Derive asymptotics for $f(z) := e^{z+z^3/3}$ using the saddle point equation $$s_0 \frac{f^\prime(s_0)}{f(s_0)} = n.$$
Hint: Plug $s_0 = \alpha_1n^{\beta_1} + \alpha_2n^{\beta_2} \pm n^{\beta_1-1-\varepsilon}$ into the saddle point equation and choose the constants so that the left-hand side becomes $n \pm cn^{- \varepsilon} + \mathcal{o}(n^{- \varepsilon})$ for some $ c > 0$ and argue that $s_0 = (1+\mathcal{o}(1))(\alpha_1 n^{\beta_1} + \alpha_2 n^{\beta_2}).$
The strategy for this problem is probably the following: We recall the following theorem (basically from the Flajolet & Sedgewick book, p.568,569; see below):
Theorem: Let $P(z) := \sum_{k \ge 1} p_kz^k$ be a polynomial with non-negative coefficients and aperiodic (in the sense that $\gcd\{k : p_k \ne 0\} = 1$. Let $f(z) := e^{P(z)}$. Then we have that
$$[z^n]f(z) \sim \frac{e^{P(s_0)}} {s_0^n \sqrt{2 \pi (n + s_0^2 P^{\prime\prime}(s_0))}}, $$
where $s_0$ is the unique solution to $s_0P^\prime(s_0) = n$.
So we note that
$$f(z) = e^{z + z^3/3}$$ $$f'(z) = (z^2+1)e^{z+z^3/3}$$ $$f''(z) = (z^4 + 2z^2+2z+1) e^{z+z^3/3}$$
Since we have an estimation of $s_0$, $$s_0 = (1+\mathcal{o}(1))(\alpha_1 n^{\beta_1} + \alpha_2 n^{\beta_2})$$ we also have an estimation of $e^{P(s_0)}$, namely
$$e^{P(s_0)} \sim e^{\alpha_1 n^{\beta_1} + \alpha_2 n^{\beta_2}}.$$
However, I do not see how to use the hint to come to the given estimation or how to find an estimation for $s_0 P^{\prime\prime}(s_0)$. Could you please help me?
The saddle point equation is $s_0^3 + s_0 = n$, whose only real root is given by $$ s_0 = n^{1/3} - \frac{1}{3}n^{ - 1/3} + \frac{1}{{81}}n^{ - 5/3} + \mathcal{O}(n^{ - 7/3} ). $$ Thus $$ P(s_0 ) = \frac{1}{3}n + \frac{2}{3}n^{1/3} + \mathcal{O}(n^{ - 1/3} ), $$ $$ s_0^2 P''(s_0 ) = 2s_0^3 = 2n - 2n^{1/3} + \mathcal{O}(n^{ - 1/3} ), $$ and $$ s_0^n = n^{n/3} \exp \left( { - \frac{1}{3}n^{1/3} +\mathcal{O}(n^{ - 1/3} )} \right). $$ Substituting these into the formula in the theorem and keeping only terms that contribute to leading order, we find $$ \left[ {z^n } \right]{\rm e}^{z + z^3 /3} \sim \frac{{\exp \left( {\frac{1}{3}n + n^{1/3} } \right)}}{{\sqrt {6\pi } \,n^{n/3 + 1/2} }} $$ as $n\to +\infty$.