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$\DeclareMathOperator\ad{ad}$The identity $$\log(e^X e^Y) = X + \frac{\ad_X}{1 - e^{-\ad_X}}Y + O(Y^2)\tag{$*$}\label{star}$$ is due to Poincaré [1], who showed that $W(t)=\log (e^X e^{tY})$ solves the ODE $$W'(t)=\frac{\ad_{W(t)}}{1-e^{-\ad_{W(t)}}}Y,\;\;W(0)=X.$$ The identity \eqref{star} is the solution to first order in $t$. It holds generally, you don't need $[X,Y]=sY$.

[1] H. Poincaré, Sur les groupes continus, Trans. Cambridge Phil. Soc., 18, 220–255 (1900).

$\DeclareMathOperator\ad{ad}$The identity $$\log(e^X e^Y) = X + \frac{\ad_X}{1 - e^{-\ad_X}}Y + O(Y^2)\tag{$*$}\label{star}$$ is due to Poincaré [1], who showed that $W(t)=\log (e^X e^{tY})$ solves the ODE $$W'(t)=\frac{\ad_{W(t)}}{1-e^{-\ad_{W(t)}}}Y,\;\;W(0)=X.$$ The identity \eqref{star} is the solution to first order in $t$. It holds generally, you don't need $[X,Y]=sY$.

[1] H. Poincaré, Sur les groupes continus, Trans. Cambridge Phil. Soc., 18, 220–255 (1900). JFM 31.0386.01

The identity $$\log(e^X e^Y) = X + \frac{\text{ad}_X}{1 - e^{-\text{ad}_X}}Y + O(Y^2)\qquad\qquad(*)$$ is due to Poincaré [1], who showed that $W(t)=\log (e^X e^{tY})$ solves the ODE $$W'(t)=\frac{\text{ad}_{W(t)}}{1-e^{-\text{ad}_{W(t)}}}Y,\;\;W(0)=X.$$ The identity $(*)$ is the solution to first order in $t$. It holds generally, you don't need $[X,Y]=sY$.

[1] H. Poincaré, Sur les groupes continus, Trans. Cambridge Phil. Soc., 18, 220–255 (1900).

$\DeclareMathOperator\ad{ad}$The identity $$\log(e^X e^Y) = X + \frac{\ad_X}{1 - e^{-\ad_X}}Y + O(Y^2)\tag{$*$}\label{star}$$ is due to Poincaré [1], who showed that $W(t)=\log (e^X e^{tY})$ solves the ODE $$W'(t)=\frac{\ad_{W(t)}}{1-e^{-\ad_{W(t)}}}Y,\;\;W(0)=X.$$ The identity \eqref{star} is the solution to first order in $t$. It holds generally, you don't need $[X,Y]=sY$.

[1] H. Poincaré, Sur les groupes continus, Trans. Cambridge Phil. Soc., 18, 220–255 (1900).

The identity $$\log(e^X e^Y) = X + \frac{\text{ad}_X}{1 - e^{-\text{ad}_X}}Y + O(Y^2)\qquad\qquad(*)$$ is due to Poincaré [1], who showed that $W(t)=\log (e^X e^{tY})$ solves the ODE $$W'(t)=\frac{\text{ad}_{W(t)}}{1-e^{-\text{ad}_{W(t)}}}(Y),\;\;W(0)=X.$$ The identity $(*)$ is the solution to first order in $t$. It holds generally, you don't need $[X,Y]=sY$.

[1] H. Poincaré, Sur les groupes continus, Trans. Cambridge Phil. Soc., 18, 220–255 (1900).

The identity $$\log(e^X e^Y) = X + \frac{\text{ad}_X}{1 - e^{-\text{ad}_X}}Y + O(Y^2)\qquad\qquad(*)$$ is due to Poincaré [1], who showed that $W(t)=\log (e^X e^{tY})$ solves the ODE $$W'(t)=\frac{\text{ad}_{W(t)}}{1-e^{-\text{ad}_{W(t)}}}Y,\;\;W(0)=X.$$ The identity $(*)$ is the solution to first order in $t$. It holds generally, you don't need $[X,Y]=sY$.

[1] H. Poincaré, Sur les groupes continus, Trans. Cambridge Phil. Soc., 18, 220–255 (1900).