Timeline for "Partition" of a smooth function in $\mathbb R^2$
Current License: CC BY-SA 3.0
16 events
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| Oct 31, 2014 at 13:34 | comment | added | GH from MO | @JochenWengenroth: Thanks for the clarification. I understand the statement now, as well as PepeToro's comments. | |
| Oct 31, 2014 at 10:09 | comment | added | PepeToro | Well there is not much difference from what I wrote. What @JochenWengenroth wrote above is correct. I hope things are clear now. | |
| Oct 31, 2014 at 9:39 | comment | added | Jochen Wengenroth | @GHfromMO Formally, you are right. The statement means: For all $f\in C^\infty(\mathbb R^2)$ with $f(0,0)=0$ there are $g_1,g_2 \in C^\infty(\mathbb R^2)$ such that $f(x,y)=g_1(xy,x)+g_2(xy,y)$ for all $(x,y)\in\mathbb R^2$. | |
| Oct 31, 2014 at 7:23 | comment | added | PepeToro | 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, 2014 at 6:12 | comment | added | PepeToro | 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, 2014 at 23:54 | comment | added | GH from MO | I also don't see what is the role of $\hat f$ in the EDIT section. I am not asking about the proof of your result, but about what the result says. | |
| Oct 30, 2014 at 23:53 | comment | added | GH from MO | It is still not clear to me what you mean. By $f_1$ we abbreviate the function $f_1(x,y)$. The function $f_1(xy,y)$ derived from it is a different object, e.g. for $f_1(x,y):=xy$ we have $f_1(xy,y)=xy^2$. It is not clear which of the two is smooth, and which of the two is used in the decomposition. It is also not clear to me what you mean by a flat function. | |
| Oct 30, 2014 at 19:10 | history | edited | PepeToro | CC BY-SA 3.0 |
added 571 characters in body
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| Oct 30, 2014 at 18:03 | comment | added | Christian Remling | The 'formal series expansion' is the Taylor series of $f$. | |
| Oct 30, 2014 at 17:32 | comment | added | GH from MO | Well, the notation $f_1=f_1(xy,x)$ etc. is confusing in itself. Also, it is not clear to me what you mean by formal series expansion. In short, it is not clear to me what you are stating. | |
| Oct 30, 2014 at 17:06 | comment | added | PepeToro | @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, 2014 at 16:40 | comment | added | GH from MO | Did you not mean $f_2(x,xy)$ instead of $f_2(xy,y)$? | |
| S Oct 30, 2014 at 16:19 | history | suggested | Mostafa Mirabi | CC BY-SA 3.0 |
cerrected spelling
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| Oct 30, 2014 at 15:58 | review | Suggested edits | |||
| S Oct 30, 2014 at 16:19 | |||||
| Oct 30, 2014 at 13:01 | history | edited | PepeToro |
edited tags
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| Oct 30, 2014 at 12:53 | history | asked | PepeToro | CC BY-SA 3.0 |