# Finite Transforms with Applications to Bessel Differential Equations of
Order Higher than Two

Gabriel López Garza  Mathematics Department, Universidad Autónoma
Metropolitana
Ciudad de México, México

###### Abstract

A finite transformation method is introduced. This method is equivalent to the
$Z$ transform method to a certain extent but generalizes it. By applying the
presented method to the Bessel functions, it is possible to solve related
ordinary differential equations of order higher than two with given initial
conditions.

###### keywords:

Finite Transforms, Bessel Functions, Operational Calculus

††journal: Integral Transforms and Special Functions

##  1 Introduction

The best-known transformation defined by sequences, among the so-called
“Finite Transforms”, is probably the $Z$ transform (for an extensive study see
[13]). Given a sequence $\\{f_{n}\\}$, the Z transform is defined by
$F(z)=\sum_{n=0}^{\infty}\frac{f_{n}}{z^{n}}$. The main difference between the
$Z$ transform with the transforms introduced in this paper is that whereas a
sequence is transformable if the series in the definition converges for at
least one value of $z$, for the transforms that we study, convergence is not
required. In section 4 it is shown how the zeta transform is a special case of
the transforms studied in this paper. Other well known examples of finite
transforms are the Sturm-Liouville Transforms (see for instance [4]).
Basically, in order to solve some differential equations (ordinary or even
partial differential equations), a sequence can be associated to certain
suitable functions. As it is known, the eigenfunctions $\\{\phi_{n}(t)\\}$ of
self-adjoint operators satisfying certain boundary conditions (i. e., the
solutions of a given Sturm-Liouville problem), provide a generalized Fourier
expansion for a function $f$ (which satisfies certain conditions depending on
the problem under consideration), given by

$f(t)=\sum_{n=0}^{\infty}a_{n}\phi_{n}(t),\quad a_{n}=\langle
f,\phi_{n}\rangle,$

where $\langle\cdot,\cdot\rangle$ is the inner product defined for each Sturm-
Liouville problem. So, in this example of finite transform, the sequence
$\\{a_{n}\\}$, may be defined as the transform, say $\mathcal{T}$, of a
function $f(t)$ by

$\mathcal{T}[f(t)]\stackrel{{\scriptstyle def}}{{=}}\\{a_{n}\\},\,n\geq 0.$

The transform so defined is very useful in solving differential equations to
some extent. Still, the method induced by this transform (for instance in [4])
does not provide a complete Mikusiński’s operational method as defined in
[11]. This is so since the convolution product induced by the inner product
has non-zero divisors of zero (for instance for any $n\neq m,$
$\langle\phi_{n},\phi_{m}\rangle=0$ given that the eigenfunctions of Sturm-
Liouville problems are orthogonal and, of course, are different from zero).
This limitation (limitation from the heuristic standpoint) is a rich source of
examples for convolution algebras [8], nevertheless. The main interest of this
paper is trying to extend the finite transform method as much as possible, if
not to complete operational calculus in the Mikusiński sense, but to extend it
to a method that allows solving many ordinary or partial differential
equations involving Bessel and other operators in various practical problems.

Examples of finite transforms are, the Legendre Transform [3], and the Lagerre
transform [10]. Still, there are many other forms of associating a sequence to
a given function besides the Sturm-Liouville transform, for instance, the
Neuman series [14, Chapter XVI]. In our approach, a transform is constructed
through a differential operator which is in some sense an extension of the
Maclaurin series for Bessel Functions as will be explained.

The transform method that is studied in this paper consists in associating to
a certain set of suitable functions a sequence $\\{a_{n}\\}\in\mathbb{C}$.
Since the set of sequences is not an algebraic field, the study is restricted
to sequences that are invertible for the Cauchy product. The fact that not
each sequence different from zero is invertible is the main cause of the
impossibility of extending our study to a complete Mikusiński’s Operational
Calculus. So our study shares more similarities with the traditional Laplace
transform Method than with the Mikusiński’s Operational Calculus as that
studied in [1]. In fact, to certain differential equations with given initial
conditions, a polynomial may be associated via the finite transformation, so
that an equation for the transform may be solved in terms of partial fraction
decomposition. The partial fraction decomposition may be associated with known
transforms or the Cauchy product of known transforms. Finally, the inverse
transform is obtained with which the problem is solved, as usual.

