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Trying to learn math :)


6h
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.
9h
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.
10h
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.
21h
revised “Partition” of a smooth function in $\mathbb R^2$
added 571 characters in body
23h
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.
1d
revised “Partition” of a smooth function in $\mathbb R^2$
edited tags
1d
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