# Boubaker's Theorem/Proof of Uniqueness

## Theorem

Let $\struct {R, +, \circ}$ be a commutative ring.

Let $\struct {D, +, \circ}$ 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 $D \sqbrk X$ be the ring of polynomial forms in $X$ over $D$.

Consider the following properties:

 $\text {(1)}: \quad$ $\ds \sum_{k \mathop = 1}^N {\map {p_n} 0}$ $=$ $\ds -2N$ $\text {(2)}: \quad$ $\ds \sum_{k \mathop = 1}^N {\map {p_n} {\alpha_k} }$ $=$ $\ds 0$ $\text {(3)}: \quad$ $\ds \valueat {\sum_{k \mathop = 1}^N \frac {\d \map {p_x} x} {\d x} } {x \mathop = 0}$ $=$ $\ds 0$ $\text {(4)}: \quad$ $\ds \valueat {\sum_{k \mathop = 1}^N \frac {\d \map { {p_n}^2} x} {\d x^2} } {x \mathop = 0}$ $=$ $\ds \frac 8 3 N \paren {N^2 - 1}$

where, for a given positive integer $n$, $p_n \in D \sqbrk X$ is a non-null polynomial such that $p_n$ has $N$ roots $\alpha_k$ in $F$.

Then the subsequence $\sequence {\map {B_{4 n} } x}$ of the Boubaker polynomials is the unique polynomial sequence of $D \sqbrk X$ which verifies simultaneously the four properties $(1) - (4)$.

## Proof

Let:

$\struct {R, +, \circ}$ be a commutative ring
$\struct {D, +, \circ}$ 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 $\map {B_{4 n} } x$, defined in $D \sqbrk X$ as:

$\ds \map {B_{4 n} } x = 4 \sum_{p \mathop = 0}^{2 n} \frac {n - p} {4 n - p} \binom {4 n - p} p \paren {-1}^p x^{2 \paren {2 n - p} }$

satisfies the properties:

 $\text {(1)}: \quad$ $\ds \sum_{k \mathop = 1}^N {\map {p_n} 0}$ $=$ $\ds -2N$ $\text {(2)}: \quad$ $\ds \sum_{k \mathop = 1}^N {\map {p_n} {\alpha_k} }$ $=$ $\ds 0$ $\text {(3)}: \quad$ $\ds \valueat {\sum_{k \mathop = 1}^N \frac {\d \map {p_x} x} {\d x} } {x \mathop = 0}$ $=$ $\ds 0$ $\text {(4)}: \quad$ $\ds \valueat {\sum_{k \mathop = 1}^N \frac {\d \map { {p_n}^2} x} {\d x^2} } {x \mathop = 0}$ $=$ $\ds \frac 8 3 N \paren {N^2 - 1}$

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

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

Let:

$\ds \map {B_{4 n} } x = \sum_{p \mathop = 0}^{2 n} a_{4 n, p} x^{2 \paren {2 n - p} }$

and:

$\ds \map {q_{4 n} } x = \sum_{p \mathop = 0}^{2 n} b_{4 n, p} x^{2 \paren {2 n - p} }$

and:

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

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

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

It has also been demonstrated that $B_{4 n}$ has exactly $4 n - 2$ real roots inside the domain $\closedint {-2} 2$.

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

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

$\bigvalueat {\d_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n} = 0$

and consequently:

$\bigvalueat {a_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n} = \bigvalueat {b_{4 n, p} } {p \mathop = 0 \,.\,.\, 2 n}$

which means:

$\map {q_{4 n} } x = \map {B_{4 n} } x$

$\blacksquare$

## Source of Name

This entry was named for Boubaker Boubaker.