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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: [mg39935] Re: [mg39914] general solution for element of series
  • From: Dr Bob <drbob at bigfoot.com>
  • Date: Wed, 12 Mar 2003 02:30:57 -0500 (EST)
  • References: <200303110738.CAA08713@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

Here are a couple of approaches:

Clear@h
h[1][x_] := 1/x
h[n_Integer][x_] /; n > 1 := 1/(x - h[n - 1][x])
h[#][x] & /@ Range[5]

or

Clear@h
h[x_, n_Integer?Positive] := Nest[1/(x - #) &, 0, n]
h[x, #] & /@ Range[5]

The second approach is probably better.

Bobby

On Tue, 11 Mar 2003 02:38:35 -0500 (EST), Michael Beqq 
<mbekkali at iastate.edu> 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
>
>
>
>
>



-- 
majort at cox-internet.com
Bobby R. Treat



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