Boubaker's Theorem
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 $D$.
Finally, consider the following properties:
| \((1):\quad\) | \(\displaystyle \sum_{k \mathop = 1}^N {p_n \left({0}\right)}\) | \(=\) | \(\displaystyle -2N\) | $\quad$ | $\quad$ | ||||||||
| \((2):\quad\) | \(\displaystyle \sum_{k \mathop = 1}^N {p_n \left({\alpha_k}\right)}\) | \(=\) | \(\displaystyle 0\) | $\quad$ | $\quad$ | ||||||||
| \((3):\quad\) | \(\displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_x \left({x}\right)} {\mathrm d x} }\right\vert_{x \mathop = 0}\) | \(=\) | \(\displaystyle 0\) | $\quad$ | $\quad$ | ||||||||
| \((4):\quad\) | \(\displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d {p_n}^2 \left({x}\right)} {\mathrm d x^2} }\right\vert_{x \mathop = 0}\) | \(=\) | \(\displaystyle \frac 8 3 N \left({N^2 - 1}\right)\) | $\quad$ | $\quad$ |
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
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 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:
- $\displaystyle B_n \left({x}\right) = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p x^{n - 2 p}$
which gives:
- $B_{4 n} \left({0}\right) = \dfrac {n - 4 n / 2} {n - n / 2} \dbinom {n - n / 2} {n / 2} = -2$
and finally:
- $(1): \quad \displaystyle \sum_{k \mathop = 1}^N B_{4 n} \left({0}\right) = \sum_{k \mathop = 1}^N -2 = - 2 N$
- 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:
- $B_{4 n} \left({\alpha_k}\right) = 0$
then the equality:
- $(2): \quad \displaystyle \sum_{k \mathop = 1}^N B_{4 n} \left({\alpha_k}\right) = 0$
holds.
- Property $(3)$
According to the closed form of the the Boubaker Polynomials:
- $\displaystyle B_n \left({x}\right) = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p x^{n - 2 p}$
We have:
- $\displaystyle \frac {\mathrm d B_{4 n} \left({x}\right)} {\mathrm d x} = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor - 1} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p \left({n - 2 p}\right) x^{n - 2 p - 1}$
The minimal power in this expansion is obtained for $p = {\left\lfloor{n / 2}\right\rfloor - 1}$, hence:
- $\dfrac {\mathrm d B_{4 n} } {\mathrm d x} \left({0}\right) = 0$
and the equality:
- $(3): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d B_{4 n} \left({x}\right)} {\mathrm d x} } \right\vert_{x \mathop = 0} = 0$
holds.
- Property $(4)$
Starting from the closed form of the the Boubaker Polynomials:
- $\displaystyle B_n \left({x}\right) = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p x^{n - 2 p}$
we have consequently:
- $\displaystyle \frac {\mathrm d^2 B_{4n} \left({x}\right)} {\mathrm d x^2} = \sum_{p \mathop = 0}^{\left\lfloor{n / 2}\right\rfloor - 2} \frac {n - 4 p} {n - p} \binom {n - p} p \left({-1}\right)^p \left({n - 2 p}\right) \left({n - 2 p - 1}\right) x^{n - 2 p - 2}$
The minimal power in this expansion is obtained for $p = \left\lfloor{n / 2}\right\rfloor - 2$, hence:
- $\displaystyle \frac {\mathrm d^2 B_{4 n} } {\mathrm d x^2} \left({0}\right) = \left({-1}\right)^p \left({n - 2 p}\right) \left({n - 2 p - 1}\right) 0$
and the equality:
- $(4): \quad \displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d^2 B_{4 n} \left({x}\right)} {\mathrm d x^2}}\right\vert_{x \mathop = 0} = \frac 8 3 N \left({N^2 - 1}\right)$
holds.
$\blacksquare$
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Proof of Uniqueness
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:
| \((1):\quad\) | \(\displaystyle \sum_{k \mathop = 1}^N {p_n \left({0}\right)}\) | \(=\) | \(\displaystyle -2N\) | $\quad$ | $\quad$ | ||||||||
| \((2):\quad\) | \(\displaystyle \sum_{k \mathop = 1}^N {p_n \left({\alpha_k}\right)}\) | \(=\) | \(\displaystyle 0\) | $\quad$ | $\quad$ | ||||||||
| \((3):\quad\) | \(\displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_x \left({x}\right)} {\mathrm d x} }\right\vert_{x \mathop = 0}\) | \(=\) | \(\displaystyle 0\) | $\quad$ | $\quad$ | ||||||||
| \((4):\quad\) | \(\displaystyle \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d {p_n}^2 \left({x}\right)} {\mathrm d x^2} }\right\vert_{x \mathop = 0}\) | \(=\) | \(\displaystyle \frac 8 3 N \left({N^2 - 1}\right)\) | $\quad$ | $\quad$ |
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:
- $q_{4 n} \left({x}\right) = B_{4 n} \left({x}\right)$
$\blacksquare$
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Source of Name
This entry was named for Boubaker Boubaker.
