0
$\begingroup$

How to solve the following equation?

$$e^{dz} - mdz = 1$$

where $z$ is the unknown variable, the others are constants, $\exp(x)$ is taking $e$ to the power $x$ . I am interested in real solutions for m > 1.

$\endgroup$
4
$\begingroup$

$z = 0$ is always a solution. Others, for $m, d \ne 0$, are

$$ z = -d^{-1} \left(1/m + W(-e^{-1/m}/m) \right) $$

where $W$ is a branch of the Lambert W function.

EDIT: I suspect you're interested in real solutions where $m$ is real. For $m < 0$, $-\exp(-1/m)/m) > 0$ so the principal branch of Lambert W can be used, but it will give $z = 0$. For $m > 0$, $-e^{-1} \le \exp(-1/m)/m < 0$ so either the $-1$ or the $0$ branch can be used; for $0 < m < 1$ the $-1$ branch gives $z=0$ while the principal branch gives a negative solution. For $m > 1$ the principal branch gives $z=0$ while the $-1$ branch gives a positive solution.

$\endgroup$
  • $\begingroup$ Could you elaborate on how to use the W function for m>1 to get the positive solution? $\endgroup$ – Serge Rogatch Apr 23 '16 at 7:47

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.