236 reputation
15
bio website
location
age
visits member for 2 years, 2 months
seen yesterday

Sep
17
awarded  Supporter
Sep
17
comment Necessary and Sufficient Conditions for $f_{xy} = f_{yx}$
You are correct. I misinterpreted what you were attempting to say. However, this is a simple restatement of the original theorem. In a sense, I'm asking that if we declare the mixed partials to be equal, then what can we deduce about $f$? Also, see my original post, edited to ask about some relevant illustrative examples.
Sep
17
revised Necessary and Sufficient Conditions for $f_{xy} = f_{yx}$
added 306 characters in body
Sep
17
comment Necessary and Sufficient Conditions for $f_{xy} = f_{yx}$
The construction of $f$ there is clever. Thank you.
Sep
17
comment Necessary and Sufficient Conditions for $f_{xy} = f_{yx}$
@Graham, nice formatting. However, the "change of order" step is only valid assuming continuity of second partials. You accidentally included a bit of your conclusion in your premise.
Sep
17
revised Necessary and Sufficient Conditions for $f_{xy} = f_{yx}$
edited title
Sep
17
asked Necessary and Sufficient Conditions for $f_{xy} = f_{yx}$
Sep
5
comment Calculating limits of functions explicitly
@Nir, Check my edits to see if it addresses your question. Remember x*y = y/(1/x) (except when x=0), so with limits you can often use this trick to get things in a form where L'Hopital's Rule applies.
Sep
5
revised Calculating limits of functions explicitly
Updated to a more explicit example
Sep
4
awarded  Yearling
Sep
3
awarded  Editor
Sep
3
revised Nested Sum Encountered in Maclaurin Expansion of $e^{-x^2}$
deleted 16 characters in body
Sep
3
comment Nested Sum Encountered in Maclaurin Expansion of $e^{-x^2}$
Thank you. The recursive definition you reformulated the sum into is quite nice. I don't, however, see the 'easy induction' proof. Would you mind suppressing the laziness urge providing a little more in the way of proof? :) Also edited question to more clearly state desire of proof.
Sep
3
answered Calculating limits of functions explicitly
Sep
3
awarded  Student
Sep
3
comment Nested Sum Encountered in Maclaurin Expansion of $e^{-x^2}$
@user166967, they are just indices of the respective sums.
Sep
3
asked Nested Sum Encountered in Maclaurin Expansion of $e^{-x^2}$
Nov
25
comment Characteristic properties for topological pushouts and pullbacks
@Bachmaninoff Basically, yes. These constructions work any any category as long as the relevant diagrams commute. Often these constructions correspond to more familiarly named constructions. As an exercise, try to see what pullbacks, pushouts, etc are in the category Set of sets.
Nov
19
comment Characteristic properties for topological pushouts and pullbacks
@Bachmaninoff Looks like you got it, sir. Note that it is crucial for the combined maps to all be continuous and that h be unique; I think this is what you were intending to say though.
Nov
18
comment Characteristic properties for topological pushouts and pullbacks
@Bachmaninoff Similary for quotient topologies, you are identifying elements in a parent topology P to get a quotient topology Q, so you should expect f:P->Q instead of f:Q->P, so we can write f(a)=f(b) and the like. Then since the image of f should be Q, we need f to be surjective (an epimorphism in Top). It is interesting to note that the subspace topology and quotient topology are dual to each other. Why? If you can convince yourself, then you probably can handle product and disjoint union characterizations yourself.