Talk:Implicit function theorem

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In view of the importance of this theorem, this article seems quite unadequate... but better than nothing! MFH 05:50, 10 Mar 2005 (UTC)

That's what the "edit this page" link is for ;) Dysprosia 07:19, 10 Mar 2005 (UTC)

By reversible matrix, we mean invertible, right? Smmurphy^(Talk) 01:37, 25 October 2006 (UTC)Reply

• I agree, this is an important theorem, and the tiny section on the theorem in the article Implicit function is not very clear. As far as the merge is concerned, I think that there is no useful information on the aforementioned page that needs to be added to this one. TSchellhous 21:47, 12 December 2006 (UTC)Reply

This is a major theorem, and deserves its own article. Dfeuer 17:09, 7 October 2007 (UTC)Reply

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)Reply

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)Reply

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)Reply

Does anyone have a (GDFL or public domain) proof available? Dfeuer 17:15, 7 October 2007 (UTC)Reply

More than 10 years later... late but if you read French you have a proof under GFDL here. Search for "Théorème de la fonction implicite". It is based on the inverse function theorem, which is also proved in the same text.

Laurent.Claessens (talk) 05:56, 14 January 2019 (UTC)Reply

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)Reply

In the example of the inverse function theorem, shouldn't the left part of (Df)(a,b) have -1's instead of 1's? Arthena^(talk) 19:48, 21 January 2008 (UTC)Reply

Also, the statement that one cannot implicitly create a differentiable function near (1,0) and (-1,0) seems pretty false. Sure you can't write an expression of y in terms of x, but clearly you can write x in terms of y. This theorem would be fairly weak if it couldn't fully characterize one of the most basic shapes in geometry. jordan (talk) 23:33, 7 May 2008 (UTC)Reply

No, you can't. Scineram (talk) 14:41, 5 May 2010 (UTC)Reply

Is the term "relation" and the respecitve link, appropriate in this context? —Preceding unsigned comment added by Stdazi (talk • contribs) 22:06, 26 January 2008 (UTC)Reply

The link does not seem right. Arthena^(talk) 23:08, 26 January 2008 (UTC

I've removed the respective link temporarily... Stdazi (talk) 11:05, 27 January 2008 (UTC)Reply

I've put it back :) A "relation" is like a "multi-valued function". For example, the (x,y) co-oridinates of a circle are a relation, even though we can't write y as a function of x because there are two values. That is the type of relation defined in relation (mathematics), and the type meant here. The intro to relation (mathematics) is misleading, so it now link straight to the definition. LachlanA (talk) 19:07, 19 February 2008 (UTC)Reply

No such thing as relationx or not about these, just toolx, doesn't matter. — Preceding unsigned comment added by Lyhendl (talk • contribs) 22:56, 31 August 2018 (UTC)Reply

I think you can't get (1,0) (-1,0) because the derivative of y is 0, and the implicit fn thm has you putting things over the derivative.

The question answeared by the implicit function theorem "can we write f(x,y)=0 as y=g(x)?" is also addressed by a theorem about so-called Curve genus [1]. An implicit function can be parameterized using rational polynoms iff curve genus equals zero. This is elaborated on in [2]. The curve genus is also used in the Riemann curve theorem [3] but I can't tell if this is subsumed with a similar "genus" in the Riemann–Roch theorem. EverGreg (talk) 09:23, 15 May 2008 (UTC)Reply

This is not correct. If a curve can be parameterized by rational polynomials, then it is an algebraic curve, so f(x,y) is a polynomial. But the implicit function theorem applies to f which are not polynomial. f need not even be analytic, nor does it have to be smooth, only C¹. For instance, if you took a continuous everywhere non-differentiable function and called its antiderivative g, then the relation g(y) - x (that is, the graph of g turned a quarter circle) is a perfectly good f, and the implicit function theorem applies.

The Riemann-Roch theorem is usually stated using the geometric genus of a smooth plane curve. This is the same thing as the arithmetic genus, which is what appears in the "Riemann curve theorem" link you cited. (I've never heard that theorem given a name before, but I'm used to hearing it stated as the birational invariance of the arithmetic genus, not as a theorem about plane curves.) Ozob (talk) 23:00, 15 May 2008 (UTC)Reply

Yes, it was not correct to say that "the curve theorem" and the implicit function theorem address the same issue. I think the motivation for stating the "curve theorem" for plane curves is that the Plücker Characteristics [4] can be determined by inspecting a computer generated plot of f(x,y)=0 where f is a polynomial. This will then tell you whether or not f can be parameterized. But now that I've learned that this can be determined from the jacobian determinant, the curve theorem strikes me as a somewhat roundabout approach. :-/ EverGreg (talk) 10:54, 16 May 2008 (UTC)Reply

