English

Method for solving an iterative functional equation $A^{2^n}(x)=F(x)$

Combinatorics 2013-02-12 v2 Classical Analysis and ODEs Functional Analysis Number Theory

Abstract

Using the notion of the composita, we obtain a method of solving iterative functional equations of the form A2n(x)=F(x)A^{2^n}(x)=F(x), where F(x)=n>0f(n)xnF(x)=\sum_{n>0} f(n)x^n, f(1)0f(1)\neq 0. We prove that if F(x)=n>0f(n)xnF(x)=\sum_{n>0} f(n)x^n has integer coefficients, then the generating function A(x)=n>0a(n)xnA(x)=\sum_{n>0}a(n)x^n, which is obtained from the iterative functional equation 4A(A(x))=F(4x)4A(A(x))=F(4x), has integer coefficients. Key words: iterative functional equation, composition of generating functions, composita.

Keywords

Cite

@article{arxiv.1302.1986,
  title  = {Method for solving an iterative functional equation $A^{2^n}(x)=F(x)$},
  author = {Dmitry Kruchinin and Vladimir Kruchinin},
  journal= {arXiv preprint arXiv:1302.1986},
  year   = {2013}
}

Comments

11 pages

R2 v1 2026-06-21T23:23:06.398Z