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I am trying to take a free-form input Function to create a list using NestList.

  Manipulate[list2 = NestList[f, .8, 10], {f, (1/3) x^3, InputField[_, String]}, 
     ControlType -> InputField]

Which generates this output:

{0.8, ((x^3)/3)[0.8], ((x^3)/3)[((x^3)/3)[0.8]], ((x^3)/ 3)[((x^3)/3)[((x^3)/3)[0.8]]], ((x^3)/ 3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[0.8]]]], ((x^3)/ 3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[0.8]]]]], ((x^3)/ 3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[0.8]]]]]], (( x^3)/3)[((x^3)/ 3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[0.8]]]]]]], (( x^3)/3)[((x^3)/ 3)[((x^3)/ 3)[((x^3)/ 3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[0.8]]]]]]]], ((x^3)/ 3)[((x^3)/ 3)[((x^3)/ 3)[((x^3)/ 3)[((x^3)/ 3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[0.8]]]]]]]]], ((x^3)/ 3)[((x^3)/ 3)[((x^3)/ 3)[((x^3)/ 3)[((x^3)/ 3)[((x^3)/ 3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[((x^3)/3)[0.8]]]]]]]]]]}

I would like to have a numeric output.

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  • $\begingroup$ The problem is in function definition, x does not mean anything in this form. Try (1/3) #^3 & $\endgroup$ – Kuba Jun 13 '13 at 19:15
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As I have mentioned in comment the problem is that (1/3)x^3 is not a proper function definition. Applying this to Your starting value will give You, as one can expect:

(* => (1/3)x^3 [.8]*)

Natural way is to use pure function form for it: (1/3)#^3&.

Manipulate[
 list2 = NestList[f, .8, 10], 
 {f, (1/3) #^3 &, InputField[_]}, ControlType -> InputField]

enter image description here

But it does not look good in InputField, it is also not convenient to type. The following solution will fix that:

Manipulate[
 list2 = NestList[f /. x -> # &, .8, 10],
 {f, (1/3) x^3, InputField[_]}, ControlType -> InputField]

enter image description here

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  • $\begingroup$ What happens if the user enters 42 + u^3? :P $\endgroup$ – m_goldberg Apr 18 '14 at 22:57
  • $\begingroup$ @m_goldberg well, nothing special, I;m not sure what do you mean :P $\endgroup$ – Kuba Apr 19 '14 at 6:45

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