PepeToro
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 Oct 31 comment “Partition” of a smooth function in $\mathbb R^2$ Well there is not much difference from what I wrote. What @JochenWengenroth wrote above is correct. I hope things are clear now. Oct 31 comment “Partition” of a smooth function in $\mathbb R^2$ By $f_1$ we abbreviate the smooth function of the variables $(xy,x)$ and by $f_2$ the smooth function of the variables $(xy,y)$. Just as shown in the expansion above. Call $w=xy$, then $\hat f_1(w,x)=\sum a_{mn}w^mx^n$ and $\hat f_2(w,y)=\sum a_{mn}w^my^n$, and make the coefficients coincide. Some of them are zero if necessary. Oct 31 comment “Partition” of a smooth function in $\mathbb R^2$ Both $f_i$ are smooth. Both are used at the decomposition. That is why I showed at least the formal level of the proof. Flat means zero Taylor expansion. The result means that a function in two variables may be partitioned as showed. The importance is that it can be done at the level of smooth functions and not only at the level of formal functions. My application is in dynamical systems for example. It may happen that $xy$ is a first integral (a constant along the trajectories of a vector field). Then it is much easier to integrate, say $f$, if we have such a result. Oct 30 revised “Partition” of a smooth function in $\mathbb R^2$ added 571 characters in body Oct 30 comment “Partition” of a smooth function in $\mathbb R^2$ @GHfromMO nope, I really meant what is written. Imagine in the formal series expansion of the form $x^iy^j$, $f_1$ contains monomials where $i\geq j$ and $f_2$ the rest. Oct 30 revised “Partition” of a smooth function in $\mathbb R^2$ edited tags Oct 30 asked “Partition” of a smooth function in $\mathbb R^2$ Sep 24 awarded Autobiographer Aug 5 awarded Editor Aug 5 revised Smooth normal forms of vector fields (the path method) edited body Aug 5 asked Smooth normal forms of vector fields (the path method) Aug 30 revised Stratification of a smooth map edited tags Aug 30 awarded Student Aug 30 asked Stratification of a smooth map