# Boubaker's Theorem/Proof of Uniqueness

## 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 $D$.

Consider the following properties:

\(\text {(1)}: \quad\) | \(\ds \sum_{k \mathop = 1}^N {p_n \left({0}\right)}\) | \(=\) | \(\ds -2N\) | |||||||||||

\(\text {(2)}: \quad\) | \(\ds \sum_{k \mathop = 1}^N {p_n \left({\alpha_k}\right)}\) | \(=\) | \(\ds 0\) | |||||||||||

\(\text {(3)}: \quad\) | \(\ds \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_x \left({x}\right)} {\mathrm d x} }\right\vert_{x \mathop = 0}\) | \(=\) | \(\ds 0\) | |||||||||||

\(\text {(4)}: \quad\) | \(\ds \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d {p_n}^2 \left({x}\right)} {\mathrm d x^2} }\right\vert_{x \mathop = 0}\) | \(=\) | \(\ds \frac 8 3 N \left({N^2 - 1}\right)\) |

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

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 $D$.

It has been demonstrated that the Boubaker Polynomials sub-sequence $B_{4 n} \left({x}\right)$, defined in $D \left[{X}\right]$ as:

- $\displaystyle B_{4 n} \left({x}\right) = 4 \sum_{p \mathop = 0}^{2 n} \frac {n - p} {4 n - p} \binom {4 n - p} p \left({-1}\right)^p x^{2 \left({2n - p}\right)}$

satisfies the properties:

\(\text {(1)}: \quad\) | \(\ds \sum_{k \mathop = 1}^N {p_n \left({0}\right)}\) | \(=\) | \(\ds -2N\) | |||||||||||

\(\text {(2)}: \quad\) | \(\ds \sum_{k \mathop = 1}^N {p_n \left({\alpha_k}\right)}\) | \(=\) | \(\ds 0\) | |||||||||||

\(\text {(3)}: \quad\) | \(\ds \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_x \left({x}\right)} {\mathrm d x} }\right\vert_{x \mathop = 0}\) | \(=\) | \(\ds 0\) | |||||||||||

\(\text {(4)}: \quad\) | \(\ds \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d {p_n}^2 \left({x}\right)} {\mathrm d x^2} }\right\vert_{x \mathop = 0}\) | \(=\) | \(\ds \frac 8 3 N \left({N^2 - 1}\right)\) |

with $\left. {\alpha_k}\right\vert_{k \mathop = 1 \,.\,.\, N}$ roots of $B_{4 n}$.

Suppose there exists another $4 n$-indexed polynomial $q_{4 n} \left({x}\right)$, with $N$ roots $\left.{\beta_k}\right\vert_{k \mathop = 1 \,.\,.\, N}$ in $F$ and which also satisfies simultaneously properties $(1)$ to $(4)$.

Let:

- $\displaystyle B_{4 n} \left({x}\right) = \sum_{p \mathop = 0}^{2 n} a_{4 n, p} x^{2 \left({2 n - p}\right)}$

and:

- $\displaystyle q_{4 n} \left({x}\right) = \sum_{p \mathop = 0}^{2 n} b_{4 n, p} x^{2 \left({2 n - p}\right)}$

and:

- $\displaystyle \mathrm d_{4 n, p} = a_{4 n, p} - b_{4 n, p}$ for $p = 0 \,.\,.\, 2 n$

then, simultaneous expressions of conditions $(1)$ and $(3)$ give:

- $ \quad \displaystyle \sum_{k \mathop = 1}^N \mathrm d_{4 n, 2 n} = 0$
- $ \quad \displaystyle \sum_{k \mathop = 1}^N \mathrm d_{4 n, 2 n - 2} = 0$

It has also been demonstrated that $ B_{4 n}$ has exactly $4 n - 2$ real roots inside the domain $\left[{-2 \,.\,.\, 2}\right]$.

So application of conditions $(3)$ and $(4)$ give $4n-2$ linear equation with variables $\left.{ d_{4n,p}}\right|_{p \mathop = 0 \,.\,.\, 2n-3}$.

Finally, since $B_{4 n}$ contains $2 n$ monomial terms (see definition), we obtain a Cramer system in variables $\left.{ d_{4 n, p}}\right\vert_{p \mathop = 0 \,.\,.\, 2 n}$, with evident solution:

- $\left.{\mathrm d_{4 n, p}}\right\vert_{p \mathop = 0 \,.\,.\, 2 n} = 0 $

and consequently:

- $\left.{a_{4 n, p} }\right\vert_{p \mathop = 0 \,.\,.\, 2 n} = \left.{b_{4 n, p} }\right\vert_{p \mathop = 0 \,.\,.\, 2 n}$

which means:

- $q_{4 n} \left({x}\right) = B_{4 n} \left({x}\right)$

$\blacksquare$

## Source of Name

This entry was named for Boubaker Boubaker.