Relation of Boubaker Polynomials to Chebyshev Polynomials

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Theorem

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

$B_n \left({2x}\right) = \dfrac {4x}n \dfrac{\mathrm d}{\mathrm d x} T_n \left({x}\right) - 2 T_n \left({x}\right)$
$B_n \left({2x}\right) = -2 T_n \left({x}\right) + 4 x U_{n-1} \left({x}\right)$

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:

$\displaystyle \sum_{n \geqslant 0} B_n \left({t}\right)x^n = \frac {1+3x^2 }{1-xt+x^2}= \left (1+3x^2 \biggl\lvert 1+x^2 \right )\left({\frac 1 {1-xt}}\right)$

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

$\displaystyle T \left({\frac 1 4 - \frac {x^2} 4 \biggl \lvert \frac 1 2 + \frac 1 2 x^2}\right)$

and:

$\displaystyle T \left({\frac 1 2 \biggl\lvert \frac 1 2 + \frac 1 2 x^2}\right)$

and by setting:

$\displaystyle T \left({1+3x^2 \biggl \lvert 1+x^2}\right) = T \left({1+3x^2 \biggl\lvert 1}\right) T \left({\frac 1 2 \biggl\lvert \frac 1 2 + \frac 1 2 x^2}\right)$

we obtain finally:

$\displaystyle B_n \left({x}\right) = U_n \left({\frac x 2}\right) + 3U_{n-2} \left({\frac x 2}\right)$

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

$B_n \left({2x}\right) = -2 T_n \left({x}\right) + 4 x U_{n-1} \left({x}\right)$
$B_n \left({2x}\right) = \dfrac {4x}n \dfrac{\mathrm d}{\mathrm d x} T_n \left({x}\right) - 2 T_n \left({x}\right)$

$\blacksquare$


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