ Range[5]

{8, 15, 22, 29, 36}

If you have a list of functions or if somehow the expression f1 + f2 + f3 evaluates to something else inside of ResourceFunction["ThroughOperator"], you can also use:

list = {f1, f2, f3};
o = ResourceFunction["ThroughOperator"][list, Plus];
o /@ Range[5]
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  • $\begingroup$ Nice answer! What 's the advantage of this ResourceFunction compared to a list of pure functions {f1,f2,f3} and Total@Through[{f1, f2, f3}[5]]? $\endgroup$
    – Ulrich Neumann
    Commented Dec 23, 2022 at 15:20
  • $\begingroup$ @UlrichNeumann Through[{f1, f2, f3}[5]] is annoying to map over lists, which is one of the main motivations I made ThroughOperator. Another reason is that Through[h[f1, f2][x]] sometimes suffers from order-of-evaluation issues that are difficult to circumvent. $\endgroup$
    – Sjoerd Smit
    Commented Jan 9, 2023 at 10:47
  • $\begingroup$ Thank you for your explanation! $\endgroup$
    – Ulrich Neumann
    Commented Jan 9, 2023 at 10:56

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