# Random stuff 2026 week 4
If $A$ and $B$ are operators on functions, what can we say when $[A,B]=AB-BA$ is a constant, namely, there exists a function $g$ such that for any function $f$, $[A,B]f=g$?
For example if $X[f](x)=xf(x)$, and $A[f](x)=f(x)+3$, then $[X,A]=3-3x$
Unlike $[D,X]=I$, where $D$ is derivative operator.
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For operator $(D+cI)$, we can construct a right-inverse $(D+cI)^{R}_{(x_{0},y_{0})}$ where define $$
(D+cI)^{R}_{(x_{0},y_{0})}[f]=e^{-cx}(\int^{x}_{x_{0}} e^{cx}f(x)dx+y_{0})
$$
Then we claim $(D+cI)(D+cI)^{R}_{(x_{0},y_{0})}=I$.
And for operator $(D+c(x)I)$, we can construct a right-inverse $(D+c(x)I)^{R}_{(x_{0},y_{0})}$ where $$
(D+c(x)I)^{R}_{(x_{0},y_{0})}[f]=e^{-\int c(x) dx}(\int_{x_{0}}^{x}e^{\int c(x)dx}f(x)dx+y_{0})
$$
This can then solve linear DEs with nonconstant coefficients, provided that we can "factor" into linear terms.
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How to linearize binary trees for computation?
Let capital letters A, B, C, ... be binary operations, and lower case x, y, z, w, ... be values.
A stacked based approach
((xAy)Bz)Cw) = xyAzBwC
((xAy)B(zCw)) = xyAzwCB
(xA((yBz)Cw)) = xyzBwCA