I have a (log-) polynomial root equation given by
$$ f(n) = \log_2\left( \text{LargestRoot} \left[ \lambda^n - 3\sum_{i=0}^{n-1} \lambda^i = 0 \right]\right) $$
The final equation of interest is then the fractional loss of $f(n)$ from 2:
$$ L(n) = \frac{2-f(n)}{2} $$
Numerical solution for small $n$ is plotted below, and shows clear log-linear asymptotic behavior. The line shown is fit using $n \in [5, 20]$. I would like an analytical solution for the asymptote but don't see how to get there. Any help would be much appreciated.
Let's first simplify the equation:
$$ \lambda^n-3\sum_{i=0}^{n-1}\lambda^i=\lambda^n-3\frac{1-\lambda^n}{1-\lambda}=0 $$
and thus
$$ \lambda^{n+1}-4\lambda^n+3=0 $$
(which has the same roots as the original equation, and an additional one at $\lambda=1$ because we multplied by $\lambda-1$). Since the largest root appears to be converging to $\lambda=4$, let's substitute $\lambda=4+\epsilon$:
\begin{eqnarray} (4+\epsilon)^{n+1}-4(4+\epsilon)^n+3 &=& 4^{n+1}+4^n(n+1)\epsilon-4^{n+1}-4^nn\epsilon+3+O\left(\epsilon^2\right) \\ &=& 4^n\epsilon+3+O\left(\epsilon^2\right) \\ &=& 0\;. \end{eqnarray}
Thus $\epsilon\approx-\frac3{4^n}$, which yields
\begin{eqnarray} f(n) &=& \log_2(4+\epsilon) \\ &=& \log_24+\log_2\left(1+\frac\epsilon4\right) \\ &\approx& 2+\frac\epsilon{4\ln2} \\ &\approx& 2-\frac3{4^{n+1}\ln2} \end{eqnarray}
and thus
\begin{eqnarray} L(n) &=& \frac{2-f(n)}2 \\ &\approx& \frac3{8\ln2}4^{-n} \end{eqnarray}
and
\begin{eqnarray} \log_{10}L(n) &\approx&\log_{10}\left(\frac3{8\ln2}\right)-n\log_{10}4 \\ &\approx& -0.27-0.60n\;, \end{eqnarray}
in agreement with your fit result.