Boubaker's Theorem

Theorem
Let $\left({R, +, \circ}\right)$ be a commutative ring.

Let $\left({D, +, \circ}\right)$ be an integral subdomain of $R$ whose zero is $0_D$ and whose unity is $1_D$.

Let $X \in R$ be transcendental over $D$.

Let $D \left[{X}\right]$ be the ring of polynomial forms in $X$ over $D$.

Finally, consider the following properties:

where, for a given positive integer $n$, $p_n \in D \left[{X}\right]$ is a non-null polynomial such that $p_n$ has $N$ roots $\alpha_k$ in $F$.

Then the subsequence $\left \langle {B_{4 n} \left({x}\right)}\right \rangle$ of the Boubaker polynomials is the unique polynomial sequence of $D \left[{X}\right]$ which verifies simultaneously the four properties $(1) - (4)$.

Proof of validity
We first prove that the Boubaker Polynomials sub-sequence $ B_{4n}(x)$, defined in $D \left[{X}\right]$ verifies properties $(1)$, $(2)$, $(3)$ and $(4)$.

Let:
 * $\left({R, +, \circ}\right)$ be a commutative ring
 * $\left({D, +, \circ}\right)$ be an integral subdomain of $R$ whose zero is $0_D$ and whose unity is $1_D$
 * $X \in R$ be transcendental over $D$.


 * Property $(1)$:

We have the closed form of the the Boubaker Polynomials:
 * $\displaystyle B_n \left({x}\right) = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p x^{n - 2 p}$

which gives:
 * $B_{4 n} \left({0}\right) = \dfrac {n - 4 n / 2} {n - n / 2} \dbinom {n - n / 2} {n / 2} = -2$

and finally:
 * $(1): \quad \displaystyle \sum_{k \mathop = 1}^N B_{4 n} \left({0}\right) = \sum_{k \mathop = 1}^N -2 = - 2 N$

We have, for given integer $n$, $ B_{4n} \in D \left[{X}\right]$ is a non-null polynomial with $N$ roots $\alpha_k$ in $F$.
 * Property $(2)$:

Since:
 * $B_{4 n} \left({\alpha_k}\right) = 0$

then the equality:
 * $(2): \quad \displaystyle \sum_{k \mathop = 1}^N B_{4 n} \left({\alpha_k}\right) = 0$

holds.

According to the closed form of the the Boubaker Polynomials:
 * Property $(3)$:
 * $\displaystyle B_n \left({x}\right) = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p x^{n - 2 p}$

We have:
 * $\displaystyle \frac {\mathrm d B_{4 n} \left({x}\right)} {\mathrm d x} = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor - 1} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p \left({n - 2 p}\right) x^{n - 2 p - 1}$

The minimal power in this expansion is obtained for $p = {\left\lfloor{n / 2}\right\rfloor - 1}$, hence:


 * $\dfrac {\mathrm d B_{4 n} } {\mathrm d x} \left({0}\right) = 0$

and the equality:


 * $(3): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d B_{4 n} \left({x}\right)} {\mathrm d x} } \right\vert_{x \mathop = 0} = 0$

holds.


 * Property $(4)$:

Starting from the closed form of the the Boubaker Polynomials:


 * $\displaystyle B_n \left({x}\right) = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p x^{n - 2 p}$

we have consequently:
 * $\displaystyle \frac {\mathrm d^2 B_{4n} \left({x}\right)} {\mathrm d x^2} = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor - 2} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p \left({n - 2 p}\right) \left({n - 2 p - 1}\right) x^{n - 2 p - 2}$

The minimal power in this expansion is obtained for $p = \left\lfloor{n / 2}\right\rfloor - 2$, hence:


 * $\displaystyle \frac {\mathrm d^2 B_{4 n} } {\mathrm d x^2} \left({0}\right) = \left({-1}\right)^p \left({n - 2 p}\right) \left({n - 2 p - 1}\right) 0$

and the equality:
 * $(4): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d^2 B_{4 n} \left({x}\right)} {\mathrm d x^2}}\right\vert_{x \mathop = 0} = \frac 8 3 N \left({N^2 - 1}\right)$

holds.