Non-rigorous reasoning in rigorous mathematics - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:46:04Z http://mathoverflow.net/feeds/question/115032 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics Non-rigorous reasoning in rigorous mathematics David Harris 2012-11-30T23:12:13Z 2012-12-05T00:24:42Z <p>I was wondering what role non-rigorous, heuristic type arguments play in rigorous math. Are there examples of rigorous, formal proofs in which a non-rigorous reasoning still plays a central part?</p> <p>Here is an example of what I am thinking of. You want to prove that some formula $f(n)$ holds, and you want to prove this by induction. Based on heuristic arguments, you conjecture what the correct formula is. Then you prove it by induction. But, if you had just given the induction proof on its own, then you would have to pluck this mysterious formula out of thin air.</p> <p>I am interested in situations in which there is a heuristic argument which is valid and can be formalized. I am more interested in cases in which there is a heuristic argument and a separate (or complementary) rigorous argument, but the heuristic argument is more enlightening and more explanatory.</p> http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics/115037#115037 Answer by Rodrigo A. Pérez for Non-rigorous reasoning in rigorous mathematics Rodrigo A. Pérez 2012-12-01T00:15:36Z 2012-12-01T00:15:36Z <p>You do "pluck the mysterious formula from thin air". That is why there are (not enough) jobs available to mathematicians: Doing Math is not something you can leave to a computer program. </p> <p>The german word ansatz describes this mental process very well. To solve a system of linear ODEs you assume the solutions are exponentials, and then proceed to find the coefficients. This <i>assumption</i> step is where intuition takes place.</p> <p>Of course this example is old, trivial, and well known, but similar insights are part of all new results. Your intuition shows you the way and THEN you formalize the proof.</p> http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics/115038#115038 Answer by Brian Rushton for Non-rigorous reasoning in rigorous mathematics Brian Rushton 2012-12-01T00:16:44Z 2012-12-01T00:16:44Z <p>I feel that almost all of math is this way. One specific example is the Riemann mapping theorem for annuli (which is equivalent to the standard Riemann mapping theorem). Riemann is said to have conceived of the idea by imagining current flowing from the inside of an annulus to the outside. The current flows and equipotentials would form an orthogonal set of coordinates which could be "stretched out" to form a perfect cylinder. Riemann's first proof of this theorem was shown to have an error, but he reportedly commented that it didn't matter, because he knew the theorem was true anyway. (Most of this comes from Jim Cannon's paper The Combinatorial Riemann Mapping Theorem). </p> http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics/115043#115043 Answer by J. H. S. for Non-rigorous reasoning in rigorous mathematics J. H. S. 2012-12-01T01:10:30Z 2012-12-01T05:43:59Z <p>I think that the following "derivation" of the Prime Number Theorem from the well-known identity</p> <p>$\sum_{d|n}\Lambda(d) = \log n$</p> <p>is a particularly prominent example of what you are asking. Indeed, it follows from the said identity that</p> <p>$\sum_{n \leq x} \sum_{d|n} \Lambda(n) = \sum_{d\leq x} \Lambda(d)\sum_{n \leq x, d|n} 1 = \sum_{d\leq x}\Lambda(d)\lfloor \frac{x}{d}\rfloor$</p> <p>and whence,</p> <p>$\sum_{n \leq x}\Lambda(n)\lfloor \frac{x}{n} \rfloor = \sum_{n \leq x} \log n \sim x \log x$.</p> <p>Now if we replaced the $\lfloor x/n \rfloor$ in the previous line by $x/n$, we would get</p> <p>$\sum_{n\leq x}\frac{\Lambda(n)}{n} \sim \log x \sim \sum_{n \leq x}\frac{1}{n}$.</p> <p>This might lead us to ascertain that the function $\Lambda$ of von Mangoldt behaves in the average like the arithmetical function that is identically equal to $1$, thus</p> <p>$\psi(x):=\sum_{n\leq x} \Lambda(n)\sim x.$ (<strong>Voilà!</strong>)</p> <p>As to the formal version of the preceding argument you may want to take a look at sections 9.9 through 9.12 of [<strong>2</strong>]. You are to find there a proof of the Prime Number Theorem (presumably due to Ingham) based on the estimate</p> <p>$\sum_{n \leq x} \psi(\frac{x}{n}) = x\log x - x+ O(\log x), \quad x \geq 1.$</p> <p>According to Prof. Balanzario (see [<strong>1</strong>, page 59]): "This demonstration ... is the correct version of our heuristic reasoning [given above]." </p> <p><strong>References</strong></p> <p>[<strong>1</strong>] E. P. Balanzario. <em>Breviario de Teoría Analítica de los Números</em>. SMM, México, 2003. </p> <p>[<strong>2</strong>] W. Rudin. <em>Functional Analysis</em>. Tata McGraw Hill Publishing Company Ltd., 1974.