The article is divided as follows: in section 2 the operations used in the set
of transformations are described and the finite transform used is defined
(subsection 2.1). In section 3 some higher order Bessel equations are solved
and equivalence to Laplace transform is shown but is worth noticing that the
transforms are defined by differential and not with integral operators as in
the case of classic Laplace transform. In the final section, the striking
similarities that have long been noted [5, p. 136] between the functions ber
and bei with the functions cos and sin, respectively, are fully explained
within the context of the transform method studied in this article.

##  2 Mathematical setting

Given two sequences $\\{a_{n}\\},\\{b_{n}\\}$ we define the product (the so-
called Cauchy product) by

$\displaystyle\\{a_{n}\\}\\{b_{n}\\}$ $\displaystyle=$
$\displaystyle\left\\{\sum_{\tau=0}^{n}a_{\tau}b_{n-\tau}\right\\}$
$\displaystyle=$ $\displaystyle\\{a_{0}b_{0},a_{0}b_{1}+a_{1}b_{0},\dots\\}.$

In [2, p. 721] it is shown that the Cauchy product of two sequences is zero if
and only if one of the sequences is the zero sequence $a_{n}=0,n\geq 0$ so
that a quotient field of sequences can be constructed. Nevertheless not every
sequence different from the zero sequence is invertible. In order to be
invertible, the first term of a given sequence is required to be different
from zero. Actually, given a sequence
$\\{a_{n}\\},a_{n}\in\mathbb{C},a_{0}\neq 0$, the multiplicative inverse
$\\{b_{n}\\}$ with respect to the Cauchy product is easily calculated
recursively, since $\\{a_{n}\\}\\{b_{n}\\}=\\{1,0,0,\dots\\}$ implies

$\displaystyle b_{0}$ $\displaystyle=$ $\displaystyle\frac{1}{a_{0}}$
$\displaystyle b_{1}$ $\displaystyle=$ $\displaystyle-\frac{a_{1}}{a_{0}^{2}}$
$\displaystyle b_{2}$ $\displaystyle=$
$\displaystyle\frac{a_{1}^{2}}{a_{0}^{3}}-\frac{a_{2}}{a_{0}^{2}}$
$\displaystyle\vdots$ $\displaystyle b_{n}$ $\displaystyle=$
$\displaystyle\frac{-1}{a_{0}}\left(a_{1}b_{n-1}+\cdots+a_{n}b_{0}\right).$

The fact that not every sequence different from zero has a multiplicative
inverse restricts the construction of an operational calculus in the
Mikusiński’s sense [11], but, as we will see, from the invertible sequences
used in this paper, it is possible to build many transforms and operate with
them as in many other transform methods used to solving differential
equations. After the last considerations the set of suitable sequences
$\mathcal{A}=\\{\\{a_{n}\\}:a_{0}\neq 0\\}$ is defined.

With the Cauchy product, it is possible to construct many operators
$T:\mathcal{A}\to\mathcal{A}$. An important example is the right shift, which
is an operator defined by the Cauchy product with the sequence
$s=\\{0,1,0,0,\dots\\}$ if $\\{a_{n}\\},n\geq 0$ is any sequence with
$a_{n}\in\mathbb{C}$, we have

$s\\{a_{0},a_{1},a_{2},\dots\\}=\\{0,1,0,0,\dots\\}\\{a_{0},a_{1},a_{2},\dots\\}=\\{0,a_{0},a_{1},a_{2},\dots\\}.$

So the right shift operator $S:\mathcal{A}\to\mathcal{A}$ is defined by
$S[\\{a_{n}\\}]\stackrel{{\scriptstyle def}}{{=}}s\\{a_{n}\\}$. The notation
$S[\\{a_{n}\\}]$ is not in use in most papers and calling the operator $s$
instead $S$ is the standard procedure (for instance in [2]), we will follow
this practice throughout this paper to avoid confusion with an established
practice.

It follows that $s^{2}=ss=\\{0,0,1,0,0,\dots\\}$ and, in general

$s^{n}=\\{0,0,\dots,0,1,0,\dots\\},$

is the sequence with zeros everywhere except in the place $n+1$ where there is
the number one. With the powers of $s$ the following notation is standard

$\displaystyle\\{a_{0},a_{1},a_{2},\dots\\}\stackrel{{\scriptstyle
def}}{{=}}a_{0}s^{0}+a_{1}s+a_{2}s^{2}+\cdots,$ (2)

where $s^{0}=\\{1,0,0,\dots\\}$ and subsequently we write simply
$a_{0}s^{0}=a_{0}$, and any constant $c$ in our calculus represents the
sequence $c\stackrel{{\scriptstyle def}}{{=}}\\{c,0,0,\dots\\}$. We recall
that the relation (2) is purely formal and does not involve the concept of
convergence at all [2, p. 723].

