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I'm trying to create a table whose elements will be a sum over previously declared elements up to a certain iteration. I've reduced what I'm trying to do to the simplest form (posted below):

M[x_] := M[x] = Table[
   If[
    n == 1, r, 2 x^n Sum[M[x][[i]], {i, 1, n-1}]
    ],
   {n, 1, 5}]

When I evaluate the above expression, it produces an exceedingly long expression that is obviously wrong. What's the mistake that I'm making here?

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  • $\begingroup$ M[x] only references itself here, so it will be an infinite recursion. When you call M[1] for instance, the RHS contains a sum over the elements of M[1]. $\endgroup$ – Marius Ladegård Meyer Mar 5 '16 at 23:14
  • $\begingroup$ I'm sorry I don't understand. Why would it be an infinite recursion? Say for instance, we're on the iteratino n=2 then it should sum from i=1 to i=1 and the expression should be 2x^n M[x][[1]] where x could now be specified. Right? $\endgroup$ – user38332 Mar 5 '16 at 23:25
  • $\begingroup$ Each item M[x] is referenced it must call itself to determine its value. Table must complete and return a value to M[x] before M[x] can be assigned a value. Since the looping is happening inside of Table the calls in Table to M[x] are causing M[x] to be called again in a never ending loop. $\endgroup$ – Edmund Mar 5 '16 at 23:31
  • 1
    $\begingroup$ Have you seen RecurrenceTable[]? $\endgroup$ – J. M.'s ennui♦ Mar 5 '16 at 23:52
  • $\begingroup$ If I wanted to set the ith element to be the sum of all previous elements would RecurrenceTable[] be able to do the job? $\endgroup$ – user38332 Mar 6 '16 at 0:18
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I think FoldList is what you are looking for.

m[x_] := FoldList[#1 + #1 2 x^#2 &, Prepend[r]@Range[2, 3]]

Then

m[y]

{r, r + 2 r y^2, r + 2 r y^2 + 2 y^3 (r + 2 r y^2)}

Just need to change the 3 to a 5.

Hope this helps.

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