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Timeline for "Universal" differential identities

Current License: CC BY-SA 3.0

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Jun 21, 2017 at 12:45 comment added Asaf Shachar @TerryTao Thanks. By the way, a-priori there could be more non-polynomial universal identities, right? i.e your argument does not rule out identities of the form $\sin(f_x)f_y=e^{f_{zz}}$. I guess there are no such non-trivial identities.
May 22, 2017 at 12:59 comment added Vít Tuček @TerryTao Right, so the answer to the question: "Are there any universal identities which are not consequences of the commutation of the mixed derivatives?" is "No, because universal identities would have to be true for polynomials and monomials are linearly independent."
May 21, 2017 at 15:50 vote accept Asaf Shachar
May 20, 2017 at 22:19 comment added Terry Tao Given any finite set of coefficients $c_{i,j}$, one can create a function $f(x,y)$ with $\partial_x^i \partial_y^j f(0,0) = c_{i,j}$ for each $(i,j)$ in the finite set, simply by forming the finite Taylor series $\sum_{i,j} \frac{c_{i,j}}{i! j!} x^i y^j$. Of course, one cannot assign two different values to $f_{xy}$ and $f_{yx}$ by this procedure, as there is only one slot for both of these in the Taylor expansion.
May 20, 2017 at 12:50 comment added Vít Tuček I don't understand. What breaks down in this proof if we don't assume that $f_{xy} = f_{yx}$? Isn't "the values of finitely many of the various derivatives are completely unconstrained" precisely the statement the OP wants to prove here?
May 18, 2017 at 5:01 comment added Terry Tao Oh, you're right of course. I was worried about having some sort of unusual cancellation in the nonlinear case but, yeah, that doesn't actually happen.
May 18, 2017 at 2:09 comment added Will Sawin Can't you simplify this by applying your "the values of finitely many of the various derivatives are completely unconstrained" argument to $f$ instead of $g$?
May 17, 2017 at 22:48 history answered Terry Tao CC BY-SA 3.0