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Who was the first to prove the mean value theorem, i.e., the existence of an intermediate point where the slope equals the average slope over the interval?

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  • $\begingroup$ Related: mathoverflow.net/questions/184358/… . So certainly Rolle didn't do it, and any proof would have had to be after 1691 (since proving the mean value theorem would have automatically proved Rolle's as a special case). There may be no well-defined answer to the question, because the answer may depend on what people think of as a rigorous proof, and historical judgments about when the real number system was formulated with mathematical precision. $\endgroup$
    – user466
    Jun 21, 2016 at 17:56
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    $\begingroup$ This article : abesenyei.web.elte.hu/publications/meanvalue.pdf : gives a long (and rather repetitive) history, but requires a good grasp of a number of different languages. $\endgroup$
    – nwr
    Jun 21, 2016 at 18:24
  • $\begingroup$ The relevant part of the pdf link posted by Nick R seems to be section 9. $\endgroup$
    – user466
    Jun 21, 2016 at 20:17
  • $\begingroup$ @NickR, would you care to elaborate a bit and format your comment as an answer? $\endgroup$ Jun 22, 2016 at 8:36
  • $\begingroup$ @MikhailKatz I have posted an answer as per your request. Please note the additional reference which was necessary to clear up an potential ambiguity in Besenyei's paper concerning authorship of the relevant proof. $\endgroup$
    – nwr
    Jun 23, 2016 at 4:18

2 Answers 2

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As per your comment requesting details of the linked paper :

Besenyei's paper begins with a history of the development of Rolle's Theorem into its general form. It follows this with the history of various generalisations of Rolle's Theorem equivalent to MVT and attributed to the likes of Cauchy, Bonnet, Serret, Dini, and Harnack.

Looking at Besenyei's paper, at first it would appear that Joseph Alfred Serret is the first to state and prove the result in its modern form, however this is not the case - please see below. From Serret's Cours de calcu infinitesimal of 1868 as Theorem I (translation) :

Let $f(x)$ be a function of $x$ which remains continuous for values of $x$ between two given limits, and which, for these values, has a well-determined derivative $f'(x)$. If $x_0$ and $X$ denote two values of $x$ between these same limits, the following $$\frac{f(X) - f(x_0)}{X-x_0} = f'(x_1)$$ will hold, with $x_1$ being a value between $x_0$ and $X$.

Besenyei's paper does not include the relevant proof.

This appears on the page denoted 11/15 (lower right hand corner) - page 83/152 of the PDF document.

However, according to this paper by Renaud Chorlay, Serret is here simply stating Bonnet's proof. (See page 7 of 16 of the PDF document.) Thus, the correct answer as to who first proved MVT in its modern form appears to be Pierre-Ossian Bonnet. The Renaud Chorlay paper does appear to give Bonnet's proof.

This conclusion appears to be contrary to the wikipedia entry for the MVT which credits Cauchy as the first to state it in its modern form. However, Cauchy's statement of MVT is actually and extension of MVT to deal with two different continuous functions $f$ and $g$ which states $$\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)}$$ and which predates the Serret/Bonnet version of MVT.


EDIT

As Mikhail has pointed out in his comment, the correct answer is in fact Cauchy, as is evident in Besenyei's paper. Cauchy gives both forms of stating the MVT.

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    $\begingroup$ (1) As Besenyei notes, Cauchy gave both forms of the theorem; (2) As Besenyei notes, Serret presented Bonnet's proof. $\endgroup$ Jun 23, 2016 at 11:32
  • $\begingroup$ @MikhailKatz I surely must be blind! Upon re-examining Besenyei's paper, sure enough, there it is as plain as day, Cauchy does indeed give both forms. Also, Besenyei's presentation of Serret's statement indented under Bonnet's entry is almost as clear. Thanks for accepting my answer, even if I got it wrong! $\endgroup$
    – nwr
    Jun 23, 2016 at 16:38
  • $\begingroup$ (The link to Chorlay’s paper seems broken.) $\endgroup$ Apr 25, 2018 at 2:47
  • $\begingroup$ Yes, it's very odd. Here is an alternative link to Chorlay's paper - perhaps an alternative version of the original link??. I shall edit the broken link as well. $\endgroup$
    – nwr
    Apr 25, 2018 at 3:37
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    $\begingroup$ It would be interesting to see how solid Cauchy's proof is, and what hypotheses he exploits exactly (differentiability, $C^1$, etc.). $\endgroup$ Apr 26, 2018 at 9:28
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The theorem is black on white in Lagrange’s Théorie des fonctions analytiques, as the initial case of “Taylor’s theorem with Lagrange remainder” (1797, §§51–53):

Hence the value of the quantity $P$ relative to $z=1$ [i.e., $\frac{f(x)\ -\ f(0)}x$] can be expressed as $f'u$, $u$ being a quantity between $0$ and $x$. Likewise (...)

