Magnus coefficient changes under longitude conjugation depend on lower-order coefficients

Write $E(u)=1+U$ and $E(\lambda )=1+V$ with $U,V\in \mathfrak{m}$. Then $E(u^{-1}\lambda u)-E(\lambda )=(1+U{)}^{-1}V(1+U)-V\mathrm{.}$ Expand $(1+U{)}^{-1}={\sum }_{a\ge 0}(-U{)}^a$. The term with no $U$ on either side is $V$ and cancels the subtracted $V$. Every remaining term is a product $(-U{)}^{a}V{U}^b$ with $a+b\ge 1$.

A coefficient of degree $m$ in such a product uses a degree-$p$ monomial of $U^a$, a degree-$s$ monomial of $V$, and a degree-$q$ monomial of $U^b$ with $p+s+q=m$ and $p+q\ge 1$. The $V$ monomial therefore has degree $s<m$ and occupies a contiguous middle block of the target word $w$, while the nonempty left or right block comes from $U$. Summing the finitely many decompositions of $w$ expresses its coefficient as an integer linear combination of coefficients of $E(\lambda )$ on proper contiguous subwords. Appending the longitude index $j$ makes each such index a proper subsequence of $(w,j)$.