**1. Introduction**

In a recent article [1] we noted that the first and second type Chebyshev polynomials can be used to separate the real from the imaginary part of the Appell polynomials. This is just one of the countless applications of these classic polynomials in Function theory. The purpose of this article is to highlight another of their applications, which is, in some way, analogous to the previous one. In fact everybody knows that the even and the odd part of a function *<sup>F</sup>*(*x*) are derived, in a trivial form, by the equations 1 2 [*F*(*x*) + *<sup>F</sup>*(−*<sup>x</sup>*)] and 12 [*F*(*x*) − *<sup>F</sup>*(−*<sup>x</sup>*)].

However, in the case of more complicated expressions of the *F* function, for example in relation to the Appell-type functions, the result is not so obvious, and it will be shown here that it can be obtained using another property of the first and second kind Chebyshev polynomials.

The article is organized as follows. In the first section we use a formula, similar to that of Euler, to separate the even part from the odd part of a binomial power containing hyperbolic functions, showing a connection with the Chebyshev polynomials considered outside their orthogonality interval. Then the results are applied to the case of the Appell polynomials and, in the following sections, to the case of the first kind Bessel functions and to the recently introduced Appell–Bessel functions [2].

It is noteworthy that the study of Appell's polynomials, and of their various extensions, has been considered in both earlier [3] and recent times [4–6]. In these articles have been shown applications to difference equations, expansions in polynomial series and has been analyzed the relative internal structure.

#### **2. Recalling the Chebyshev Polynomials**

It is well known that the first kind Chebyshev polynomias *Tn*(*x*) can be defined outside the interval [−1, 1], by using their expression in terms of hyperbolic functions [7].

Putting *x* = cosh *θ*, and *f*(*θ*) = cosh *θ* + sinh *θ*, we can write:

$$[f(\theta)]^n = (\cosh \theta + \sinh \theta)^n = \sum\_{h=0}^{\lfloor n/2 \rfloor} \binom{n}{2h} [\cosh \theta]^{n-2h} ([\cosh \theta]^2 - 1)^h +$$

$$+ \sinh \theta \sum\_{h=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2h+1} [\cosh \theta]^{n-2h-1} ([\cosh \theta]^2 - 1)^h = \tag{1}$$

$$= \mathcal{E} [f(\theta)]^n + \mathcal{O} [f(\theta)]^n \,\_t$$

where E[ *f*(*θ*)]*n* and O[ *f*(*θ*)]*n* denote the even and odd part of the [ *f*(*θ*)]*n* function. Therefore, using the explicit expression of Chebyshev polynomias, we find

$$[f(\theta)]^n = (\cosh \theta + \sinh \theta)^n = T\_n(\cosh \theta) + \sinh \theta \,\, lL\_{n-1}(\cosh \theta) \,\, \, \tag{2}$$

and

$$\mathcal{E}[f(\theta)]^n = T\_n(\cosh \theta) \,, \qquad \mathcal{O}[f(\theta)]^n = \sinh \theta \,\,\mathrm{d}I\_{n-1}(\cosh \theta) \,. \tag{3}$$

Equations (2) and (3) can be interpreted as an Euler-type formula, owing the analogy with the classical one:

$$[\exp(\mathrm{i}\,\theta)]^n = (\cos\theta + \mathrm{i}\,\sin\theta)^n = \sum\_{h=0}^{\lfloor n/2 \rfloor} \binom{n}{2h} [\cos\theta]^{n-2h} ([\cos\theta]^2 - 1)^h +$$

$$+ \mathrm{i}\,\sin\theta \sum\_{h=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2h+1} [\cos\theta]^{n-2h-1} ([\cos\theta]^2 - 1)^h = \tag{4}$$

.

$$=\mathfrak{R}[(\exp(\mathrm{i}\,\theta))^n] + \mathrm{i}\,\mathfrak{J}[(\exp(\mathrm{i}\,\theta))^n] \,\,\_2\theta$$

#### *Consequences of the Euler-Type Formula*

Using the expansions proven in [8] (but in the case of hyperbolic functions), we find:

**Theorem 1.** *The Taylor expansions hold:*

*e<sup>x</sup> τ* cosh(*y τ*) = ∞ ∑ *<sup>n</sup>*=0 *C* ˜ *<sup>n</sup>*(*<sup>x</sup>*, *y*) *τnn*! , *e<sup>x</sup> τ* sinh(*y τ*) = ∞ ∑ *<sup>n</sup>*=0 *S* ˜ *<sup>n</sup>*(*<sup>x</sup>*, *y*) *τnn*! , (5)