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)Reply

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)Reply

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)Reply

The current expression for partial derivatives seems wrong. At least, the sum indexing and vector dimensions seem to be inconsistent. Can anyone fix this and provide some proof or reference? My attempted fix apparently was undone. --Eskilj (talk) 18:28, 26 June 2018 (UTC)Reply

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)Reply

@D.Lazard: 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)Reply

The theorem is true if one takes ${\displaystyle V=\mathbb {R} ^{n}}$. In fact, if it is true for a given V, it remains true if V increases. This results from continuity and the condition ${\displaystyle g(a)=b.}$ It is necessary to specify V only if one omits the condition ${\displaystyle g(a)=b,}$ which is not the case here. In fact, specifying V is equivalent with the condition ${\displaystyle g(a)=b.}$ D.Lazard (talk) 14:46, 23 November 2018 (UTC)Reply

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)Reply

@D.Lazard: 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)Reply

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)Reply

In the sentence "From here we know that ${\displaystyle -{\tfrac {\partial _{x}F}{\partial _{y}F}}}$ 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)Reply

I don't see it either. Can someone clarify this? JBW (talk) 19:52, 1 September 2023 (UTC)Reply

This conclusion seems absolutely unfounded to me. We're talking about this:

Since ${\displaystyle F}$ is continuously differentiable and from the assumption we have

$${\displaystyle |\partial _{x}F|<\infty ,|\partial _{y}F|<\infty ,\partial _{y}F\neq 0.}$$

From this we know that ${\displaystyle {\tfrac {\partial _{x}F}{\partial _{y}F}}}$ is continuous and bounded on both ends. From here we know that ${\displaystyle -{\tfrac {\partial _{x}F}{\partial _{y}F}}}$ is Lipschitz continuous in both ${\displaystyle x}$ and ${\displaystyle y}$.

There's no way that "continuous and bounded" imply "Lipschitz continuous", so I don't see how to fix this step in the reasoning. Please, someone, look up a correct proof and put it in this article! John Baez (talk) 17:53, 23 February 2025 (UTC)Reply

I should mention that you can also find a proof at Inverse_function_theorem#Implicit_function_theorem. But of course the proof here should be corrected by someone (me!?). —- Taku (talk) 14:20, 6 March 2025 (UTC)Reply

This is ridiculous to not using here the basic theorem that a continuously differentiable function is Lipschitz continuous. Also the whole proof was ridiculous of pedantry. I rewrote it into a few elementary lines. D.Lazard (talk) 20:21, 2 April 2025 (UTC)Reply

If I am not mistaken, the problem seems that f is C^1 (continuously differentiable) so the derivative is merely continuous; not necessarily continuously differentiable. So, Lipschitzness isn't clear. I am not too sure if the proof works without C^2. -- Taku (talk) 07:56, 4 April 2025 (UTC)Reply

I agree, and I restored the {{dubious}} tag removed by my rewrite of the proof. One must look on standard textbooks to see the standard answer to this question. D.Lazard (talk) 09:10, 4 April 2025 (UTC)Reply

It must be noted that the condition "⁠${\displaystyle f}$⁠ is continuously differentiable" is stronger than "the partial derivatives of ⁠${\displaystyle f}$⁠ are continuous". For the moment, the continuity of the differential is not used in the proof. So, it is possible that C^2 is not needed, if one knows how using fully the fact that the differential is continuous. D.Lazard (talk) 16:10, 4 April 2025 (UTC)Reply

Actually, if I remember right, the continuous differentiability is equivalent to that partial derivatives exist and are continuous. So the current proof seems to use C^2. We do have the Peano existence theorem, but it only gives the existence. (Maybe we can get the uniqueness in some other way?) —- Taku (talk) 12:52, 5 April 2025 (UTC)Reply

The set V is needed for the correct formulation of the statement but it is missing in the statement. The formulation states that f(x, g(x))=0 for all x \ in U but misses the fact that all (x,y) U x V with f(x, x) = 0 satisfy y=g(x). Also this sentence is not correct: "... there exists a unique function g:U\to R^m such that g(a)=b and f(x ,g(x))=0 for all x \in U". The function is only unique if g(x) is restricted to V.

OK, I see that there is the section "Definitions" which mentions V. However, my expectation is that "Statement of the theorem" states the theorem fully. "Definitions" may define terms used in the formulation of the statement. I think this is not a good way to present such an important theorem.

The Banach space version of the theorem is stated correctly. Mc22222 (talk) 23:21, 15 March 2025 (UTC)Reply