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Recursion without starting values

Harry Potter
Posted 10 years ago

r[0] := r0; r[1] :=r1; r[n + 1] := ar[n] + br[n - 1]; Table[r[n + 1], {n, -1, 12}]

all I get is only...

{r0, r1, r[2], r[3], r[4], r[5], r[6], r[7],
  r[8], r[9], r[10], r[11], r[12], r[13]}

but i need r[2], r[3], r[4] ... r[13] depending on r0 and r1 for example: 

r[2] = a b r0+ (a^2 + b) r1

thx for helping
POSTED BY: Harry Potter
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Harry Potter
Harry Potter
Posted 10 years ago

That was exactly what I needed. Thanks. Ta! ^^
POSTED BY: Harry Potter

Hi,

what about this?

r[0] = r0; r[1] = r1; r[n_] := a r[n - 1] + b r[n - 2]; Table[r[x], {x, 0, 12}]

Cheers,

Marco
POSTED BY: Marco Thiel

Hi Harry Potter,

this is a simple difference equation which can be solved by Mathematica:

ClearAll["Global`*"]
sol = First @ RSolve[{r[n + 1] == a r[n] + b r[n - 1], r[0] == r0, r[1] == r1}, r[n], n];
r[n_, a_, b_] = Simplify[r[n] /. sol];
Table[r[n, a, b], {n, 0, 13}]

In case this is a homework problem the solution alone is by no means sufficient, you need to supply the intermediate calculations anyway. This can be done in an elegant way in analogy to the calculation of a closed form for Fiboncci numbers ("matrix form").

Regards -- Henrik
POSTED BY: Henrik Schachner
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