Notice that $s=\\{0,1,0,\dots\\}$ is not invertible with respect to the Cauchy
product, but the sequence $\\{1,0,0,\dots\\}+s=\\{1,1,0,0,\dots\\}$, actually
is, and the following formulas are derived easily [2]

$\displaystyle\\{1\\}$ $\displaystyle=$ $\displaystyle\frac{1}{1-s},$ (3)
$\displaystyle\\{r^{n}\\}$ $\displaystyle=$ $\displaystyle\frac{1}{1-rs},$ (4)
$\displaystyle\left\\{\cos\frac{\pi}{2}n\right\\}$ $\displaystyle=$
$\displaystyle\\{1,0,-1,0,\dots\\}=\frac{1}{1+s^{2}},$ (5)
$\displaystyle\left\\{\sin\frac{\pi}{2}n\right\\}$ $\displaystyle=$
$\displaystyle\\{0,1,0,-1,\dots\\}=\frac{s}{1+s^{2}}.$ (6)

That is, the sequence $1-s$ is the multiplicative inverse of the sequence
$\\{1\\}=\\{1,1,1,\dots\\}$ for the Cauchy product or, more properly speaking,
the sequence $1-s$ is a representative of the class of inverse sequences of
$\\{1\\}$, and similar meaning has the symbol $1/G(s)$ for the identities from
(4) to (6), as well.

The left shift $l$ of a sequence $\\{y_{n}\\}$ is defined by

$\displaystyle l\\{y_{n}\\}=l\\{y_{0},y_{1},y_{2},\dots\\}$
$\displaystyle\stackrel{{\scriptstyle def}}{{=}}$
$\displaystyle\\{y_{1},y_{2},y_{3},\dots\\}=\\{y_{n+1}\\}$ $\displaystyle
l^{m}\\{y_{n}\\}$ $\displaystyle\stackrel{{\scriptstyle def}}{{=}}$
$\displaystyle\\{y_{n+m}\\}.$ (7)

The following formula [2, p. 724] will be relevant for solving differential
equations related to Bessel and other operators in this paper:

$\displaystyle
s^{m}(l^{m}\\{y_{n}\\})=s^{m}\\{y_{n+m}\\}=\\{y_{n}\\}-y_{0}-sy_{1}-\cdots-s^{m-1}y_{m-1}.$
(8)

Notice that formula (8) is similar to the Laplace transform formula for the
derivative of order $m$ of a given function.

###  2.1 Finite Transform definition

Given a series $g(t)=a_{0}f_{0}(t)+a_{1}f_{1}(t)+a_{2}f_{2}(t)+\cdots$, where
$a_{n}\in\mathbb{C}$ and $f_{n}(t)$ are given functions, we define the
transform $\mathcal{T}[g(t)]=G(s)$ by the formula

$\displaystyle\mathcal{T}[g(t)]$ $\displaystyle\stackrel{{\scriptstyle
def}}{{=}}$ $\displaystyle\\{a_{n}\\}=\\{a_{0},a_{1},a_{2},\dots\\}$ (9)
$\displaystyle=$ $\displaystyle a_{0}+a_{1}s+a_{2}s^{2}+\cdots=G(s),$ (10)

if and only if there exists a operator $L$ such that
$L[f_{n}(t)]=f_{n-1}(t),n\geq 1$. The operator $L$ corresponds to the left
shift $s$ and it is called the concrete realization of the shift for the
sequence $\\{f_{n}(t)\\},n\geq 0$.

The $a_{n}$ will be given by different instances of differential operators
applied to given functions as seen in the following examples. Notice that (10)
is a purely formal expression in powers of $s$ and this notation is not
germane with the question of convergence.

##  3 Examples of applications

  1. 1.

Transform induced by Bessel functions. For Bessel functions of order $\nu\geq
0$ we consider the monomials of the form
$f_{n,\nu}(t)=\frac{(t/2)^{2n+\nu}}{\Gamma(\nu+n+1)n!}$. If
$L_{\nu}\stackrel{{\scriptstyle
def}}{{=}}\frac{1}{t}DtD-\frac{\nu^{2}}{t^{2}}$ is a differential operator
where $D$ denotes the derivative with respect to $t$, then a direct
calculation shows that

$\displaystyle L_{\nu}f_{n,\nu}(t)$ $\displaystyle=$ $\displaystyle
f_{n-1,\nu}(t),\text{ for }n\neq 0,$ (11) $\displaystyle L_{\nu}f_{0,\nu}(t)$
$\displaystyle=$ $\displaystyle 0,\text{ for }n=0,$ (12)

in this way $L_{\nu}$ corresponds to the left shift $s$ in this concrete
realization. In fact, if $f(t)$ is a function for which

