The LambertW Function

Friday 9 April 2004 at 2:35 pm | In Articles | 1 Comment

Many equations cannot be solved exactly without using special functions. For example, to solve 2^x=3 requires the use of the \ln function (or similar). This function is sometimes defined in terms of an integral from which their properties can be deduced. Thus \ln is defined by \displaystyle \ln x=\int ^x _1 \dfrac{dt}{t}\text{ for } x>0 and it is then clear that, for example, \ln 1=0

There are many equations that can only be solved in terms of newly-defined functions. One such function that isn’t all that well known is the LambertW function where w\left(x\right) is defined as a solution (for w) of we^w=x. This allows you to solve equations like 2^x=x^8 which was asked about on the S.O.S. Mathematics CyberBoard

To solve 2^x=x^8 let w=-\ln x so that x=e^{-w}. Then

2^x=x^8 \Longrightarrow 2^{e^{-w}}=e^{-8w} \Longrightarrow e^{-w}\ln 2=-8w \Longrightarrow -\frac{1}{8}\ln 2=we^w

Thus w=w(\frac{1}{8}\ln 2) and so x=e^{-w}=-8\dfrac{w(\frac{1}{8}\ln 2)}{\ln 2} which is our answer.

Using tables or software this gives 1.100.

But hang on, is that the only solution? No, because 2^x<x ^8 for small values of x and 2^x grows much faster than x^8 so 2^x>x^8 for large values of x. Since both x \mapsto 2^x and x \mapsto x^8 are continuous on \mathbb{R} there is another value of x for which 2^x=x^8. A quick fiddle with a calculator gives x\approx43.5.

Research into the LambertW function to find out how this other solution can be given in terms of this function.

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  1. Hi,
    I want to solve an equation with the MATLAB programing language. It’s here:

    I = c1*I + c2 + A1*(exp(q*(V+I*Rs)/(k*T))-1) + A2*(exp (q*(V+I*Rs)/(2*k*T))-1)

    (I must find “I” )

    Could you possibly help me? Is there any special command to solve it in MATLAB tools?
    I must note that the ‘solve’ command didn’t work for it.
    I’m looking forward to hearing the response. Please mail to me! ;)

    Best wishes,

    Comment by Somayeh — Monday 18 February 2008 8:51 am #

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