3
$\begingroup$

I am trying to define a recursive function with a subscript. Something similar would be $f_{n+1}(x) = \int 3 x f_{n}(x)dx$. I've tried lists, For function, NestList, and RecurrenceTable. What do you recommend?

$\endgroup$
2

2 Answers 2

3
$\begingroup$

Assume some $f_1(x)$ and then define a univariate recursion in terms of n:

f[1] := x + 1;
f[n_Integer /; n >= 2] := f[n] = Integrate[3 x f[n - 1], x]

For example:

f[5]

27/560 ((35 x^8)/8 + (16 x^9)/9)

$\endgroup$
3
$\begingroup$

Continuing with corey979's approach

Clear["Global`*"]

f[1] := x + 1;
f[n_Integer /; n >= 2] := f[n] = Integrate[3 x f[n - 1], x] // Simplify

Generating a sequence from the recursion

seq = f /@ Range[6] // FullSimplify

(* {1 + x, (3 x^2)/2 + x^3, 3/40 x^4 (15 + 8 x), 9/560 x^6 (35 + 16 x), 
     (3 x^8 (315 + 128 x))/4480, (9 x^10 (693 + 256 x))/98560} *)

Using FindSequenceFunction directly doesn't work. To simplify the problem, break the sequence into its component parts

seq2 = Simplify[seq/(x^(2 (# - 1)) & /@ Range[6])]

(* {1 + x, 3/2 + x, 9/8 + (3 x)/5, 9/16 + (9 x)/35, 27/128 + (3 x)/35, 
 81/1280 + (9 x)/385} *)

seq3 = Coefficient[#, x] & /@ seq2

(* {1, 1, 3/5, 9/35, 3/35, 9/385} *)

seq4 = seq2 - x*seq3

(* {1, 3/2, 9/8, 9/16, 27/128, 81/1280} *)

The original sequence is then represented by

f2[n_, x_] = (FindSequenceFunction[seq4, n] + 
      FindSequenceFunction[seq3, n]*x) x^(2 n - 2) // FunctionExpand // 
  FullSimplify

(* (6^(-1 + n) x^(-2 + 2 n) *
    (Sqrt[π] x Gamma[n] + 2 Gamma[1/2 + n]))/(Sqrt[π] Gamma[2 n]) *)

Verifying that f == f2 over a much broader range than the original sequence

And @@ Simplify[Table[f[n] == f2[n, x], {n, 1, 40}]]

(* True *)

f2 is not limited to integer values of n

Plot3D[f2[n, x], {n, 1/2, 10}, {x, 0, 2},
 AxesLabel -> (Style[#, 14] & /@ {"n", "x", "f"}),
 ClippingStyle -> None,
 ImageSize -> Medium]

enter image description here

EDIT: More succinctly, to go from seq to f2

f2[n_, x_] = (Total[FindSequenceFunction[#, n] & /@ 
  Transpose[List @@@ (seq // Expand)]] // FunctionExpand // FullSimplify)

(* (6^(-1 + n) (x^2)^(-1 + n) * 
  (Sqrt[π] x Gamma[n] + 2 Gamma[1/2 + n]))/(Sqrt[π] Gamma[2 n]) *)
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.