$f(t)=a_{0}f_{0,\nu}(t)+a_{1}f_{1,\nu}(t)+a_{2}f_{2,\nu}(t)+\cdots,a_{n}\in\mathbb{C},$

then we define the Bessel Transform $\mathcal{T}_{B_{\nu}}$ by

$\mathcal{T}_{B_{\nu}}[f(t)]=\\{a_{0},a_{1},a_{2},\dots\\}$

and, since

$L_{\nu}f(t)=a_{1}f_{0,\nu}(t)+a_{2}f_{1,\nu}(t)+\cdots,$

by (11) and (12), we have

$\mathcal{T}_{B_{\nu}}[L_{\nu}f(t)]=\\{a_{1},a_{2},a_{3}\dots\\}$

so that applying $L_{\nu}$ to a function $f(t)$ corresponds to the left shift
applied to $\mathcal{T}[f(t)]$, as we claimed.

Examples of transforms for fixed $\nu\geq 0$ are

$\displaystyle\mathcal{T}_{B_{\nu}}[J_{\nu}(t)]$ $\displaystyle=$
$\displaystyle\\{1,-1,1,-1,\dots\\},$ (13)
$\displaystyle\mathcal{T}_{B_{\nu}}[I_{\nu}(t)]$ $\displaystyle=$
$\displaystyle\\{1,1,1,\dots\\},$ (14)
$\displaystyle\mathcal{T}_{B_{\nu}}[\text{Ber}_{\nu}(t)]$ $\displaystyle=$
$\displaystyle\left\\{\cos\frac{(3\nu+2n)\pi}{4}\right\\},$ (15)
$\displaystyle\mathcal{T}_{B_{\nu}}[\text{Bei}_{\nu}(t)]$ $\displaystyle=$
$\displaystyle\left\\{\sin\frac{(3\nu+2n)\pi}{4}\right\\}.$ (16)

where $J_{\nu}(t)$ are the well known Bessel functions of order $\nu\geq 0$
and $I_{\nu}(t)$ are the modified Bessel functions of the first kind.
Particular interesting cases of (15) and (16) occur when $\nu=0$ for which

$\displaystyle\mathcal{T}_{B_{0}}[\text{Ber}(t)]$ $\displaystyle=$
$\displaystyle\\{1,0,-1,0,\dots\\}$ (17)
$\displaystyle\mathcal{T}_{B_{0}}[\text{Bei}(t)]$ $\displaystyle=$
$\displaystyle\\{0,-1,0,1,\dots\\}.$ (18)

Formulas (13) to (18) are well-known (see for instance [12] p. 140 for
formulas (15) and (16)). For general functions, the Transform
$\mathcal{T}_{B_{\nu}}$ can be calculated by iterating the operator $L_{\nu}$.
Let $L_{\nu}^{m}=L_{\nu}(L_{\nu}^{m-1}),m\geq 2$ be the order $m$ operator,
then given a function $f(t)$ for which the limit

$\displaystyle\lim_{t\to 0}(L^{m}f(t))=a_{m}$ (19)

does exist we will denote $\lim_{t\to
0}(L_{\nu}^{m}f(t))=L^{m}_{\nu}f(0)=a_{m}$, so for the class of functions for
which that limit exists for any $m\in\mathbb{N}$ we define

$\mathcal{T}_{B_{\nu}}[f(t)]=\\{f(0),L_{\nu}f(0),L_{\nu}^{2}f(0),\dots\\}.$

Note the similarity of the series
$f(x)=f(0)+a_{1}f_{1,\nu}(t)+a_{2}f_{2,\nu}(t)+\cdots$ with the Maclaurin
series. The coincidence is not developed further, for our purposes, the
existence of the transform $\mathcal{T}_{B_{\nu}}$ requires only the existence
of the limit (19). There are many functions for which the transform can be
computed by applying the $L_{\nu}$ operator repeatedly, for example,
$\sqrt{\frac{2}{\pi t}}\sinh t$, $\sqrt{\frac{2}{\pi x}}\cosh t$,
$\sqrt{\frac{2}{\pi t}}\left(\sin t+\frac{\cos t}{t}\right)$, and many others.
The reader can verify directly by applying the operator that the transform
$\mathcal{T}_{B_{\nu}}$ exists, but a direct proof follows easily from the
formulas for Bessel functions of order equal to one half and odd integer [12,
pg. 138 and p. 140]. For example, with the formula 24.58 in [12] we have
$I_{1/2}(t)=\sqrt{\frac{2}{\pi t}}\sinh t$, and therefore

$\mathcal{T}_{B_{1/2}}\left[\sqrt{\frac{2}{\pi t}}\sinh
t\right]=\\{1,1,1,\dots\\}.$

  2. 2.

