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Let $f$ be a function of $n$. I would like to define a function $A$ with module which takes $f[n]$ as input. So for example:

f[n_]:=f[n]=(n^2+1)/n;
A[f[n]_]:=A[f[n]]=Module[{d},
d[n_]=d[n]=Numerator[f[n-1]];
Return[d[n]]
]

My code above does not work. I want f[n] still be a function of n inside the module but it seems I did it wrong. Any help is appreciated.

Chen M Ling
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    Instead of a function defined with a pattern like f[n_], you can try f = Function[n, (n^2 + 1)/n]. Then pass both f and n to A: A[f_Function, n_] := ... – Marius Ladegård Meyer Jun 15 '16 at 06:08
  • @MariusLadegårdMeyer. I think I cannot do that because actually f is another output from another function. But thanks for the answer :D – Chen M Ling Jun 15 '16 at 08:32

1 Answers1

2

Do you mean:

ClearAll[A]
Attributes@A = HoldAll;
A[f[n_]] := A[f[n]] = Module[{d}, d = Numerator@f[n - 1]]

I've removed the redundant Return. Module is also irrelevant here actually. Anyway, I suppose you need Module in your real problem so don't take it away.

Or you need f to be arbitrary, too? Then:

ClearAll[A]
Attributes@A = HoldAll;
A[f_[n_]] := A[f[n]] = Module[{d}, d = Numerator@f[n - 1]]

f[n_] := f[n] = (n^2 + 1)/n;
g[n_] := g[n] = (n^2 + 2)/n;

A[g[3]]
(* 3 *)
A[f[3]]
(* 5 *)

A // DownValues
(* {HoldPattern[A[f[3]]] :> 5, HoldPattern[A[g[3]]] :> 3, 
    HoldPattern[A[f_[n_]]] :> (A[f[n]] = Module[{d}, d = Numerator[f[n - 1]]])} *)
xzczd
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  • +1 for the nice examples, but if interpretation is correct, and I think it is, this should probably be closed as a duplicate. – Mr.Wizard Jun 15 '16 at 06:36
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    @Mr.Wizard Yeah, I believe it's a duplicate, too, but posts about functions are really hard to search 囧. Hope someone can find the duplicate post. – xzczd Jun 15 '16 at 06:40
  • Oh. This works! But when I removed Attributes@A = HoldAll; it did not work. What is that for? – Chen M Ling Jun 15 '16 at 08:30
  • Oh when I make $d$ a function in $n$ too, it doeas not work. A[f_[n_]] := A[f[n]] = Module[{d}, d[n_]:=d[n] = Numerator@f[n - 1]] doesn't work.

    Sorry, I do not know how to format a code in a comment.

     – Chen M Ling Jun 15 '16 at 08:48
  • @ChenMLing To understand the meaning of Attributes@A = HoldAll;, just check the document of Attributes and HoldAll, you may also feel this post interesting. As to the d[n] part, I don't quite understand what you mean. What output are you expecting? – xzczd Jun 15 '16 at 10:37
  • @xzczd I am expecting the output will be a function of n as well, so it will Return[d[n]]. I have edited my question :) – Chen M Ling Jun 15 '16 at 12:14
  • @ChenMLing Are you expecting to create a function d[n] that can be used outside e.g. something like A[f[n]]; d[3] (*output: 5*)? Then you should not localize d with Module. Or you just want a expression rather than function? Then you don't need to define function inside the Module. BTW formatting code in comment is the same as formatting in post. – xzczd Jun 15 '16 at 12:19
  • @xzczd, my real code takes function of n as input and gives function of n as output. What I have learnt is to use Module to write big function, and return what we are expected. I don't know how to do it without Module. – Chen M Ling Jun 15 '16 at 12:28
  • @ChenMLing Module is a function for localization i.e. isolating the possible influence from outside. Writing big function does often require using Module etc. but it doesn't mean you can use it blindly. Check documentation of Module for more details. To be blunter: don't use Module if you don't understand what you're getting from it. In your case what you need is simply: d[n_] := d[n] = Numerator@f[n - 1]. – xzczd Jun 15 '16 at 12:38
  • @xzczd. OK, understand :) Thank you very much x :) – Chen M Ling Jun 15 '16 at 13:14