FAIR and interactive data graphics from a scientific knowledge graph

This  level-5 vital article is rated C-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Mathematics High‑priority This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics High This article has been rated as High-priority on the project's priority scale.

Using the flawed mainstream calculus and a patch I, John Gabriel, introduced to prove it:

https://drive.google.com/open?id=0B-mOEooW03iLZG1pNlVIX2RTR0E

Using the first and only rigorous formulation of calculus in human history:

https://drive.google.com/open?id=0B-mOEooW03iLblJNLWJUeGxqV0E

2601:646:8B82:E709:EC2A:A37E:89CF:63E5 (talk) 19:33, 2 February 2017 (UTC)[reply]

Dear John, Wikipedia should be a source of (widely) accepted knowledge, not a platform to spread revolutionary ideas. Limits and infinitesimals are totally accepted in mathematics. That's why your "new formulation of calculus" is very misleading for someone who is not a professional mathematician. From a mathematicians point of view, your approach just looks wrong and informal.

Kegelkugel (talk) 20:04, 2 February 2017 (UTC)[reply]

Your comments are unsubstantiated assertions. I am a professional mathematician and quite possibly the greatest mathematician today. That limits and infinitesimals are accepted in mainstream mathematics does not mean that these are valid concepts.

www.academia.edu/98758410/Was_Newtons_and_Leibnizs_method_of_setting_h_0_valid_The_answer_is_YES

My approach may "look wrong" to ignorant people like you, but once again the problem is your bad perspective or lack of it. 64.99.242.121 (talk) 11:53, 23 March 2023 (UTC)[reply]

You're not a professional mathematician of any kind. 81.230.146.243 (talk) 12:05, 5 July 2023 (UTC)[reply]

Your approach is wrong. 84.216.157.21 (talk) 10:24, 8 September 2023 (UTC)[reply]

Limits and infinitesimals are not generally accepted outside of mainstream academia. There are no real mathematicians in mainstream academia, only mythmaticians. To someone not in mainstream academia, the new calculus makes perfect sense. That's something you can't say about mainstream calculus whose results are generally correct, but the formulation is flawed. As for accepted knowledge, it is delusional to think that only it should be published. If one is not part of the academic cabal, the chances of publishing anything are zero. My ideas are not revolutionary, they are common sense. So common sense in fact, that most with a high school education can learn the new calculus without taking a lifetime of learning.

All this aside, you cannot refute that my constructive proof of the MVT was never realised by anyone before me. It hasn't been published anywhere because I am generally not liked by stupid people in the mainstream. That's a loss for future real mathematicians. 2601:646:8B82:E709:2D34:EBFE:E7A6:6A34 (talk) 21:05, 5 February 2017 (UTC)[reply]

The section has:

Define g(x) = f(x) − rx, where r is a constant. Since f is continuous on [a, b] and differentiable on (a, b), the same is true for g. We now want to choose r so that g satisfies the conditions of Rolle's theorem.

You can't just choose an r so that g satisfies the conditions of Rolle's theorem, because:

a. f(b) is not necessarily equal to f(a) in the MVT.

b. By choosing r in the way you do to "satisfy" Rolle's theorem, that is, f(a)-ra=f(b)-rb, you are in fact using the very c you set out to prove exists. Hint: you can't use c in your proof because that's what you need to prove.

Big Fail! You have been failing since the inception of Wikipedia. Mythmaticians the last 400 years have failed. I will not give you the answer because you are dishonest, ignorant and incompetent. J-O-H-N Gabriel - discoverer of the new calculus. 116.252.221.146 (talk) 06:11, 24 November 2014 (UTC)[reply]

The c in the proof is from the function g, not f.

We define g such that it gives it a c, and this c is then used to prove the MVT for f. 84.216.157.141 (talk) 09:45, 9 September 2023 (UTC)[reply]

"You can't just choose an r so that g satisfies the conditions of Rolle's theorem, because:

a. f(b) is not necessarily equal to f(a) in the MVT."

No one said it has to be, g(x) is NOT f(x). We define g(x) to be this function such that g(a) = g(b), and r is arbitrary, we have not made a statement yet of its purpose beforehand. Assuming any property we want from it is not circular, as long as it is a well defined property. Assuming a property in a matter that f'(c)=(f(b)-f(a))/(b-a) would be circular, since I am considering f(x) to be a general function, so this couldn't just be a definition, but would adding a specific property I never proved, of f(x).

"b. By choosing r in the way you do to "satisfy" Rolle's theorem, that is, f(a)-ra=f(b)-rb, you are in fact using the very c you set out to prove exists. Hint: you can't use c in your proof because that's what you need to prove. "

Wrong. What we are trying to prove is that there is a c in between interval [a,b] such that f'(c)=(f(b)-f(a))/(b-a). Remember, we have said no property of r, there is not said one thing of r in the statement. We have not yet stated its purpose, this is how we DEFINE g(x) and r, and yet not limit f(x) in any way. Whether or not it is well defined is a different topic. And it is as will be shown below.

