Lambert W-1 function
06-17-2021, 10:21 PM
Post: #3
 Albert Chan Senior Member Posts: 2,142 Joined: Jul 2018
RE: Lambert W-1 function
A good guess for e^W(x,-1) is 2*x^2 ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ , where x = [-1/e, 0]

We can setup Newton's method to get e^W(x,-1):

> X := −0.1
> 2*X*X ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ → 0.02
> (X+Ans)/(1+LN(Ans)), then Enter, Enter, Enter ...

2.74723104353ᴇ−2
2.79535734165ᴇ−2
2.79551995963ᴇ−2
2.79551996147ᴇ−2 ﻿ ﻿ ﻿ // = e^W(-0.1,-1)

> LN(Ans), or X/Ans
−3.57715206396 ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ // = W(-0.1,-1)

Or, Newton's method straight for W(x,-1), solving f(w) = w + ln(w/x) = 0

> LN(2*X*X) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ → −3.91202300543
> LN(e*X/Ans)*Ans/(Ans+1), then Enter, Enter, Enter ...

−3.58237415547
−3.57715353929
−3.57715206396 ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ // = W(-0.1,-1)
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 Messages In This Thread Lambert W-1 function - robmio - 06-04-2021, 01:36 PM RE: Lambert W-1 function - Stevetuc - 06-04-2021, 01:50 PM RE: Lambert W-1 function - Albert Chan - 06-17-2021 10:21 PM

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