[math-fun] When (-1)-choose-(-1) is 1/2
Following up on issues raised in the thread "Atrocious Mma 8&9, WolframAlpha bug", here are some thoughts on (-1)-choose-(-1): Under the Euler-measure-as-generalized-cardinality philosophy, (-1)-choose-(-1) could be interpreted as the "number" of ways, given a polyhedral set S of Euler measure -1, to choose a polyhedral subset S' of Euler measure -1. (Here and hereafter, "number" should be understood in a very loose way. E.g., the set of triples r,s,t with 0<r<s<t<1 is an open 3-cell and therefore has Euler measure (-1)^3 = -1, so we say that there are -1 such triples. For an introduction to this mode of thinking, mostly due to Schanuel, see http://arxiv.org/abs/math/0203289.) In 1 dimension (the easiest place to pursue these notions), the simplest kind of polyhedral set of Euler measure -1 is just an open interval. Let's take S = (0,1), for definiteness. Its polyhedral subsets S' of Euler measure -1 are unions of k points and k+1 open intervals (which can abut the k points), for some non-negative k. We stratify the collection of all such subsets by the number of "breakpoints". With 0 breakpoints, the only possibility is S' = (0,1) itself. With 1 breakpoint, we have subsets of the form (0,t) and (t,1) (with t strictly between 0 and 1). How many such values of t are there? -1 of them (in the Euler measure sense). So there are -2 subsets of (0,1) with Euler measure -1 that have 1 breakpoint. With 2 breakpoints, s and t (with s < t), the possibilities for S' are (s,t), (0,s] \union (t,1), (0,s) \union [t,1), (0,s) \union (s,t], and [s,t) \union (t,1). How many pairs s,t with 0<s<t<1 are there? +1 of them. So there are +5 subsets of (0,1) with Euler measure -1 that have 2 breakpoints. The sequence continues 1, -2, 5, -13, 35, -96, ..., and to count all the subsets S' we're interested in, we just have to add up these numbers. It turns out that the power series 1 - 2q + 5q^2 - 13q^3 + 35q^4 - 96q^5 + ... can be "evaluated" at q=1 by analytic continuation, because it is the expansion of 2/(1 + 3 q + Sqrt[1 + 2 q - 3 q^2]); substituting q=1, we get 1/2 as our answer. I suspect that in this theory (or rather the theory into which such flimflam will eventually evolve), the generalized Euler measure of the set of Euler-measure-k polyhedral subsets S' of an Euler-measure-n polyhedral set S depends not just on n and k but also on other properties of S. So it's wrongheaded to think of the above calculation as determining "the" right value of (-1)-choose-(-1); there will be other right values (such as 0) in other contexts. Jim Propp
I once thought that the best way to continue the triangle upward was to make the row above [0* 1 0*] be (1/2 -1/2)* 1/2 1/2 (-1/2 1/2)* which is symmetric, and mostly periodic except for the small glitch in the center. Subsequent upfill gives a compromise between the two main candidates, trefoil and biwedge: We get a trefoil, but with the upper two pie slices multiplied by 1/2. There may also be a sign adjustment --- I haven't been following this discussion very closely. It does preserve the property that the sum of two (horizontal) neighbors is the term below, and the left-right symmetry. I'd decided to not mention the "third way", but JP provided the perfect opening. Rich ----- Quoting James Propp <jamespropp@gmail.com>:
Following up on issues raised in the thread "Atrocious Mma 8&9, WolframAlpha bug", here are some thoughts on (-1)-choose-(-1):
Under the Euler-measure-as-generalized-cardinality philosophy, (-1)-choose-(-1) could be interpreted as the "number" of ways, given a polyhedral set S of Euler measure -1, to choose a polyhedral subset S' of Euler measure -1. (Here and hereafter, "number" should be understood in a very loose way. E.g., the set of triples r,s,t with 0<r<s<t<1 is an open 3-cell and therefore has Euler measure (-1)^3 = -1, so we say that there are -1 such triples. For an introduction to this mode of thinking, mostly due to Schanuel, see http://arxiv.org/abs/math/0203289.)
In 1 dimension (the easiest place to pursue these notions), the simplest kind of polyhedral set of Euler measure -1 is just an open interval. Let's take S = (0,1), for definiteness. Its polyhedral subsets S' of Euler measure -1 are unions of k points and k+1 open intervals (which can abut the k points), for some non-negative k. We stratify the collection of all such subsets by the number of "breakpoints".
With 0 breakpoints, the only possibility is S' = (0,1) itself.
With 1 breakpoint, we have subsets of the form (0,t) and (t,1) (with t strictly between 0 and 1). How many such values of t are there? -1 of them (in the Euler measure sense). So there are -2 subsets of (0,1) with Euler measure -1 that have 1 breakpoint.
