Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the following Schröder functional equation:


where $f,h: ℂ→ℂ$, here $f$ is not analytic and $h$ is analytic and $c∈ℝ$.

My question is: How we can solve this equation (the form of $f$ and its domain of definition). We can take $h$ as $h(s)=1-s$ or $h(s)=s-1$ and for both cases we can take $c=-1$.

Some motivations are available here:

share|cite|improve this question
up vote 1 down vote accepted

I solve such questions using the Carleman-matrix-concept (which you also find mentioned in the wikipedia article). Carleman-matrices contain in its rows the coefficients of a function in its power series representation, and of its powers. So the Carleman-matrix, say C contains the coefficients of $f(x)^0$ (which is the constant 1), $f(x)$, $f(x)^2$ ... in its rows.
In your case where $f(x)=1 - 1 x$ we have $$ C= \begin{bmatrix} 1&.&.&. \\ 1&-1&.&. \\ 1&-2&1&. \\ 1&-3&3&1 \\ \end{bmatrix} \qquad \text{ for } \qquad \begin{array} {} f(x)^0 &=&1 \\ f(x)&=&1-x \\ f(x)^2&=&1-2x+x^2 \\ f(x)^3&=&1-3x+3x^2-x^3 \\ \end{array} $$ such that with a "vandermonde"-type column-vector $V(x) = [1,x,x^2,x^3,...]$ of appropriate size we shall have $$ C \cdot V(x) = V(f(x)) $$

Then if you find a diagonalization of C such that $M^{-1}\cdot D \cdot M = C $ where $D$ is diagonal and $M$ and $M^{-1}$ are triangular, then $M$ in its second row contains the coefficient of the Schröder-function and that in $M^{-1}$ in its second row its inverse. In our case we find that $$ M^{-1} =\begin{bmatrix} 1 & . & . & . \\ 1/2 & 1/2 & . & . \\ 1/6 & 1/2 & 1/3 & . \\ 0 & 1/4 & 1/2 & 1/4 \end{bmatrix} $$ , with $D=\operatorname{diag}([1,-1,1,-1])$ and $$ M = \begin{bmatrix} 1 & . & . & . \\ -1 & 2 & . & . \\ 1 & -3 & 3 & . \\ -1 & 4 & -6 & 4 \end{bmatrix}$$ is a possible solution. (Here $M^{-1}$ can be recognized as the set of coefficients of integrals of the bernoulli-polynomials when we extend the dimension of the matrix infinitely)
In general, the eigenvector-matrix $M$ can be understood as limit of the n'th power of $C$ scaled by the reciprocal of the n'th power of $f'(0)$ when n goes to infinity, and so the Schröder-function as limit of the n'th iterate of $f(x)$ divided by $f'(0)^n$ where $n \to \infty \qquad $ - but can furtherly be scaled by an arbitrary constant factor $\gamma \ne 0$

Remark: Your example which requires only matrix size of $n \times n= 2 \times 2$ is much easier, but then one wouldn't see the general principle (and the relation to the bernoulli-polynomials, so I used the bigger matrix size with n=4 here.

share|cite|improve this answer
@ Gottfried Helms : This is a great answer. Thank you very much. But the function $f$ is not analytic. – ZE1 Dec 31 '12 at 11:13
@rh1: hmm, what do you mean? Unfortunately I mixed your given $h(s)$ by my usual $f(x)$ and so this overlaps your $f(s)$, sorry perhaps I should adapt this. By my example, the function $f(s)$ in your sense is $f(s)=-1+2s$ and $f^{-1}(s)=1/2(1+s)$ so $h(s)=f^{-1}(c \cdot f(s))= 1/2(1+(-1(-1+2s)))=1/2(2-2s)=1-s$ for any complex $s$. The $m$'th iterates are then expressible simply by $m$'th powers of $c$ - and, taking care of the problem of non-unique solutions of fractional or complex powers of $c=-1$ one can define even fractional- and complex-valued iterations heights $m$ – Gottfried Helms Dec 31 '12 at 12:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.