Introduction

In this post we will talk about the Abel Functional Equation, and to some extent, the Schroeder Functional Equation and the Boettcher Functional Equation:

\( A(f(x)) = A(x) + 1 \)

\( S(f(x)) = c S(x) \)

\( B(f(x)) = B(x)^c \)

Upon inspection we find that each is the exponential of the previous function. In other words: \( S(x) = c^{A(x)} \) and \( B(x) = b^{S(x)} \). So in theory solving any of these functional equations should produce the same results. In practice, however, different methods are used to solve each of these functional equations. As noted elsewhere in this forum, all of these work by the general principle of converting unknown iterates to known iterates, since the functions \( x \mapsto x+1 \), \( x \mapsto cx \) and \( x \mapsto x^c \) all have obvious continuous iterates. The general form of each of these functional equations (FE) is:

\( F \circ G = H \circ F \)

where G and H are known and F is the function being solved for. One thing to note about Schroeder's functional equation is that it is equivalent to using PDM/Bell/Carleman matrices and using matrix diagonalization to evaluate non-integer matrix powers. This can clearly be seen by taking the PDM of the equation:

\( \text{Pdm}[S]\text{Pdm}[f] = \text{Pdm}_x[cx]\text{Pdm}[S] \)

\( \text{Pdm}[f] = \text{Pdm}^{-1}[S]\text{Pdm}_x[cx]\text{Pdm}[S] \)

\( \text{Pdm}[f] = \mathbf{U}^{-1} \text{Pdm}_x[cx] \mathbf{U} \)

and since \( \text{Pdm}_x[cx] \) is a diagonal matrix, this is the diagonalization of \( \text{Pdm}[f] \). The first place I have found this relationship (between diagonalization and Schroeder's FE) is in the work of Aldrovandi on Bell (PDM) matrices. That is all we're going to talk about Schroeder's FE. The rest of this post is going to be about Abel's FE.

Approximate Solutions to the Abel FE

If we start with the assumption that the Abel function has a series expansion, then we can define its coefficients as follows, with an equivalent definition of f(x):

\( A(x) = \sum_{k=0}^{\infty} A_k (x-x_0)^k \)

\( f(x) = \sum_{k=0}^{\infty} f_k (x-x_0)^k \)

and when we form a sequence of the derivatives of the Abel Functional Equation, we get an infinite set of equations in an infinite number of variables:

\(

\begin{array}{rl}

A(f(x_0)) & = A(x_0) + 1 \\

A'(f(x_0))f'(x_0) & = A'(x_0) \\

A''(f(x_0))f'(x_0)^2 + A'(f(x_0))f''(x_0) & = A''(x_0) \\

\cdots & = \cdots

\end{array}

\)

the first thing to notice is that it's impossible to solve for A(x) when \( x_0 \) is a fixed-point of f(x). So keeping this in mind, we can continue to express these equations in terms of the coefficients in order to derive a more managable form:\begin{array}{rl}

A(f(x_0)) & = A(x_0) + 1 \\

A'(f(x_0))f'(x_0) & = A'(x_0) \\

A''(f(x_0))f'(x_0)^2 + A'(f(x_0))f''(x_0) & = A''(x_0) \\

\cdots & = \cdots

\end{array}

\)

\(

\begin{array}{rl}

A(f(x)) - A(x) & = 1 \\

\frac{\partial^n}{\partial^n x}\left[ A(f(x)) - A(x) \right]_{x=x_0} & = \delta_{n0} \\

\frac{\partial^n}{\partial^n x}\left[ \sum_{k=0}^{\infty} A_k (f(x)-x_0)^k - \sum_{k=0}^{\infty} A_k (x-x_0)^k \right]_{x=x_0} & = \delta_{n0} \\

\sum_{k=0}^{\infty} A_k \frac{\partial^n}{\partial^n x}\left[ (f(x)-x_0)^k - (x-x_0)^k \right]_{x=x_0} & = \delta_{n0}

\end{array}

\)

\begin{array}{rl}

A(f(x)) - A(x) & = 1 \\

\frac{\partial^n}{\partial^n x}\left[ A(f(x)) - A(x) \right]_{x=x_0} & = \delta_{n0} \\

\frac{\partial^n}{\partial^n x}\left[ \sum_{k=0}^{\infty} A_k (f(x)-x_0)^k - \sum_{k=0}^{\infty} A_k (x-x_0)^k \right]_{x=x_0} & = \delta_{n0} \\

\sum_{k=0}^{\infty} A_k \frac{\partial^n}{\partial^n x}\left[ (f(x)-x_0)^k - (x-x_0)^k \right]_{x=x_0} & = \delta_{n0}