The idea is I have my functions f(x) which I can not make any definitions on besides the original statement, otherwise would not be general. I define some arbitrary function, g(x), such that g(x)=f(x)-rx for r. Of course, I don't know r, and so can add an arbitrary condition on g(x) such that I can now choose an r. That is, let g(a) = g(b). There is no mention of g(x) in the MVT statement, this is how we define g(x) by definition, which will satisfy the property irrelevant of f(x). For this to be the case, I then am able to derive r. Again, these properties is how it is defined, and makes no claim of the nature of f(x) in any statement. From this, I derive r = (f(b)-f(a))/(b-a) which is always well defined since b=/=a and regardless of my function f(x), will find g(a) = g(b). Whenever I plug in x=a or x=b, will necessarily, based off of this definition of g(x) and r, get g(a) = g(b).

So there is nothing wrong here, as I can define some random g(x), however I want, as long as the definition is always applicable for any f(x)(THEN I would be assuming a property of f(x) for g(x) to then apply) in terms of some r, and these are definitions, not assumptions on the nature of f(x).

________________________________________

Your comments are noted as flawed. You haven't refuted anything that I've said. — Preceding unsigned comment added by 172.92.105.191 (talk) 00:07, 14 October 2022 (UTC)[reply]

"Wrong. What we are trying to prove is that there is a c in between interval [a,b] such that f'(c)=(f(b)-f(a))/(b-a). Remember, we have said no property of r, there is not said one thing of r in the statement. We have not yet stated its purpose, this is how we DEFINE g(x) and r, and yet not limit f(x) in any way. Whether or not it is well defined is a different topic. And it is as will be shown below."

You're wrong. If you say f(a)-ra=f(b)-rb, then you have already assumed that r has that property. Your flawed thinking and conclusions are nothing new. 2A02:587:AF28:900:2947:A928:1446:5D1A (talk) 16:42, 11 August 2024 (UTC)[reply]

And you do not get to "define anything" according to your whimsical notions. Definitions must be well formed. 2A02:587:AF28:900:2947:A928:1446:5D1A (talk) 16:43, 11 August 2024 (UTC)[reply]

Hi, can anyone here help me out? I'm looking for applications of the MVT for integration? It's certainly easy to understand, but why do it? Why do we have the theorem? THANKS! -TN

You won't find anything like that here. Try the new calculus of John Gabriel. There is no separate theorem for integration, only a variant of the mean value theorem which has never been understood the last 400 years. 116.252.221.146 (talk) 06:06, 24 November 2014 (UTC)[reply]

https://www.stumblingrobot.com/2015/10/23/application-of-the-mean-value-theorem-for-integrals/

https://math.stackexchange.com/questions/2275527/application-of-mvt-for-integrals

https://www.jstor.org/stable/24340031

Just some examples. 84.216.157.141 (talk) 09:39, 9 September 2023 (UTC)[reply]

The fact that you are asking such a question is evidence that the mainstream have never understood that the mean value theorem IS the fundamental theorem of calculus. 2A02:587:AF28:900:2947:A928:1446:5D1A (talk) 16:45, 11 August 2024 (UTC)[reply]

This mvt things badly needs a picture to clear things up.

The agreed policy is that all words in a species' official common name should be capitalised, other than those following a hyphen if they refer to a part of the animal: "Bald Eagle", "Red-necked Phalarope", "Wilson's Storm-Petrel".

The biology convention appears to be applicable to math as well. Pizza Puzzle

...and were the mean value theorem an animal, I'd agree with you. -- The Anome 14:13 2 Jul 2003 (UTC)

Well, so much for logic...do you have some conventional ruling which you feel somehow overrids that and the convention that proper names should be capitalized? How is it easier to read Joseph-Louis LaGrange; instead of simply LaGrange. Students of the Mean Value Thoerem do not need to be on a first name basis with Joe. Pizza Puzzle

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)

The illustration says "there exist", where evidently it means "there exists". Could someone correct this? -- I don't know what software created the illustration. Thanks. Michael Hardy 00:22, 15 Nov 2003 (UTC)

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)

'Twould be nice to have something in here about how the MVT generalizes to a function f:Rⁿ --> R^m... -GTBacchus 23:38, 31 October 2005 (UTC)[reply]

moved to Talk:Mean value theorem (divided differences) Oleg Alexandrov (talk) 17:13, 20 November 2005 (UTC)[reply]

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 .

In the same way, if you divide by g'(c), you must have for all x in ]a ; b[.

