# Given $f'(x) = 2f(x)^2$ find the recurrence formula for $a_n$ in $f(x) =\sum a_n x^n$.

Let $f(x) = \sum_{n=0}^\infty a_n x^n$ with radius convergence of $R>0$. We know that $f'(x) = 2f(x)^2$. Find the recurrence formula of $a_n$.

I don't know if it makes a difference but $f(x)$ should be $f:(-1,1)\to\mathbb{R}$ and the relation with the derivative should be valid for all $x\in (-1,1)$.

So it's given that: $f'(x)=2f(x)^2$. Hence:

$$\sum_{n=0}^\infty na_n x^{n-1} = 2\left(\sum_{n=0}^\infty a_n x^n \right)^2$$

Now, we have a product of summations:

$$\sum_{n=1}^\infty na_n x^{n-1} = 2 \sum_{n=0}^\infty \left(\sum_{k=0}^n a_k a_{n-k} \right)x^n$$

How to proceed?

• The summation on the derivative should start from $1$, not $0$, since the constant term vanished by differentiation. – Spine Feast Dec 14 '14 at 13:33
• Right, I've edited it. – AlonAlon Dec 14 '14 at 13:37

Well at first $$\sum_{n=0}^\infty na_nx^{n-1}=\sum_{n=0}^\infty (n+1)a_{n+1}x^n=\sum_{n=0}^\infty 2\left(\sum_{k=0}^n a_k a_{n-k} \right)x^n$$ Now group coefficients with $x^n$ so you get $(n+1)a_{n+1}-2\sum_{k=0}^na_ka_{n-k}=0$,since if the coefficients are non-zero than the equation doesn't hold for all $x$.So $$a_{n+1}=\frac{2\sum_{k=0}^na_ka_{n-k}}{n+1}$$
• Did you guys lose the $2$ somewhere? – Spine Feast Dec 14 '14 at 13:41
• You're right here too ($2$). Edited. Thanks. – AlonAlon Dec 14 '14 at 13:42