
Jun 29th, 2012, 06:31 PM
#1
Thread Starter
New Member

Feb 11th, 2013, 09:40 AM
#2
New Member

Mar 26th, 2014, 11:52 PM
#3
New Member
Re: Lambert W function for Excel, work on real and complex number
Thank you also.
One minor quibble: there seems to be an issue with the results that come from input values between 0 and 0.0212. As soon as it hits 0.0212 it's like hitting a wall, beyond that the results are wildly off:
0.0218 > 0.02229
0.0217 > 0.02219
0.0216 > 0.02208
0.0215 > 0.02198
0.0214 > 0.02187
0.0213 > 0.02177
0.0212 > 5.51714
0.0211 > 5.57716
...and so on, all input values from there to zero are similarly off.
Can anyone see why? Unfortunately I just lack the coding skills to pick it apart  gave it a go, but just no experience there whatsoever :(
Any help much appreciated.
Last edited by Chipmunker; Mar 27th, 2014 at 01:40 PM.

Mar 30th, 2014, 01:05 PM
#4
Thread Starter
New Member
Re: Lambert W function for Excel, work on real and complex number
See: http://www.had2know.com/academics/la...alculator.html
"If x is within the sub interval (1/e, 0), the calculator returns two values."
And for x = 0.0211, the calculator returns 5.577162 and 0.02156, both values are valid.
In Excel, the function return only 1 of them, I will check if i can return the more suitable one.

Mar 30th, 2014, 03:00 PM
#5
New Member
Re: Lambert W function for Excel, work on real and complex number
Aha, that all makes sense. Thank you.
Please don't spend any more time on my account  now the dual correct answers are apparent I can work with that.
Thanks again.

Feb 4th, 2015, 04:12 AM
#6
Registered User
Re: Lambert W function for Excel, work on real and complex number
I searched high and low for access top a vba routine for Lambert W.
This was the best  BUT it sometimes gives the ve branch solutions instead of the +ve.
It is also complex which i do not need.
For others I will post my solution here.
It returns the +ve branch solution only, is accurate to 9dp for most of the useful space and is reasoanbly faast
Here it is and hope others find it useful.
Bob J.
Code:
Public Function myLambertW(x As Double) As Double
'
' Function provided on the web seems to have starting problems and uses complex Nos where I only need real
' This version solves for the upper branch of the solution only
'
' It uses standard Newton iteration
'
' The starting value is log(x+1) which is within 40% of target from 0.33 < x < 10^100
'
' Results are accurate to about 9dp from 0.3678 < x < 10^100
'
' Values for second Branch are not computed ie 0.36788... < x < 0, W(x) < 1
'
' There is some potential to reduce iterations for smaller +ve x values
' but may need to increase as x approaches 1
'
' RB Jordan 3/2/2015
'
Dim xTry As Double
Dim Iter As Integer
If x < Exp(1) Then
myLambert = CVErr(xlErrValue)
Exit Function
End If
xTry = Log(1 + x)
For Iter = 1 To 9
xTry = xTry  (xTry  x / Exp(xTry)) / (1 + xTry)
Next Iter
myLambertW = xTry
End Function
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