Transforms method for Bessel operators. After separation of variables of the
Plum equation $\Delta^{2}u-\gamma\Delta u-\frac{4\gamma}{r^{2}}u=\Lambda u$,
where $\Delta$ is the laplacian in polar coordinates [6], the resulting fourth
order equation
$(ty^{\prime\prime})^{\prime\prime}-((9t^{-1}+8\mu^{-1}t)y^{\prime})^{\prime}=\Lambda
ty$ may be solved by using the transforms of example 1. In fact, last equation
may be written in the form equivalent to [1, eq. 45]

$\displaystyle\left[L^{2}_{2}-\frac{8}{\mu}L_{2}-\left(\lambda^{2}+\frac{8}{\mu}\right)\lambda^{2}\right]y(t)=0.$
(20)

Applying $\mathcal{T}_{B_{2}}$ to equation (20) we obtain

$\displaystyle\\{Y_{n+2}\\}-\frac{8}{\mu}\\{Y_{n+1}\\}-\left(\lambda^{2}+\frac{8}{\mu}\right)\lambda^{2}\\{Y_{n}\\}=0,$
(21)

where $\\{Y_{n}\\}=\mathcal{T}_{B_{2}}[y(t)]$,
$\mathcal{T}_{B_{2}}[L^{2}_{2}y(t)]$, and $\mathcal{T}_{B_{2}}[L_{2}y(t)]$ are
obtained after using formula (7). Multiplying (21) by $s^{2}$ and applying
formula (8) we have

$\displaystyle\\{Y_{n}\\}-Y_{0}-sY_{1}-s\frac{8}{\mu}(\\{Y_{n}\\}-Y_{0})-s^{2}\left(\lambda^{2}+\frac{8}{\mu}\right)\lambda^{2}\\{Y_{n}\\}=0,$

and hence

$\displaystyle\\{Y_{n}\\}=\frac{Y_{0}+s\left(Y_{1}-\frac{8}{\mu}Y_{0}\right)}{(1+\lambda^{2}s)\left(1-\left(\frac{8}{\mu}+\lambda^{2}\right)s\right)}$
(22)

and after partial fraction decomposition in (22)

$\displaystyle\\{Y_{n}\\}=\frac{1}{2(4+\lambda^{2}\mu)}\left(\frac{8Y_{0}-\mu
Y_{1}+Y_{0}\lambda^{2}\mu}{1+\lambda^{2}s}+\frac{Y_{1}\mu+Y_{0}\lambda^{2}\mu}{1-(8/\mu+\lambda^{2})s}\right).$
(23)

By using a general form of formulas (13) and (14), i. e.

$\displaystyle\mathcal{T}_{B_{\nu}}[J_{\nu}(\sqrt{\lambda}t)]$
$\displaystyle=$
$\displaystyle\lambda^{\nu/2}\\{\lambda^{k}\\}=\frac{\lambda^{\nu/2}}{1-\lambda
s}$ (24) $\displaystyle\mathcal{T}_{B_{\nu}}[I_{\nu}(\sqrt{\lambda}t)]$
$\displaystyle=$
$\displaystyle\lambda^{\nu/2}\\{(-\lambda)^{k}\\}=\frac{\lambda^{\nu/2}}{1+\lambda
s}$ (25)

we have

$\displaystyle\mathcal{T}_{B_{2}}[J_{2}(\lambda t)]$ $\displaystyle=$
$\displaystyle\lambda^{2}\\{1,-1,1,-1,\dots\\}=\frac{1}{1-(-\lambda^{2})s}$
$\displaystyle\mathcal{T}_{B_{2}}\left[I_{2}\left(\sqrt{\frac{8}{\mu}+\lambda^{2}}\,t\right)\right]$
$\displaystyle=$
$\displaystyle\left(\frac{8}{\mu}+\lambda^{2}\right)\\{1,1,1,1,\dots\\}=\frac{1}{1-\left(\frac{8}{\mu}+\lambda^{2}\right)s}.$

With the last two equations, it is possible to take inverse transform in
equation (23) to solve the initial value problem (20) with initial conditions
$y(0)=Y_{0}$, and $\lim_{t\to 0}L_{2}y(t)=Y_{1}$. The reader may notice that
with the identities $J_{2}(u)=-J_{0}(u)+\frac{2}{u}J_{1}(u)$ and
$I_{2}(u)=I_{0}(u)-\frac{2}{u}I_{1}(u)$ we obtain the same solution as that
given by formulas in [7] and in [1].

