Series of inverse function $A(s) = \sum_{k>0}a_ks^k$ and $A(s)+A(s)^3=s$.
I want calculate $a_5$. What ways to do it most efficiently?
 A: It is pretty obvious that $s\mapsto A(s)$ is odd. If you really only need $a_5$ the simplest thing is to write
$$A(s)=s+a_3s^3+a_5s^5+?s^7\ ,$$
whereby the question mark represents a full power series. Then
$$A^3(s)=s^3(1+a_3s^2+?s^4)^3$$
and therefore
$$0=A(s)+A^3(s)-s=s+a_3s^3+a_5s^5+?s^7+s^3\bigl(1+3(a_3s^2+?s^4)+?s^4\bigr)-s\ .$$
Comparing coefficients gives $a_3=-1$, $\>a_5=3$.
A: $A(s) = \sum_{k>0}a_ks^k$ and $A(s)+A(s)^{3} =s$
We know (Cauchy product): $$A(s)^{2} = \sum_{n>0}^{\infty} \left( \sum_{i=0}^{n}a_{i} a_{n-i} \right) s^n$$
And
$$A(s)^{3} = \sum_{n>0}^{\infty} \left( \sum_{j=0}^{n}a_{n-j} \left( \sum_{i=0}^{j}a_{i} a_{j-i} \right) \right) s^n$$
Hence:
$$ \sum_{n>0}^{\infty}a_ns^n+\sum_{n>0}^{\infty} \left( \sum_{j=0}^{n}a_{n-j} \left( \sum_{i=0}^{j}a_{i} a_{j-i} \right) \right) s^n = s$$
For $n=0$:$$a_{0}+{a_0}^{3}=0$$
For n=1:
$$a_{1}+a_{1}.0+0=1$$
By raising the $n$ you get the $a_{n}$
A: Another possibility: the Lagrange inversion formula
$A$ is the inverse of $B(t)=t+t^3$, so
$$A(s)=\sum_{n=1}^\infty\lim_{t\to 0}\frac{s^n}{n!}\frac{d^{n-1}}{dt^{n-1}}\left(\frac{t}{t+t^3}\right)^n.$$
A: We can elaborate on the Lagrange inversion concept.
Suppose we have $$A(s) = \sum_{n\ge 0} a_n s^n$$
and $A(s)+A(s)^3=s$ and we seek $a_n.$

Using the Cauchy Residue Theorem to prepare for Lagrange inversion we have that
$$a_n = 
\frac{1}{2\pi i}
\int_{|s|=\epsilon}
\frac{1}{s^{n+1}} A(s) \; ds.$$
Now put $A(s)=w$ so that $w+w^3 = s$ and
$$ds = 1 + 3w^2 \;dw.$$
This yields
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{(w+w^3)^{n+1}} w \times (1+3w^2) \; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n}} 
\frac{1}{(1+w^2)^{n+1}} \times (3w^2+3-2) \; dw.$$
The first component here is
$$3 \times\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n}} 
\frac{1}{(1+w^2)^{n}} \; dw
= 3\times[w^{n-1}] \frac{1}{(1+w^2)^{n}}.$$
This is zero when $n$ is even
and when $n$ is odd it yields
$$3\times (-1)^{(n-1)/2}\times {(n-1)/2+n-1\choose n-1}
= 3 (-1)^{(n-1)/2} {3/2n-3/2\choose n-1}.$$
The second component is
$$-2\times \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n}} 
\frac{1}{(1+w^2)^{n+1}} \; dw
= -2\times [w^{n-1}] \frac{1}{(1+w^2)^{n+1}}.$$
This is again zero when $n$ is even 
and when $n$ is odd it yields
$$-2\times (-1)^{(n-1)/2} {(n-1)/2+n\choose n}
= -2 (-1)^{(n-1)/2} {3/2n-1/2\choose n}.$$
Combining these two yields
$$(-1)^{(n-1)/2}
\left(3 {3/2n-3/2\choose n-1}
-2 {3/2n-1/2\choose n}\right)$$
when $n$ is odd and zero otherwise.

If desired this can be simplified to
$$(-1)^{(n-1)/2}
\left(3-2 \frac{3/2n-1/2}{n}\right)
{3/2n-3/2\choose n-1}
= (-1)^{(n-1)/2}
\frac{1}{n}{3/2n-3/2\choose n-1}.$$
A: Here we use a somewhat simpler but equivalent variant of the Lagrange Inversion Formula. (See Theorem A.2 in Analytic Combinatorics by P. Flajolet and R. Sedgewick for the equivalence of the variants).

Lagrange inversion formula: 
Let $B(s), A(s)\in s\mathbb{C}[s]$ be inverses: $B(A(s))=s$. If $B(s)=\frac{s}{\phi(s)}$ and 
  $A(s)=s\phi\left(A(s)\right)$, then
  \begin{align*}
[s^n]A(s)=\frac{1}{n}\left[s^{n-1}\right]\left(\phi(s)\right)^n\tag{1}
\end{align*}

The functional relation
\begin{align*}
(A(s)) + (A(s))^{3} = s
\end{align*}
can be written as $B\left(A(s)\right)=s$ with
\begin{align*}
B(s)&=s+s^3\\
&=s(1+s^2)
\end{align*}

We can write $$\phi(s)=\frac{s}{B(s)}=\frac{1}{1+s^2}$$ 
and obtain from (1)
\begin{align*}
[s^5]A(s)&=\frac{1}{5}\left[s^{4}\right]\left(\phi(s)\right)^5\tag{2}\\
&=\frac{1}{5}\left[s^{4}\right]\frac{1}{(1+s^2)^{5}}\\
&=\frac{1}{5}\left[s^{4}\right]\sum_{k=0}^{\infty}\binom{-5}{k}s^{2k}\tag{3}\\
&=\frac{1}{5}\binom{-5}{2}\\
&=\frac{1}{5}\cdot\frac{(-5)(-6)}{2!}\\
&=3
\end{align*}

Comment:


*

*In (2) we apply the Lagrange inversion formula (1)

*In (3) we use the binomial series expansion
A: Derive $A+A^3=s$ on $s$ and evaluate at $0$, giving with $A_0=0$,
$$A'+3A^2A'=1\implies A'_0=1$$
$$A''+6AA'^2+3A^2A''=0\implies A''_0=0$$
$$A'''+6A'^3+18AA'A''+3A^2A'''=0\implies A'''_0=-6$$
$$A''''+36A'^2A''+18AA''^2+24AA'A'''+3A^2A''''=0\implies A''''_0=0$$
$$A'''''+90A'A''^2+60A'^2A'''+60AA''A'''+30AA'A''''+3A^2A'''''=0\implies A'''''_0=360$$
Hence by Taylor
$$A(s)=s-\frac6{3!}s^3+\frac{360}{5!}s^5\cdots$$
