A calculation yields the following result:

$ \text{kh}\to \frac{2 \sqrt{3} \sqrt{\nu +n-1} \sqrt{\text{R}^2 n^3+3 \nu +12 \nu n+15 n-3}+6 \nu -6 (2 \nu +1) n-6}{\text{R} (\nu +1) n^2}$

Knowing that $ n \gg 1$ , I can approximate this solution (by hand) to:

$ \text{kh}\to \frac{2 \left(\sqrt{3 \left(\text{R}^2 \text{n}^2+12 \nu +15\right)}-3 (2 \nu +1)\right)}{\text{R} (\nu +1) \text{n}}$

A log-log plot of the exact solution and the approximate solution looks like:

**It is clear from this graph, that the the curve has a minimum at $ n\gg1$ but $ n \ll \infty$ . Knowing that $ n \gg 1$ and $ R \ll 1$ (and, if it helps, $ 0\leq\nu\leq1/2$ ), how do I generate this approximate solution with Mathematica?**

My initial attempts include doing a Taylor series expansion around $ n=\infty$ , as well all as expansions around both $ n=\infty$ and $ R=0$ , but even considering many terms in the expansion (e.g. 9), still does not come very close to my approximate solution.

Here is the Mathematica code to get the original solution (the plots above were generated using: $ R\to0.008, \nu\to1/2$ ):

` eqk1=-((1 + \[Nu]) + (1 + \[Nu])/12 R kh n) Cn + (n - (1 - \[Nu])) Dn == 0; eqk2 = (R^2/12 n^2 + (1 + \[Nu])/12 R kh n + 2 (1 + \[Nu])) Cn - ((1 + \[Nu]) (n + 1/12 R kh n^2)) Dn == 0; matkh = Normal@CoefficientArrays[{eqk1, eqk2}, {Dn, Cn}]; khnAll = Solve[Det[matkh[[2]]] == 0, kh][[1]][[1]] // FullSimplify `

The closest I’ve come to generating the approximate solution is by squaring the exact solution, expanding around infinity, and then taking a square root, i.e.

` FullSimplify[Sqrt[Normal[Series[(kh /. khnAll)^2, {n, \[Infinity], 2}]]]] `

This leads to a graph that looks like:

Any help would be greatly appreciated.