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We know the solution(commutative case of Spitzer's Identity) of the equation $b=1+\text{P}(ba)$ when the operator $\text{P}$ satisfies Rota-Baxter eqution $\text{P}(x)\text{P}(y)=\text{P}(x\text{P}(y))+\text{P}(\text{P}(x)y)+u\text{P}(xy)$.(see Theorem 1.3.12 in the book "An introduction to Rota-Baxter Algebra" of Li Guo)

Is there anyway to find the solution of the equation $b=1+\text{P}(ba)$ when the operator $\text{P}$ satisfies $ \text{P}(x)\text{P}(y)=\text{P}(x\text{P}(y))+\text{P}(\text{P}(x)y)+uxy$?

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  • $\begingroup$ Why do you put $P$ in text mode? it's not standard $\endgroup$ – YCor Mar 15 at 16:14

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