# 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.

$\blacksquare$

This article needs proofreading.Please check it for mathematical errors.If you believe there are none, please remove `{{Proofread}}` from the code.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Proofread}}` from the code. |

### 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:

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

$\blacksquare$

This needs considerable tedious hard slog to complete it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Finish}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Source of Name

This entry was named for Boubaker Boubaker.