0
$\begingroup$

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

$xe^x=k$

or

$e^x+x=k$

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

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

and

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

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

$$z=k-e^{e^z}$$

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 )

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.