Talk:Mean value theorem

message: Shouldn't the theorem text in the image be cropped? The theorem is stated in the article, and in greater detail. Dysprosia 11:42, 29 Aug 2003 (UTC)
thread_title: Image text

message: I moved the calculus table down to the See also section. Both the table and diagram wouldn't fit side by side at many window sizes, and there wasn't room for it right under the main picture or in any other section. Besides this, it makes sense in see also — it is, after all, a list of related links. Deco 02:33, 28 Apr 2005 (UTC)
thread_title:  Calculus table move 

message: 'Twould be nice to have something in here about how the MVT generalizes to a function f:Rn --> Rm... -GTBacchus 23:38, 31 October 2005 (UTC)
thread_title:  Generalization 

message: : moved to Talk:Mean value theorem (divided differences) Oleg Alexandrov (talk) 17:13, 20 November 2005 (UTC)
thread_title:  The section on divided differences in the article 

message: Hello, I'm french and my english is really too bad but I think, there is an error in this theorem : if you divide by g(b) - g(a), necessary, you must have g(b) \ne g(a). 
In the same way, if you divide by g'(c), you must have g'(x) \ne 0  for all x in ]a ; b[.
 
HB in wikipedia.fr
: You are right, fixed! Oleg Alexandrov (talk) 23:07, 25 February 2006 (UTC)
thread_title:  Cauchy's mean value theorem  

message: The word "integrable" links to the article on integration - but the word "integrable" does not occur in the linked article.  If anyone knows of an article in which the term "integrable" is defined, maybe you should link to that instead.
:The word appears in both Riemann integral and Lebesgue integration
:"integrable" deserves its own short article in a dictionary/encyclopedia. Could you please review this and consider to create the article? Currently there is a disambiguation page for integrability.
:IMHO meaning of "integrable"  is this:
::Integrable function $f$ is a function that
::* is suitable for integration (for example, measurable) and
::* the integral of $f$ (over its domain) is well-defined (the integral "exists") and finite
::The meaning of both conditions depends on the particular integral under consideration, see for example Riemann integral and Lebesgue integration. The first condition is sometimes neglected (by mistake or by agreement), sometimes implicitly implied by the existence of the integral.
::There is also a more general notion of "locally integrable" function, where the second condition is required only for suitable (IMHO compact) subdomains of the domain of $f$
::90.180.192.165 (talk) 17:27, 27 November 2012 (UTC)
thread_title:  Integrable? 

message: The example given in the introductory section appears to be totally wrong (or I'm just not getting it).  Currently it states:
This theorem can be understood concretely by applying it to motion: if a car travels one hundred miles in one hour, so that its average speed during that time was 50 miles per hour, then at some time its instantaneous speed must have been exactly 120 miles per hour.
Does it not actually mean that "its average speed during that time was 100 mph", and "at some time its instantaneous speed must have been  exactly 100 mph"?  I came here to discover what the MVT is, so I'm no expert, but I was pretty sure I got it before this example was given, tossing me back into doubt and despair. Is this in need of correction?
—Preceding unsigned comment added by 128.101.140.175 (talk) 20:03, 11 September 2008 (UTC)
replies: text: You were victim of a vandalised version of this article, now corrected. Always be aware that this can happen before you believe what you read. Also please add questions at the end of a talk page. Marc van Leeuwen (talk) 06:06, 12 September 2008 (UTC)
thread_title:  Error 

message: The article says:
* Only continuity of ƒ, not differentiability, is needed at the endpoints of the interval I and therefore superfluous if I is open
: 
That doesn't make sense.  "Superfluous if I is open" means the theorem is true on open intervals.  What is the statement of the theorem on open intervals?  There isn't any.  The fraction (ƒ(b) − ƒ(a))/(b − a) relies on the endpoints a and b. Michael Hardy (talk) 23:35, 26 November 2008 (UTC)
replies: text: OK, different proposition.  But the remark remains confused as it stands.  I will edit further. Michael Hardy (talk) 23:39, 26 November 2008 (UTC)
thread_title:  Confused remark 

