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
- References:
- general solution for element of series
- From: "Michael Beqq" <mbekkali@iastate.edu>
- general solution for element of series