I am attempting to calculate the power series coefficients of the solution to the differential equation $$f'(x)^2+f(x)^4=1\qquad f(0)=0.\tag{1}$$
I am trying to do so, because $f(x)$ is the inverse function of $$g(x)=\int_0^x\frac{dt}{\sqrt{1-t^4}},$$ i.e. $f\circ g(x)=g\circ f(x)=x$. I already know that $f$ can be represented in terms of the Jacobi elliptic function $\mathrm{sn}$ as $$f(x)=\mathrm{sn}(x|-1),$$ and I was hoping to understand a little more about elliptic functions like this through the lens of power series. I have made a little progress, I think.
First off, let $f'(x)^2=\sum_{n\ge0}a_nx^n$ and $f(x)^4=\sum_{n\ge0}b_nx^n$. Then $$\sum_{n\ge0}(a_n+b_n)x^n=1$$ so $a_0+b_0=1$ and $a_n=-b_n$ for $n>0$. Then let $f(x)=\sum_{n\ge0}f_nx^n$. Our aim is to find $f_n$, or at least evaluate the first few $f_n$. from $(1)$, it's pretty easy to see that $f_0=0$ and $f_1=1$. Then we see, from the Cauchy product, $$f'(x)^2=\sum_{n\ge0}x^n\sum_{k=0}^{n}(k+1)(n-k+1)f_{k+1}f_{n-k+1}$$ and $$f(x)^4=\sum_{n\ge0}x^n\sum_{k=0}^{n}q_kq_{n-k}$$ where $$q_n=\sum_{r=0}^{n}f_rf_{n-r}.$$ Thus for $n>0$, $$\sum_{k=0}^{n}\left(q_kq_{n-k}+(k+1)(n-k+1)f_{k+1}f_{n-k+1}\right)=0.$$ Is there an easier way to calculate these coefficients?
The exponential generating function of OEIS sequence A104203 is your $\,f(x).\,$ This function is known as the Gauss lemniscate sine. It satisfies the differential equations $$ f\,'(x)^2 + f(x)^4 = 1 \quad \text{ and }\quad f''(x) = -2f(x)^3. $$ The second differential equation implies that the sequence satisfies the recursion $$ a_{n+2} = -2 \sum_{i+j+k = n} \frac{n!}{i!\,j!\,k!}\,a_i\,a_j\,a_k $$ The first few terms are $$ a_2 = -2\,a_0^3,\;\; a_3 = -6\,a_0^2a_1,\;\; a_4 = 12\,a_0(a_0^4-a_1^2),\;\; a_5 = -12\,a_1(-9\,a_0^4+a_1^2). $$ When $\,a_0=0,\,a_1=1,\,$ we get the sequence $\,A104203: 0, 1, 0, 0, 0, -12, 0, 0, 0,3024,\dots.$