I believe that lambert_w calculates the principal branch, which is denoted
W_0. What you are looking for is W_{-1}. Studying the paper by Corless, I
don't think there's a readily usable relationship between W_0 and W_{-1}.
However, using the paper of Chapeau-Blondeau and Monir, one can easily
construct a simple numerical implementation of W_{-1}:
W1(z):=block([w1,w2,p:-sqrt(2*float(%e)*z+2)],w1:((((((-221/8505*p)+769/1728
0)*p-43/540)*p+11/72)*p-1/3)*p+1)*p-1,while(w1#w2) do
(w2:w1,w1:float(log(z/w1))),w1);
Do give it a try, I tested it by verifying that W1(z)*exp(W1(z)) gives back
z for -1/%e<z<0, so it seems to do the trick.
Viktor
-----Original Message-----
From: maxima-bounces at math.utexas.edu [mailto:maxima-bounces at math.utexas.edu]
On Behalf Of Oliver Kullmann
Sent: Thursday, May 22, 2008 7:37 PM
To: Barton Willis
Cc: maxima at math.utexas.edu
Subject: Re: [Maxima] Lambert W function?
Perhaps I've missed it, but what
branch of the Lambert-W-function does
lambert_w represent?
Apparently only this one selected branch
is available? I need the function in
the interval [-1/e, 0[, and I need
exactly the other real-valued branch than
the one computed by lambert_w. How
to get that?
Oliver
On Tue, May 20, 2008 at 01:11:24PM -0500, Barton Willis wrote:
> The function "lambert_w" is new to 5.15.0 (thanks due to Raymond and
> Stavros).
> The solve doesn't know about lambert_w. So far, lambert_w evaluates for
> doubles (but not big floats):
>
> (%i5) lambert_w(5.60);
> (%o5) 1.392014569377103
> (%i6) diff(lambert_w(x),x);
> (%o6) %e^(-lambert_w(x))/(lambert_w(x)+1)
> (%i9) taylor(lambert_w(x),x,0,4);
> (%o9) x-x^2+(3*x^3)/2-(8*x^4)/3+...
>
> Barton
>
--
Dr. Oliver Kullmann
Computer Science Department
Swansea University
Faraday Building, Singleton Park
Swansea SA2 8PP, UK
http://cs.swan.ac.uk/~csoliver/
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