Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Many famous results were discovered through non-rigorous proofs, with correct proofs being found only later and with greater difficulty. One that is well known is Euler's 1737 proof that

$1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots =\frac{\pi^2}{6}$

in which he pretends that the power series for $\frac{\sin\sqrt{x}}{\sqrt{x}}$ is an infinite polynomial and factorizes it from knowledge of its roots.

Another example, of a different type, is the Jordan curve theorem. In this case, the theorem seems obvious, and Jordan gets credit for realizing that it requires proof. However, the proof was harder than he thought, and the first rigorous proof was found some decades later than Jordan's attempt. Many of the basic theorems of topology are like this.

Then of course there is Ramanujan, who is in a class of his own when it comes to discovering theorems without proving them.

I'd be interested to see other examples, and in your thoughts on what the examples reveal about the connection between discovery and proof.

Clarification. When I posed the question I was hoping for some explanations for the gap between discovery and proof to emerge, without any hinting from me. Since this hasn't happened much yet, let me suggest some possible explanations that I had in mind:

Physical intuition. This lies behind results such as the Jordan curve theorem, Riemann mapping theorem, Fourier analysis.

Lack of foundations. This accounts for the late arrival of rigor in calculus, topology, and (?) algebraic geometry.

Complexity. Hard results cannot proved correctly the first time, only via a series of partially correct, or incomplete, proofs. Example: Fermat's last theorem.

I hope this gives a better idea of what I was looking for. Feel free to edit your answers if you have anything to add.

share|improve this question
5  
I was thinking also of stuff like Witten. –  Steve Huntsman Jun 11 '10 at 1:01
11  
In Tom Hales account of Jordan's proof, he states that there is essentially no problem with Jordan's original proof, and that claims to the contrary are themselves wrong or based on misunderstandings. As far as I can tell, he is correct, and there is no reason to impugn Jordan's original proof. (See "Jordan's proof of the Jordan curve theorem" at math.pitt.edu/~thales/papers ) –  Emerton Jun 11 '10 at 2:49
1  
@Emerton. I stand corrected. Maybe Jordan's proof should be in the same category as Heegner's: thought to be incorrect, but essentially correct when properly understood. –  John Stillwell Jun 11 '10 at 3:09
10  
A further remark: I think that is important to distinguish between polishing an argument, or perhaps interpreting it in terms of contemporary language and formalism, which will almost always be required when reading arguments (especially subtle ones) from 100 or more years ago, and genuinely incomplete arguments. As an example of the latter, one can think of Riemann's arguments with the Dirichlet principle, where this result was simply taken as an axiom. Additional work was genuinely required to validate the Dirichlet principle, and thus complete Riemann's arguments. –  Emerton Jun 11 '10 at 5:59
3  
I would argue that (although it came after the drive for rigor had already started thanks to Cantor, Weierstrass, et al.) the dawn of modern statistical and quantum physics had a great deal to do with the consolidation of rigor throughout mathematics. Indeed, ergodic theory and functional analysis owe a great deal to these disciplines, and neither could have existed in the time of (say) Euler because the approach to mathematics was different. –  Steve Huntsman Jun 11 '10 at 12:32
show 9 more comments

36 Answers

The (sharp) bound on the number of non-repelling cycles of a rational map of a Riemann sphere, sometimes called Fatou-Shishikura inequality, is such an example. It says that a rational map $f: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$ of degree $d \geq 2$ has at most $2d-2$ non-repelling (i.e., attracting or neutral) cycles.

This bound was first stated (without proof or even any particular motivation) by Lucjan Emil Boettcher, in his paper ''Zasady rachunku iteracyjnego (czesc pierwsza i czesc druga) [Principles of iterational calculus (part one and two)]", Prace Matematyczno Fizyczne, vol. X (1899-1900), pp. 65-86, 86-101. In 1920 it was formulated independently by Pierre Fatou. He managed to prove a weaker estimate, by $4d-4$. Later Adrien Douady and John Hamal Hubbard proved the conjectured estimate in the case when $f$ is polynomial, and finally Mitsuhiro Shishikura proved it in the general case, using the theory of quasiconformal surgery. (Shishikura, Mitsuhiro: Surgery of complex analytic dynamical systems. In: Dynamical systems and nonlinear oscillations (Kyoto, 1985), 93-105, World Sci. Adv. Ser. Dynam. Systems, 1, World Sci. Publishing, Singapore, 1986). Subsequently, another proof was given by Adam L. Epstein: http://arxiv.org/pdf/math/9902158.pdf

share|improve this answer
add comment

According to Atiyah (Responses to: A. Jaffe and F. Quinn, ``Theoretical mathematics: toward a cultural synthesis of mathematics and theoretical physics'' Bull. Amer. Math. Soc. (N.S.) 29(1993), no. 1, 1--13; MR1202292 (94h:00007)) Hodge's proofs on what is now called Hodge Theory (representation of deRham cohomology classes by harmonic forms) were incorrect, because Hodge was not an analyst, though the theory was correct.

share|improve this answer
add comment

Results in complexity theory such as $P \neq NP$.

Philosophically, it makes sense that there is a difference between verification and search, and no-one has discovered a counterexample.

(Note, that this result is not strictly speaking known to be correct. However, it is believed to be correct and routinely used as if it were simply true. No one, as far as I can tell, would ever begin a proof in complexity theory by assuming $P = NP$. )

share|improve this answer
show 2 more comments

Renormalizations in QFT

Renormalizations as discarding perturbative corrections to masses and charges were not easily accepted, even by their inventors, because of being obviously anti-mathematic. It remains to be a prescription, lucky in some rare cases and wrong in the others.

In Physics we use a perturbation theory where the perturbation is supposed to be small but it is "big" in QFT. First we write down a non perturbed Hamiltonian, let's say:

$\hat H_0 = -\frac{\hbar^2}{2m_e}\frac{d^2}{dx^2} + \hat{V}_0 (x)$ (1)

Everything in it is quite physical including the electron mass. Then we "develop" our theory and include, as we think, a small interaction that has also a kinetic and a potential term:

$\hat H_{int} = -\epsilon\frac{d^2}{dx^2} + \hat{V}_1 (x)$ (2)

The kinetic term shifts the particle mass, it is obvious. But our mass is already good in (1) and any its shifting worsens agreement with experiment. Discarding this correction "restores" the right kinetic part of the Hamiltonian, and taking $\hat{V}_1$ into account improves agreement with experiment. So the discarding practice became a part of QFT calculations.

Appearance of a kinetic perturbative term is due to our misunderstanding interactions. Some part of interactions cannot be treated perturbatively but should be present in the zeroth-order approximation. Discarding is a very bad practice. For (2) it may luckily work, but for other our guesses of interactions it can be more complicated and be just "non renormalizable".

Although shown on a simplest example, the renormalizations in QFT have nothing else in their meaning but repairing a wrongly guessed Hamiltonian via repairing the corresponding solutions. Normally it is difficult to see explicitly that some part of guessed interaction, namely a "self-action" term, is of a kinetic nature. That is why presently they "explain" renormalizations differently.

A correct theory development should not include kinetic perturbative terms. Then the perturbative series will be reasonable, in my opinion.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.