Talk:Mean value theorem

Help! MVT for Integration?

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

Picture

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

Name

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

Image text

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)

Calculus table move

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)

Generalization

'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)

The section on divided differences in the article

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

Cauchy's mean value theorem

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)

I think it should also be mentioned that as a consequence the geometric form of the theorem must allow for an exception when $f'(x)=g'(x)=0$ for some x: in this case the conclusion of Cauchy's theorem is satisfied for $c=x$, but no tangent is (probably) defined at this singular point of the parametric curve. As an example take $(f(x),g(x))=(x^3,x^2+1)$ on the interval [-1,1]. Marc van Leeuwen (talk) 07:29, 20 March 2008 (UTC)

Integrable?

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.

Error

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)

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)

Confused remark

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&nbap;b. Michael Hardy (talk) 23:35, 26 November 2008 (UTC)

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

The slight generalization.

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)

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)

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

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

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 $+\infty$ or $-\infty$ 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)

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)

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

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

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]

$\begin{align} f(x) & = 0 \qquad & 0 \le x < 0.5 \\ f(x) & = 2x-1 \qquad & 0.5 \le x \le 1 \\ \end{align}$

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)

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)

Vector-valued generalization

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)

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)

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)

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)

Problem with proof?

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)

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)

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)

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)

"M v t" vs. "M-v t": spelling

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)

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)

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)

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)

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)

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)

Gauss's mean value theorem

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)

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)

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)

Smooth vs. Differentiable

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)

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)

Positive feedback

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)

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)

Wrong Picture!

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)

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)

Government of Beijing celebrates the mean value theorem

WTF???

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

Proof of ( * )

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

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