Talk:Implicit function theorem

message: In view of the importance of this theorem, this article seems quite unadequate... but better than nothing! MFH 05:50, 10 Mar 2005 (UTC)
replies: text: That's what the "edit this page" link is for ;) Dysprosia 07:19, 10 Mar 2005 (UTC)

message: This is a major theorem, and deserves its own article.  Dfeuer 17:09, 7 October 2007 (UTC)
thread_title:  Split from [[Implicit function]] 

message: What is the relationship between the implicit function theorem and the inverse function theorem?  That should probably be mentioned in both articles Dfeuer 17:10, 7 October 2007 (UTC)
replies: text: The intended link might be that both theorems require the jacobian to be invertible, but it's still unclear to me. :-/ EverGreg (talk) 09:27, 15 May 2008 (UTC)
replies: text: Ah, didn't read all of it. They'r both special cases of the constant rank theorem. I added this info on the other page. EverGreg (talk) 12:25, 15 May 2008 (UTC)
thread_title:  Relationship with [[Inverse function theorem]] 

message: There is a sentence here that says "Thus, here, the Y in the statement of the theorem is just the number 2b".
But there is no "Y" in the statement of the theorem.  — Preceding unsigned comment added by 128.227.159.177 (talk) 20:08, 15 November 2017 (UTC)
replies: text: No, you can't. Scineram (talk) 14:41, 5 May 2010 (UTC)
thread_title: Example

message: Suggestion: in the section statement of the theorem, just state the theorem as concisely as possible.  There is no further need for motivation in this section.  This would help greatly if you just want to look up the theorem, as opposed to learn about it. --345Kai (talk) 21:34, 29 March 2009 (UTC)
replies: text: I strongly agree with this.  This section should be split into two sections about the concise statement of the theorem and an intuitive explanation of it.  I will see what I can do. Erispre (talk) 01:43, 20 January 2013 (UTC)

text: The section still does not seem to be fixed with the high standard of math articles across the wiki.178.194.227.250 (talk) 13:23, 26 July 2014 (UTC)
thread_title:  Section: statement of the theorem 

message: Statement of the theorem: a new constant c appears. The discussion until that point was on f=0.  — Preceding unsigned comment added by Alv (talk • contribs) 09:05, 20 August 2011 (UTC)
thread_title:  Why not incorporate constant c into function f? 

message: The Inverse function theorem requires a restriction on the co-domain. Why wouldn't the implicit function theorem require such an assumption too? Moreover I noticed that the versions found in the literature also have this restriction. There seem to be global versions too, but don't they need additional assumptions? Odulon (talk) 14:17, 23 November 2018 (UTC)
replies: text: The theorem is true if one takes V=\mathbb R^n. In fact, if it is true for a given , it remains true if  increases. This results from continuity and the condition g(a)=b. It is necessary to specify  only if one omits the condition g(a)=b, which is not the case here. In fact, specifying  is equivalent with the condition g(a)=b. D.Lazard (talk) 14:46, 23 November 2018 (UTC)
thread_title:  Restricting the co-domain 

message: In the "Implicit functions from non-differentiable functions" section a cited source states without proof that the existence of implicit function implies the locally one-to-one property.  A counterexample to this is provided by explicit function f(x,y)=x^2. The implicit function would be x(y)=0, but the explicit function doesn't satisfy the locally one-to-one property.
I wrote this talk post in hope that in the future someone doesn't re-edit the mistake into the article. 193.40.13.174 (talk) 10:27, 19 October 2022 (UTC)
thread_title:  Converse is not true 

message: I see that in the article you have interpreted "almost always" in a measure-theoretic sense. Before making my recent edit to the article, which you have reverted, I thought very carefully, and decided that the average reader, as opposed to a mathematician with knowledge of measure theory, might be likely to take it as meaning something like "for almost all functions". I am perfectly happy with a statement in the article which makes it clear that it refers to almost everywhere on a particular functions, but, while you and I know enough to understand it in that sense, the overwhelming majority of Wikipedia readers don't. Just providing a wikilink to an article defining the term doesn't adequately address the question, as only a very small proportion of readers will follow that link.  I'm not sure what would be a good way of phrasing it, but maybe something like "almost everywhere on the function"?  (In my opinion the single worst thing about Wikipedia's coverage of mathematical topics is that far too often it is aimed at graduate mathematicians, and is often insufficiently accessible to the vast majority of readers of the encyclopaedia.) JBW (talk) 10:51, 14 August 2023 (UTC)
replies: text: I agree that the link is too technical, but I have not found a better one. On the other hand, I find "very often" much too weak. So, I suggest to unlink "almost always", and possibly to add some explanation, such as "almost always (it suffices that some auxiliary function be nonzero at a given point)". If parentheses break reading too much, they may be replaced with an explantory footnote or a separate sentence such as "Here, almost always means always except if some auxiliary function is zero at a given point". D.Lazard (talk) 09:08, 15 August 2023 (UTC)
thread_title: Almost always

message: In the sentence "From here we know that -\tfrac{\partial_x F}{\partial_y F} is Lipschitz continuous in both x and y." I don't see why it's Lipshitz continuous. 
2607:FB91:1010:AB6A:2850:D949:74B3:FAE1 (talk) 16:28, 30 August 2023 (UTC)
replies: text: I don't see it either. Can someone clarify this? JBW (talk) 19:52, 1 September 2023 (UTC)
thread_title: Question about Proof