  3. 3.

Transform induced by Maclaurin series. Consider the monomials of the form
$f_{n}(t)=t^{n}/n!$, so that, since $\frac{d}{dt}t^{n}/n!=t^{n-1}/(n-1)!$, the
left shift in this concrete realization is the derivative with respect to $t$.
Given a function $f(t)$ for which $\lim_{t\to 0}f^{(n)}(t)=f^{(n)}(0),\,n\geq
0$ does exist, the discrete transform of $f$, say $\mathcal{T}_{M}[f(t)]=F(s)$
is, by definition, the sequence

$\displaystyle\mathcal{T}_{M}[f(t)]$ $\displaystyle\stackrel{{\scriptstyle
def}}{{=}}$
$\displaystyle\\{f(0),f^{\prime}(0),f^{(2)}(0),\dots,f^{(n)}(0),\dots\\}$
$\displaystyle=$ $\displaystyle f(0)+f^{\prime}(0)s+f^{(2)}(0)s^{2}+\cdots,$

and, formally,

$\mathcal{T}_{M}[f^{\prime}(t)]=\\{f^{\prime}(0),f^{(2)}(0),f^{(3)}(0),\dots\\}=f^{\prime}(0)+f^{(2)}(0)s+f^{(3)}(0)s^{2}+\cdots.$

In general if
$\mathcal{T}_{M}[y(t)]=\\{a_{0},a_{1},a_{2},\dots\\}=\\{a_{n}\\},n\geq 0$,
then
$\mathcal{T}_{M}[y^{\prime}(t)]=\\{a_{1},a_{2},\dots\\}=\\{a_{n+1}\\},n\geq
0$. Consequently
$\mathcal{T}_{M}[y^{(m)}(t)]=\\{a_{m},a_{m+1},\dots\\}=\\{a_{n+m}\\},n\geq 0$,
so that the derivative of order $m$ corresponds to the right shift of $m$
places.

Observe that, for instance, the sequence $\\{1,1,1,\dots\\}$ corresponds to
the exponential function, and by formula (3),
$\mathcal{T}_{M}[\,e^{t}\,]=\\{1,1,1,\dots\\}=\frac{1}{1-s}$. Also, it is easy
to see, according to formula (5), that $\mathcal{T}_{M}[\,\cos
t\,]=\frac{1}{1+s^{2}}$. And, moreover, by formula (6),
$\mathcal{T}_{M}[\,\sin t\,]=\frac{s}{1+s^{2}}.$

With the formula (8) now it is possible to solve non-homogeneous differential
equations with constant coefficients, as an example we solve

$\displaystyle y^{\prime\prime}-3y^{\prime}+2y=e^{3t};$ (26) $\displaystyle
y(0)=1,\;y^{\prime}(0)=0.$ (27)

We have by (7) that if $\mathcal{T}_{M}[y(t)]=\\{y_{n}\\},n\geq 0$ then in
this realization, necessarily the derivative satisfies

$\displaystyle\mathcal{T}_{M}[y^{(m)}(t)]=l^{m}\\{y_{n}\\}=\\{y_{n+m}\\}.$
(28)

Taking transforms in (26) and setting
$\mathcal{T}_{M}[y(t)]=\\{y_{n}\\}=Y(s)$, we have by formula (28)

$\displaystyle\\{y_{n+2}\\}-3\\{y_{n+1}\\}+2\\{y_{n}\\}=\frac{1}{1-3s}.$ (29)

Multiplying (29) by $s^{2}$, taking into account the initial conditions (27)
so that $\mathcal{T}_{M}[y(0)]=y_{0}=1$,
$\mathcal{T}_{M}[y^{\prime}(0)]=y_{1}=0$, hence by formula (8), we obtain
after simplification

$\displaystyle\\{y_{n}\\}(1-3s+2s^{2})$ $\displaystyle=$
$\displaystyle\frac{s^{2}}{1-3s}+1-3s$ (30) $\displaystyle\\{y_{n}\\}$
$\displaystyle=$ $\displaystyle\frac{1-6s+10s^{2}}{(1-3s)(s-1)(2s-1)}$ (31)
$\displaystyle Y(s)$ $\displaystyle=$
$\displaystyle\frac{\frac{1}{2}}{1-3s}+\frac{\frac{5}{2}}{1-s}-\frac{2}{1-2s}$
(32) $\displaystyle y(t)$ $\displaystyle=$
$\displaystyle\mathcal{T}_{M}^{-1}[Y(s)]=\frac{1}{2}e^{3t}+\frac{5}{2}e^{t}-2e^{2t}.$
(33)

