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Example 1

F(0,t)=t
F(n+1,t)=F(n,2*t)

Example 2

F(0,y,t)=t
F(n+1,y,t)=F(n,y+1,t*y)

When I tried to manually solve above equation for close form solution and my findings are as follows(correct me if m wrong)

F(n,t)=2^n*t

F(n,y,t)=t*y*(y+1)*....*(y+n)

When I tried in Mathematica using RSolve, I got the following results

RSolve[{F[n + 1, t] == F[n, 2 t], F[0, t] == t}, F[n, t], n]

No Result

RSolve[{F[n+1,y,t]==F[n,y+1,y*t], F[0,y,t]==t}, F[n, t], n]

No Result

Any other-way to get the expected result?

Looking forward for suggestion and correction ..

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  • $\begingroup$ I had tried using RSolve and got the results as shown above, however, I did not mention it in the previous version of the question. i have updated the question. Thank you $\endgroup$ – Tom Apr 28 '16 at 14:02
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This is a "computer assisted" solution rather than certified closed form but it may satisfy your needs.

F1[n_, t_] := F1[n, t] = If[n >= 1, F1[n - 1, 2 t], t];
Table[F1[n, t], {n, 0, 10}]
FindSequenceFunction[%]

output :

{t, 2 t, 4 t, 8 t, 16 t, 32 t, 64 t, 128 t, 256 t, 512 t, 1024 t}
2^(-1 + #1) t &

and

F2[n_, y_, t_] := F2[n, y, t] = If[n >= 1, F2[n - 1, y + 1, t y], t]
Table[F2[n, y, t], {n, 0, 10}]
FindSequenceFunction[%]

output:

{t, t y, t y (1 + y), t y (1 + y) (2 + y), 
 t y (1 + y) (2 + y) (3 + y), t y (1 + y) (2 + y) (3 + y) (4 + y), 
 t y (1 + y) (2 + y) (3 + y) (4 + y) (5 + y), 
 t y (1 + y) (2 + y) (3 + y) (4 + y) (5 + y) (6 + y), 
 t y (1 + y) (2 + y) (3 + y) (4 + y) (5 + y) (6 + y) (7 + y), 
 t y (1 + y) (2 + y) (3 + y) (4 + y) (5 + y) (6 + y) (7 + y) (8 + y), 
 t y (1 + y) (2 + y) (3 + y) (4 + y) (5 + y) (6 + y) (7 + y) (8 + 
    y) (9 + y)}
t Pochhammer[y, -1 + #1] &
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