A note on Sonnenschein summability matrices

TL;DR

This paper presents a straightforward method for calculating the column sums of Sonnenschein summability matrices, simplifying an aspect of their analysis.

Contribution

It introduces a simple computational approach for column sums of Sonnenschein matrices, enhancing understanding and usability.

Findings

Efficient computation method for column sums

Simplifies analysis of Sonnenschein matrices

Potential applications in summability theory

Abstract

In this note, we give a simple method for computing the column sums of the Sonnenschein summability matrices.

Equations

$$f(z)=\frac{\alpha +\left(1-\alpha -\beta \right)z}{1-\beta z},z\in \mathbb{C}\backslash \{\frac{1}{\beta }\},$$

$${f_{n,k}}=\sum\limits_{v=0}^{k}{\left(\begin{array}[]{l}n\\ v\end{array}\right){{\left({1-\alpha-\beta}\right)}^{v}}{\alpha^{n-v}}\left(\begin{array}[]{l}n+k-v-1\\ \,\,\,\,\,\,\,\,\,\,k-v\end{array}\right){\beta^{k-v}}}.$$

$$\sum _{n=0}^{\infty }{\left[f\left(z\right)\right]}^n=\frac{1}{1-f(z)}.$$

$$\frac{1}{1-f(z)}=\frac{1}{1-\alpha }+\frac{\left(1-\beta \right)z}{\left(1-\alpha \right)\left(1-z\right)}=\frac{1}{1-\alpha }+\frac{1-\beta }{1-\alpha }\sum _{k=1}^{\infty }{z}^k.$$

$$h(z)=si{n}^2(\frac{\pi z}{2}),(z\in \mathbb{C})$$

$$\begin{array}[]{l}\frac{1}{{1-h(z)}}={\sec^{2}}\left({\frac{{\pi z}}{2}}\right)=\frac{2}{\pi}\frac{{d\tan\left({\frac{{\pi z}}{2}}\right)}}{{dz}}\\ \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{2}{\pi}\frac{{d\left({\sum\limits_{n=0}^{\infty}{\frac{{{{(-1)}^{n}}2({2^{2n+2}}-1)}}{{\left({2n+2}\right)!}}{B_{2n+2}}~{}{\pi^{2n+1}}}{z^{2n+1}}}\right)}}{{dz}}\\ \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\sum\limits_{n=0}^{\infty}{\frac{{{{(-1)}^{n}}\left({8n+4}\right)({2^{2n+2}}-1)}}{{\left({2n+2}\right)!}}{B_{2n+2}}~{}{\pi^{2n}}}{z^{2n}},\end{array}$$

$$B_n=\sum _{k=0}^n\frac{1}{k+1}\sum _{r=0}^k{(-1)}^r\left({}_r^k\right)r^n.(n=0,1,2,...)$$

$$\frac{{(-1)}^n\left(8n+4\right)(2^{2n+2}-1)}{\left(2n+2\right)!}{B}_{2n+2}{\pi }^{2n}.$$

Full text

1 Introduction

Let $f(z)$ be an analytic function in $|z|<r,r\geq 1$ with $f(1)=1.$ The matrix $S=(f_{n,k}),$ where $(f_{n,k})$ are defined by ${\left[{f\left(z\right)}\right]^{n}}=\sum_{k=0}^{\infty}{{f_{n,k}}{z^{k}}}$ is called a Sonnenschein matrix [4]. The special choice

[TABLE]

where $\alpha$ and $\beta$ are complex numbers, gives the Karamata matrix $K[\alpha,\beta]$ and its coefficients as a Sonnenschein matrix are given by [3]

[TABLE]

Recently in [2], the authors have calculated the row and column sums of Karamata matrices in a relatively complicated way. In this note we give a new and simple method for computing the column sums of these matrices which can be also applied to another Sonnenschein matrices.

Start with a general $f.$ It is clear that if $|f(0)|<1$ then $|f(z)|<1$ in a neighborhood of $z=0.$ Thus

[TABLE]

Now the sum we want is the coefficient of $z^{k}$ of the right-hand side of the above equation. For the case of the Karamata matrices we have $f(0)=\alpha,$ so we assume first that $|\alpha|<1.$ Now

[TABLE]

The coefficient of $z^{k}$ is then easily found. Indeed, the sum of the first column is $\frac{1}{{1-\alpha}}$, and the sum of all other columns are $\frac{{1-\beta}}{{1-\alpha}}.$

The point is that this method may apply to other choices of the function $f$ and other Sonnenschein matrices. As another example consider the function

[TABLE]

which is holomorphic function and its coefficients as a Sonnenschein matrix are as followings:

If $n=0$, then ${a_{0,k}}=\delta_{0k},$ for all $k;$ 2. 2.

If $n\neq 0$ and $k=0$, then ${a_{n,k}}=\frac{1}{{{4^{n}}}}\left({{}_{\,\,n}^{2n}}\right)+\sum\limits_{r=0}^{n-1}{\frac{{{{(-1)}^{n+r}}}}{{{2^{2n-1}}}}\left({{}_{2r}^{2n}}\right)}$; 3. 3.

If $n\neq 0$ and $k\neq~{}0$, then ~{}~{}{a_{n,k}}=\left\{\begin{array}[]{l}\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k is odd$,\\ \\ {\pi^{2k}}\sum\limits_{r=0}^{n-1}{\frac{{{{\left({-1}\right)}^{n+r+k}}{{\left({n-r}\right)}^{2k}}}}{{{2^{2n-1}}\left({2k}\right)!}}\left({{}_{2r}^{2n}}\right)}\,\,\,\,\,\,\,\,k$ is even.\end{array}\right.

We have $h(0)=0$, thus

[TABLE]

where $B_{n}$ is the sequence of Bernoulli numbers (see [1], pp. 274 - 275), defined by

[TABLE]

Therefore, the sum of odd columns are [math], and the sum of $2n-$th columns ($n=0,1,2,...$) is

[TABLE]

Acknowledgements

The authors would like to thank professor Mourad Ismail [University of Central Florida, Orlando] for his technical assistance during the preparation of the this note.

References