1
$\begingroup$

I'm looking at a paper involving a constrained optimisation problem using Lagrange multipliers, in which the following Lagrangian function appears:

$\gamma = \max \sum_{i=1}^nr_i(1-e^{-a_ix_i})+\lambda(1-\sum_{i=1}^{n}x_i)$

For the partial derivative of $\gamma$ with respect to $x_1$, the authors write:

$\frac{\partial\gamma}{\partial x_1} = a_1x_1+ \ln(\lambda)-\ln(r_1a_1)$

As a thoroughgoing newbie to all this, I thought that the partial derivative with respect to $x_1$ would be:

$\frac{\partial\gamma}{\partial x_1} = -a_1r_1e^{-a_1x_1} +\lambda$

...because I thought the partial derivative with respect to $x$ of $e^{kx}$ is $k\cdot e^{kx} $

Clearly, I'm missing something pretty fundamental (!), so I'd be really grateful if anyone could explain how the partial derivative above appears! Thanks in advance!

$\endgroup$
  • 1
    $\begingroup$ You are totally correct (may be, just a sign error we don't care about since the expression will be set equal to zero). I really wonder how they made the loarithm appearing ! Typo's, once more I bet. $\endgroup$ – Claude Leibovici Dec 27 '15 at 13:14
  • $\begingroup$ You are correct and the authors are not just wrong, but horribly wrong (you have the wrong signs, though). Assuming that you understood correctly their article, I suggest not reading it anymore. $\endgroup$ – Alex M. Dec 27 '15 at 13:16
1
$\begingroup$

My first reaction was : "one more typo". And, I was wrong !!

But let $$F=\sum _{i=1}^n r_i \left(1-e^{-a_i x_i}\right)+\lambda \left(1-\sum _{i=1}^n x_i\right)$$ which gives $$\frac{dF}{dx_i}=a_i r_i e^{-a_i x_i}-\lambda$$ what you properly found. But they now set the derivative equal equal to $0$. Then, and only then, $$a_i r_i e^{-a_i x_i}=\lambda$$ Now, take the logarithms. But this does not represent the derivative of anything.

$\endgroup$
  • $\begingroup$ Thank you Claude and @AlexM. for your help (and for pointing out I had the signs wrong!). I'm so sorry for my simpleton questions (I'm slowly learning!), but just to check I understand: your answer here, Claude, doesn't mean that the authors were right after all, does it? Since the Lagrange multiplier method involves setting the partial derivatives equal to 0, will they get the right results for the optimisation problem? Please forgive my slow brain, and thanks for the help! $\endgroup$ – Sprog Dec 27 '15 at 14:26
  • $\begingroup$ @Sprog. If they write that this is the derivative, they are criminals ! The expression with the logarithm is true when the derivative is zero but nowhere else. And, by the way, you are very welcome. Cheers. $\endgroup$ – Claude Leibovici Dec 27 '15 at 14:29
  • $\begingroup$ Ah, I see - thanks! $\endgroup$ – Sprog Dec 27 '15 at 14:50

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.