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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?

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Notice that you should be able to get that asymptotic expansion using Newton's method (with $(n/A)^{1/k}$ as your initial guess). Perhaps there are results that ensure that Newton's method works in your situation, but I don't know offhand... –  Matt Young Feb 17 '10 at 2:24
    
The technique here is essentially the same as the technique used to prove Hensel's lemma, but I haven't quite figured out if it is directly applicable. –  Qiaochu Yuan Feb 17 '10 at 3:28
    
Don Zagier has a very general method to compute asymptotic expansions; it fluctuates from one paper to another with different implementations depending on different needs. The easiest link is mpim-bonn.mpg.de/preprints/send?bid=3138 (appendix there). –  Wadim Zudilin Jun 22 '10 at 14:25
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5 Answers

This is the well-known theory of Puiseux series (substitute $T=1/n$ to match the notation in that website). The convergence of the series is mentioned there too. So in your dissertation, I would recommend that you cite one of the references in the linked-to page instead of including a proof. Good luck with your thesis!

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If you have a good initial guess (such as your case), then you don't need either undetermined coefficients or Newton's method (although they both work), you can in fact get a procedure which will give you as many coefficients as you want. Harald's first step is excellent, so we'll start from that. Given $F(\epsilon,w) = \epsilon^2w+\epsilon w^2+w^3-1 = 0$, we can derive a differential equation for $w(\epsilon)$. With the help of Maple's gfun[algeqtodiffeq], giving $$ 9+ \left( 6{\epsilon}^{5}+7{\epsilon}^{2} \right) w ( \epsilon ) + \left( 27-6{\epsilon}^{6}-7{\epsilon}^{ 3} \right) {\frac {d}{d\epsilon}}w(\epsilon) + \left( -3{\epsilon}^{7}-14{\epsilon}^{4}-27\epsilon \right) {\frac {d^{2}}{d{\epsilon}^{2}}}w ( \epsilon ) $$ with initial conditions $w (0) =1,w^{'''}(0) =-4/9 $ [the DE is singular at $\epsilon=0$ so the initial conditions are non-standard). Nevertheless, from there one can continue and use gfun[diffeqtorec] to get a recurrence for the coefficients of the series. The result is $$ \left( 6-3n-3{n}^{2} \right) u(n) + \left( -98-77n-14{n}^{2} \right) u (n+3) + \left( - 648-27{n}^{2}-270n \right) u (n+6 ) $$ with starting conditions $u(0) =1, u(1) =-1/3,u (2) =-2/9,u (3) ={\frac {7}{81}},u(4) =0,u(5) ={\frac {14}{729}}$. These are sufficient to allow you to 'unwind' the recurrence as much as you want.

Bjorn's answer gives the underlying 'analytic' answer that justifies that the above gives a convergent result. Actually, the procedure above works in other cases as well, but then the result is only valid in a sector, and computing the size of that sector can be fiendishly difficult.

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Possibly a more generic answer than the others: Use a bit of scaling. You are looking for a root $z$ near $n^{1/3}$, so put $z=n^{1/3}w$ and rewrite your equation, using $\epsilon=n^{-1/3}$, as $$F(\epsilon,w)=\epsilon^2w+\epsilon w^2+w^3-1=0.$$ Note that $\partial F(\epsilon,w)/\partial w=3\ne0$ at $(\epsilon,w)=(0,1)$ and employ the analytic version of the implicit function theorem to see that $w$ is an analytic function of $\epsilon$.

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Well, you essentially answered that yourself: set up the corresponding system for the coefficients, note that it is always solvable and that the partial sum $S_m$ of the series satisfies the equation with an error $O(n^{-p(m)})$ with $p(m)\to+\infty$ as $m\to+\infty$ (you should get something like $m/k$ for the exact value of $p(m)$), so it differs from the true root by $O(n^{-p(m)})$ for large $n$, which is exactly what the asymptotic series is about.

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If you're worried about convergence... You can rewrite your equation as $g(z) = {1 \over f(1/z)} = 1/n$ for $z$ near ${1 \over n}$. $g(z)$ is thus $z^d$ divided by a polynomial with a nonzero constant term. Here $d$ is the degree of $f(z)$. Hence $g(z) = h(z)^d$ for some $h(z)$ with $h(0) = 0$ and $h'(0) \neq 0$. So you're solving $h(z) = n^{-{1 \over d}}$ which has a convergent power series solution in $n^{-{1 \over d}}$ by inverting $h(z)$. Then the series you originally wanted will be the reciprocal of this, so will start with $cn^{1 \over d}$ and then the exponents decrease by multiples of ${1 \over d}$.

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