Partitions over permutations

I was thinking more about the cosine approximation to the Gaussian

exp(−z²) ≈ (1 + cos(sin(z) + z))/2

that I wrote about last week. The two expressions above are close along the real axis but not along the imaginary axis.

If ziy, the right side grows much faster than the left, behaving like exp(exp(y)).

This led to me looking up the power series for the double exponential function exp(exp(y)). This is an interesting series because the coefficient of xn is

e Bn / n!

where Bn is the nth Bell number, which equals the number of ways to partition a set of n labeled items [1]. And of course n! is the number of ways to permute a set of n labeled items. So the nth coefficient in the power series for exp(exp(y)) is the ratio of the number of partitions to permutations for a set of n labeled things, multiplied by e.

The number of ways to partition a set of n things grows quickly as n increases, almost as quickly as the number of permutations, and so the series for the double exponential function converges very slowly.

Computing

SymPy has a function bell for computing Bell numbers, so you could compute the ratio of partitions to permutations as follows.

from sympy import bell, factorial
f = lambda n: bell(n)/factorial(n)

This returns a number of type sympy.core.numbers.Rational and so the result is exact. You can cast it to float for convenience.

Asymptotics

If we look at only the terms in the asymptotic series for log Bn and log n! that grow with n we have

log Bn ~ n log n − n log log n
log n! ~ n log n − ½ log n

and so

log( Bnn! ) ~ ½ log n − n log log n

There’s also an asymptotic series for log( Bnn! ) involving the Lambert W function:

log( Bnn! ) ~ n/r − 1 − n log r

where r = W(n).

Related posts

[1] It’s important that the items are labeled. Partition numbers are the number of partitions of an unlabeled set. Partition numbers are much smaller than Bell numbers.

2 thoughts on “Partitions over permutations

  1. Michael Kinyon

    One could make a case that exp(exp(x)-1) is slightly more natural than exp(exp(x)) (but only slightly). Besides its value being 1 at x=0, you get rid of the extra e in the series coefficients.

  2. Michael Kinyon

    Er, what I should have added (but for some reason, forgot) is that exp(exp(x)-1) is the exponential generating function for the Bell numbers.

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