## f(f(x))=exp(x) Generates the Multinomials

To find an analytic function f(x) such that f(f(x)) = exp(x), one
approach would be to consider the power series expansion
f(x) = c0 + c1 x + c2 x^2 + c3 x^3 + c4 x^4 + ...
Substituting this expression into itself for x, and collecting terms
by powers of x, we can equate the resulting coefficients to the
corresponding coefficients of exp(x). This gives a set of conditions
on the derivatives of f at 0 and c0. Letting fn(x) denote the nth
derivative of f, we have the conditions
f0(c0) = 1
f1(0) f1(c0) = 1
f2(0) f1(c0) + f1(0)^2 f2(c0) = 1
f3(0) f1(c0) + 2 f2(0) f1(0) f2(c0) + f1(0)^3 f3(c0) = 1
f4(0) f1(c0) + 3 f2(0)^2 f2(c0) + 4 f1(0) f3(0) f2(c0)
+ 6 f1(0)^2 f2(0) f3(c0) + f1(0)^2 f4(c0) = 1
and so on. In general, the kth condition consists of p(k) terms,
where p(k) denotes the number of partitions of k. For any given
partition we can define the parameters e_j such that
k = 1 e_1 + 2 e_2 + 3 e_3 + ... + k e_k
and then define the sum
q = e_1 + e_2 + ... + e_k
With these parameters we can write the kth condition on the derivatives
of f(x) as follows:
_ _
| k! k |
SUM | ---------------------- fq(c0) PROD fj(0)^e_j | = 1
| k j=1 |
| PROD [(j!)^e_j (e_j)!] |
|_ j=1 _|
where the sum is evaluated over all the partitions of k. The
coefficient of
fq(c0) f1(0)^e1 f2(0)^e2 ... fk(0)^ek
is the multinomial coefficient denoted as M_3 in Abramowitz and
Stegun's "Handbook of Mathematical Functions". Thus, in a peculiar
sense, the function f(x) implicitly defined by f(f(x))=exp(x) can
be regarded (at least formally) as the "generating function" of
this family of coefficients. This raises the question of what
other sequences of numbers have implicit generating functions of
this kind.
It's also interesting that, whereas we customarily define an analytic
function around a certain point using the Taylor series expansion by
essentially specifying the all derivatives of the function at that
point, in this case we are (formally) defining the function f(x) by
specifying conditions on the derivatives at TWO distinct points, x=0
and x=c0=f(0).

Return to MathPages Main Menu