*where*

$$\begin{aligned} \mathsf{C}\_{n}(\mathbf{x}, \mathbf{y}) &= \sum\_{j=0}^{\lfloor n/2 \rfloor} \binom{h}{2j} \mathbf{x}^{n-2j} \mathbf{y}^{2j} \\ \mathsf{S}\_{n}(\mathbf{x}, \mathbf{y}) &= \sum\_{j=0}^{\lfloor (n-1)/2 \rfloor} \binom{h}{2j+1} \mathbf{x}^{n-2j-1} \mathbf{y}^{2j+1} . \end{aligned} \tag{6}$$

**Proof.** The result follows by using the product of series: **I.** Cauchy product involving an even function.

$$\sum\_{k=0}^{\infty} c\_k \frac{\tau^k}{k!} \sum\_{k=0}^{\infty} d\_k \frac{\tau^{2k}}{(2k)!} = \sum\_{k=0}^{\infty} \left[ \sum\_{h=0}^{\left[\frac{k}{2}\right]} \binom{k}{2h} c\_{k-2h} d\_h \right] \frac{\tau^k}{k!} \tag{7}$$

**II.** Cauchy product involving an odd function.

$$\sum\_{k=0}^{\infty} a\_k \frac{\tau^k}{k!} \sum\_{k=0}^{\infty} b\_k \frac{\tau^{2k+1}}{(2k+1)!} = \sum\_{k=0}^{\infty} \left[ \sum\_{h=0}^{\left[\frac{k-1}{2}\right]} \binom{k}{2h+1} a\_{k-2h-1} b\_h \right] \frac{\tau^k}{k!} \,. \tag{8}$$

As a consequence of Equations (5) and (6) we find:

$$e^{\mathbf{x}\cdot\mathbf{\tau}}\left[\cosh(y\,\mathbf{\tau})+\sinh(y\,\mathbf{\tau})\right] = e^{(x+y)\cdot\mathbf{\tau}} = \sum\_{n=0}^{\infty}\left[\tilde{\mathbb{C}}\_{n}(\mathbf{x},y)+\tilde{\mathbb{S}}\_{n}(\mathbf{x},y)\right]\frac{\tau^{n}}{n!},\tag{9}$$

and putting *x* = cosh *θ*, *y* = sinh *θ*

$$\begin{aligned} \exp\left[\left(\cosh\theta + \sinh\theta\right)\tau\right] &= \sum\_{n=0}^{\infty} (\cosh\theta + \sinh\theta)^n \frac{\tau^n}{n!} = \\ &= \sum\_{n=0}^{\infty} \left[\tilde{\mathcal{C}}\_n(\cosh\theta, \sinh\theta) + \tilde{S}\_n(\cosh\theta, \sinh\theta)\right] \frac{\tau^n}{n!} .\end{aligned} \tag{10}$$

Therefore, we conclude that

$$\begin{aligned} \vec{\text{C}}\_{\text{ll}}(\cosh \theta, \sinh \theta) &= T\_{\text{ll}}(\cosh \theta) \\\\ T\_{\text{ll}}(\cosh \theta, \sinh \theta) &= \sinh \theta \,\mathcal{U}\_{\text{ll}-1}(\cosh \theta) \end{aligned} \tag{11}$$

and furthermore:

$$\mathcal{C}\_{\mathbf{n}}(\mathbf{x}, \sqrt{\mathbf{x}^2 - 1}) = T\_{\mathbf{n}}(\mathbf{x}) \,, \qquad \mathcal{S}\_{\mathbf{n}}(\mathbf{x}, \sqrt{\mathbf{x}^2 - 1}) = \sqrt{\mathbf{x}^2 - 1} \, \mathcal{U}\_{\mathbf{n} - 1}(\mathbf{x}) \, . \tag{12}$$

#### **3. The Even and Odd Part of Appell Polynomials**

*S*

˜

In this section we show how to represent the even and odd part of Appell polynomials. Consider the Appell polynomials [9–11], defined by the generating function [12]

$$A(t) \; e^{\mathbf{x} \cdot t} = \sum\_{n=0}^{\infty} a\_n(\mathbf{x}) \frac{t^n}{n!} \; , \tag{13}$$

where

$$A(t) = \sum\_{n=0}^{\infty} a\_k \frac{t^k}{k!} \,. \tag{14}$$

Putting *y* := √*x*<sup>2</sup> − 1, we have

$$A(t)\,e^{(x+y)\,t} = \sum\_{n=0}^{\infty} A\_{ll}(x,y)\,\frac{t^n}{n!},\tag{15}$$

where

$$A\_n(\mathbf{x}, \mathbf{y}) = A\_n(\mathbf{x}, \sqrt{\mathbf{x}^2 - 1}) = a\_n(\mathbf{x} + \sqrt{\mathbf{x}^2 - 1}). \tag{16}$$

