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MathGroup Archive 2003

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Re: general solution for element of series

  • To: mathgroup at
  • Subject: [mg39991] Re: general solution for element of series
  • From: Paul Abbott <paul at>
  • Date: Fri, 14 Mar 2003 04:45:21 -0500 (EST)
  • Organization: The University of Western Australia
  • References: <b4k3uo$8hn$>
  • Sender: owner-wri-mathgroup at

In article <b4k3uo$8hn$1 at>,
 "Michael Beqq" <mbekkali at> 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;  

Since it is a continued fraction, you can use

 H[x_,j_]:=FromContinuedFraction[Join[{0}, Table[(-1)^i x, {i, 0, j-1}]]]


Paul Abbott                                   Phone: +61 8 9380 2734
School of Physics, M013                         Fax: +61 8 9380 1014
The University of Western Australia      (CRICOS Provider No 00126G)         
35 Stirling Highway
Crawley WA 6009                      mailto:paul at 

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