I'm solving an NDSolve equation, like:

enito = 10;
sm = Range[enito];

For [a = 1, a < (enito + 1), a++, 

sm[[a]] = 
  NDSolve[{y''[x] + a*y[x] == 0, y[0] == 1, y'[0] == 1}, 
   y, {x, 0, 30}]]

And then I do:

valor = Range[enito];

For[a = 1, a < (enito + 1), a++, 

 valor[[a]] = 
With[{exp = Through[({y} /. First[sm[[a]]])[#]]}, exp &]];

Up to this point, I think everything is alright. Now, I'm interested in construct a list of functions with valor[[a]]. For example if I define:

F[x_] := x^2

I'm interested in have something like:

gr[[a]][x_] := valor[[a]]/F[x]

But I can't get anything similar. What can I use for obtain an list with the 10 elements of gr[[a]], making them a function (using F[x] for each one of them)? i.e, for have a list (each element, a different a) of functions of x?

I'd appreciate a lot all the answers. Thank you.

  • 2
    $\begingroup$ Minimalistic problems are nice. Make it clear, what is your goal. At the moment ti seems only the last line is relevant. $\endgroup$
    – Johu
    Jun 9, 2016 at 6:52
  • $\begingroup$ Okay, I'll try: I'm just searching for a way of get a list of functions. I'm interested in evaluate each a solution for the NDsolve, divided by x^2, and evaluated at a given x*. $\endgroup$ Jun 9, 2016 at 6:55
  • $\begingroup$ This does give you a list of functions, without me understanding any of your NDSolve stuff. fList = Table[a/#^2 & /. a -> aa, {aa, Range[1.22, 1.88, 0.11]}] $\endgroup$
    – Johu
    Jun 9, 2016 at 7:04
  • $\begingroup$ And if my function F[x_] is numerical, and I cannot get an analytical expression for it? $\endgroup$ Jun 9, 2016 at 7:09
  • $\begingroup$ If you would replace the last line with gr[a_,x_] := valor[[a]]/F[x], you would not be happy? $\endgroup$
    – Johu
    Jun 9, 2016 at 7:09

1 Answer 1


A simple solution is:


Can this help?

  • $\begingroup$ That was another way of solving my problem, thanks. $\endgroup$ Jun 9, 2016 at 7:17
  • 1
    $\begingroup$ But unlike mine this actually does, what is asked in the title of the question. You perhaps should accept this answer. $\endgroup$
    – Johu
    Jun 9, 2016 at 7:19

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