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Suppose $\alpha\in(0,\frac12)$ and $\beta\in(0,\infty)$ are fixed. Initially I have $N\in\mathbb N\backslash\{0\}$ and $n\in\{0,\ldots,N\}$. I'd like to know, as a function $n$, the solution of the cubic equation
$$ (N+2\beta)x^3-(N+n+3\beta)x^2+(n+\beta+N\alpha-N\alpha^2)x+n\alpha^2-n\alpha=0 $$ in the limit $N\to\infty$.

Dividing the polynomial by $N$ gives $$ (1+2\epsilon_2)x^3-(1+\epsilon_1+3\epsilon_2)x^2+(\epsilon_1+\epsilon_2+\alpha-\alpha^2)x+\epsilon_1(\alpha^2-\alpha) = 0, $$ where $\epsilon_2 = \beta/N \to 0$. For fixed $n$, $\epsilon_1 = n/N \to 0$ so it seems in this case a multivariate generalization of perturbation theory might be appropriate. Does such a thing exist?

Bonus addition for the curious: the solution $q$ lies in $[\alpha,1-\alpha]$ and possess a symmetry about $\frac N2$. So the solution for $n = O(N)$ is given by the symmetry condition $q(n) = 1 - q(N-n)$. Also near $n=\frac N2$ the solution is well approximate by $q=\frac n N$.

Plotted below is the solution for $N=1001, \alpha = \frac12, \beta = \frac14$ in black and $q= \frac n N$ in red:

enter image description here

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You can differentiate the equation with respect to the parameters (as in this answer) to get a multivariate Taylor expansion for the roots. – joriki Feb 10 '12 at 7:10
Can you clarify something for me? In your question you say that $n\in\{0,\dots,N\}$ and then take the limit when $N\to\infty$. Surely, this can be interpreted to mean that $n$ can, potentially, be of $O(N)$. My question is: can we actually assume that $n\ll N$? Because if we can, we'll get nice leading order behaviour (even though it won't depend on $n$). Otherwise, I do not think you can do much with with this cubic except just solving it directly. – Aleksey Pichugin Mar 1 '12 at 9:57
@AlekseyPichugin -- I added some additional notes that should answer your question. joriki - thank you for the link, I haven't had time to go through it in detail yet. – Chris Ferrie Mar 1 '12 at 14:33
Yes, my fault, I can now see that you get another nice scaling in the case when $n\sim N$. OK, in this case, what is the problem? Surely, both the expansion for $n\ll N$ and for $n\sim N$ are easy enough to obtain. Would this be of interest to you? – Aleksey Pichugin Mar 1 '12 at 19:49
@AlekseyPichugin -- Yes, that is essentially my questions. – Chris Ferrie Mar 1 '12 at 20:19

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