My question is related to this one: Approximating for $a \gg b$

I have a rather complex symbolic expression of interest. The expression is comprised of about ten different symbolic quantities. I wish to make an approximation regarding order of magnitude. I provided a minimal example. Let us assume that I have the following expression:

$$ expr=\frac{e^{\beta\,\epsilon_1}V}{\cosh(\alpha\,\epsilon_1)(e^{\beta\,\epsilon_1}+e^{\beta\,\epsilon_2})} $$

In the limit of $\beta\epsilon_1\gg\beta\epsilon_2$ this trivially reduces to $expr\approx \frac{V}{\cosh(\alpha\,\epsilon_1)}$. If on the other hand the ratio of $\epsilon_1$ and $\epsilon_2$ is unknown, but we take the limit of $\alpha\epsilon_1\gg1$, the expression may be approximated as $expr\approx \frac{2e^{\beta\,\epsilon_1}V}{e^{\alpha\,\epsilon_1}(e^{\beta\,\epsilon_1}+e^{\beta\,\epsilon_2})}$. Thus the expression can have various approximations in different regimes of the physical parameters, where the form the expression depends on the relation between these parameters.

As explained my actual is much more complex, and it is hard to determine the correct approximation by inspection. As MMA knows how to compute limits, I thought that it might be able to do this type of approximations. It comes down to two questions:

  1. Can I ask MMA to compute a limit where, the variable taken to infinity is the product of two parameters, while keeping each of them separately finite? I mean something that would resemble Limit[expr, \[Alpha]*\[Epsilon]1 -> Infinity]]

  2. For the expression given in the example, even if this limit can be taken by MMA, the result will be zero. However, this isn't my desire as I want a functional expression, not a number. Therefore I assume the taking the limit won't do the trick. Can I make such approximations using Series, or perhaps in another way?


After playing with the solution found in: Approximating for $a \gg b$, I think that it insufficient for my case as if I write:

Normal[expr /. \[Epsilon]1 -> \[Epsilon]1 + O[\[Epsilon]2]]
(E^(\[Beta] \[Epsilon]1) v Sech[\[Alpha] \[Epsilon]1])/(1 + \E^(\[Beta] \[Epsilon]1))

I'm able approximate only the exponents, but if I try the other limit I proposed:

Normal[expr /. \[Alpha] -> \[Alpha] + O[\[Epsilon]1]]

The result is:

$$ v \text{$\epsilon $1}^2 \left(\frac{\frac{\beta ^2}{2}-\frac{\alpha ^2}{2}}{e^{\beta \text{$\epsilon $2}}+1}-\frac{\beta ^2}{\left(e^{\beta \text{$\epsilon $2}}+1\right)^2}-\frac{\beta ^2 \left(e^{\beta \text{$\epsilon $2}}-1\right)}{2 \left(e^{\beta \text{$\epsilon $2}}+1\right)^3}\right)+v \text{$\epsilon $1} \left(\frac{\beta }{e^{\beta \text{$\epsilon $2}}+1}-\frac{\beta }{\left(e^{\beta \text{$\epsilon $2}}+1\right)^2}\right)+\frac{v}{e^{\beta \text{$\epsilon $2}}+1} $$

Namely when taking the series expansion, MMA doesn't know to keep $\epsilon_1$ within the exponents while expanding Cosh.

  • $\begingroup$ even your simple example is not strictly correct, you need Be1 >> Be2 $\endgroup$ – george2079 Apr 5 '17 at 15:37
  • $\begingroup$ True, I edited the question accordingly $\endgroup$ – Yair M Apr 5 '17 at 15:39
  • 1
    $\begingroup$ Try Series[yourExpression /. e2 -> x/e1, {x, 0, 1}]. It doesn't get you all the way there, but it's close. $\endgroup$ – march Apr 5 '17 at 16:01
  • $\begingroup$ My actual expression has more parameters, so I need something more robust $\endgroup$ – Yair M Apr 5 '17 at 16:03

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