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Given $$ f(x)=x-x^2$$ I have to expand this as an infinite Taylor series ($x'=f(x)$), $$T_s(x)= x + s\beta_1(f) + s^2(2!)^{-1}\beta_2(f)+s^3(3!)^{-1}\beta_3(f) + \dots $$ where $\beta_1(f)=f$, $\beta_2(f)=f'f$, $\beta_j(f)=\beta_2(\beta_{j-1}(f)) $.

I tried to come up with a rule for that, but calculating $\beta_i$ for $i=1,2,3,\dots$ didn't help.

Can someone give me a hint, how to expand this properly? Thank you.

Greetings, Bernd

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  • $\begingroup$ Solving by separation of variables, then expanding the solution as a power series, looks feasible. Does that suit your needs? $\endgroup$ Nov 6 '16 at 19:28
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So you want to compute $$ x(s)=\sum_{k=0}^\infty β_k\frac{s^k}{k!}\implies x'(s)=\sum_{k=0}^\infty β_{k+1}\frac{s^k}{k!} $$ and per Cauchy product $$ x(s)^2=\sum_{k=0}^\infty \sum_{m=0}^k\binom{k}{m}β_mβ_{k-m}\frac{s^k}{k!} $$ and thus comparing coefficients on both sides of the equation $$ β_{k+1}=β_k-\sum_{m=0}^k\binom{k}{m}β_mβ_{k-m} $$ which gives a recursive formula for the computation of the Taylor coefficients.

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