With 2 breakpoints, s and t (with s < t), the possibilities for S' are (s,t), (0,s] \union (t,1), (0,s) \union [t,1), (0,s) \union (s,t], and [s,t) \union (t,1). How many pairs s,t with 0<s<t<1 are there? +1 of them. So there are +5 subsets of (0,1) with Euler measure -1 that have 2 breakpoints.
The sequence continues 1, -2, 5, -13, 35, -96, ..., and to count all the subsets S' we're interested in, we just have to add up these numbers. It turns out that the power series 1 - 2q + 5q^2 - 13q^3 + 35q^4 - 96q^5 + ... can be "evaluated" at q=1 by analytic continuation, because it is the expansion of 2/(1 + 3 q + Sqrt[1 + 2 q - 3 q^2]); substituting q=1, we get 1/2 as our answer.
I suspect that in this theory (or rather the theory into which such flimflam will eventually evolve), the generalized Euler measure of the set of Euler-measure-k polyhedral subsets S' of an Euler-measure-n polyhedral set S depends not just on n and k but also on other properties of S. So it's wrongheaded to think of the above calculation as determining "the" right value of (-1)-choose-(-1); there will be other right values (such as 0) in other contexts.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
For some thoughts about the enterprise of computing quantities one hasn't rigorously defined, see http://mathoverflow.net/questions/135536/procedure-based-as-opposed-to-defin... (which I hope will elicit interesting responses over the next few days). Jim Propp On Sun, Jun 30, 2013 at 11:06 PM, James Propp <jamespropp@gmail.com> wrote:
Following up on issues raised in the thread "Atrocious Mma 8&9, WolframAlpha bug", here are some thoughts on (-1)-choose-(-1):
Under the Euler-measure-as-generalized-cardinality philosophy, (-1)-choose-(-1) could be interpreted as the "number" of ways, given a polyhedral set S of Euler measure -1, to choose a polyhedral subset S' of Euler measure -1. (Here and hereafter, "number" should be understood in a very loose way. E.g., the set of triples r,s,t with 0<r<s<t<1 is an open 3-cell and therefore has Euler measure (-1)^3 = -1, so we say that there are -1 such triples. For an introduction to this mode of thinking, mostly due to Schanuel, see http://arxiv.org/abs/math/0203289.)
In 1 dimension (the easiest place to pursue these notions), the simplest kind of polyhedral set of Euler measure -1 is just an open interval. Let's take S = (0,1), for definiteness. Its polyhedral subsets S' of Euler measure -1 are unions of k points and k+1 open intervals (which can abut the k points), for some non-negative k. We stratify the collection of all such subsets by the number of "breakpoints".
With 0 breakpoints, the only possibility is S' = (0,1) itself.
With 1 breakpoint, we have subsets of the form (0,t) and (t,1) (with t strictly between 0 and 1). How many such values of t are there? -1 of them (in the Euler measure sense). So there are -2 subsets of (0,1) with Euler measure -1 that have 1 breakpoint.
With 2 breakpoints, s and t (with s < t), the possibilities for S' are (s,t), (0,s] \union (t,1), (0,s) \union [t,1), (0,s) \union (s,t], and [s,t) \union (t,1). How many pairs s,t with 0<s<t<1 are there? +1 of them. So there are +5 subsets of (0,1) with Euler measure -1 that have 2 breakpoints.
The sequence continues 1, -2, 5, -13, 35, -96, ..., and to count all the subsets S' we're interested in, we just have to add up these numbers. It turns out that the power series 1 - 2q + 5q^2 - 13q^3 + 35q^4 - 96q^5 + ... can be "evaluated" at q=1 by analytic continuation, because it is the expansion of 2/(1 + 3 q + Sqrt[1 + 2 q - 3 q^2]); substituting q=1, we get 1/2 as our answer.
I suspect that in this theory (or rather the theory into which such flimflam will eventually evolve), the generalized Euler measure of the set of Euler-measure-k polyhedral subsets S' of an Euler-measure-n polyhedral set S depends not just on n and k but also on other properties of S. So it's wrongheaded to think of the above calculation as determining "the" right value of (-1)-choose-(-1); there will be other right values (such as 0) in other contexts.
Jim Propp
I'm impressed by the large fraction of small-size polyominoes and polyhexes that individually tile the plane (using arbitrary congruent copies, which may be flipped over). All polyominoes through the 35 hexominoes tile the plane, plus all but 3 of the 108 distinct heptominoes. And over 94% of the369 octominoes. All polyhexes through the 22 pentahexes tile the plane, plus all but 4 of the 82 distinct hexahexes. And over 91% of the 333 distinct heptahexes. So: Is there some heuristic reason to expect such a high fraction of small-size polyominoes and polyhexes to tile the plane? This phenomenon seems very counterintuitive to me. --Dan
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