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64

Computer experiments (in the early 1960s!) led Birch and Swinnerton-Dyer to the formulation of their conjecture, which stimulated the development of much of arithmetic geometry.


58

If the Riemann zeta function had only trivial zeroes, then after multiplying by the gamma factor, it would become a zero-free entire function of finite order. Every zero-free entire function of finite order is of the form $\exp(poly(z))$. This cannot happen for the zeta function, as can be seen by considering behavior as $s\to\infty$ along the reals, for ...


42

It seems to be the case that the polynomial $P_n(x) =\sum_{m=1}^n m!S(n,m)x^m$ has only real zeros. (I know it is true that $\sum_{m=1}^n S(n,m)x^m$ has only real zeros.) If this is true, then the value of $m$ maximizing $m!S(n,m)$ is within 1 of $P'_n(1)/P_n(1)$ by a theorem of J. N. Darroch, Ann. Math. Stat. 35 (1964), 1317-1321. See also J. Pitman, ...


33

The Prime Number Theorem was conjectured by Gauss from looking (very hard, one can presume...) at a table of the primes $\leq10^6$. It is not with too much effort that one can read his Disquisitiones as a set of tricks to determine primality with as little work as possible, and one can understand the motivation: he was his own computer, in a way :P (I don't ...


31

Here's a variation on the arguments using the explicit formula. If there were no zeros in the critical strip then the explicit formula would say $$\psi_0(x) = x - \frac{\zeta'(0)}{\zeta(0)} - \frac12 \log(1-x^{-2}),$$ where $\psi_0(x)$ is the same as the usual $\psi(x)$ except when $x$ takes integer values $\psi_0(x)$ is the average of the left and right ...


28

An intuitive reason is indicated by Bukh in another answer. The reason there are nontrivial zeros is because the zeta-function is known to grow in a way that it wouldn't grow if there were no nontrivial zeros. Here is some additional (but not complete) detail. Let's pass from the zeta-function to the completed zeta-function $$ Z(s) = ...


28

Rather than the problem of why the zeta function has non-trivial zeros, let me address Gowers's question of why the error term in the prime number theorem needs to be large. The short answer that I propose is: because the integers are so well distributed. To make this precise, I shall prove a general result on semigroups, showing that either the ...


26

A lot of Rich Schwartz's work begins with computer experiments. Perhaps most notably is his proof that a triangle whose largest angle is less than 100 degrees has a periodic billiard trajectory.


24

Another thing you might like to check out is Herb Wilf's very nice article in the Princeton Companion to Mathematics, in which he talks about the interplay between theory and experiment, and the many forms it can take. To give one of his examples (or rather classes of examples), if you generate a sequence of integers, you can plug it into Sloane's database ...


22

(I've just typed a comment but somehow it got lost. I'll type an answer instead). Bloom is right, no zeros implies too nice a formula, except the right approximation to $\pi(x)$ is $\int_2^x dt/\log t$, not $x / \log x$. The existence of zeros implies that the error cannot be better than $O(x^{1/2})$. On the other hand, the zeta function of ...


17

The proof of the four color theorem, the proof of the nonexistence of a projective plane of order 10, the proof of the Kepler conjecture, the search for Mersenne primes, the discovery of the Lorenz attractor and the Feigenbaum constant among others are examples. These are in the wikipedia article on experimental mathematics with other examples. There is also ...


16

I have one personal experience with experimental mathematics. It started with a computer assisted simple proof of a result that every five dimensional polytope contains a 2-simensional face which is a triangle or a quadrangle. It turns out that the negation of this theorem implies a certain number of inequalities for the so called flag numbers which together ...


16

This reference contains the best result of this kind currently known for $\mu(n)$: Tadej Kotnik and Herman te Riele The Mertens Conjecture revisited. Algorithmic number theory, 156--167, Lecture Notes in Comput. Sci., 4076, Springer, Berlin, 2006. They prove that $\limsup_{x \rightarrow +\infty}M(x)/\sqrt{x} \geq 1.218$ and that $\liminf_{x \rightarrow ...


15

This is part of topological dynamics (a close cousin of ergodic theory, aka measurable dynamics, in which the underlying space on which the dynamics takes place is a topological space rather than a measure space). The relationship with combinatorics is roughly as follows: topological dynamics is to colouring Ramsey theorems (such as van der Waerden) as ...


15

You probably know this, but: Set $s(n) = \mu(1) + \cdots + \mu(n)$. Suppose, for the sake of contradiction, that $s(n) = O(n^{1/2 - \epsilon})$. Then $$\sum s(n) \left( n^{-s} - (n+1)^{-s} \right)$$ would converge for $Re(s) > 1/2-\epsilon$. This would give an analytic extension of $1/\zeta(s)$ to this half plane, contradicting that $\zeta$ has zeroes on ...


