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[math-fun] Most influential mathematician?
by Henry Baker 02 Mar '06

02 Mar '06

Does anyone on this list have a suggestion of who is the most influential mathematician today? I'm looking for someone of the stature of Hilbert; i.e., someone whom most people inside & outside of mathematics would acknowledge as being influential. Of course, this presumes that anyone outside of mathematics knows or cares about mathematicians! For example, during WWII, Einstein was considered the most influential of physicists by people inside & outside of the physics community. Thanks in advance for any suggestions. Henry Baker

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Re: [math-fun] Most influential mathematician?
by dasimov＠earthlink.net 02 Mar '06

02 Mar '06

Henry writes: << Does anyone on this list have a suggestion of who is the most influential mathematician today? I'm looking for someone of the stature of Hilbert; i.e., someone whom most people inside & outside of mathematics would acknowledge as being influential. Of course, this presumes that anyone outside of mathematics knows or cares about mathematicians! . . . >> As you suggest, there is little overlap between the most influential mathematicians and the ones most known to the general population. So it may be necessary to narrow down what you're asking for some more. You said most influential *today*, but not that the mathematician need be alive today. I'm not sure what you meant regarding this. Shiing-Shen Chern, the differential geometer (1911-2004) is said to be well-known to the general population of China. Cf. <http://en.wikipedia.org/wiki/Chern>. Martin Gardner (mostly self-taught in math) and Ian Stewart are widely known as popularizers of math; many mathematicians attribute their choice of profession to Gardner's columns in Sci. Am. Within math per se, I think of John Milnor and Jean-Pierre Serre as probably the most influential mathematicians alive today. (I'd pick Milnor over Serre, because Milnor's scope is probably wider and he's written more math texts, all of which are gems of exposition, and accessible survey articles.) Excluding Euclid, Pythagoras, Archimedes, and Al-Khwarizmi for being in different categories, I'd say that Carl Friedrich Gauss is the clear winner, if the question is which mathematician living or dead, has had the most impact on modern civilization AND within math per se. Cf. <http://en.wikipedia.org/wiki/List_of_topics_named_after_Carl_Friedrich_Gauss> for evidence of this. (Full disclosure: I would've said Gauss anyway, but recently learned from the Mathematics Genealogy Project that Gauss is a direct mathematical ancestor of mine, i.e., via a sequence of thesis advisors.) --Dan

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Re: [math-fun] Stirling Numbers?
by dasimov＠earthlink.net 01 Mar '06

01 Mar '06

Thanks, everyone, for the interesting asymptotic expressions for B(n). (Would only that they were expressed purely in terms of n, rather than a root of some weird equation in n.) --Dan

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Re: [math-fun] Stirling Numbers?
by N. J. A. Sloane 01 Mar '06

01 Mar '06

Dan, the entry for the Bell numbers is A000110 there you will find: %F A000110 a(n) is asymptotic to n!*(2 Pi r^2 exp(r))^(-1/2) exp(exp(r)-1) / r^n, where r is the p\ ositive root of r exp(r) = n. - see e.g. the Odlyzko reference. Neil

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Re: [math-fun] Polyomino question
by Michael Reid 28 Feb '06

28 Feb '06

> What is the smallest polyomino that contains all polyominos of size n? > For n = 4, there is a unique size 6 polyomino that works: > OOOO > OO XX XXXX also works. it is clear that size 6 is the smallest possible, because the square tetromino and the straight one can overlap in at most two squares. > For n = 5, the best I can do is 9: > O > OOOOO > OOO > I don't know if this is unique; I would be very surprised if 8 is possible. X XXXXX XXX also works. it is easy to show that size 8 is indeed impossible. note that the "I" and "W" pentominoes can overlap in at most 2 squares, so at least 8 squares are required to accommodate even those two. there are only a handful of ways to overlap them, covering as few as 8 squares, and none of the resulting shapes contain the "X" pentomino: W W W W W W W W WW WW WW WW BBIII IBBII IIBBI IIIBB BBIII IBBII IIBBI IIIBB WW WW WW WW a simple back-of-the-envelope type calculations shows that the only 9-ominoes that work are the two above. what i did was: add a single square to each of the above octominoes in a way that would also accommodate the X pentomino, and check those shapes. there were 15 resulting 9-ominoes (one occurred twice) and only 2 contain all pentominoes. i also tried all ways of overlapping the I and W pentominoes in a single square (9 ways) but none of the shapes contains the X pentomino. i realize that this approach is prone to error, so it would be nice to have independent confirmation. i haven't examined the hexomino case yet, but it seems unlikely to succumb to such a simple analysis. mike

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[math-fun] Re: partitions of a set
by James Propp 28 Feb '06

