I am reading The Magnus expansion and some of its applications by Blanes et al. and I have a question about one equation regarding the Floquet-Magnus expansion.
I. Defintions
I use the same definitions used in Blanes' paper. For the nested commutator, we have: $$ \text{ad}_A B=[A,B], \quad \text{ad}_A^j B=[A,\text{ad}_A^{j-1}B], \quad \text{ad}_A^0 B=B.\tag{1.1}$$ The exponential operator of $\text{ad}_A$ is defined as: $$ \text{Ad}_A = \exp \text{ad}_A,\quad \text{Ad}_A(B)=e^BAe^{-B}=\sum_{k=0}^\infty \frac{1}{k!} \text{ad}_A^k B. \tag{1.2}$$
II. Magnus expansion
Here, I only provided a quick summary of the Magnus expansion. We start with the differential equation $$ Y'(t)= A(t)Y(t), \tag{2.1}$$ where $Y(t),A(t)$ are operators acting on a Hilbert space $\mathcal{H}$. The prime in $A'(t)$ denotes the derivative with respect to $t$. The goal is to find a solution of the form $Y(t)=e^{\Omega(t)}$. Using this ansatz, we have: $$ Y'(t)= \text{d}\exp_{\Omega} \left[\Omega'(t)\right] e^{\Omega(t)}, \tag{2.2}$$ where $$\text{d}\exp_{\Omega} \left[B\right]=\sum_{k=0}^\infty \frac{1}{(k+1)!} \text{ad}_\Omega^k B = \frac{\exp(\text{ad}_\Omega)-I}{\text{ad}_\Omega}B, \tag{2.3}$$ and $I$ is the identity operator. The proof of equation (2.2) is shown in Blanes' paper. From equation (2.1) we obtain: $$ \text{d}\exp_{\Omega}\Omega'(t) = A(t) \implies \Omega'(t) = \text{d}\exp_{\Omega}^{-1}A(t),\tag{2.3}$$ where $$ \text{d}\exp^{-1}_{\Omega} B = \frac{\text{ad}_\Omega}{e^{\text{ad}_\Omega}-1}B, \tag{2.4}$$ is the inverse of the map $\text{d}\exp_\Omega$. So, we have arrived at a differential equation for $\Omega(t)$ which can be solved iteratively (the solutions are not relevant for this question).
III. Floquet Magnus expansion and Question
Floquet systems involve differential equations like (2.1) where $A(t)=A(t+T)$ is periodic with period $T$. Floquet theorem ensures we can find solutions of the form $$ Y(t)=P(t)e^{tF},\tag{3.1}$$ where $P(t)=P(t+T)$ is periodic and $F$ is time independent. Looking for solutions for $P(t)$ of the form $P(t)=e^{\Lambda(t)}$ we obtain: $$ Y'(t)=P'(t)e^{tF} +P(t)Fe^{tF} = A(t)P(t)e^{tF} \implies P'(t) =A(t)P(t)-P(t)F.\tag{3.2}$$ Using (2.2) we get $$ \text{d}\exp_{\Lambda}[\Lambda'(t)]= A-e^\Lambda Fe^{-\Lambda}. \tag{3.3}$$ Hence, we get $$ \Lambda'(t)= \text{d}\exp_{\Lambda}^{-1}\left[ A-e^\Lambda Fe^{-\Lambda}\right]. \tag{3.3}$$ In Blanes' paper, they write $$ \Lambda'(t) = \sum_{k=0}^\infty \frac{B_k}{k!}\text{ad}_\Lambda^k \left[A+(-1)^{k+1}F\right], \tag{3.4}$$ where $B_k$ is the $k^\text{th}$ Bernoulli number.
My question is: are equations (3.3) and (3.4) equivalent? How can I show it? Or did I make a mistake when writing (3.3)?
I have tried to show the two expressions are equivalent by using the series expansion of $\text{d}\exp_\Lambda^{-1} B$: $$ \text{d}\exp_\Lambda^{-1} B = \sum_{k=0}^\infty \frac{B_k}{k!}\text{ad}_\Lambda^k B ,\tag{3.5}$$ the equality $$ e^{\Lambda}Fe^{-\Lambda} = \sum_{k=0}^\infty \frac{1}{k!}\text{ad}_\Lambda^k F, \tag{3.6}$$ and the fact that $\text{ad}_\Lambda^k$ is a linear operator. However, in trying to show (3.3) and (3.4) are the same, I only arrive at $$ \Lambda'(t) = \text{d}\exp^{-1}_{\Lambda}(A-e^\Lambda F e^{-\Lambda})=\sum_{k=0}^\infty \frac{B_k}{k!}\left( \text{ad}_\Lambda^k A - \sum_{n=0}^\infty \frac{1}{n!} \text{ad}_\Lambda^{k+n} F\right),$$ which does not take me to far.
