Re: general solution for element of series

*To*: mathgroup at smc.vnet.net*Subject*: [mg39922] Re: general solution for element of series*From*: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>*Date*: Wed, 12 Mar 2003 02:28:03 -0500 (EST)*Organization*: Universitaet Leipzig*References*: <b4k3uo$8hn$1@smc.vnet.net>*Reply-to*: kuska at informatik.uni-leipzig.de*Sender*: owner-wri-mathgroup at wolfram.com

Hi, just define H[x_, 1] := 1/x H[x_, n_Integer] := H[x, n] = 1/(x - H[x, n - 1]) Regards Jens Michael Beqq wrote: > > Suppose I have expressions of x generated by some function f for any given > j: > > (j=1) => z=1/x=H[x,1] > (j=2) => z=1/(x-(1/x))=H[x,2] > (j=3) => z=1/(x-(1/(x-(1/x))))=H[x,3] > (j=4) => z=1/(x-1/(x-(1/(x-(1/x)))))=H[x,4] > ........... > (j=j) => z= H(x,j) > > I would like to know how I can find the function that generates this > sequence for some particular element j, that is, for any j=1,....,J I can > express z as a function of x and j; for example, if j=2 I would like to get > z=1/(x-(1/x)), while if j=3 I would like to get z=1/(x-(1/(x-(1/x)))). > There must be a way to do that, at least I hope so, because there exist > operators in Mathematica 4.0 (full version) that allow one to get any > element, z, of the sequence if one knows the generating function, f, and the > element's number, j (in short, I need to find the functional form of > H[x,j]). > > I hope I was clear enough :) > > Please help. Michael Beqq