Constructing a polynomial with arbitrary Mahler Measure

Let p(z) \in \mathbb{C}[z], and p(z) = a(z-\alpha_1)(z-\alpha_2)\cdots(z-\alpha_n).

Here is a definition of Mahler measure.

M: \mathbb{C}[z] \to \mathbb{R} \text{, with } M(p) = |a| \prod_{i=1}^n \max\{1, |\alpha_i|\}

Lehmer conjectured,

Lehmer’s Mahler-measure Conjecture:
There exists a real number \mu such that \forall p(x) \in \mathbb{Z}[x], M(p) = 1 or 1 < \mu \leq M(p) . Lehmer suggested

M(p_\mu) = \mu = 1.176280818...

where p_\mu is given as

p_\mu = x^{10} - x^9 + x^7 - x^6 + x^5 - x^4 + x^3 - x^2 + 1.

Now, let us “try” to construct a polynomial p(x), such that 1 < M(p) < \mu.

First, consider 1 < \alpha < \mu. Hence, (x-\alpha)(x-\frac{1}{\alpha}) = x^2 - (\alpha+\frac{1}{\alpha}) + 1; however, \alpha + \frac{1}{\alpha} \not \in \mathbb{Z}, so we need to calibrate the polynomial with more terms.

Let

p_N(x) = (x^2-(\alpha+\frac{1}{\alpha})x + 1) \prod_{i=1}^N (x- \beta_i)(x-\overline{\beta_i})

where \forall i, \beta_i, \overline{\beta_i} \in \mathbb{C} and \beta_i \overline{\beta_i} = {|\beta_i|}^2 = 1.

Thus, (x-\beta_i)(x-\overline{\beta_i}) = x^2-2\Re(\beta_i)x+1.

Let us substitute (\alpha+\frac{1}{\alpha}) by b_0 and 2\Re(\beta_i) by b_i; then, we may rewrite the polynomial p(x) as

p_{N}(x) = \prod_{i=0}^N (x^2-b_ix+1)

Here are the first few cases;

n = 0, then p_0(x) = x^2-b_0x+1

n = 1, then p_1(x) = x^4 - (b_0+b_1) x^3 + (b_0b_1+2)x^2 - (b_0+b_1)x + 1

n = 2, then p_2(x) = x^6 - (b_0+b_1+b_2)x^5 + (b_0b_1+b_0b_2+b_1b_2+3)x^4-(2b_0+2b_1+2b_2+b_0b_1b_2)x^3 + (b_0b_1+b_0b_2+b_1b_2+3)x^2 - (b_0+b_1+b_2)x + 1

Obviously, for given n, x^k and x^{2n-k} share the same coefficient.

Since the polynomial is getting messier, we need a new notation; let \pi_k^n denote the sum of product of k-combination from set of \{b_i\}_{i=0}^{n-1}. For instance, \pi_2^3 = b_0b_1 + b_0b_2 + b_1b_2.

Now, in stead of writing down the whole family of polynomials into a table. Let us define a matrix X where each entry x_{k,j}, (the k-th row and j-th column) represents the coefficient of x^{j-1} from p_{k-1}(x). The entry is left blank if the coefficient is 0, just to avoid confusion.

Featured image

The coefficients seem pretty random though, there is a specific pattern lurking.

 Consider k \leq 2i. Then, x^k term in p_i(x) is only determined by x^{k-2}, x^{k-1} and x^k terms from p_{i-1}(x).

For example, consider x_{5,5}; it is 10 + 3\pi_2^5 + \pi_4^5. According to the idea presented above, this term is created by

(4+\pi_2^4)x^2 \cdot x^2 + (-3\pi_1^4 - \pi_3^4)x^3 \cdot (-b_4) x + (6 + 2\pi_2^4 + \pi_4^4)x^4 \cdot 1

Since the degree of x is all 4, let us not consider about the x^4 for now. Then,

4 + 6 = 10

\pi_2^4 + (-b_4)(-3\pi_1^4) + 2\pi_2^4 = 3\pi_2^4 + 3b_4\pi_1^3 = 3 \pi_2^5

and

(-b_4)(-\pi_3^4) + \pi_4^4 = \pi_4^5.

Thus, it nicely transforms to the term, as desired.

With this nice property, I induced a following equation.

Lee Equation:

 Featured image

As we move back to the matrix, let us assume that p_{n-1}(x) has Mahler measure less than \mu. Then, every term along the n-th row is composed of integers; obviously it leads to the conclusion that for all k \leq n, \pi_k^n are integers.

If Lehmer’s conjecture turns out to be true, then it implies for every n, there exists no sets of \{b_i\}_{i=0}^{n-1} that renders \pi_k^n \in \mathbb{Z} for every k.

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Constructing a polynomial with arbitrary Mahler Measure

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