An unusual series that produces pi was discovered by Jonas Castillo Toloza in 2007; the series consists of the reciprocals of the triangular numbers and, as such, could be detected in Pascal’s triangle.

http://www.cut-the-knot.org/arithmetic/algebra/TriPiInPascal.shtml ]]>

Joel, you’ve given a list of evaluations of the binomial probability distribution *at the mean* (or the integers either side of the mean), and the Gaussian approximation to that distribution at the mean. And sure, these values at the mean get closer together as n increases.

Unfortunately, though, that doesn’t help us obtain a good approximation for the product of all the numbers in each row of Pascal’s triangle, which is the aim of the calculations in this blog post. For n=1,..,10, the ratio between the actual product and the product of the numbers you get from the Gaussian approximation takes the values: 1.06747, 1.28577, 1.36777, 1.3446, 1.237, 1.07058, 0.87397, 0.674019, 0.49155, 0.339207. The limit of this ratio for large n turns out to be zero.

So, alas, this is one problem where the approximation you’ve described is not applicable.

]]>Greg- I used the Table function in Mathematica to test De Moivre’s Approximator against Jacob Bernoulli’s Exact Function. Here are my results for n ranging from 1 through 10.

Bernoulli 0.5, De Moivre 0.798 | Bernoulli 0.5, De Moivre 0.564 | Bernoulli 0.375, De Moivre 0.461 | Bernoulli 0.375, De Moivre 0.399 | Bern 0.312, De M 0.357 | Bern 0.312, De M 0.326, | Bern 0.273, De M 0.302 | Bern 0.273, De M 0.282 | Bern 0.246, De M 0.266 | Bern 0.246, De M 0.252 |

The way I do it the approximation improves with increasing n. My goal is to set a lower bound on the number of genes having an influence on any given biological manifestation. So, for example, if the bell curve for adult height solves to a 12 element binomial (13 columns) I think that implies a minimum of 12 genes with an influence on cell production rate.

]]>Let me mention something one might try to do with the bijective proofs given above. Often bijective proofs can be copied in different contexts to gain new information.

I’ve been using

to mean, not the binomial coefficient, but a certain set whose size is that binomial coefficient: the set of -element subsets of But this set corresponds to a basis of the th exterior power of , usually called In quantum physics, this is the Hilbert space for identical fermions each having degrees of freedom.

I’ve been using

to mean, not the number of balls in a -dimensional pyramid with edges of length but a certain set whose size is that number: the set of -element multisubsets of But this set corresponds to a basis of the th symmetric power of usually called This is the Hilbert space for bosons each having degrees of freedom.

So, when we get our hands on a specific bijection

when we also get an isomorphism of Hilbert spaces

relating bosons and fermions. Is this good for something? I don’t know. But this is the kind of thing one can do.

]]>Joel, approximating the binomial coefficients with a Gaussian sounds like a good idea! I tried it out myself initially. But if you work through the calculations, the product of the Gaussian approximations for each row of Pascal’s triangle turns out *not* to be asymptotic to the product of the actual binomial coefficients.

I believe this is because the approximation works very well around the mean, but not so well at the tails, and while that doesn’t matter much if you’re summing the terms (which are small at the tails), when you’re taking their product it leads to large discrepancies.

]]>One more remark I should have added to my last comment: while the restriction on the “coordinate representation” of a multisubset is that the coordinates are positive integers in ascending order, the restriction on the “count representation” is that the counts are non-negative integers that sum to the dimension of the coordinate representation. In other words, the must lie in an affine hyperplane with the appropriate fixed value for their dot product with . And it’s easy to see that the formula:

where:

will yield non-negative values for , and a dot product with of .

So what the bijection is doing is taking the -dimensional pyramid, changing the coordinates to embed it in the appropriate affine hyperplane, and then treating the new coordinates of each object in the pyramid as counts for the various values in the coordinates of each object of the -dimensional pyramid.

]]>There are two representations of a multisubset of with length . One lists all the elements in ascending order, giving a -tuple:

The other specifies the *number of elements* of each possible value, giving a -tuple:

The bijection between triangular numbers:

gives the “count representation” of one triangular number from a change of basis and origin of the “coordinate representation” of the other triangular number:

where and .

]]>Another way to describe is by specifying the number of coordinates that take a given value.

If , then the number of coordinates of that are equal to a certain value, , is:

where we define and in addition to the actual coordinates , .

]]>Using Graham’s path method, I figured out an explicit formula for the bijection:

If and , the coordinates of are given by:

where is the count of the number of coordinates of that are equal to .

Or to put this another way, is one more than the number of coordinates of that are less than or equal to .

]]>Ah, and if you *flip* Graham’s paths diagonally (that is, reverse the coordinates of each endpoint of a line segment in his two-dimensional diagram) you get a bijection that goes all the way through, from !

Actually, you can’t quite just flip the coordinates, if you want to start both paths at (1,0). But that’s just matter of tweaking the choice of coordinates; geometrically, the paths are flipped.

I don’t have time to draw a diagram of this right now, but it’s very easy to draw these paths by hand, and doing that for the first dimension-changing example:

… makes it easy to see what’s going on.

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