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
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.