Exponential Generating Function for Boubaker Polynomials

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Theorem

The Boubaker polynomials, defined as:

$B_n \left({x}\right) = \begin{cases} 1 & : n = 0 \\ x & : n = 1 \\ x^2+2 & : n = 2 \\ x B_{n-1} \left({x}\right) - B_{n-2} \left({x}\right) & : n > 2 \end{cases}$


have as an exponential generating function:

$\displaystyle f_{B_n, \operatorname{EXP}} \left({x, t}\right) = \sum_{n=0}^\infty B_n \left({x}\right) \frac{t^n}{n!} = 4 x e^{t \frac x 2} \frac {\sin \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right)} {\sqrt {1 - \left({\frac x 2}\right)^2}} - 2 e^{t \frac x 2} \cos \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right) - 3$


Proof

From the definition of the Boubaker polynomials, we have:

$ \begin{cases} B_0 \left({x}\right) = 1 \\ B_1 \left({x}\right) = x \\ \end{cases}$


We have also, for $T_n$, the Chebyshev polynomials of the first kind and $U_n$, the Chebyshev polynomials of the second kind:

$ \begin{cases} T_0 \left({x}\right) = 1 \\ T_1 \left({x}\right) = x \\ \end{cases}$

and :

$ \begin{cases} U_{-1} \left({x}\right) = 0 \\ U_0 \left({x}\right) = 1 \\ U_1 \left({x}\right) = 2x \\ \end{cases}$


Then, from the definition of the exponential generating functions of the Chebyshev polynomials of the first kind:

$\displaystyle f_{T_n, \operatorname{EXP}} \left({x, t}\right) = e^{t \frac x 2} \cos \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right) $

and the Chebyshev polynomials of the second kind:

$\displaystyle f_{U_n, \operatorname{EXP}} \left({x, t}\right) = e^{t \frac x 2} \frac {\sin \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right)} {\sqrt {1 - \left({\frac x 2}\right)^2}} $

and from the relation:

$B_n \left({x}\right) = 4 x U_{n-1}\left({\frac x 2}\right) -2 T_n \left({\frac x 2}\right)$

which is valid only for $n>1$, we have:

$\displaystyle B_0 \left({x}\right) \frac{t^0}{0!}+ B_1 \left({x}\right) \frac{t^1}{1!} + \sum_{n=2}^\infty B_n \left({x}\right) \frac{t^n}{n!} = 4x \sum_{n=2}^\infty U_{n-1} \left({x}\right) \frac{t^n}{n!} - 2 \sum_{n=2}^\infty T_n \left({x}\right) \frac{t^n}{n!}+B_0 \left({x}\right) \frac{t^0}{0!}+ B_1 \left({x}\right) \frac{t^1}{1!} -4x \sum_{n=0}^1 U_{n-1} \left({x}\right) \frac{t^n}{n!} - 2 \sum_{n=0}^1 T_n \left({x}\right) \frac{t^n}{n!} $


Finally, by simplifying the two sides and considering the exponential generating functions and the values of the first terms of $T_n$ and $U_n$, we obtain:

$\displaystyle \sum_{n=0}^\infty B_n \left({x}\right) \frac{t^n}{n!} = 4xe^{t \frac x 2} \frac {\sin \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right)} {\sqrt {1 - \left({\frac x 2}\right)^2}} - 2 e^{t \frac x 2} \cos \left({t \sqrt {1 - \left({\frac x 2}\right)^2}}\right) - 3$

Hence the result.

$\blacksquare$