Though the question was not to me, the matrix form of
\( f(e^x)=f(x)+1 \) or by composition \( f\circ\exp = \tau_1\circ f \) is
\( E\cdot F = F\cdot T_1 \) where E is the Carleman matrix of exp and so on. \( T_1 \) is the upper triangular pascal matrix. But this does not help here/gives no way to solve the equation. If we however reduce F to its first column \( (f_0,f_1,\dots)^T \) on the left side, we can reduce \( T_1 \) to the first column on the right side \( t_1=(1,1,0,\dots)^T \). Then \( F\cdot t_1=(1+f_0,f_1,f_2,\dots)^T=\vec{f}+(1,0,\dots)^T \) and the fate is on its way \( E\cdot\vec{f}=\vec{f}+(1,0,\dots)^T \) and \( (E-I)\vec{f}=(1,0,\dots)^T \).
For the truncated matrices I think for no \( F \) is \( E=F\cdot T_1 \cdot F^{-1} \) because the Jordan normal form of \( E \) is a Diagonal matrix with pairwise different eigenvalues while the Jordan normal form of \( T_1 \) has only the eigenvalue 1.
\( f(e^x)=f(x)+1 \) or by composition \( f\circ\exp = \tau_1\circ f \) is
\( E\cdot F = F\cdot T_1 \) where E is the Carleman matrix of exp and so on. \( T_1 \) is the upper triangular pascal matrix. But this does not help here/gives no way to solve the equation. If we however reduce F to its first column \( (f_0,f_1,\dots)^T \) on the left side, we can reduce \( T_1 \) to the first column on the right side \( t_1=(1,1,0,\dots)^T \). Then \( F\cdot t_1=(1+f_0,f_1,f_2,\dots)^T=\vec{f}+(1,0,\dots)^T \) and the fate is on its way \( E\cdot\vec{f}=\vec{f}+(1,0,\dots)^T \) and \( (E-I)\vec{f}=(1,0,\dots)^T \).
For the truncated matrices I think for no \( F \) is \( E=F\cdot T_1 \cdot F^{-1} \) because the Jordan normal form of \( E \) is a Diagonal matrix with pairwise different eigenvalues while the Jordan normal form of \( T_1 \) has only the eigenvalue 1.
