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Here is a simple example to illustrate my problem:

expr = {a + b*Exp[-r t], a - b*Exp[-r t]}
(* {a + b E^(-r t), a - b E^(-r t)} *)

g[t_, b_] := Total[Evaluate[expr], {2}]

g[10, {b1, b2}]
(* {a + b E^(-r t), a - b E^(-r t)} *)

It looks is like Total[] undoes the Evaluate[] statement. Indeed:

g[10, 1]
(* {a + b E^(-r t), a - b E^(-r t)} *)

What I wanted and expected was the same output as when doing

h[t_, b_] := Evaluate[expr]
    
Total[h[10, {b1, b2}], {2}]  
(* {2 a + b1 E^(-10 r) + b2 E^(-10 r), 
     2 a - b1 E^(-10 r) - b2 E^(-10 r)} *)

I would like to understand why it can't be done the way I did, and how to do it instead (in the function definition). I am using Mathematica 13.3 on Windows.

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  • 2
    $\begingroup$ To make your test cases a bit simpler, look at : expr = x; f1[x_] := expr; f2[x_] := Evaluate[expr]; f3[x_] := f4[Evaluate[expr]]; {f1[0], f2[0], f3[0]} $\endgroup$
    – Domen
    Commented Oct 12, 2023 at 15:17
  • $\begingroup$ Right. I get {x, 0, f4[x]}, and it is the last output I don't understand. $\endgroup$
    – Sooner
    Commented Oct 12, 2023 at 15:20
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    $\begingroup$ Actually, it is the second output that should pop-out to you :) You may want to look at Definition[f1], Definition[f2] and Definition[f3]. You will see that in f2, expr evaluated to x already at the time you defined the function, while in the other two cases, expr remained intact, as expected. x in expr will then not be automatically replaced with the function argument x! There are various solutions, for example: g[t_, bb_] := Total[expr /. b -> bb, {2}] or Clear[expr]; expr[b_] := {a + b*Exp[-r t], a - b*Exp[-r t]}; g[t_, b_] := Total[expr[b], {2}] $\endgroup$
    – Domen
    Commented Oct 12, 2023 at 15:27
  • $\begingroup$ No, that is precisely why first and second do not pop out to me. They gave what I expected. For example for f2, the defining statement is already equivalent to f2[x_]: = x at the outset, so it is clear that f2[0]=0. I would have thought that in the same way, the definition for f3 is f3[x]:=f4[x], since I asked for the inside of f4 to be evaluated immediately. But as Bob Hanlon says below, it appears Evaluate is buried too deeply - which IMO somewhat defeats the purpose of it. $\endgroup$
    – Sooner
    Commented Oct 12, 2023 at 15:42
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    $\begingroup$ If you read the documentation for Evaluate, it states "Evaluate works only on the first level, directly inside a held function". This seems perfectly reasonable to me, because otherwise the evaluation process would need to parse the expression tree entirely looking for any Evaluates before moving on. $\endgroup$
    – lericr
    Commented Oct 12, 2023 at 16:16

1 Answer 1

Reset to default
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$Version

(* "13.3.1 for Mac OS X ARM (64-bit) (July 24, 2023)" *)

Clear["Global`*"]

expr = {a + b*Exp[-r t], a - b*Exp[-r t]};

g[t_, b_] := Total[Evaluate[expr], {2}]

The Evaluate is buried too deep to be evaluated when g is defined

?g

enter image description here

Consequently, the b and t in expr are not the same as the localized b and t in the subsequent evaluation of g. You would need

g[t2_, b2_] := Total[Evaluate[expr /. {t -> t2, b -> b2}], {2}]

?g

enter image description here

g[10, {b1, b2}]

(* {2 a + b1 E^(-10 r) + b2 E^(-10 r), 2 a - b1 E^(-10 r) - b2 E^(-10 r)} *)

But then the Evaluate is not necessary.

Clear[g]

g[t2_, b2_] := Total[expr /. {t -> t2, b -> b2}, {2}]

g[10, {b1, b2}]

(* {2 a + b1 E^(-10 r) + b2 E^(-10 r), 2 a - b1 E^(-10 r) - b2 E^(-10 r)} *)
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