Clearly (32) is obtained after partial fraction decomposition of (31), and
(33) is obtained from (4), since in this realization
$\mathcal{T}_{M}[e^{rt}]=\\{r^{n}\\}=\frac{1}{1-rs}$. Of course, as usual in
transform methods, if $\mathcal{T}_{M}[y(t)]=Y(s)$ then we define
$y(t)\stackrel{{\scriptstyle def}}{{=}}\mathcal{T}_{M}^{-1}[Y(s)]$.

##  4 The $Z$ transform case

A correspondence between the $Z$ transform and the Maclaurin transform studied
in this article is established now. For a function $G\in\mathcal{C}^{\infty}$
a sequence $\\{g_{n}\\}$ may be defined by the formula

$\displaystyle g_{n}\stackrel{{\scriptstyle def}}{{=}}\lim_{x\to
0}\frac{1}{n!}\frac{d^{n}}{dx^{n}}G\left(\frac{1}{x}\right)$ (34)

if and only if the limit does exist. With the sequence $\\{g_{n}\\}$, the zeta
transform of the sequence is defined as

$\displaystyle\mathcal{T}_{Z}[\\{g_{n}\\}]=F(z)=g_{0}+\frac{g_{1}}{z}+\frac{g_{2}}{z^{2}}+\cdots=\sum_{n=0}^{\infty}\frac{g_{n}}{z^{n}},$
(35)

if the series (35) converges for at least one $z\in\mathbb{C}$. Properties of
the zeta transform are well known (see for instance [13]). Notice that the
correspondence

$\displaystyle z^{-n}\leftrightarrow s^{n}$ (36)

establishes a one-to-one correspondence, hence an equivalence between the zeta
transform and Maclaurin transform. The principal difference between these
transforms is that the left shift $s$ is associated with a differential
operator in all other transforms studied in this paper, (and in particular for
the Maclaurin transform $\mathcal{T}_{M}$, in such a way that $s$ corresponds
to the derivative), meanwhile by derivating $z^{-n}$ is not possible to find
the ${g_{n}}$ from a given Laurent series
$\sum_{n=0}^{\infty}\frac{g_{n}}{z^{n}}$ and evaluating at $z=0$, but it is
possible by complex integration.

Another difference is, as already mentioned, the convergence of a given series
is not determinant for the existence of the Maclaurin transforms. For
instance, the sequence $\\{1,1,1,\dots,\\}$ has $Z$ transform

$\displaystyle\mathcal{T}_{Z}[\\{1,1,\dots\\}]=F(z)=1+\frac{1}{z}+\frac{1}{z^{2}}+\cdots=\frac{z}{z-1}$
(37)

which exists only if $|z|>1,$ but

$\displaystyle\mathcal{T}_{M}[\\{1,1,\dots\\}]=\frac{1}{1-s}$ (38)

is well defined. So, meanwhile in (37) the $Z$ transform $F(z)$ only exists
for $|z|>1$, formula (38) indicates that the multiplicative inverse respect to
the Cauchy product of the sequence $\\{1,1,1,\dots\,1\dots\\}$ is the sequence
$\\{1,-1,0,0,\dots,0,\dots\\}=1-s$, as the reader may easily corroborate.

Example [Equivalence with $Z$ transform]. As an illustration of the
equivalence between the $Z$ transform and the Maclaurin transform we solve an
initial value problem:

$\displaystyle y_{k+1}-3y_{k}$ $\displaystyle=$ $\displaystyle 4$ (39)
$\displaystyle y_{0}$ $\displaystyle=$ $\displaystyle 1.$

which is already solved by using $Z$ transforms in [9, Example 3.42]. By the
properties of the $Z$ transform [13, Chapter 3, section 3.7] equation and
initial condition in (39) is transformed in

$\displaystyle Y(z)=\mathcal{T}_{Z}[\\{y_{n}\\}]$ $\displaystyle=$
$\displaystyle\frac{-2z}{z-1}+\frac{3z}{z-3}$ (40) $\displaystyle=$
$\displaystyle-2\sum_{n=0}^{\infty}\frac{1}{z^{n}}+3\sum_{n=0}^{\infty}\frac{3}{z^{n}}$
(41)

where (41) is obtained by expanding in Laurent series (40). In considering the
correspondence (36) and applying it to (41) it is possible to find the
Maclaurin transform of (39) given by