message: I'm very tried, so maybe I'm asking a stupid question, but I thought that the mean-value theorem is false for a vector-valued function. In fact, I added a counterexample a couple of days ago. (I think I'm just missing something trivial, but just can't figure what it is.) -- Taku (talk) 00:14, 19 April 2009 (UTC)
replies: text: Note that the vecor-valued generalization claimed is NOT an exact analog of the MVT for real-valued functions. I have added a proof for the version claimed, so it should be easier to understand now. To prove the estimation if Df is bounded one needs that the norm of an integral of a vector-valued function is less than or equal to the integral of the norm of that function, but I felt this isn't the right place to provide a proof of that fact. If required it can be added of course.--Chris Barista (talk) 14:02, 19 April 2009 (UTC)
replies: text: Never mind. I've made the generalization an independent section and now there are proofs for all claims in that section. I also tried to provide some intuition why the mean value theorem in its usual form doesn't apply to vector-valued functions.--Chris Barista (talk) 16:43, 19 April 2009 (UTC)

text: First of all, the short explanation you put is very nice: if you can explain, then that's much better than a mere counterexample. Second, I'm still not sure about the vector-valued version. Isn't that version essentially the same as the fundamental theorem of calculus, written in a compact form for multi-variables. This is not surprising since we're assuming that a function is continuously differentiable. The point of the mean value theorem, as I understand, is that it applies to the case (i) when the function isn't differentiable, and (ii) the derivative actually attains a value so that the equation is satisfied. This point is much more than technicality. The theorem is often used to prove that the continuos differentiability or differentiability in a tricky setting. If you can apply the FTC, the FTC is usually better since one can obtain a better estimate. (Ok, we should stress this point in the article.) -- Taku (talk) 22:18, 20 April 2009 (UTC)
thread_title:  Vector-valued generalization 

message: Unless I'm very much mistaken, the proof of the First mean value theorem for integration given in this article doesn't actually prove what it claims. The statement it's claiming to prove is:
: ... there exists a number x in (a, b) such that...
And what it actually proves is:
: ... there exists x in [a, b] such that...
Obviously the former condition is more restrictive than the latter, so the proof is not actually proving the former statement. Unfortunately, one cannot simply change the closed interval to an open one because the conclusion of the intermediate value theorem, upon which this one is based, refers to a closed interval, not an open one. - dcljr (talk) 00:23, 23 June 2009 (UTC)
replies: text: You're right.  I've altered the statement of the theorem to refer to the closed interval rather than the open interval.  I've also modified the proof.  The intermediate value theorem won't do the trick unless the sup and inf are actually attained.  The fact that a continuous function on a closed bounded interval attains its extreme values is a separate result that needs to be mentioned.  We should probably also link to those results. Michael Hardy (talk) 01:07, 23 June 2009 (UTC)
replies: text: But what is the exact statement of the theorem? Does it refer to a closed interval? open one? (Actually I have never seen this version of mean value theorem outside Wikipedia.) -- Taku (talk) 01:55, 23 June 2009 (UTC)

text: I'm not sure if I've seen it elsewhere or not.  I'll look around........ Michael Hardy (talk) 02:02, 23 June 2009 (UTC)
thread_title:  Problem with proof? 

message: I believe the correct spelling of this should be the latter, i.e. "mean-value theorem".  After all, it's not a very mean theorem ;-)  We have a redirect from "m-v t" to "m v t", whereas Wolfram has a redirect from "m v t" to "m-v t". Papppfaffe (talk) 09:48, 25 June 2009 (UTC)
replies: text: You have a point but I have never seen "mean-value theorem". Maybe this isn't a big deal since there is only one mean value theorem, no ambiguity is a possibility. -- Taku (talk) 10:32, 25 June 2009 (UTC)
replies: text: Yes, according to English compound#Compound adjectives, hyphens are generally not used for compound adjectives in case no ambiguity is possible.  Now the question is of course whether one should think ambiguity is possible here.  Looking at Special:PrefixIndex/mean, the general consensus appears to be "no," so let me rest my case. Papppfaffe (talk) 08:09, 26 June 2009 (UTC)
replies: text:  I think the spelling should be without the hyphen.  I supposed I am influenced by the books I am familiar with, in this case the word mean is being use as a synonym for average, and I do not think it is appropriate to hyphenate average value.  I think more telling than Wolfram is that Spivak, Hughes-Hallet et al, Stewart all opt for the non-hyphenated version. Thenub314 (talk) 18:36, 26 June 2009 (UTC)

