NANACAOct 19, 2015

A Further Property of Functions in Class ${\bf B}^{\boldsymbol(m)}$

arXiv:1510.05501
Originality Synthesis-oriented
AI Analysis

This is an incremental theoretical result for researchers working on asymptotic analysis and numerical integration of functions with special asymptotic expansions.

The paper proves that if a function f belongs to class B^{(m)} and g belongs to A^{(s)} with g(x)>0, then the composition f∘g also belongs to B^{(m)}, extending the applicability of the D^{(m)}-transformation to integrals with complicated arguments. The result is demonstrated with two numerical examples using the D^{(3)} transformation.

We say that a function $α(x)$ belongs to the set ${\bf A}^{(γ)}$ if it has an asymptotic expansion of the form $α(x)\sim \sum^\infty_{i=0}α_ix^{γ-i}$ as $x\to\infty$, which can be differentiated term by term infinitely many times. A function $f(x)$ is in the class ${\bf B}^{(m)}$ if it satisfies a linear homogeneous differential equation of the form $f(x)=\sum^m_{k=1}p_k(x)f^{(k)}(x)$, with $p_k\in {\bf A}^{(i_k)}$, $i_k$ being integers satisfying $i_k\leq k$. These functions have been shown to have many interesting properties, and their integrals $\int^\infty_0 f(x)\,dx$, whether convergent or divergent, can be evaluated very efficiently via the Levin--Sidi $D^{(m)}$-transformation. (In case of divergence, they are defined in some summability sense, such as Abel summability or Hadamard finite part or a mixture of these two.) In this note, we show that if $f(x)$ is in ${\bf B}^{(m)}$, then so is $(f\circ g)(x)=f(g(x))$, where $g(x)>0$ for all large $x$ and $g\in {\bf A}^{(s)}$, $s$ being a positive integer. This enlarges the scope of the $D^{(m)}$-transformation considerably to include functions of complicated arguments. We demonstrate the validity of our result with an application of the $D^{(3)}$ transformation to two integrals $I[f]$ and $I[f\circ g]$, for some $f\in{\bf B}^{(3)}$ and $g\in{\bf A}^{(2)}$.

Foundations

The foundational work for this paper's niche, ranked by how specifically the neighbourhood builds on it — not by global fame.

Your Notes