I know how to solve equations using Lambert's W function like




But how can I solve this two kinds of equations involving natural log ?

$e^x \ln(x)=k$



I can't figure how to rewrite them in one way where I can apply $W$

  • $\begingroup$ I don't think you can get an explicit solution. Probably, only numerical methods would do the job (easily). $\endgroup$ – Claude Leibovici Mar 20 '15 at 14:32
  • $\begingroup$ Yes, @ClaudeLeibovici but I'm interested in closed-forms involving only special functions, they are too easy to be solved with approximations. $\endgroup$ – Renato Faraone Mar 20 '15 at 14:34
  • 1
    $\begingroup$ They would be very special functions ! $\endgroup$ – Claude Leibovici Mar 20 '15 at 14:36

If you place $z=\ln(x)$ the equation can be rewritten as:


Applying lagrange inversion you will find:

$$z=k+\sum_{n=1} \frac{ (-1)^n}{n!} \left[\left(\frac{d}{du}\right)^{n -1}e^{ne^u}\right]_{u=k}$$

now remembering the Rodrigues representation of Touchard polynomials: ( see http://en.wikipedia.org/wiki/Touchard_polynomials )

$$T_n(u)=e^{-e^{u}} \left(\frac{d}{du}\right)^n e^{e^{u}} $$

you can find a formal solution:

$$z(k)=k+\sum_{n=1} \frac{ (-1)^n}{n!} e^{ne^k} T_{n-1}(k+\ln(n))$$

Therefore replacing $z=\ln(x)$ we find:

$$\ln(x(k))=k+\sum_{n=1} \frac{ (-1)^n}{n!} e^{ne^k} T_{n-1}(k+\ln(n))$$

If a link with Lambert W function or its one parameter generalization exists I don't know.

References For details about transcendental equations which can be solved by Lagrange series of mixed exponential/hypergeometric polynomials (that is to say transseries) obtained by suitable Rodrigues formulas see:

"Generalization of Lambert W function, Bessel polynomials and transcendental equations" Giorgio Mugnaini

http://arxiv.org/abs/1501.00138 (pdf: http://arxiv.org/pdf/1501.00138v3 )


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