Prove the series has positive integer coefficients

Question

How can I show that the Maclaurin series for $$ \mu(x) = (x^4+12x^3+14x^2-12x+1)^{-1/4} \\ = 1+3\,x+19\,{x}^{2}+147\,{x}^{3}+1251\,{x}^{4}+11193\,{x}^{5}+103279\, {x}^{6}+973167\,{x}^{7}+9311071\,{x}^{8}+\cdots $$ has positive integer coefficients? (I have others to do, too, but this one will be a start.)

possibilities
(a) The coefficients $Q(n)$ satisfy the recurrence $$ (n+1)Q(n)+(12n+21)Q(n+1)+(14n+35)Q(n+2)+(-12n-39)Q(n+3)+(n+4)Q(n+4) = 0 $$ (b) $\mu(x)$ satisfies the differential equation $$ (x^3+9x^2+7x-3)\mu(x)+(x^4+12x^3+14x^2-12x+1)\mu'(x)=0 $$ (c) factorization of $x^4+12x^3+14x^2-12x+1$ is $$ \left( x-\sqrt {5}+3+\sqrt {15-6\,\sqrt {5}} \right) \left( x-\sqrt {5}+3-\sqrt {15-6\,\sqrt {5}} \right) \left( x+\sqrt {5}+3-\sqrt {15+ 6\,\sqrt {5}} \right) \left( x+\sqrt {5}+3+\sqrt {15+6\,\sqrt {5}} \right) $$ (d) $(1-8X)^{-1/4}$ has positive integer coefficients, But if $X$ is defined by $1-8X=x^4+12x^3+14x^2-12x+1$, then $X$ does not have integer coefficients.

(e) Can we compute the series for $\log\mu(x)$ $$3\,x+{\frac {29}{2}}{x}^{2}+99\,{x}^{3}+{\frac {3121}{4}}{x}^{4}+{ \frac {32943}{5}}{x}^{5}+{\frac {348029}{6}}{x}^{6}+\dots $$ and then recognize that its exponential has integer coefficients?

Answer 1

I can prove the integer property, but not (yet) positivity.

Lemma: The coefficients of Taylor series of $\mu(x)$ are integers.

Proof. We can write $$x^4+12x^3+14x^2-12x+1=\left(1-x\right)^4\Bigl(1-8\nu(x)\Bigr),$$ where $\nu(x)$ is given by $$\nu(x)=\frac{2}{1-x}-\frac{7}{\left(1-x\right)^2}+\frac{7}{\left(1-x\right)^3}-\frac{2}{\left(1-x\right)^4}.$$ The coefficients of Taylor series of $\nu(x)$ are clearly integer and we have $\nu(0)=0$. Since the coefficients of $(1-x)^{-1}$ and $(1-8x)^{-\frac14}$ are integers, the statement follows. $\qquad\square$

Another option is to rewrite $\mu(x)$ in terms of $y=\frac{x}{1-x}$: $$\mu(x)\mapsto \frac{1+y}{\left[1-8y\left(1-y^2\right)\left(1+2y\right)\right]^{\frac14}}.$$ It is interesting to note that even the coefficients of different powers of $y$ seem to be positive.

Answer 2

Here is a proof for positivitiy. Use the recurrence $$ (n+1)Q(n)+(12n+21)Q(n+1)+(14n+35)Q(n+2)+(-12n-39)Q(n+3)+(n+4)Q(n+4) = 0 $$ $Q(0)=1,Q(1)=3,Q(2)=19,Q(3)=147$. Prove by induction that $Q(n) \ge 3 Q(n-1)$ for $n \ge 1$. This is true for the first few terms. Assume true up to $Q(n+3)$, then prove it for $Q(n+4)$ as follows: $$ (n+4)Q(n+4) = (12n+39)Q(n+3)-(14n+35)Q(n+2)-(12n+21)Q(n+1)-(n+1)Q(n) \\ \ge \left[(12n+39)-\frac{14n+35}{3}-\frac{12n+21}{9}-\frac{n+1}{27}\right]Q(n+3) \\ =\left[\frac{161}{27} n + \frac{692}{27}\right] Q(n+3) \gt (3n+12)Q(n+3)=(n+4)3Q(n+3) $$ and therefore $Q(n+4) > 3 Q(n+3)$.