text: Based on the etymology of the term I've read (source unknown), the Central Limit Theorem is not a theorem about a "central limit", it's a limit theorem that is central, hence the hyphen would be incorrect. - dcljr (talk) 22:15, 4 August 2009 (UTC)

text: In this case I think it's preferable to follow dominant usage in the mathematical community, rather than "correct" usage. I'd only consider correctness a deciding factor if both were prevalent in textbooks and publications. Dcoetzee 22:21, 4 August 2009 (UTC)
thread_title:  "M v t" vs. "M-v t": spelling 

message: Could someone with more knowledge then me add a section about Gauss's mean value theorem, see Mathworld --ojs (talk) 20:26, 23 August 2009 (UTC)
replies: text: Isn't it the same as Cauchy's integral formula? (Just make a change of variable to get this from the usual formulation.) In other words, unrelated. -- Taku (talk) 21:19, 23 August 2009 (UTC)
replies: text: I don't know if it is the same, I would then like to see the derivation since Gauss's formula is in the complex plane, but Cauchy's in the real plane. --ojs (talk) 16:16, 4 September 2009 (UTC)
thread_title:  Gauss's mean value theorem 

message: In the header the hypothesis for the theorem is that the function is smooth.  However, in the formal statement and the proof, all that is mentioned is differentiability.  I've changed the statement in the header. —Preceding unsigned comment added by Last Octagon (talk • contribs) 21:05, 18 September 2009
: The term "smooth" is often used to indicate some amount of differentiability.  I think it's more appropriate here in the informal header section (which is just supposed to give a rough idea of the theorem), as it makes the article appeal to a wider audience (people can have a rough intuitive idea of what "smooth" is but might not know "differentiable").  As such, I've put "smooth" back in, but left "differentiable" after it in brackets, so all audiences are catered for... Tcnuk (talk) 09:15, 19 September 2009 (UTC)
replies: text: No. The term smooth is the second of two conditions required for differentiability. The first is continuity. Smoothness occurs in an interval where only one tangent line can be constructed at each point (except for inflection points). Differentiability means continuous and smooth. There is no "smooth versus differentiable".  2601:646:8B82:E709:4B5:21D:20FC:91F3 (talk) 08:52, 6 January 2017 (UTC)
replies: text: The problem with this is  that 'smooth' is a precise mathematical term meaning that a function is infinitely differentiable, which doesn't have to be the case here.  —Preceding unsigned comment added by 217.171.129.70 (talk) 07:03, 16 November 2009 (UTC)
thread_title:  Smooth vs. Differentiable 

message: I just wanted to say that as a second semester Calculus student I love the intuitive way the first two paragraphs describe the theorem! This is quite unlike most wikipedia articles about math topics (granted, many topics simply are very hard to explain layman's terms). I think its very good idea to put a rough and potentially incomplete but easy to grasp summary of a topic somewhere near the introduction of an article. -HannesJvV- (talk) 21:27, 23 September 2009 (UTC)
thread_title:  Positive feedback 

message: I think that the picture given in the Cauchy's mean value theoram section does not depict a function. Reason? draw a vertical line and it intersects the graph of the function at more than one point! Hope somebody changes this!  —Preceding unsigned comment added by Suryamp (talk • contribs) 10:34, 16 January 2010 (UTC)
replies: text: The picture is generated from two separate functions, f(t) and g(t). If you drew two separate plots, one of t and f(t), and one of t and g(t), you would find that f and g both pass the vertical line test. Strad (talk) 18:26, 16 January 2010 (UTC)
thread_title:  Wrong Picture! 

