Good day to everyone!

I have two (just for simple example) functions and they compose a map:

s[k_]:={#[[1]],#[[2]]-k2 #[[1]]^2}&

If I use this map in NestList all works nice:


But now I want to change the first argument t of m[t] at each (or with some step) iteration. For this I was hoping to use FoldList, but this does not work:


How can I use Composition in FoldList?


I'm not sure can I add some more wishes or should I post another question.

To speed-up iterations, I can Compile functions:

sc=Compile[{k2,{x,_Real,1}},{x[[1]],x[[2]]-k2 x[[1]]^2}]

Using With technique that @kglr advised in comment I can write:

FoldList[With[{xx = #2},RightComposition[mc[.2+xx,#]&,sc[2,#]&]],

This works. But what confuses me is that #2 is in purple and Composition does not work when replaced RightComposition. What is the difference in this case between Composition and RightComposition? Am I using Compile correctly?

  • 2
    $\begingroup$ you probably meant s[k2_]:=...? $\endgroup$
    – kglr
    Apr 16, 2021 at 12:35
  • 1
    $\begingroup$ FoldList[With[{xx = #2}, Composition[m[.2 + xx], s[2]]@#] &,...]? $\endgroup$
    – kglr
    Apr 16, 2021 at 12:49
  • $\begingroup$ @kglr, yes, that works! You can post this as answer, and porbably add explanation of why direct using #2 fails. $\endgroup$
    – macros
    Apr 16, 2021 at 14:24
  • $\begingroup$ @kglr, I edited question, could you give me some hints, or answers? $\endgroup$
    – macros
    Apr 16, 2021 at 15:49
  • 1
    $\begingroup$ Also, you haven't correctly understood @kglr 's comment. If I rewrite it as FoldList[Function[{foldarg1, foldarg2}, With[{xx = foldarg2}, Composition[m[.2 + xx], s[2]]@foldarg1]], ...], is it easier to understand? $\endgroup$
    – xzczd
    Apr 18, 2021 at 2:30


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