
Puzzle 3. In the limit as \(n \to \infty\), what is the probability that the shortest cycle in a randomly chosen permutation of an \(n\)element set has length \(\gt k\)?
We mainly need to count the permutations of an \(n\)element set whose shortest cycle has length \(\gt k\). So, we define a functor
\[ \textrm{Perm}_{\gt k} \colon S \to \textrm{FinSet} \]from the groupoid of finite sets and bijections, \(S\), to the category of finite sets, \(\textrm{FinSet}\), sending any finite set \(X\) to the finite set
\[ \textrm{Perm}_{\gt k}(X) = \{ \textrm{permutations of } X \textrm{ whose shortest cycle has length } \gt k \} \]A functor from \(S\) to \(\textrm{FinSet}\) is called a finite species, and this is what my course was about. We think of a finite species \(G : S \to \textrm{FinSet}\) as a map sending any finite set \(X\) to the set of '\(G\)structures' we can put on \(X\). Any finite species \(G\) has a generating function
\[ G(x) = \sum_{n \ge 0} \frac{G(n)}{n!} x^n \]where in \(G(n)\) we are using \(n\) as our notation for some \(n\)element set. Our basic strategy will be to compute the generating function of the species \(\textrm{Perm}_{\gt k}\) and use that to count permutations of an \(n\)element set whose shortest cycle has length \(k\). But in fact, to save work, we will just work out the asymptotics as \(n \to \infty\).
We can compose species: a \(G \circ H\) structure on \(X\) consists of a way of writing \(X\) as a union of disjoint parts, putting a \(G\)structure on the set of parts, and putting an \(H\)structure on each part. We have
\[ G \circ H = G \circ H \]where at right we're composing generating functions.
So, if we can write \(\textrm{Perm}_{\gt k}\) as a composite of two species whose generating functions we know, we can work out \(\textrm{Perm}_{\gt k}\).
In fact \(\textrm{Perm}_{\gt k}\) is the composite of two species called \(\textrm{Exp}\) and \(\textrm{Cyc}_{\gt k}\). The species \(\textrm{Exp}\) is called 'being a finite set': we can put an \(\textrm{Exp}\)structure on any finite set in exactly one way. The species \(\textrm{Cyc}_{\gt k}\) is called 'being a cyclically ordered set of cardinality \(\gt k\)': we can put a \(\textrm{Cyc}_{\gt k}\)structure on a finite set only if its cardinality is \(\gt k\), and then a \(\textrm{Cyc}_{\gt k}\)structure on it is just a cyclic ordering.
We have
\[ \textrm{Perm}_{\gt k} \cong \textrm{Exp} \circ \textrm{Cyc}_{\gt k} \]because a permutation with cycles of length \(\gt k\) on a finite set \(X\) is the same as a way of chopping \(X\) into subsets of cardinality \(\gt k\) and cyclically ordering each one. Thus, we have
\[ \textrm{Perm}_{\gt k} = \textrm{Exp} \circ \textrm{Cyc}_{\gt k} \]and we just need to compute the two generating functions on the righthand side.
The generating function of \(\textrm{Exp}\) is easy to compute:
\[ \textrm{Exp}(x) = \sum_{n \ge 0} \frac{1}{n!} x^n = \exp(x) \]and this explains the name \(\textrm{Exp}\). The generating function of \(\textrm{Cyc}_{\gt k}\) is
\[ \textrm{Cyc}_{\gt k}(x) = \sum_{n \gt k} \frac{(n1)!}{n!} x^n = \sum_{n \gt k} \frac{x^n}{n} \]since there are \((n1)!\) cyclic orderings on an \(n\)element set. We have
\[ \sum_{n \ge 1} \frac{x^n}{n} = \ln\left(\frac{1}{1x}\right) \](just integrate the geometric series) so we get
\[ \textrm{Cyc}_{\gt k}(x) = \ln\left(\frac{1}{1x}\right)  \left(x + \frac{x^2}{2} + \cdots + \frac{x^k}{k}\right)\]Since
\[ \textrm{Perm}_{\gt k} = \textrm{Exp} \circ \textrm{Cyc}_{\gt k} \]we get
\[ \textrm{Perm}_{\gt k}(x) = e^{ \ln\left(\frac{1}{1x}\right)  \left(x + \frac{x^2}{2} + \cdots + \frac{x^k}{k}\right) } \]or simply
\[ \textrm{Perm}_{\gt k}(x) = \frac{e^{\left(x + \frac{x^2}{2} + \cdots + \frac{x^k}{k}\right)}}{1x} \]This is really a kind of formula for \(\textrm{Perm}_{\gt k}(n)\), the number of permutations of an \(n\)element set with shortest cycle of length \(\gt k\). After all, by the definition of generating function, this equation says
\[ \sum_{n \ge 0} \frac{\textrm{Perm}_{\gt k}(n)}{n!} x^n = \frac{e^{\left(x + \frac{x^2}{2} + \cdots + \frac{x^k}{k}\right)}}{1x} \]Even better, the coefficient
\[ \frac{\textrm{Perm}_{\gt k}(n)}{n!} \]is the probability we're after: the probability that the shortest cycle in a randomly chosen permutation of an \(n\)element set has length \(\gt k\).
