quantum field theory - Derivation of the Renormalization Group from Renormalized Coupling?

At page 303 in the book Quantum Field Theory for the Gifted Amateur by Blundell and Lancaster, they argue that the renormalization group equation for the coupling $\lambda$ in $\phi^4$ theory can be derived as follows. Firstly, we introduce the renormalized coupling $$\lambda_R(s_0) = \lambda +\lambda^2 C \ln \left( \frac{s_0}{\Lambda^2}\right) + \ldots , $$ and rewrite the matrix element in terms of $\lambda_R$: $$ iM = - \lambda_R(s_0) -C \ln \left( \frac{s }{s_0}\right) \lambda_R^2(s_0) + \ldots \tag{1} $$ For a different scale $s_1$, we find $$\lambda_R(s_1) = \lambda +\lambda^2 C \ln \left( \frac{s_1}{\Lambda^2}\right) + \ldots , $$ $$ iM = - \lambda_R(s_1) -C \ln \left( \frac{s }{s_1}\right) \lambda_R^2(s_1) + \ldots \tag{2} $$ They then argue that by "subtracting" it follows that $$\lambda_R(s_1) = \lambda_R(s_0) +C \ln \left( \frac{s_1 }{s_0}\right) \lambda_R^2(s_0) + \ldots \tag{3}$$ I'm failing to see why this should be true. If we subtract Eq. 2 from Eq. 1 we find a term of the form $$ -C \ln \left( \frac{s }{s_1}\right) \lambda_R^2(s_1)+C \ln \left( \frac{s_0}{\Lambda^2}\right)\lambda_R^2(s_0) $$ which is not equal to the one in Eq. 3 unless $\lambda_R^2(s_0) = \lambda_R^2(s_1)$.

Am I missing something or is the derivation shown here wrong?

Answer

After subtraction you get $$ \lambda_R(s_1)=\lambda_R(s_0)+C\ln\left(\frac{s}{s_0}\right)\lambda_R^2(s_0)-C\ln\left(\frac{s}{s_1}\right)\lambda_R^2(s_1)+O\left(\lambda_R^3\right). $$ This is an implicit equation for $\lambda_R(s_1)$ that can be solved iteratively. Then, $$ \lambda_R(s_1)=\lambda_R(s_0)+C\ln\left(\frac{s}{s_0}\right)\lambda_R^2(s_0)-C\ln\left(\frac{s}{s_1}\right)\left[\lambda_R(s_0)+C\ln\left(\frac{s}{s_0}\right)\lambda_R^2(s_0)+\dots\right]^2+\ldots. $$ From this, neglecting powers higher than 2 of the coupling, you are able to recover the book's result.