HB in wikipedia.fr

You are right, fixed! Oleg Alexandrov (talk) 23:07, 25 February 2006 (UTC)[reply]

I think it should also be mentioned that as a consequence the geometric form of the theorem must allow for an exception when for some x: in this case the conclusion of Cauchy's theorem is satisfied for , but no tangent is (probably) defined at this singular point of the parametric curve. As an example take on the interval [-1,1]. Marc van Leeuwen (talk) 07:29, 20 March 2008 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

OK, different proposition. But the remark remains confused as it stands. I will edit further. Michael Hardy (talk) 23:39, 26 November 2008 (UTC)[reply]

Does anyone know of a reference (or a least a proof) of the Mean Value theorem in the case that we allow infinite derivatives. I haven't thought about this much but it seems to me that one can construct functions that have an infinite derivative on very "large" sets. I checked a few of my introductory analysis texts, and did a few google searches but I haven't had any luck finding this statement. Before I did anymore leg work, I thought someone here might know. Thenub314 (talk) 09:32, 20 January 2009 (UTC)[reply]

Hi. I don't really understand your question. Is it related to a specific part of the article? I didn't see it. Is your question about the very classical formula at the head of the article? In this case I think it is wrong. Or about the integral formulas? --Bdmy (talk) 10:40, 20 January 2009 (UTC)[reply]

Not a problem, I figured out how to prove the statement I was after, which is effectively the same as the usual proof. I was referring to the sentence:

The mean value theorem is still valid in a slightly more general setting. One only needs to assume that f : [a, b] → R is continuous on [a, b], and that for every x in (a, b) the limit

exists as a finite number or equals +∞ or −∞.

Well, I missed that sentence when I read rapidly through the article. In the meantime I realized just what you say, that (for example) Rolle's theorem is clear if the derivative exists in the extended sense, simply because the derivative can't be or where the function assumes a maximal value. When I said it was probably wrong, I had in mind trivial examples with half-derivatives with opposite infinite values on each side. But just as you, I don't have a reference to give. The case of Cantor function that you mention is not clear to me, because there are many points where the derivative does not exist in the previous sense. I don't know if there is a function having this extended derivative everywhere and infinite on an uncountable set. --Bdmy (talk) 12:16, 20 January 2009 (UTC)[reply]

There is, it is not the cantor function, but a related function which has infinite derivative on an uncountable set. There is a construction in Boas & Boas "A Primer of real functions" (pp 164-165). He uses it to give an example where you have two functions with everywhere equal (not necessarily finite) derivatives, that do not differ by a constant.Thenub314 (talk)

Which bothered me momentarily when I started thinking about a function on [0,1] whose derivative is positive or +∞, and it is +∞ on the Cantor set. But you get very odd behavior allowing functions whose derivative is infinite, the sum and difference of such functions may no longer have derivatives at every point etc. But I believe the statement is correct, at least I have a (trivial) proof of it. But it would still be nice to have a reference, since it is not so standard to allow the function to have infinite derivatives. Most likely it is an exercise somewhere. Thenub314 (talk) 11:28, 20 January 2009 (UTC)[reply]

Hi. The following generalization is not true:

The mean value theorem is still valid in a slightly more general setting. One only needs to assume that f : [a, b] → R is continuous on [a, b], and that for every x in (a, b) the limit

exists as a finite number or equals +∞ or −∞. If finite, that limit equals f' (x).

The limit cannot be infinite. Take for example the following function on interval [0, 1]

According to the theorem, there should be c such that f' (c) = 1. There isn't any, f' (x) equals to 0, 2, and +∞ only. —Preceding unsigned comment added by 78.128.165.70 (talk) 15:03, 3 April 2009 (UTC)[reply]

Hi. In your example the required limit does not exist at x = 0.5 and it does not show that the generalization is not true. --Bdmy (talk) 19:07, 3 April 2009 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

Maybe this is beating a dead horse, but just for comparison we have central limit theorem as opposed to "central-limit theorem". -- 22:24, 26 June 2009 (UTC)

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

I second that.........I am seeing a few other similar comments in other technical articles discussions within Wiki. Most articles still need 'dumb-down' introductions to be added though. —Preceding unsigned comment added by 99.147.240.11 (talk) 14:37, 6 September 2010 (UTC)[reply]

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)[reply]

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)[reply]

WTF???

There is no reference to this picture in the text. How does this picture help with the understanding of the Mean value theorem?

I have a question about how did you get the third equality in proving (*) ?

Mohammad 99 (talk) 13:45, 21 February 2011 (UTC)[reply]

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) — Preceding unsigned comment added by Wishcow (talk • contribs)

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)[reply]

I have added a new reference to an article by Hobson in 1909 and accordingly also removed the text claiming "This variant was proved by Hiroshi Okamura in 1947." Since precisely the same formulation was known in 1909 it cannot have been proved by Okamura in 1947, and MathSciNet also turned up no results for Okamura and mean value theorems. There had been a 'citation needed' tag on the claim since 2007 so I think it is high time to remove it. — Preceding unsigned comment added by 130.60.188.209 (talk) 11:47, 12 December 2014 (UTC)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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's_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)[reply]

You're right, I didn't notice that u and v were different symbols. That being said: Complex differentiable does actually mean holomorphic (and also analytic). These three terms mean different things, but they are equivalent in the case of complex-valued functions in the complex plane. So yes, holomorphicity is needed if you want the derivative of f to be a single complex number (two real degrees of freedom for the value of the derivative at any given point), as opposed to a real-differentiable vector field in the plane (four real degrees of freedom). MasterHigure (talk) 09:08, 27 May 2022 (UTC)[reply]

By differentiability, I didn't mean complex-differentiability (which as you said means holomorphic) and for the existence of 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)[reply]

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)[reply]