Answer 3

For the periodicity mod $5$, note that $$(1 - x^4)^4 \equiv (1 + 3 x + 4 x^2 + 2 x^3)^4 (1-12 x+14 x^2+12 x^3+x^4) \mod 5 $$ and thus in the field $\mathbb Z_5((x))$ of formal Laurent series in $x$ over the integers mod $5$ we have $$ \eqalign{&\dfrac{1}{1-12 x+14 x^2+12 x^3+x^4} = \left(\dfrac{1+3x+4x^2+2x^3}{1-x^4}\right)^4\cr &= \left(1+3x+4x^2+2x^3 + (1+3x+4x^2+2x^3) x^4 + (1+3x+4x^2+2x^3) x^8 + \ldots \right)^4\cr} $$

Given there is a solution over the integers, taking the coefficients mod $5$ gives us a solution in $\mathbb Z_5((x))$. Of course there are four fourth roots, but this is the only solution whose constant coefficient is $1$ (the others have $2, 3$ or $4$).

Similarly, to show the coefficients are odd, $$(1-x)^4 \equiv 1 + x \equiv 1 -12 x + 14 x^2 + 12 x^3 + x^4 \mod 2$$ so that in $\mathbb Z_2((x))$ $$\dfrac{1}{1-12 x+14 x^2+12 x^3+x^4} = \dfrac{1}{(1-x)^4} = (1 + x + x^2 + x^3 + \ldots)^{4}$$

Combine the solutions mod $5$ and mod $2$ and we see that mod $10$ the coefficients repeat $1,3,9,7$.

The pattern mod $3$ also seems quite interesting (but not periodic): I'll leave that as an exercise. Solution tomorrow if nobody else posts it.

EDIT: OK, time's up. Mod $3$ we have $$\eqalign{(1 - x^2 + x^4)^{-1/4} &\equiv 1+{x}^{2}+{x}^{6}+{x}^{8}+{x}^{18}+{x}^{20}+{x}^{24}+{x}^{26} \ldots\cr &= \prod_{j=0}^\infty \left(1 + x^{2\cdot 3^j}\right) = \sum_{k \in S} x^{k}\cr}$$ where $S$ is the set of positive integers whose base-3 representation contains no $1$'s. This is because $$(1+x^{m})^3 \equiv 1 + x^{3m}$$ so that $$ \dfrac{1+x^{2\cdot 3^{n+1}}}{1+x^2} \equiv \prod_{j=0}^n \dfrac{1+x^{2\cdot 3^{j+1}}}{1+x^{2\cdot 3^j}} \equiv \prod_{j=0}^n \left(1 + x^{2\cdot 3^j}\right)^2$$ Taking $n \to \infty$ we get $$ \dfrac{1}{1+x^2} \equiv \prod_{j=0}^\infty \left(1 + x^{2\cdot 3^j}\right)^2$$ and then $$ \dfrac{1}{1-x^2 + x^4} \equiv \dfrac{1}{(1+x^2)^2} \equiv \prod_{j=0}^\infty \left(1 + x^{2\cdot 3^j}\right)^4 $$

Answer 4

Here is an "almost" solution for integer coefficients.

The zeros of $x^4+12x^3+14x^2-12x+1$ are algebraic integers. So are their reciprocals. Now in the series $$ (1-xa)^{-1/4} = 1+\frac{a}{4} x+\frac{5 a^2}{32} x^2 +\frac{15a^3}{128} x^3 + \cdots $$ where $a$ is an algebraic integer, the coefficients are algebraic integers (except for power-of-2 denominators). Multiply four of these $$ (x^4+12x^3+14x^2-12x+1)^{-1/4} = \prod_a (1-xa)^{-1/4} $$ where the product is over all $a$ such that $1/a$ is a zero of $x^4+12x^3+14x^2-12x+1$. Its coefficients are also algebraic integers, except possibly for power-of-2 denominators. But this is invariant under permutation of the roots, so the coefficients are rational numbers. (We can also see this more directly.) Therefore the coefficients are rational integers, except possibly for power-of-2 denominators.

After doing this, I thought I would be able to finish by using 2-adic numbers, to analyze these denominators. But it seems this polynomial has no 2-adic zeros. So I would have to go to an algebraic extension of the 2-adics...