I am trying to minimize $f(x,y)= \exp(-xy)+\alpha \exp\left(-\frac{1}{x}\right)$ with reespect to $x\in[0,\gamma]$ for a given $y\geq 0$. Here $\alpha$ and $\gamma$ are positive constants.

Differnetiating $f(x,y)$ and solving does not provide any closed-form solution. I need to find a simple solution which is in some way related to the optimal value.

So, I tried replacing $e^{-1/x}$ with $\ln(x)$ as they both have similar behavior for certian range of $x$. I scaled $\ln(x)$ in a way that they both have same value at $\gamma$.

\begin{equation} g(x,y)= \exp(-xy)+\alpha e^{-\frac{1}{\gamma}} \ln(x) / \ln(\gamma) \end{equation}

g(x,y) has a closed-form solution in terms of LambertW function. For cerain range of $x$, using plots, I see it to be very close to the optimal value of $f(x.y)$.

Now, I am trying to mathmatically show that they are indeed close.

Is there any good way to show this?

I tried the following

1) Derivative of $f(x.y)$ is close to zero at $x^{*}$ that minimizes $g(x.y)$

2) Show that both $f(x.y)$ an $g(x.y)$ are close for that range of $x^*$

  • $\begingroup$ is there anything known about $\alpha$ or $\gamma$? I guess the result depends on their values as well $\endgroup$ – Jane Apr 29 at 4:50
  • $\begingroup$ They both are positive constants. There are no other restrictions on them in general. I am trying to find the range of $\alpha$ for which this closeness works. $\endgroup$ – sar1729 Apr 29 at 5:02

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