13

A fairly classical (and pretty dated at that) example is the study of solitons which was to a large extent triggered by sloving numerically the so-called Fermi--Pasta--Ulam chain and then of its continuous limit, the Korteweg--de Vries equation. It would not be much of exaggeration to say that the whole modern theory of integrable systems grew out of this. ...


13

Donald Knuth comments in one of his papers on the key role of the use of computers in making his paper possible. D. E. Knuth, A class of projective planes. Transactions of the American Mathematical Society 115 (1965) 541–549. Basically what happened was that Knuth used a computer to study an example and once he saw how that example worked he was able to ...


13

You asked for a classification of types of experimental mathematics. I would like to mention just one class that interests me greatly (which is the class that predominates in polymath5). There are many examples of mathematical statements that become easier to prove when you strengthen them, because you have much more of a clue of what sort of proof is ...


12

I'm surprised that no one has mentioned the Mandelbrot set yet, arguably the most famous new mathematical object of the last 30 years, at least among the general public. Benoit Mandelbrot discovered it in 1979 as a result of computer experiments. He says that when he first saw it he was so surprised by its appearance that he thought it must be the result of ...


12

Also V.I. Arnol'd is quite enthusiastic about the experimental mathematics. In particular, he wrote two books on the subject in Russian, Experimental mathematics, Fazis, Moscow, 2005, and Experimental observation of mathematical facts, MCCME, Moscow, 2006, where, I guess, one could find a number of examples of the kind you ask for. Unfortunately, to the best ...


11

The Monstrous Moonshine relationships between the Monster group (the largest sporadic finite simple group) and modular functions were investigated after observations of several numerical coincidences: Several of the lowest coefficients of the Fourier expansion of the $j$ modular function are small sums of the smallest dimensions of irreducible ...


11

May I offer an attempt at a very lowbrow answer? This answer might naturally strike anyone who knows the functional equation, who looks at a color coded graph of zeta function such as http://en.wikipedia.org/wiki/File:Complex_zeta.jpg, and who knows a little about the long-range behavior of exponential functions and $\Gamma(z)$. Nothing I will say is ...


11

For $t$ fixed, the count is proportional to $\lambda^n$, where $\lambda = 2 \cos \frac\pi{2t+2}$ is the principal eigenvalue of the adjacency matrix of the path with $2t+1$ vertices. The all-positive (Perron-Frobenius) eigenvector corresponding to $\lambda$ is $$\bigg(\sin \frac{\pi}{2t+2}, \sin \frac{2\pi}{2t+2},\sin \frac{2\pi}{2t+2},\dots,sin ...


11

The correct asymptotic is $C \cdot (\log N)^{1/2}$. (c.f Selberg-Delange method).


11

The idea (from the Selberg-Delange) method to doing this problem is the following steps: 1) Let $F(s) = \sum_{n\ge 1} \frac{1}{n^s d(n)} = \prod_{p} \left(1 + \sum_{k=1}^{\infty} \frac{1}{(k+1) p^{ks}} \right)$. The latter is by multiplicativity of $d(n)$. 2) If we look, instead at $G(s) = \prod_p \left( 1 + \frac{1}{2 p^s} \right)$ we can see that ...


11

Update 2: Original answer below. I've put together graphs showing more than just champions, using every $N\leq 10^4$ and also every $N\equiv 0\pmod{100}$ up to $10^5$. This is for the original version, not the variant, but I'd expect that to be essentially the same. I am starting to be somewhat skeptical of my $1/3$ estimate. I'll collect some data ...


9

This looks like the Stirling numbers of the second kind (up to the $m!$ factor). This and this papers are specifically devoted to the maximal Striling numbers. It seems that for large $n$ the relevant asymptotic expansion is $$k! S(n,k)= (e^r-1)^k \frac{n!}{r^n}(2\pi k B)^{-1/2}\left(1-\frac{6r^2\theta^2 +6r\theta+1}{12re^r}+O(n^{-2})\right),$$ where ...


9

(edited) The answer can be extracted from a paper of Ramanujan, "Some formulae in the analytic theory of numbers", no. 17 in his collected papers. There he gives, among other things, the formula $\sum_{n\leq X} \frac{1}{d(n)} \sim \frac{X}{\sqrt{\log{X}}}\pi^{-\frac{1}{2}}\prod_{p}\sqrt{p^2-p}\log{\frac{p}{p-1}}$. The answer to the original question can ...


9

John Conway initially thought that his cellular automata game of Life could not lead to an unlimited number of active cells starting from a finite number of cells. Computer experimentation by Bill Gosper led to the discovery in 1970 of the Gosper Glider Gun, a finite collection of cells which leads to a recurring state which continues to emit "spaceships" ...


9

Thistlethwaite's discovery of links with trivial Jones polynomial (MR1831681) should be mentioned here. He apparently discovered them "during the course of a routine computer ennumeration." Later he and Eliahou and Kauffman (MR1928648) were able to recognize these links as part of an infinite family of links with trivial Jones polynomial. Further, it's my ...



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