A Combinatorial Proof of log(e^{z}) = z

Doron ZEILBERGER ^{1}

As he was rushing to his Complex Analysis graduate class, my colleague, Hector Sussmann, asked me whether I can find a formal-power-series-proof of log(e^{z}) = z, where

e^{z} := ∑_{k=0}^{∞} \frac{z^{k}}{k!} ,

and

log(1 + z) := ∑_{k=1}^{∞}(−1)^{k−1} \frac{z^{k}}{k} .

Of course, Hector was aware that one way is to use ‘calculus’, of the formal kind (of course, he knows that I don’t believe in any other versions!), as outlined, for example, in my classroom note

http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/lag.pdf .

Indeed (log(e^{z}))′ = (e^{z})′/e^{z} = e^{z}/e^{z} = 1 and ‘integrating’ gives the desired result, but he asked for a direct calculus-free proof, even of the good (i.e. formal) kind.

Here goes

log(e^{z}) = log(1 + ∑_{k=1}^{∞} \frac{z^{k}}{k!} ) = ∑_{k=1}^{∞}(−1)^{k−1} \frac{1}{k} (∑_{a=1}^{∞} \frac{z^{a}}{a!} )^{k}

= ∑_{k=1}^{∞}(−1)^{k−1} \frac{1}{k} ∏_{i=1}^{k}( ∑_{a_{i}=1}^{∞} \frac{z_{i}^{a}}{a_{i}!})

= ∑_{k=1}^{∞}(−1)^{k−1} \frac{1}{k} ∑_{n=1}^{∞} ( ∑_{{(a_{1},...,a_{k})∈N^{k} \atop a_{1}+...+a_{k}=n}} \frac{1}{a_{1}! ⋯ a_{k}!}) z^{n}

= ∑_{k=1}^{∞}(−1)^{k−1} \frac{1}{k} ∑_{n=1}^{∞} k!S(n, k)\frac{z^{n}}{n!} ,

where S(n, k), the Stirling numbers of the second kind, are the number of set partitions of {1, 2, …, n} into k sets. It is well-known and trivial to see that

S(n, k) = S(n − 1, k − 1) + kS(n − 1, k) ,

1 Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen

Rd., Piscataway, NJ 08854-8019, USA. zeilberg at math dot rutgers dot edu , http://www.math.rutgers.edu/~zeilberg/.

Sept 27, 2005. Exclusively published, as a class-room note in http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/hector.pdf . Supported in part by the NSF.

(element n may be either a loner, in which case the rest are partitioned into k − 1 sets or it joins one of the k existing sets). Continuing, we get, by changing order of summation,

= ∑_{n=1}^{∞} (∑_{k=1}^{n}(−1)^{k−1}(k − 1)!S(n, k)) \frac{z^{n}}{n!}

The coefficient of z^{n}/n! is

∑_{k=1}^{n}(−1)^{k−1}(k − 1)![S(n − 1, k − 1) + kS(n − 1, k)] =

∑_{k=1}^{n}(−1)^{k−1}(k − 1)!S(n − 1, k − 1) + ∑_{k=1}^{n−1}(−1)^{k−1}k!S(n − 1, k) =

∑_{k=1}^{n−1}(−1)^{k}k!S(n − 1, k) − ∑_{k=1}^{n−1}(−1)^{k}k!S(n − 1, k) = 0 ,

if n > 1. Of course, when n = 1 it is 1.