# Relation of Boubaker Polynomials to Chebyshev Polynomials

## Theorem

The Boubaker polynomials are related to Chebyshev polynomials through the equations:

$\map {B_n} {2 x} = \dfrac {4 x} n \dfrac \d {\d x} \map {T_n} x - 2 \map {T_n} x$
$\map {B_n} {2 x} = -2 \map {T_n} x + 4 x \map {U_{n - 1} } x$

where:

$T_n$ denotes the Chebyshev polynomials of the first kind
$U_n$ denotes the Chebyshev polynomials of the second kind.

## Proof

(Using Riordan Matrix)

The ordinary generating function of the Boubaker Polynomials can be expressed in terms of Riordan matrices:

$\ds \sum_{n \geqslant 0} \map {B_n} t x^n = \frac {1 + 3x^2} {1 - x t + x^2} = \paren {1 + 3 x^2 \mid 1 + x^2} \paren {\dfrac 1 {1 - x t} }$

By considering the Riordan matrices of Chebyshev Polynomials of the first and second kind $T_n$ and $U_n$, respectively:

$\map T {\dfrac 1 4 - \dfrac {x^2} 4 \mid \dfrac 1 2 + \dfrac 1 2 x^2}$

and:

$\map T {\dfrac 1 2 \mid\dfrac 1 2 + \dfrac 1 2 x^2}$

and by setting:

$\map T {1 + 3x^2 \mid 1 + x^2} = \map T {1 + 3x^2 \mid 1} \map T {\dfrac 1 2 \mid \dfrac 1 2 + \dfrac 1 2 x^2}$

we obtain finally:

$\map {B_n} x = \map {U_n} {\dfrac x 2} + 3 \map {U_{n - 2} } {\dfrac x 2}$

which gives, due the relation between Chebyshev Polynomials of the first and second kind $T_n$ and $U_n$ and their derivatives:

$\map {B_n} {2 x} = -2 \map {T_n} x + 4 x \map {U_{n - 1} } x$
$\map {B_n} {2 x} = \dfrac {4 x} n \dfrac \d {\d x} \map {T_n} x - 2 \map {T_n} x$

$\blacksquare$