By using the Cauchy product we find:

$$A\_n(\mathbf{x}, \mathbf{y}) = \sum\_{k=0}^n \binom{n}{k} a\_{n-k} \left(\mathbf{x} + \mathbf{y}\right)^k = \mathcal{E}\left[A\_n(\mathbf{x}, \mathbf{y})\right] + \mathcal{O}\left[A\_n(\mathbf{x}, \mathbf{y})\right],\tag{17}$$

and putting *x* = cosh *θ*, *y* = √*x*<sup>2</sup> − 1 = sinh *θ*,

$$A\_n(\mathbf{x}, \mathbf{y}) = \sum\_{k=0}^n \binom{n}{k} a\_{n-k} \left(\cosh \theta + \sinh \theta\right)^k = \sum\_{k=0}^n \binom{n}{k} a\_{n-k} \left[\tilde{\mathcal{L}}\_k(\mathbf{x}, \mathbf{y}) + \tilde{S}\_k(\mathbf{x}, \mathbf{y})\right] = 0$$

$$= \sum\_{k=0}^n \binom{n}{k} a\_{n-k} \left[T\_k(\mathbf{x}) + \mathbf{y} \,\mathcal{U}\_{k-1}(\mathbf{x})\right].$$

Therefore, by recalling (2) and (3), we conclude that

$$\begin{aligned} \mathcal{E}\left[A\_{\boldsymbol{n}}(\mathbf{x}, \sqrt{\mathbf{x}^2 - 1})\right] &= \sum\_{k=0}^{n} \binom{n}{k} a\_{n-k} \ \operatorname{T}\_k(\mathbf{x}) \ , \\\\ \mathcal{O}\left[A\_{\boldsymbol{n}}(\mathbf{x}, \sqrt{\mathbf{x}^2 - 1})\right] &= \sqrt{\mathbf{x}^2 - 1} \ \sum\_{k=0}^{n} \binom{n}{k} a\_{n-k} \ \operatorname{U}\_{k-1}(\mathbf{x}) \ . \end{aligned} \tag{18}$$

**Remark 1.** *Note that Equation (18) can be applied in general, since, when* |*x*| > 1*, the position x* + √*x*<sup>2</sup> − 1 = *u is equivalent to x* = *<sup>u</sup>*2−1 2*u , and* √*x*<sup>2</sup> − 1 = *<sup>u</sup>*2−1 2*u , so that Equation (13) becomes:*

$$A(t)\;e^{\mu\;t} = \sum\_{n=0}^{\infty} a\_n(\mu)\frac{t^n}{n!}.\tag{19}$$

*Therefore, we can conclude with the theorem.*

**Theorem 2.** *The even and odd part of the Appell polynomials <sup>α</sup>n*(*u*) *defined by the generating function* (19) *can be represented, in terms of the first and second kind Chebyshev polynomials, by the equations:*

$$\begin{aligned} \mathcal{E}\left[a\_n(u)\right] &= \sum\_{k=0}^n \binom{n}{k} \, a\_{n-k} \, T\_k\left(\frac{u^2+1}{2u}\right), \\\\ \mathcal{O}[a\_n(u)] &= \frac{u^2-1}{2u} \sum\_{k=0}^n \binom{n}{k} \, a\_{n-k} \, \mathcal{U}\_{k-1}\left(\frac{u^2+1}{2u}\right) \,. \end{aligned} \tag{20}$$

#### **4. 1st Kind Bessel Functions**

We consider here the first kind of Bessel functions with integer order [13], defined by the generating function [12]:

$$e^{\frac{\pi}{2}\left(t-\frac{1}{T}\right)} = \sum\_{n=-\infty}^{\infty} J\_n(x) \, t^n. \tag{21}$$

Putting, for shortness:

$$J\_0(\mathbf{x}) := \frac{1}{2} J\_0(\mathbf{x})\,, \qquad J\_k(\mathbf{x}) := J\_k(\mathbf{x})\,, \quad (k \ge 1)\,. \tag{22}$$

> Equation (21) writes:

$$e^{\frac{\pi}{2}\left(t-\frac{1}{T}\right)} = \sum\_{n=0}^{\infty} \bar{f}\_n(\mathbf{x}) \, t^n + \sum\_{n=0}^{\infty} (-1)^n \, \bar{f}\_n(\mathbf{x}) \, t^{-n} \,. \tag{23}$$