28 Feb '06

> Is there a formula for LP(n) ? > What about an asymptotic formula? Rota's proof of Dubinski's formula, as described in the book "Finite Operator Calculus", is a gem. This little book is one of the reasons I went into combinatorics. Jim Propp

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[math-fun] Stirling Numbers?
by Schroeppel, Richard 28 Feb '06

28 Feb '06

This message is from Dan Asimov, about 10 days ago. The mailing list software didn't like it for some reason. I did some heavy editing to unHTML it, so I may have mangled it. --Rich ---------------------------------------------- Suppose you want to count all possible partitions of the set X_n = {1,2,...,n}. I.e., you want to know how many [collections of mutually disjoint nonempty subsets of X_n] have union = X_n. Ordinary partitions are of course equivalence classes of these "labeled partitions" (my term). ; E.g., for X_5: ordinary partitions: # of corresponding labeled partitions -------------------- -------------------------- 1+1+1+1+1 = 1 1+1+1+2 C(5,2) = 10 1+2+2 5 * C(4,2)/2 = 15 1+1+3 C(5,3) = 10 1+4 C(5,1) = 5 2+3 C(5,2) = 10 5 C(5,5) = 1 ----- total: 52 Is there a formula for LP(n) ? What about an asymptotic formula? --Dan

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[math-fun] squaring a series for pi
by Schroeppel, Richard 27 Feb '06

27 Feb '06

Bill Cordwell has called my attention to a paper by Robin Chapman that collects 14 proofs that zeta(2) [:= sum(1/n^2)] = (pi^2)/6. http://www.maths.ex.ac.uk/~rjc/etc/zeta2.ps (or .pdf) Proof #13 is what I asked for in HAKMEM: The proof squares the double-ended series ... + 1/-3 - 1/-1 + 1/1 - 1/3 + 1/5 - ... (easily seen to be just 2*(pi/4) when summed from the center) and shows equality to the double-ended series ... + 1/9 + 1/1 + 1/1 + 1/9 + 1/25 + ... (which is absolutely convergent) and is equal to (pi^2)/4. From there, it's a quick wave of the hands to get zeta(2) = (pi^2)/6. Chapman credits the proof to the Borweins' book "Pi and the AGM", and, indeed, it is exercise 11 in section 11.3, on page 381. The proof only requires some algebra, the logN estimate for a harmonic sum, and limits. The first move is to clip the series, summing each series from the -(2K+1) term through the 2K+1 term. The term squares cancel, leaving the problem of showing that the cross terms in the squared series sum to a value that approaches 0 as K->infinity. Since the clipped series are finite, there are no nasty worries about conditional convergence. It's a little complicated, but a nice first principles proof. Even the logN estimate is dispensable, being replaceable with an upper bound of, say, sqrtN for the harmonic sum. Rich

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[math-fun] Re: Polyomino question
by Dean Hickerson 27 Feb '06

27 Feb '06

Franklin T. Adams-Watters asked: > What is the smallest polyomino that contains all polyominos of size n? ... > 2n-2 is a simple lower bound, since fitting the general "line" polyomino > requires n in a single row (or column), while the general "snake" > polyomino has only 2 squares in any given row. This bound can be improved by considering 'linear' polyominoes of other slopes. For 0 <= s <= 1, let P(s,n) be an n-omino which approximates a line of slope s. E.g. the straight line polyomino is P(0,n), and the snake is P(1,n). P(1/2,n) looks like a subset of this: ooo ooo ooo ooo ooo ooo ... P(2/3,n) could be a subset of either of these; I don't know which will give the best lower bound: oooo ooo o oo oooo ooo o oo oooo ooo o oo oooo ooo o oo oooo ooo ... ... P(1/2,n) can overlap P(0,n) in at most 3 squares, and overlap P(1,n) in at most 6 squares, so any polyomino that contains P(0,n), P(1,n), and P(1/2,n) has at least n + (n-2) + (n-3-6) = 3n-11 squares. If we also include P(1/3,n), we need at least 4n-31 squares, if I've maximized the intersections correctly: slopes maximum intersection 0 1 2 0 1/2 3 1 1/2 6 0 1/3 4 1 1/3 4 1/2 1/3 12 I'll leave it to someone else to figure out the best choice for P(s,n), a general formula for the maximum intersection of P(s,n) and P(t,n), and the choice of slopes which gives the best lower bound for n-ominoes. Dean Hickerson dean(a)math.ucdavis.edu

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Re: [math-fun] Polyomino question
by Michael Reid 27 Feb '06

27 Feb '06

> What is the smallest polyomino that contains all polyominos of size n? interesting question. a relevant reference: http://www.xprt.net/~munizao/polycover/ (but i'm not sure how far he's taken it.) mike

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