From the Geometric Series to Stirling Numbers

The geometric series

with |x| < 1 can be written in the form

Differentiating with respect to x gives

The numerator inside the parentheses is a polynomial of degree 1 in n with coefficients that are functions of x.  Letting p₁(n,x) denote this polynomial, the above identity can be written as

In general, given a relation of the form

where k is a positive integer and p_k(n,x) is a polynomial of degree k in n with coefficients that are functions of x, we can differentiate with respect to x to give

The numerator of the coefficient of xⁿ in this summation is a polynomial of degree k+1 in n, so we will denote it by p_k+1.  Thus we can define an infinite family of polynomials by means of the recurrence

The first several polynomials of this form are

Notice that these expressions contain the sub-polynomials

We will denote the kth such polynomial by s_k(n), the roots of which are the integers 1 through k.  Thus for all k > 1 we see that the first term of p_k vanishes for n equal to 0, 1,.., k-1, whereas the second term vanishes for n equal to 0, 1,.., k-2, and for n = k-1 the second term equals -(k!)x.  This leads us to split the summation in equation (1) into two parts, as shown below

The first summation is simply -k!, independent of x, so we have

If we add k to each appearance of n in the summand, we can change the indices so they range from 0 to infinity.  Thus we have

Letting P_k(n,x) denote p_k(n+k,x), we can write

where

The coefficients of x in these expressions are actually the same polynomials as the "constant" coefficients, except n is replaced by n+1.  Therefore, in terms of the polynomials

we can write the general expression for P_k(n,x) in the form

Obviously we have r_k(n) = (-1)^ks_k(-n).  The coefficients of the first several polynomials s_k are shown below.

Letting S_k,j denote the jth number in the kth row, the magnitude of S_k,j is the sum of all products of j of the first k positive integers (with the convention that the null product is 1).  Obviously we have S_k,k = k!, and we have the generating function

It follows that the numbers S_k,j satisfy the recurrence relation

which corresponds to the fact that the s_k polynomials satisfy the relation s_k(n+1) = ns_k-1(n).  The numbers S_k,j are essentially Stirling numbers of the first kind.  These numbers were named for James Stirling (1692-1770), a Scotsman who is also remembered for the approximate formula for factorials

that appeared in his book Methodus differentialis on series and interpolation.  Stirling was a follower of Newton, and elaborated Newton's scheme for classifying the cubics.  He also worked on the problem of determining the earth's shape, agreeing with Newton that the earth bulges at the equator (contrary to Cassini's claim).  Later, Stirling became manager of the Scottish Mining Company, a demanding job which made it difficult for him to continue with the study of mathematics and physics.  For example, he felt compelled to let lapse a correspondence with Euler, since his work at the mining company prevented him from keeping up with the discussion of Euler's "many ingenious results", "for after deliberations have been interrupted ... patience is required before the mind can be brought to think about the same things once again".  In terms of the usual nomenclature and indexing conventions for the Stirling number denoted by S_k^(j) we have

Speaking of Euler, there is a clear relation between Stirling numbers and the Eulerian numbers.  As explained in the note on Eulerian Numbers, we have the identity

where E_k,j are the Eulerian Numbers.  Thus we have

These identities can be used to determine the coefficients of a polynomial p_k(n) of degree k such that

For example, with k = 2 we seek constant coefficients A, B, C such that

Splitting up the summation by powers of n, and making use of the first three Eulerian identities, we have

Simplifying this relation, we find that it is satisfied if and only if

Since we seek a solution that is valid for arbitrary values of x, each of the quantities in parentheses must vanish, so we have thee equations in three unknowns, which can be solved to give

In the same way we can determine that the polynomials p_k(n) for any degree k are just the Stirling polynomials s_k(n) discussed previously, i.e., we have

This is a nice generalization of the geometric series.  To illustrate, with k = 4 we have the identity

Our derivation was explicitly based on the case k = 2, but it's interesting to examine the system of equations for higher degrees.  With k = 3 the system of equations is

With k = 4 the system of equations is

As a last example, we show the system of equations with k = 5

The left-most column of the coefficient matrix is a row of the Eulerian numbers, the two right-most columns of the coefficient matrix are rows of the binomial coefficients, and the right-hand column vector is a row of Stirling numbers (of the first kind).  Notice that the rows of each of these kinds of numbers sum to factorials if we take their absolute values.  On the other hand, with alternating signs each row of binomial or Stirling numbers sums to zero. 

It's interesting to review the basic recurrences defining the binomial coefficients, the two kinds of Stirling numbers, and the Eulerian numbers.  These are summarized below, with indices signifying left and right diagonal slices beginning with A(1,1) = 1.

We also have a series closely related to equation (3), namely

where, as before, we have r_k(n) = (n+1)(n+2)…(n+k). For example, with k = 4 we have the identity

Combining this with the corresponding identities based on equation (3), we can deduce identities with coefficients involving just the alternate terms of the Stirling polynomials.

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