\end{array}

\)

At this point, it should be obvious that this is a matrix equation: \( \mathbf{a}\mathbf{R} = \mathbf{b} \) where the matrix elements of R are:

\( R_{k(n+1)}[f](x_0) = \frac{\partial^n}{\partial^n x}\left[ (f(x)-x_0)^k - (x-x_0)^k \right]_{x=x_0} \)

and a is a vector of A(x) coefficients, and b is the vector (1, 0, 0, ...). However, even in this form the matrix describes an unsolvable set of equations, since \( A_0 \) plays no part in any equation, thus the point \( (x_0, A_0) \) must be given in order for the Abel function to be uniquely defined. Once this point is given, then we can use equations (n=0..4) and coefficients (k=1..5) for example, and have a 5 by 5 system of equations. Initially the definition of R seems very similar to a PDM, only it involves a subtraction. After we did the shift to allow assigning \( A_0 \), there is even less resemblance between R and a PDM, but it doesn't mean we can't derive a meaningful relationship:\(

\mathbf{R}[f](x_0) = \mathbf{J}(\text{Pdm}[f](x_0) - \mathbf{I})

\)

where \( J_{jk} = k! \delta_{(j+1)k} \). This matrix is the core of my solution to the Abel equation, and the super-logarithm. Now for some notation. Since we will be using this matrix later we will define two noations:\mathbf{R}[f](x_0) = \mathbf{J}(\text{Pdm}[f](x_0) - \mathbf{I})

\)

- \( \mathbf{R}[f](x_0)_n \) means the truncated matrix of R with elements \( R_{kj} \) where \( 1 < (j,k) < n \).

- \( \mathbf{R}[f](x_0)_{nm} \) means the Cramer's rule matrix of R where \( R_{kj}[f](x_0)_{nm} = R_{kj}[f](x_0)(1 - \delta_{mk}) + \delta_{j1}\delta_{mk} \).

These definitions allow us to express the solution to a single coefficient of the infinite matrix equation (if the limit exists) through the standard Cramer's Rule as:

\(

A_k = \lim_{n \rightarrow \infty} \frac{\mathbf{R}[f](x_0)_{nk}}{\mathbf{R}[f](x_0)_{n0}}

\)

A_k = \lim_{n \rightarrow \infty} \frac{\mathbf{R}[f](x_0)_{nk}}{\mathbf{R}[f](x_0)_{n0}}

\)

Possibly Exact Solutions to the Abel FE for \( e^x \)

In this section we will consider the special case of the Abel function for exp(x), or in other words, the natural super-logarithm. For this special case to be approachable, we will need a theorem concerning the conversion between two kinds of series. In this section we will use the term Puiseux seies for series of the form:

\(

f(x) = \sum_{k=0}^{\infty} \frac{1}{k!} \log(x)^k (f \circ \exp)^{(k)}(0)

\)

f(x) = \sum_{k=0}^{\infty} \frac{1}{k!} \log(x)^k (f \circ \exp)^{(k)}(0)

\)

Using this definition of Puiseux, we will derive a relationship between common Taylor series (about x=1) and Puiseux series (about log(x)=0). We start with the definition of the Stirling numbers of the second kind:

\(

e^{\left( e^x - 1 \right)}

= \sum^{\infty}_{k=0} \frac{\left( e^x - 1 \right)^k}{k!}

= \sum^{\infty}_{k=0} \sum^{\infty}_{n=0} \frac{x^n}{n!} \left{{n \atop k}\right}

\)

e^{\left( e^x - 1 \right)}

= \sum^{\infty}_{k=0} \frac{\left( e^x - 1 \right)^k}{k!}

= \sum^{\infty}_{k=0} \sum^{\infty}_{n=0} \frac{x^n}{n!} \left{{n \atop k}\right}

\)

Theorem:

For any analytic function:

\(

f(e^x) = \sum^{\infty}_{n=0} \frac{x^n}{n!} \sum^{n}_{k=0} \left{{n \atop k}\right} f^{(k)}(1)

\)

where \( \left{{n \atop k}\right} \) are Stirling numbers of the second kind.f(e^x) = \sum^{\infty}_{n=0} \frac{x^n}{n!} \sum^{n}_{k=0} \left{{n \atop k}\right} f^{(k)}(1)

\)

Proof:

The double generating function of Stirling numbers of the second kind is well known [Generatingfunctionology].

