Let $f(z) = z + z^2 + z^3$. Then for large $n$, $f(z) = n$ has a real solution near $n^{1/3}$, which we call $r(n)$. This appears to have an asymptotic series in descending powers of $n^{1/3}$, which begins $$ r(n) = n^{1/3} - {1 \over 3} - {2 \over 9} n^{-1/3} + {7 \over 81} n^{-2/3} + O(n^{-1}). $$ In order to derive this series I use a method of undetermined coefficients. If we assume such a series exists, of the form $$ r(n) = An^{1/3} + B + Cn^{-1/3} + Dn^{-2/3} + O(n^{-1}), $$ then applying $f$ to both sides gives $$ n = A^3 n + (3 A^2 B + A^2) n^{2/3} + (3 A^2 C + 3 A B^2 + A + 2AB)n^{1/3} + (B^2 + 2AC + 3A^2 D + B + B^3 + 6 ABC + O(n^{-1/3}). $$ Thus we have $A^3 = 1$ and so $A = 1$; $3A^2 B + A^2 = 0$ and so $B = -1/3$; similarly $C = -2/9, D = 7/81$.

More generally, let $f$ be a polynomial with leading term $Ax^k$. Then the largest real root of $f$ appears to have an asymptotic series of the form

$$ A^{-1/k} n^{1/k} + c_0 + c_1 n^{-1/k} + c_2 n^{-2/k} + \cdots $$

which can be computed by a similar method to the one I gave above.

**My question**, which I would like an answer to in order to make a minor point in my dissertation: how can I prove that the solutions of equations of the form $f(z) = n$ have such series?