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MathGroup Archive 2003

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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


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