0

Assume a continued fraction such as $\begin{equation} \cfrac{1}{1+z+\cfrac{1}{2+z+\cfrac{1}{3+z+\dots}}} \end{equation}.$

  1. Is it possible to find a series expansion around $z=\infty$? I know that around $z=0$ it is just to truncate the continued-fraction to the smallest fraction?
  2. If so, what would be the extension to a matrix continued-fraction, i.e.? $\begin{equation}\cfrac{1}{1+A(z)+\cfrac{1}{2+A(z)+\cfrac{1}{3+A(z)+\dots}}},\end{equation}$ where $A(z)$ is a $n\times n$ matrix and $1/A(z)$ should be interpreted as $A(z)^{-1}$.
amWhy
  • 210,739
J.Agusti
  • 105
  • 1
    For your first question, substitute $z = \frac 1w$ and find the Taylor series about $w = 0$. That is what a "series expansion about $z = \infty$" means. – Paul Sinclair Aug 21 '24 at 11:35

1 Answers1

5

The (first) given continued fraction has a closed form in terms of modified Bessel functions: $$ \cfrac1{1+z+\cfrac1{2+z+\cfrac1{3+z+\dots}}}=f(z):=\color{blue}{\frac{I_{1+z}(2)}{I_z(2)}}=\frac{\sum_{n=0}^\infty\frac1{n!}\prod_{k=0}^{n+1}\frac1{k+z}}{\sum_{n=0}^\infty\frac1{n!}\prod_{k=0}^n\frac1{k+z}} $$ (everywhere it converges). As a function of $z\in\mathbb{C}$, this has an infinite number of poles with unbounded magnitude (consider negative real $z$). This implies the negative answer to both questions.

An easier argument: if the Taylor expansion of $g(z)=f(1/z)$ around $z=0$ exists, then the limit $\lim\limits_{n\to\infty}g(-1/n)=\lim\limits_{n\to\infty}I_{n-1}(2)/I_n(2)$ must be finite (here $n$ is an integer), which is not the case.

Still, there is an asymptotic series $f(z)\asymp\sum_{n=1}^{(\infty)}c_n z^{-n}$ as $z\to+\infty$. It can be obtained algorithmically (using the CF directly, or the series-quotient representation above), and the first few values of $c_n$ are $$ 1,-1,0,3,-8,9,19,-141,448,-759,-951,14355,-71810,238617,-421622,\\-1283181,17699702,-110524503,494858579,-1480719213,\dots $$ I don't know what (closed or some compact form of) these values are.

metamorphy
  • 43,591
  • 1
    A related sequence is the sequence of Bessel numbers. – Gary Aug 22 '24 at 04:47
  • That is a nice answer and I never thought it could be expressed in terms of Bessel functions. In my case, $0<z$ is always positive. Would that change the analysis you did? And regarding the matrix case, is it obvious that the analysis of complex also apply to the matrix case? – J.Agusti Aug 22 '24 at 06:11
  • @J.Agusti: No, the positivity of $z$ doesn't help at all. When a (convergent) Taylor expansion exists, it has a positive radius of convergence. Such a series in $1/z$ would converge in $|z|>R$ for some $R$, but we have poles there. The same for the matrix case: we need a convergent series. Perhaps there are other analytic-continuation techniques to extend the CF definition to matrices, but these have nothing to do with the power series in $A^{-1}$. – metamorphy Aug 22 '24 at 06:47
  • 2
    The $c_n$ are tabulated at https://oeis.org/A295289 – Gerry Myerson Aug 22 '24 at 07:17
  • @metamorphy I like the easier argument, thanks! How did you figure out the closed form of the continued fraction? Is there some procedure to find them? I have another continued fraction, much more complicated, with $z^2$ terms which I cannot find the closed form yet. – J.Agusti Aug 22 '24 at 07:31
  • @J.Agusti: The closed form of the CF can be obtained from the recurrence relations for Bessel functions (after some fiddling with convergence issues) or by computing the exponential generating function of the convergents of the CF (the unknown-result-style approach I use in some of my answers); the recurrence for the convergents yields an ODE for the EGF, which reduces to Bessel's equation. As for your more complicated CF, you may want to post a dedicated question ;) – metamorphy Aug 23 '24 at 03:30
  • @metamorphy But for this case, you still need to 'guess' that the CF is related to Bessel functions or to know that the convergence yields the Bessel ODE. But there are so many functions which admit CF representation (look for example Gauss's continued fraction). For example, I have a CF G(x) that for G(x=0) I know is related to 1-Gamma[0,1/s], where G[a,z] is the incomplete gamma function. It is, for example, not clear that G(x) will also be given in terms of an incomplete gamma function. – J.Agusti Aug 23 '24 at 07:23
  • @metamorphy I posted two related questions which might need some of your feedback, Post1 and post2 – J.Agusti Aug 26 '24 at 13:10