52. Whence there results finally this new theorem, remarkable in its simplicity and its generality, that denoting by $u$ a quantity unknown but comprised between the limits $0$ and $x$, one can successively develop any function of $x$ and of arbitrary other quantities according to the powers of $x$, thus (...) \begin{align} f(z+x) &= fz + x\,f'(z+u),\\ &=fz + x\,f'z + \frac{x^2}2\,f''(z+u),\\ &=fz + x\,f'z + \frac{x^2}2\,f''z + \frac{x^3}{2.3}\,f'''(z+u),\\ &\ \ \&\mathrm c. \end{align}

Lagrange also spells out the “Rolle” case in his next book (1798, pp. 166, 175):

Consideration of the maxima and minima of parabolic lines led Stirling to a method for counting and bounding real roots in degrees 3 and 4, which Euler generalized in his Differential calculus [§298]. This method boils down, basically, to Rolle’s (... :)

between two consecutive real roots of the equation $\mathrm F\,x=0$, there always falls a real root of the equation $\mathrm F'x=0$.

So I question the accepted attribution to Cauchy (Wikipedia 2007, 2016) — though of course it all hinges on: “first to prove” under what assumptions, and how rigorously? For discussion of that: e.g.

Edit as requested:

Briefly, Dugac contends (and repeats in a later paper and book) that 1º) Lagrange first singled out, stated, and proved — perhaps inconclusively — the mean-value property for analytic functions; 2º) he was to an extent foreshadowed by Cavalieri (1635, p. 19); 3º) the first conclusive or “rigorous” proof in print is by Dini (1878, p. 71). To me this is all rather noncontroversial: the book I learned it from called it “Lagrange’s theorem”. So do quite a few others. Current textbooks (Thomas, Stewart, Petrovic,...) credit it to Lagrange. So do Bottazzini (1986, p. 53), Grabiner (1981, p. 122), Edwards (1979, p. 313), Cajori (1910, pp. 308-311), the Encyklopädie (1899, II A 2 §§7, 11), and so on all the way to Lacroix (1810: vol. 1, pp. liv, 385-386; 1800: vol. 3, pp. vii, 384). I am beginning to wonder if Besenyei might be the only author who doesn’t attribute the theorem to Lagrange?

That said, it is true that a wide shift of emphasis happened between the way Lagrange thought of his theorem and the way we use (and prove) it now. Besides already containing most of the above story, the French Encyclopédie article (1912, II 3 §§11, 22) says this well in footnote 116:

From Rolle’s theorem114) one readily deduces115) the mean value formula which plays a basic role throughout Calculus116) and is also, like Rolle’s theorem, basically a mere translation into precise language of the intuitive fact that B. Cavalieri had brought to the geometers’ attention.


114) On the first rigorous proofs of Rolle’s theorem, see A. Pringsheim (1900, p. 454); see also U. Dini (1878, p. 75); A. Harnack (1881); M. Pasch (1882); P. Mansion (1887); J. Tannery (1886); A. Demoulin (1902).

115) In his lectures at École polytechnique, O. Bonnet fully rigorously demonstrated this formula on the basis of Rolle’s theorem. He also gave a proof of Rolle’s theorem which, at least as exposed in J. A. Serret (1868, pp. 17-19), is far from immune to objections. (...)

116) One should note that this fundamental role stems mainly from the mode of exposition of Calculus generally adopted today. For J. L. Lagrange, who took a different point of view, this formula was merely a consequence of Taylor’s formula (1797, p. 49).

To summarize, my take for now (based on these references and subject to your corrections):

1635:  Cavalieri states MVT hence Rolle. (Proof rejected by Guldin, 1640.)
1690:  Rolle proves Rolle for polynomials.
1755:  Euler states Rolle for differentiable functions. (Sketch of proof accepted at the time.)
1797:  Lagrange proves MVT (hence Rolle) for analytic functions. (Proof accepted at the time.)
1823:  Cauchy proves MVT (hence Rolle) for C1 functions. (Proof essentially same as Lagrange.)
1868:  Bonnet rearranges proof of MVT to rely on Rolle as a lemma.
1878:  Dini proves Rolle (hence MVT now) for differentiable functions. (Our current proof.)

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  • $\begingroup$ +1 for an interesting post. Note that the theorem as stated in Lagrange is apparently incorrect as stated even for an analytic function, since the point $u$ apriori depends on whether you are in the first line, the second line, or the third line. If you get a chance could you provide a summary of the Dugac/Grabiner/Persson commentary? $\endgroup$ Apr 22, 2018 at 7:08
  • $\begingroup$ Bonnet's proof does not work for a differentiable (not $C^1$) function whereas according to Besenyei, Cauchy's does. So it does not seem controversial to attribute the first correct proof to Cauchy. $\endgroup$ Apr 24, 2018 at 9:56
  • $\begingroup$ All calculus textbooks use the term "Rolle's theorem" even though everybody knows Rolle did not prove what we call Rolle's theorem today. I think the same remark can be made about Lagrange's theorem. $\endgroup$ Apr 24, 2018 at 9:57
  • $\begingroup$ Of course. Lagrange stated a theorem and gave a proof. As with Pythagoras, Rolle, Cauchy-Schwarz, Stokes, etc., setting and generality later evolved. $\endgroup$ Apr 24, 2018 at 13:57
  • $\begingroup$ I hear what you are saying but a literal application of Stigler's law of eponymy would have the opposite effect, namely that this result would be named after the person who popularized it more than Lagrange did, namely Cauchy. $\endgroup$ Apr 24, 2018 at 14:07

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