Re: sum of recursive fn: solving for n

*To*: mathgroup at smc.vnet.net*Subject*: [mg22117] Re: [mg22108] sum of recursive fn: solving for n*From*: BobHanlon at aol.com*Date*: Wed, 16 Feb 2000 02:34:33 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

f[1] = 2; f[x_] := 2*f[x - 1]; For an undefined n, the recursion for f[n] does not stop since it doesn't know when it gets to f[1]. Consequently, attempting to evaluate your Sum causes a recursion limit error. A brute force method is Select[Table[{n, Sum[f[x], {x, 1, n}]}, {n, 1, 10}], #[[2]] == 62 &][[1, 1]] 5 Alternatively, Needs["DiscreteMath`RSolve`"]; Clear[f, n]; RSolve[{f[x] == 2*f[x - 1], f[1] == 2}, f[x], x] {{f[x] -> 2^x}} Solve[Sum[(f[x] /. %[[1]]), {x, 1, n}] == 62, n] Solve::"ifun": "Inverse functions are being used by \!\(Solve\), so some \ solutions may not be found." {{n -> 5}} Bob Hanlon In a message dated 2/14/2000 3:25:09 AM, reply at newsgroup.please writes: >what am i doing wrong here? > >f[x_] := (f[x-1])*2 >f[1] =2 >Solve[Sum[f[x], {x, 1,n}] ==62, n] >