Note that, using the notation

$$\varepsilon^{\frac{x}{2}\left(t-\frac{1}{\tau}\right)} = \mathcal{E}\left[\varepsilon^{\frac{x}{2}\left(t-\frac{1}{\tau}\right)}\right] + \mathcal{O}\left[\varepsilon^{\frac{x}{2}\left(t-\frac{1}{\tau}\right)}\right],\tag{24}$$

the even and odd part must be understood with respect to both *t* and *x*, and we find:

$$\begin{aligned} \mathcal{E}\left[e^{\frac{\tilde{\mathbf{x}}}{2}\left(t-\frac{1}{T}\right)}\right] &= \sum\_{n=0}^{\infty} f\_{2n}(\mathbf{x}) \left(t^{2n} + t^{-2n}\right), \\\\ \mathcal{O}\left[e^{\frac{\tilde{\mathbf{x}}}{2}\left(t-\frac{1}{T}\right)}\right] &= \sum\_{n=0}^{\infty} f\_{2n+1}(\mathbf{x}) \left(t^{2n+1} - t^{-(2n+1)}\right). \end{aligned} \tag{25}$$

#### *Representation by Chebyshev Polynomials*

Inverting the equation *τ* = 12 *t* − 1*t* , we have:

$$t = \tau + \sqrt{\tau^2 + 1}. \tag{26}$$

**Theorem 3.** *The generating function* (21) *can be represented in terms of Chebyshev polynomials by:*

$$\varepsilon^{\mathbf{x}\cdot\mathbf{r}} = 2\sum\_{n=0}^{\infty} f\_{2n}(\mathbf{x}) \, T\_{2n}(\mathbf{r}) + 2\sqrt{\pi^2 + 1} \sum\_{n=0}^{\infty} f\_{2n+1}(\mathbf{x}) \, \mathcal{U}\_{2n}(\mathbf{r}) \,. \tag{27}$$

*so that the even and odd part of the Bessel functions defined by Equation* (21) *are given by*

$$\begin{split} \mathcal{E}\left[\sum\_{k=-\infty}^{\infty} J\_k(\mathbf{x}) \, t^k \right] &= 2 \sum\_{k=0}^{\infty} J\_{2k}(\mathbf{x}) \, T\_{2k}(\mathbf{r}) \, , \\ \mathcal{O}\left[\sum\_{k=-\infty}^{\infty} J\_k(\mathbf{x}) \, t^k \right] &= 2\sqrt{\tau^2 + 1} \sum\_{k=0}^{\infty} J\_{2k+1} \, \mathcal{U}\_{2k}(\mathbf{r}) \, . \end{split} \tag{28}$$

**Proof.** Equation (21) writes:

$$\varepsilon^{\mathbf{x}\cdot\mathbf{r}} = \sum\_{n=0}^{\infty} \tilde{f}\_{2n}(\mathbf{x}) \left( t^{2n} + t^{-2n} \right) + \sum\_{n=0}^{\infty} f\_{2n+1}(\mathbf{x}) \left( t^{2n+1} - t^{-(2n+1)} \right), \tag{29}$$

so that

$$\left(t^{2n} + t^{-2n}\right) = 2\operatorname{Tr}\_{2n}(\tau) \,, \qquad \left(t^{2n+1} - t^{-(2n+1)}\right) = 2\sqrt{\tau^2 + 1} \, \operatorname{l}l\_{2n}(\tau) \, . \tag{30}$$

and the result is proven.

## **5. Appel–Bessel Functions**

Several mixed-type (or hybrid) functions have been recently considered. The starting point of these type of special functions can be found in [14,15]. In this section we consider the Appel–Bessel functions introduced in [2].

**Definition 1.** *The Appel–Bessel functions [2] are defined by generating function:*

$$G(\mathbf{x}, t) = A \left[ \frac{\mathbf{y}}{2} (t - \frac{1}{T}) \right] \exp\left[ \frac{\mathbf{y}}{2} (t - \frac{1}{T}) \right] = \sum\_{k = -\infty}^{\infty} \left[ A\_k \mathbf{f}\_k(\mathbf{x}) \right] t^k,\tag{31}$$

*where*

$$A(\tau) = \sum\_{k=0}^{\infty} a\_k \ \frac{\tau^k}{k!} \ . \quad (a\_0 \neq 0) \,. \tag{32}$$

*Since*

$$G(\mathbf{x}, t) = \sum\_{k = -\infty}^{\infty} \left[ {}\_{A}I\_{k}(\mathbf{x}) \right] t^{k} = \sum\_{k = -\infty}^{\infty} \left[ {}\_{A}I\_{-k}(\mathbf{x}) \right] t^{-k},\tag{33}$$