$\displaystyle Y(s)=\mathcal{T}_{M}[\\{y_{n}\\}]$ $\displaystyle=$
$\displaystyle\frac{-2}{1-s}+\frac{3}{1-3s}$
$\displaystyle\\{y_{k}\\}=\mathcal{T}^{-1}_{M}[Y(s)]$ $\displaystyle=$
$\displaystyle-2\\{1,1,\dots\\}+3\\{1,3,3^{2},\dots\\}$

So the difference problem (39) has the solution $y_{k}=-2+3^{k+1}$, as in [9].

Of course, problem (39) can be solved directly with the Maclaurin transform,
as shown below, last argument was made only to emphasize the correspondence
(36). To solve (39) directly with the Maclaurin transform
$Y(s)=\mathcal{T}_{M}[\\{y_{n}\\}]$ we have by the equation in problem (39)

$\\{y_{k+1}\\}-3\\{y_{k}\\}=4\\{1,1,\dots\\},$

multiplying the last equation by $s$, applying formula (8), and taking
$y_{0}=1$, which corresponds to the initial value in problem (39), it is
obtain

$\displaystyle s\\{y_{k}\\}-1-3s\\{y_{k}\\}$ $\displaystyle=$ $\displaystyle
4s\\{1,1,\dots\\}$ $\displaystyle(1-3s)Y(s)$ $\displaystyle=$
$\displaystyle\frac{4s}{1-s}+1$ $\displaystyle Y(s)$ $\displaystyle=$
$\displaystyle\frac{4s}{(1-s)(1-3s)}+\frac{1}{1-3s},$

consequently, after partial fraction decomposition

$Y(s)=\frac{-2}{1-s}+\frac{2}{1-3s}+\frac{1}{1-3s}=\frac{-2}{1-s}+\frac{3}{1-3s}.$

So by taking inverse transform and using formula (4)

$\\{y_{k}\\}=-2\\{1,1,\dots\\}+3\\{3^{k}\\},k>0,$

so, $y_{k}=-2+3^{k+1},k>0$ and $y_{0}=1$ which coincides with the $Z$
transform solution given before.

##  5 Conclusions

The striking similarity between ber and bei functions with cos and sin
functions respectively has been noticed since time ago (see for instance [5,
p. 136]) but, to the best of my knowledge, it has not been completely
understood. In the approach of this article, Ber and Bei functions and cos and
sin are in correspondence with the same transform (or sequence), Ber and Bei
for $L_{\nu}$ operator, and cos and sin for $d/dt$ operator respectively.
Table (1) shows the exact correspondence between these functions.

Operator | function | Transform | associated sequence
---|---|---|---
$L_{\nu}$ | Ber$(t)$ | $\mathcal{T}_{B_{\nu}}[\text{Ber}_{\nu}\,t]=\frac{1}{1+s^{2}}$ | $\\{1,0,-1,0,\dots\\}$
$L_{\nu}$ | Bei$(t)$ | $\mathcal{T}_{B_{\nu}}[\text{Bei}_{\nu}\,t]=\frac{s}{1+s^{2}}$ | $\\{0,-1,0,1,\dots\\}$
$L_{\nu}$ | $I_{\nu}(t)$ | $\mathcal{T}_{B_{\nu}}[I_{\nu}t]=\frac{1}{1-s}$ | $\\{1,1,\dots\\}$
$D$ | $\cos t$ | $\mathcal{T}_{M}[\cos t]=\frac{1}{1+s^{2}}$ | $\\{1,0,-1,0,\dots\\}$
$D$ | $\sin t$ | $\mathcal{T}_{M}[\sin t]=\frac{s}{1+s^{2}}$ | $\\{0,-1,0,1,\dots\\}$
$D$ | $e^{t}$ | $\mathcal{T}_{M}[e^{t}]=\frac{1}{1-s}$ | $\\{1,1,\dots\\}$
Table 1:  Correspondence between sin, cos, and Bessel functions according to
their respective operator.

So, for instance, Table (1) shows that with sequence $\\{1,1,\dots\\}$ is in
correspondence with two different functions, the Bessel function $I_{\nu}$ and
the exponential function. So that they have the same transforms even when they
were obtained with different transform methods. Of course, the operator
associated with each function is $L_{\nu}$ and $D$, respectively. So the
coincidence of two different methods is fully explained within the context of
the transform method studied in this article.

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