message: I have a question about how did you get the third equality in proving (*) ?
Mohammad 99 (talk) 13:45, 21 February 2011 (UTC)
thread_title:  Proof of ( * ) 

message: Notice that in the section 'Generalization for determinants' it is stated 'we get Lagrange's mean value theorem'. In no place in the article is the mean value theorem described as Lagrange's mean value theorem (except for a redirect). Either this should be removed, or the connection to Lagrange should be explained (or the lack of connection). In a lot of textbooks this theorem is indeed named Lagrange's theorem. Wishcow 06:42, 14 December 2013 (UTC)
thread_title:  Role of Lagrange 

message: I just removed a reference "On the second mean value theorem of integral". Mathematics, edited by theMath. Soc., Vol. 1 (1947). which does not appear to actually exist. The only google results for it were Wikipedia, and I could not find anything matching it in MathSciNet. 130.60.188.209 (talk) 11:38, 12 December 2014 (UTC)
thread_title:  Nonexistent reference 

message: In the last paragraph of "First Mean Value Theorem for Definite Integrals" it's claimed that there exists c in the open (!) interval (a,b).  The proof uses only c in the closed (!) interval [a,b].  I think, it can be proven for the open interval as well, but don't know how.  Can anyone correct this?
Kegelkugel (talk) 14:20, 23 April 2016 (UTC)
replies: text: Yes, I see the problem too. For now, I have added a note saying the proof is a proof of the weak version. Eventually the proof should be modified to prove the claimed statement (maybe I can do it when I find free time.) Shouldn’t it be hard, right?? —- Taku (talk) 09:43, 21 June 2022 (UTC)
thread_title:  First Mean Value Theorem for Definite Integrals: proof is incorrect 

message: Is there really a need to split into the real part and the imaginary part? Isn't it equivalent to just say f'(u)=(f(b)-f(a))/(b-a)? 84.211.101.246 (talk) 15:23, 24 May 2022 (UTC)
replies: text: Because the mean value theorem requires that we can compare numbers and we can’t compare complex numbers? The complex-valued function version seems to be just a special case of a vector-valued version to me (and I don’t know why you need holomorphic) —- Taku (talk) 18:37, 24 May 2022 (UTC)
replies: text: Sure you can compare complex numbers, when the comparison is equality rather than greater-than. And holomorphicity is needed for f' to make sense. It is also needed to distinguish it from a vector-valued function in that it makes sure the u used in the real part is the same as the u used in the complex part. Ultimately, though, you have a complex number f'(u) and you have a complex number (f(b) - f(a))/(b-a), and the article says their real parts are equal and their imaginary parts are equal. I wonder why you can't just say the two complex numbers are equal. MasterHigure (talk) 06:50, 25 May 2022 (UTC)

text: By "compare" I meant inequality since the usual proof of Rolle's theorem (a special case of the mean value theorem) uses inequalities. (I have just noticed there is a discussion of generalization to other fields at Rolle%27s_theorem#Generalizations_to_other_fields.) As for holomorphicity, no holomorphicity means the Cauchy–Riemann equations are satisfied; the differentiability is enough for f' to be defined. You said "u used in the real part is the same as the u used in the complex part. " but that's not the statement in the article; the vector you find for the real part is different from one from the imaginary part. (Not surprising since this is exactly the issue for vector-valued functions.) -- Taku (talk) 07:23, 26 May 2022 (UTC)

text: By differentiability, I didn't mean complex-differentiability (which as you said means holomorphic) and for the existence of f' when f is viewed as a function in two variables, you don't need complex-differentiability. Having said that, I now understand the difference from the several-var version; you divide by the complex number b-a, which is different from the dot product (so, apparently, holomorphicity is needed). I would love to see the proof, though (so I can see how holomorphicity is used). -- Taku (talk) 10:53, 27 May 2022 (UTC)
thread_title:  Complex generalisation 

message: There was an example in the lede that applied this theorem to average speed and instantaneous speed that was removed at some point 
It’s been a while since I did calculus, so I forgot  the exact name after I read a news article about average speed trapping (I initially confused it with the midpoint theorem).
Given that average speed trapping is probably the most common scenario where most people will encounter the application of this theorem, it might be worthwhile referencing this in the article. Park3r (talk) 12:16, 31 July 2022 (UTC)
thread_title:  Practical application