The Taylor series expansion of f(x) about (x=1) is just:

\( f(x) = \sum^{\infty}_{k=0} f^{(k)}(1) \frac{\left( x - 1 \right)^k}{k!} \)

Expanding \( f(e^x) \) accordingly gives:

\( f(e^x) = \sum^{\infty}_{k=0} f^{(k)}(1) \frac{(e^x-1)^k}{k!} \)

and \( \frac{\left( e^x - 1 \right)^k}{k!} \) is the generating function for \( \left{{n \atop k}\right} \), so:

\( f(e^x) = \sum^{\infty}_{k=0} f^{(k)}(1) \sum^{\infty}_{n=0} \frac{x^n}{n!} \left{{n \atop k}\right} \)

by switching the order of summation, the theorem is obtained.

Corollary:

Given a Taylor series of f(x) about (x=1), the associated Puiseux series is:

\(

\begin{array}{|c|}

\hline \\

f(x) = \sum^{\infty}_{n=0} \frac{\log(x)^n}{n!} \sum^{n}_{k=0} \left{{n \atop k}\right} f^{(k)}(1) \\

\hline

\end{array}

\)

Proof: Trivial, substitute \( x \rightarrow \log(x) \) in the theorem.\begin{array}{|c|}

\hline \\

f(x) = \sum^{\infty}_{n=0} \frac{\log(x)^n}{n!} \sum^{n}_{k=0} \left{{n \atop k}\right} f^{(k)}(1) \\

\hline

\end{array}

\)

Corollary:

Given a Puiseux series of f(x):

\( f(x) = \sum^{\infty}_{n=0} \frac{\log(x)^n}{n!} (f \circ \exp)^{(n)}(0) \)

the associated Taylor series is:

\(

\begin{array}{|c|}

\hline \\

f(x) = \sum^{\infty}_{n=0} \frac{(x - 1)^n}{n!} \sum^{n}_{k=0} \left[{n \atop k}\right] (f \circ \exp)^{(k)}(0) \\

\hline

\end{array}

\)

Proof: Trivial, the Stirling numbers of the first kind \( \left[{n \atop k}\right] \) are basically inverses of \( \left{{n \atop k}\right} \).\begin{array}{|c|}

\hline \\

f(x) = \sum^{\infty}_{n=0} \frac{(x - 1)^n}{n!} \sum^{n}_{k=0} \left[{n \atop k}\right] (f \circ \exp)^{(k)}(0) \\

\hline

\end{array}

\)

What this allows us to do is look at the Abel FE in a new light. Using the Taylor-Puiseux conversion detailed above, we can rewrite the Abel FE of exp(x) as:

\(

\begin{array}{rl}

A(e^x) & = A(x) + 1 \\

\sum^{\infty}_{n=0} \frac{x^n}{n!} \sum^{n}_{k=0} \left{{n \atop k}\right} A^{(k)}(1) & = \sum^{\infty}_{n=0} \frac{(x - 1)^n}{n!} A^{(n)}(1) + 1 \\

\sum^{\infty}_{n=0} \frac{(x-1)^n}{n!} \sum_{k=n}^{\infty} \left({k \atop n}\right) \sum_{j=0}^k \left{{k \atop j}\right} A^{(j)}(1) & = \sum^{\infty}_{n=0} \frac{(x - 1)^n}{n!} A^{(n)}(1) + 1

\end{array}

\)

which after replacing the derivatives with \( A_k \) notation (with \( x_0=1 \)), leads us to the recurrence equation:\begin{array}{rl}

A(e^x) & = A(x) + 1 \\

\sum^{\infty}_{n=0} \frac{x^n}{n!} \sum^{n}_{k=0} \left{{n \atop k}\right} A^{(k)}(1) & = \sum^{\infty}_{n=0} \frac{(x - 1)^n}{n!} A^{(n)}(1) + 1 \\

\sum^{\infty}_{n=0} \frac{(x-1)^n}{n!} \sum_{k=n}^{\infty} \left({k \atop n}\right) \sum_{j=0}^k \left{{k \atop j}\right} A^{(j)}(1) & = \sum^{\infty}_{n=0} \frac{(x - 1)^n}{n!} A^{(n)}(1) + 1

\end{array}

\)

\(

A_n = \sum_{k=n}^{\infty} \left({k \atop n}\right) \sum_{j=0}^k \left{{k \atop j}\right} A_j

\)

whether this is useful or not remains to be seen. Since this is original research, I still don't know about the convergence of this form, but from initial investigations, I have found it to be quite unpredictable, and not very convergent at all. Perhaps the inverse of this equation might have better convergence properties since this form seems to diverge so badly. I just wanted to put it out there to show that I have been thinking since the last time I updated my website. A_n = \sum_{k=n}^{\infty} \left({k \atop n}\right) \sum_{j=0}^k \left{{k \atop j}\right} A_j

\)

Andrew Robbins

http://tetration.itgo.com