Unfortunately, working out these coefficients is hard.
Fortunately, we just want to know the limit of these coefficients as \(n \to \infty\). And for this we can use a little result from complex analysis. I'll prove it later:
Lemma. Suppose
\[ f(z) = \sum_{n \ge 0} c_n z^n \]is a meromorphic function defined on a disc of radius \(\gt 1\) centered at the origin, which is analytic on this disc except for a simple pole at \(z = 1\). Then \( \lim_{n \to \infty} c_n\) equals the residue of \(f\) at \(z = 1\).
If you want to understand this lemma, I urge that you consider an example: the function \(1/(1z)\). The coefficients \(c_n\) of its power series are all \(1\), and it has a pole at \(z = 1\) with residue \(1\). This is why we need to take the residue of \(f\) in the lemma. In fact, proving the lemma consists of reducing the general problem to this special case.
Given this lemma, the quantity we're trying to compute:
\[ \lim_{n \to \infty} \frac{\textrm{Perm}_{\gt k}(n)}{n!} \]is the residue of the function
\[ \frac{e^{\left(z + \frac{z^2}{2} + \cdots + \frac{z^k}{k}\right)}}{z1} \]at the point \(z = 1\). This function is just minus the generating function of \(\textrm{Perm}_{\gt k}\), but I've changed \(x\) to \(z\) to emphasize that now we're thinking of it as a function on the complex plane. It's actually analytic everywhere except \(z = 1\), and since \(1/(z1)\) has a simple pole of residue \(1\) at this point, the residue of
\[ \frac{e^{\left(z + \frac{z^2}{2} + \cdots + \frac{z^k}{k}\right)}}{z1} \]at \(z = 1\) is just the numerator evaluated at \(z = 1\), namely
\[ e^{\left(1 + \frac{1}{2} + \cdots + \frac{1}{k}\right)} \]So this number is the probability that the shortest cycle in a randomly chosen permutation of an \(n\)element set has length \(\gt k\), in the \(n \to \infty\) limit.
This is beautiful, I think. And it leads immediately to another beautiful fact.
Puzzle 4. Give a simpler asymptotic formula for the above probability as \(k \to \infty\).
Here we use the fact that
\[ 1 + \frac{1}{2} + \cdots + \frac{1}{k} = \ln k + \gamma + o(1) \]where \(\gamma \approx 0.57721\) is Euler's constant. So, as \(k \to \infty\), the probability we've just computed is
\[ e^{\left(1 + \frac{1}{2} + \cdots + \frac{1}{k}\right)} \sim e^{\left(\ln k + \gamma \right)} = \frac{1}{e^\gamma \, k}\]When Euler discovered his constant \(\gamma\) in his 1776 paper "On a memorable number naturally occurring in the summation of the harmonic series", he voiced his suspicion that \(e^\gamma\) would be interesting... and here is some evidence that it is!
By the way, I learned about these questions here:
They were first answered in 1952:
To wrap up, let me prove that lemma. It's just a special case of a much more general useful result on the asymptotic behavior of the coefficients of a power series:
But once you've seen the proof of this special case, it's easy to generalize.
Lemma. Suppose
\[ f(z) = \sum_{n \ge 0} c_n z^n \]is a meromorphic function defined on a disc of radius \(\gt 1\) centered at the origin, which is analytic on this disc except for a simple pole at \(z = 1\). Then \( \lim_{n \to \infty} c_n\) equals the residue of \(f\) at \(z = 1\).
Proof. The idea is to 'subtract off the singularity'. We can write
\[ f(z) = g(z) + \frac{a}{1z} \]where \(g\) is analytic on a disc of radius \(\gt 1\) and \(a\) is the residue of \(f\) at \(z = 1\). If we write
\[ g(z) = \sum_{n \ge 0} d_n z^n \]then we have
\[ c_n = d_n + a \]for all \(n\). But the Cauchy–Hadamard theorem says that
\[ \limsup_{n \to \infty} d_n^{1/n} = 1/R \]where \(R\) is the radius of convergence of \(\sum_{n \ge 0} d_n z^n\), and since \(R \gt 1\) we have \(d_n \to 0\) as \(n \to \infty\). Thus
\[ c_n \to a \]as was to be shown. ■
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