</p> http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics/115044#115044 Answer by R Hahn for Non-rigorous reasoning in rigorous mathematics R Hahn 2012-12-01T01:49:59Z 2012-12-01T01:49:59Z <p>In mathematical statistics people often have experience about some method that works well in practice even though it "shouldn't" in all generality. The game is then to ask what conditions need to be satisfied to explain why the method works. </p> <p>Here is an example which I was not personally involved in, so I can only speculate. <a href="http://arxiv.org/pdf/0708.0983.pdf" rel="nofollow">This paper by Bickel and Li</a> considers local polynomial regression methods and shows that they works as well as possible (in the sense of asymptotic optimality) when the data it is being used on has low dimensional structure. The idea is that people were finding that certain regression techniques were giving reasonable generalization performance in prediction problems even when the data was high dimensional so they figured that maybe the data wasn't actually high dimensional in some relevant aspect. But <em>which</em> relevant aspect, that's the challenging part.</p> <p>To my mind, figuring out how to explicitly articulate the minimal conditions under which some obvious" fact is true is where the discover and understanding come in. It is a very different process than what a student does on problem set, where the statement and all the relevant conditions are laid out and the main job is deriving the stated implication.</p> <p>Put another way: research has degrees of freedom on both ends -- you can find/create the answer and the question as pairs, rather than being handed the one and being asked to complete the set. This perspective of course doesn't cover all cases -- notably, that of people chasing down famous open problems. But it is a way in which one can develop a rigorous understanding from non"-rigorous reasoning. When one first starts thinking vaguely about a problem there is nothing there about which to be rigorous.</p> http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics/115045#115045 Answer by Reladenine Vakalwe for Non-rigorous reasoning in rigorous mathematics Reladenine Vakalwe 2012-12-01T02:04:29Z 2012-12-01T02:04:29Z <p>Unless I am misunderstanding, the Weil conjectures fit into this framework. I believe it took about two decades for the Grothendieck school to formalize Weil's heuristic that his conjectures follow from a Lefschetz fixed point formula for varieties over finite fields. Rather than try to flesh this post out, let me point to the Brian Osserman's article for the PCM: <a href="http://www.math.ucdavis.edu/~osserman/math/pcm.pdf" rel="nofollow">http://www.math.ucdavis.edu/~osserman/math/pcm.pdf</a>. The Wikipedia account of the history also seems to be not bad (but I haven't really read it in detail): <a href="http://en.wikipedia.org/wiki/Weil_conjectures" rel="nofollow">http://en.wikipedia.org/wiki/Weil_conjectures</a>. I also seem to remember learning about the history and some of the mathematics for the first time from an article by Steven Kleiman, but cannot remember the precise reference. </p> http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics/115053#115053 Answer by none for Non-rigorous reasoning in rigorous mathematics none 2012-12-01T05:38:17Z 2012-12-01T05:38:17Z <p>In elementary calculus, the way I was taught to integrate by partial fractions was to guess the numerators of the fractions, then back-substitute to check the correctness of the guess. Very often the guess would be right for no obvious reason. When it was wrong, the discrepancy immediately suggested (non-rigorously) what the next guess should be, and the second guess was almost always correct.</p> http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics/115073#115073 Answer by Todd Trimble for Non-rigorous reasoning in rigorous mathematics Todd Trimble 2012-12-01T14:12:48Z 2012-12-01T14:12:48Z <p>Have you reads Proofs and Refutations by Lakatos? It's all about the dynamic tension between "heuristics" (don't get mad at me, Andrew Stacey!) and rigorous proof, centering particularly on a classroom situation where they are discussing Euler's formula V - E + F = 2. The difference between the first attempted "proofs" or thought experiments and the final rigorous proof involving homology is pretty stark; the first proof is however memorable and explanatory.</p> <p>Even in pre-Robinson days before they were made rigorous, you could say that the original intuitions of infinitesimals in calculus were effective and explanatory (even today, I am told, among certain physicists and engineers who might never learn the rigorous foundations). If you read the introduction to Models of Smooth Infinitesimal Analysis by Moerdijk and Reyes, you will see examples of intuitive reasoning with infinitesimals among geometers like Lie and E. Cartan which were certainly convincing to them, but which had to undergo some distortion to meet the demands of Weierstrassian rigor -- at least that was so until recent years when the types of reasoning with nilpotent infinitesimals in smooth analysis were clarified and made rigorous through sheaf theory and its internal logic. </p> http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics/115075#115075 Answer by Pablo Lessa for Non-rigorous reasoning in rigorous mathematics Pablo Lessa 2012-12-01T14:52:59Z 2012-12-01T15:42:10Z <p>There is a well known connection between parabolic and elliptic partial differential equations and Brownian motion. By now it very well explored formally (e.g. the probabilistic proof of Hörmander's theorem due to Malliavin) but it used to be the case that people would get their intuition from Brownian motion and then prove a theorem by completely different means.