Boubaker's Theorem

Theorem

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

Let $\left({D, +, \circ}\right)$ be an integral domain such that $D$ is a subring 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:

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

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

$(3): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {dp_{n}(x)}{dx}}\right|_{x \mathop = 0} = 0$

$(4): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {dp_{n}^{2}(x)}{dx^{2}}}\right|_{x \mathop = 0} = \frac 8 3 N \left({N^2-1}\right)$

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

Proof

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 domain such that $D$ is a subring 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:

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

which gives:

$\displaystyle B_{4n} \left({0}\right) = \frac {n-4n/2} {n-n/2} \binom {n-n/2} { n/2 }=-2$

and finally:

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

Property $(2)$

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

Since:

$\displaystyle B_{4n} (\alpha_k)=0$

then the equality:

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

holds.

Property $(3)$

According to the closed form of the the Boubaker Polynomials:

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

We have:

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

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

$\displaystyle \frac {dB_{4n}} {dx} (0)=0$

and the equality:

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

holds.

Property $(4)$

Starting from the closed form of the the Boubaker Polynomials:

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

we have consequently:

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

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

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

and the equality:

$(4): \quad \displaystyle \left.{\sum_{k=1}^N \frac {d^2 B_{4n} (x)} {dx^2}}\right|_{x \mathop = 0} = \frac 8 3 N(N^2-1) $

holds.

$\blacksquare$

Proof of Uniqueness

Let:

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

$\left({D, +, \circ}\right)$ be an integral domain such that $D$ is a subring 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 $B_{4n} \left({x}\right)$, defined in $D \left[{X}\right]$ as:

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

satisfies the properties:

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

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

$(3): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_n \left({x}\right)}{\mathrm d x}}\right|_{x \mathop = 0} = 0$

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

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

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

Let:

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

and:

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

and:

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

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

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

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

It has also been demonstrated that $ B_{4n}$ has exactly $4n-2$ 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_{4n}$ contains $2 n$ monomial terms (see definition), we obtain a Cramer system in variables $\left.{ d_{4n,p}}\right|_{p \mathop = 0 \,.\,.\, 2n}$, with evident solution:

$\displaystyle \left.{\mathrm d_{4n, p}}\right|_{p \mathop = 0 \,.\,.\, 2n} = 0 $

and consequently:

$\displaystyle \left. { a_{4n, p}}\right|_{p \mathop = 0 \,.\,.\, 2n} = \left.{ b_{4n,p}}\right|_{p \mathop = 0 \,.\,.\, 2n}$

which means:

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

$\blacksquare$

Source of Name

This entry was named for Boubaker Boubaker.