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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?

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  • $\begingroup$ The summation on the derivative should start from $1$, not $0$, since the constant term vanished by differentiation. $\endgroup$ – Spine Feast Dec 14 '14 at 13:33
  • $\begingroup$ Right, I've edited it. $\endgroup$ – AlonAlon Dec 14 '14 at 13:37
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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 $$a_{n+1}=\frac{2\sum_{k=0}^na_ka_{n-k}}{n+1}$$

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    $\begingroup$ Did you guys lose the $2$ somewhere? $\endgroup$ – Spine Feast Dec 14 '14 at 13:41
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    $\begingroup$ You're right here too ($2$). Edited. Thanks. $\endgroup$ – AlonAlon Dec 14 '14 at 13:42
  • $\begingroup$ Yes I copied the equation from his post,I was too lazy to write it myself :P,Edited $\endgroup$ – kingW3 Dec 14 '14 at 13:45

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