*we find*

$$\left[ \left. \right| J\_{-k}(\mathbf{x}) \right] = (-1)^{k} \left[ \left. \right| J\_{k}(\mathbf{x}) \right]. \tag{34}$$

*Furthermore, from*

$$G(\mathbf{x}, -t) = G(-\mathbf{x}, t) = \sum\_{k = -\infty}^{\infty} \left[ {}\_{A}I\_{k}(\mathbf{x}) \right] (-1)^{k} \, ^{k}t = \sum\_{k = -\infty}^{\infty} \left[ {}\_{A}I\_{k}(-\mathbf{x}) \right] \, ^{k} \mathbf{r} \tag{35}$$

*we find*

$$\left[ \, \_A J\_k(-\mathbf{x}) \right] = (-1)^k \left[ \_A J\_k(\mathbf{x}) \right]. \tag{36}$$

*That is, the same symmetry properties of the ordinary* 1*st kind Bessel functions still hold for the Appell–Bessel functions.*

#### *5.1. Representation of the Appell–Bessel Functions*

Even in this case we put, for shortness:

$$[A]\_0(\mathbf{x})\left]:=\frac{1}{2}\begin{bmatrix}\_A\end{bmatrix}\_0(\mathbf{x})\right],\qquad [\_A\ddot{J}\_k(\mathbf{x})]:=\begin{bmatrix}\_A\mathbf{J}\_k(\mathbf{x})\end{bmatrix},\quad \left(k\ge 1\right).\tag{37}$$

**Theorem 4.** *The generating function* (31) *can be represented in terms of Chebyshev polynomials by:*

$$A(\mathbf{x}\,\tau)\,\varepsilon^{\mathbf{r}\,\tau} = 2\sum\_{n=0}^{\infty} \left[ {}\_{A}I\_{2n}(\mathbf{x}) \right] T\_{2n}(\mathbf{r}) + 2\sqrt{\tau^2 + 1} \sum\_{n=0}^{\infty} \left[ {}\_{A}I\_{2n+1}(\mathbf{x}) \right] \cup I\_{2n}(\mathbf{r}),\tag{38}$$

*so that the even and odd part of the Appell-Bessel functions are given by*

$$\begin{split} \mathcal{E}\left[\sum\_{k=-\infty}^{\infty} \left[ {}\_{A}f\_{k}(\mathbf{x}) \right] t^{k} \right] &= 2 \sum\_{k=0}^{\infty} \left[ {}\_{A}\tilde{l}\_{2k}(\mathbf{x}) \right] T\_{2k}(\mathbf{r}) \,, \\ \mathcal{O}\left[\sum\_{k=-\infty}^{\infty} \left[ {}\_{A}f\_{k}(\mathbf{x}) \right] t^{k} \right] &= 2\sqrt{\tau^{2}+1} \sum\_{k=0}^{\infty} \left[ {}\_{A}f\_{2k+1} \right] \mathcal{U}\_{2k}(\mathbf{r}) \,. \end{split} \tag{39}$$

**Proof.** Using the symmetry properties (34) and (36), the same technique applied in Section 3 gives the result in the present case.

#### *5.2. Connection with the Appel–Bessel Functions*

**Theorem 5.** *The following equation holds:*

$$\sum\_{k=0}^{\infty} \sum\_{h=0}^{k} \binom{k}{h} a\_{k-h} \frac{\left(\mathbf{x}\,\,\tau\right)^{k}}{k!} = \tag{40}$$
 
$$\mathbf{x} = 2 \sum\_{n=0}^{\infty} \left[ {}\_{A}\tilde{l}\_{2n}(\mathbf{x}) \right] T\_{2n}(\mathbf{r}) + 2\sqrt{\tau^2 + 1} \sum\_{n=0}^{\infty} \left[ {}\_{A}l\_{2n+1}(\mathbf{x}) \right] l L\_{2n}(\mathbf{r}) \,.$$

**Proof.** By using the Cauchy product we find:

$$A(\mathbf{x}\,\tau)\,e^{\mathbf{x}\,\tau} = \sum\_{k=0}^{\infty} a\_k \frac{(\mathbf{x}\,\tau)^k}{k!} \sum\_{k=0}^{\infty} \frac{(\mathbf{x}\,\tau)^k}{k!} = \sum\_{k=0}^{\infty} \sum\_{h=0}^k \binom{k}{h} a\_{k-h} \frac{(\mathbf{x}\,\tau)^k}{k!}.\tag{41}$$

Therefore, the result follows by comparing Equations (38) and (41).