</p> <p>One example is the following quote from Nash's 1958 paper "Continuity of solutions of parabolic and elliptic equations":</p> <blockquote> <p>The methods here were inspired by physical intuition, but the ritual of mathematical exposition tends to hide this natural basis. For parabolic equations, diffusion, Brownian movement, and flow of heat or electrical charge all provide helpful interpretations.</p> </blockquote> <p>As a side note. One of the people who contributed the most to establishing the formal connection between Brownian motion and parabolic/elliptic equations was Joseph Doob. He had done his Phd. thesis on harmonic analysis, but couldn't find a job anywhere (this was a couple of years after the Great Depression) until he got offered a post at a probability department. He started working on formal (i.e. Kolmogorov) foundations of probability and ended up establishing the connection between harmonic functions and Martingales. He's one of my favorite mathematicians and I think his contributions are underrated.</p> http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics/115308#115308 Answer by Ronnie Brown for Non-rigorous reasoning in rigorous mathematics Ronnie Brown 2012-12-03T16:04:47Z 2012-12-04T14:28:01Z <p>In 1965 I had the idea that the proof of the Seifert-van Kampen theorem for the fundamental groupoid generalised to two dimensions, and higher, but lacked the gadget corresponding a 2-dimensional fudamental groupoid using squares composed in two directions. So this was an idea of a proof in search of a theorem. So I tried for 9 years to define this for a topological space. Finally, in 1974, Philip Higgins and I realised that we could do this for a pair of spaces, i.e. a space $X$ and subspace $A$, mapping a square to $X$ with edges mapped into $A$ and taking homotopy classes of these maps with vertices fixed in the homotopies. Fortunately, lots of work on related algebra had been done in the meanwhile, so the main stuff rolled out, and got published in 1978. </p> <p>Unfortunately, the use of groupoids and double groupoids seemed to arouse hostility. so this and the work in all dimensions was, a colleaue remarked, pursued in the teeth of opposition! </p> <p>So that is another possible affect of intuition. to say some work is ridiculous! It's a hard life! But has been lots of fun pursuing a line of intuition and trying to make it really work. I was lucky in my collaborators, too. </p> <p>Later: There is an example in J.E. Littlewood's "A mathematican's miscellany" where a picture contains the essential argument. In higher category theory, there is quite a lot of use of manipulating diagrams, and this is regarded, rightly, as rigorous. </p> http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics/115413#115413 Answer by Karl Fabian for Non-rigorous reasoning in rigorous mathematics Karl Fabian 2012-12-04T16:01:57Z 2012-12-04T16:01:57Z <p>Every geometric problem that has a two-dimensional representation is solved by almost every mathematician by first drawing a diagram, then deriving the correct formal description from this diagram, and then continuing to solve the problem in the algebraic description.</p> <p>These certainly are ubiquitous "situations in which there is a heuristic argument which is valid and can be formalized."</p> <p>Also the heuristic argument is "separate (or complementary)" to the "rigorous argument, but the heuristic argument is more enlightening and more explanatory."</p> http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics/115453#115453 Answer by P Vanchinathan for Non-rigorous reasoning in rigorous mathematics P Vanchinathan 2012-12-05T00:24:42Z 2012-12-05T00:24:42Z <p>Close to he requirement in the original question: Waring's problem which generalizes Lagranges's four-square theorem. Every positive integer can be expresses as a sum of 9 cubes, a sum of 19 fourth powers etc. For the $k$-th powers the number of summands required, denoted $g(k)$, was a heuristic guess, $g(k) = 2^k + [ (\frac32)^k ]- 2$ and some variations of this. Though Hilbert proved $g(k)$ is finite before 1910 actual specific values were proved decades later. The reason I know this is because one of the persons who 'nailed the last nail into the coffin" of this problem in 1980s was working where I started my PhD.</p> <p>The number of summands is very high for low numbers because (heuristically) you have only 1 and $2^k$ to use. For 4-th powers 79 is the culprit, needing fifteen 1's and four 16's.</p> <p>So this lead to related another natural question: as numbers needing that many summands are small in size they may be a finite number of exceptions. Define $G(k)$ as the number of summands needed for expressing every sufficiently large integer as sum of $k$-th powers (i.e. treat 79 as an exception for the case of fourth powers). $G(4)$ is known to be 16. </p>