# Boubaker's Theorem

## 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 $D$.

Finally, consider the following properties:

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

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

### Proof of validity

We first prove that the Boubaker Polynomials subsequence $\sequence {\map {B_{4 n} } x}$, defined in $D \sqbrk X$ verifies properties $(1)$, $(2)$, $(3)$ and $(4)$.

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 $D$.

Property $(1)$

We have the closed form of the the Boubaker Polynomials:

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

which gives:

$\map {B_{4 n} } 0 = \dfrac {n - 4 n / 2} {n - n / 2} \dbinom {n - n / 2} {n / 2} = -2$

and finally:

$(1): \quad \ds \sum_{k \mathop = 1}^N \map {B_{4 n} } 0 = \sum_{k \mathop = 1}^N -2 = -2 N$

Property $(2)$

We have, for given integer $n$, $B_{4n} \in D \sqbrk X$ is a non-null polynomial with $N$ roots $\alpha_k$ in $F$.

Since:

$\map {B_{4 n} } {\alpha_k} = 0$

then the equality:

$(2): \quad \ds \sum_{k \mathop = 1}^N \map {B_{4 n} } {\alpha_k} = 0$

holds.

Property $(3)$

According to the closed form of the the Boubaker Polynomials:

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

We have:

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

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

$\map {\dfrac {\d B_{4 n} } {\d x} } 0 = 0$

and the equality:

$(3): \quad \ds \valueat {\sum_{k \mathop = 1}^N \frac {\d \map {B_{4 n} } x} {\d x} } {x \mathop = 0} = 0$

holds.

Property $(4)$

Starting from the closed form of the the Boubaker Polynomials:

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

we have consequently:

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

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

$\map {\dfrac {\d^2 B_{4 n} } {\d x^2} } 0 = \paren {-1}^p \paren {n - 2 p} \paren {n - 2 p - 1}0$

and the equality:

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

holds.

$\blacksquare$

### Proof of Uniqueness

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.