Geometric Moments and Rational Lambert Values: Proofs and Further Results

A separate positive-measure estimate.

We seek positive moment weights comparable to $(k+1{)}^2(k+2)/2$, with constants depending only on $q$. In the determinant expansion these constants give factors exponential in $N$, so they do not change its cubic exponent of $q$. For fixed $0<q<1$ write $P=(q;q{)}_{\mathrm{\infty }}$, $Q=(\sqrt{q};q{)}_{\mathrm{\infty }}$, $T=(-1;q{)}_{\mathrm{\infty }}^2$, and

The choice $w=q^{m+1}$ collects all dependence on the moment index. Indeed, the three finite products in the $t$th remainder summand become

Their substitution gives $v_m^{\ast }=P^4{G}_q(q^{m+1})$ term by term. For fixed $q$ and $0<r<1$, the $t$th summand on $|w|\le r$ satisfies

Indeed, $(q;q{)}_t\ge P$, each denominator product in $w$ has modulus at least $(r;q{)}_{\mathrm{\infty }}>0$, and the numerator product has modulus at most $(-r^2;q{)}_{\mathrm{\infty }}$. The bound is independent of $t$, so the sum converges uniformly and absolutely on each such disk. The individual products converge there as well, since their tails are bounded by geometric series in $q$. Thus $G_q$ is holomorphic for $|w|<1$, and its Taylor series may be evaluated at $w=q^{m+1}$ to obtain the moment expansion. For formal series, the corresponding substitution is justified coefficient by coefficient, as in the short paper. The $q$-binomial theorem [28]

follows by comparing coefficients in $(1-w)R(w)=(1-Aw)R(qw)$ with $R(0)=1$; the series converges for $|w|<1$ since its coefficients are at most $P^{-1}$. They are nonnegative, and at least $(A;q{)}_{\mathrm{\infty }}$ if $A<1$. For $A=1$ the series equals $1$. Factor the numerator of the $t$th term using

After pairing each positive-argument factor with one denominator, the $t$th summand of $G_q$ becomes

The $q$-binomial identity makes the two ratios nonnegative coefficientwise; the remaining product also has nonnegative coefficients. When $t=0$, the first ratio is $1$ and every coefficient of the second is at least $Q$. The last fraction has coefficients at least those of $(1-w{)}^{-3}$. Thus the $t=0$ term alone bounds ${\gamma }_k$ below by $Q(\genfrac{}{}{0ex}{}{k+3}{3})$.

For the upper bound, if $R$ has nonnegative coefficients and $R(1)\le C<\mathrm{\infty }$, convolution with a nondecreasing sequence $d_k$ is bounded by $C{d}_k$. At $t=0$, bound the second ratio coefficientwise by $P^{-1}(1-w{)}^{-1}$. After extracting $(1-w{)}^{-3}$ from the last fraction, its remaining factor has value at $w=1$ at most $T{P}^{-3}$. This gives $T{P}^{-4}(\genfrac{}{}{0ex}{}{k+3}{3})$. For $t\ge 1$, both ratios are bounded by $P^{-1}(1-w{)}^{-1}$. Extracting the one factor $(1-w{)}^{-1}$ from the last fraction leaves a factor with value at $1$ at most $T{P}^{-3}$; also $(q;q{)}_t^{-1}\le P^{-1}$. The $t$th summand is therefore bounded coefficientwise by $T{P}^{-6}{w}^t(1-w{)}^{-3}$. Summing $t\ge 1$ gives $T{P}^{-6}(\genfrac{}{}{0ex}{}{k+2}{3})$ as the bound for its $k$th coefficient. Consequently

Thus $\sum _{k\ge 0}{P}^4{\gamma }_k{q}^k{\delta }_{q^k}$ is a finite positive measure on $[0,1]$, with infinitely many distinct support points and moments $v_m^{\ast }(q)$. In standard terminology, $(v_m^{\ast }(q){)}_{m\ge 0}$ is a Hausdorff moment sequence. This statement fixes $q$; the measure is not a representing measure for the coefficient sequence $s_m(p)$ considered in the next subsection. A nonzero polynomial of degree less than $N$ cannot vanish at all of its first $N$ atoms; the associated Gram matrix is positive definite. Truncate the measure to its first $K+1$ atoms. At fixed rank $N$, each matrix entry converges as $K\to \mathrm{\infty }$, and the determinant converges because it is a polynomial in those entries. Finite Cauchy–Binet and monotone convergence of the nonnegative tuple sums therefore give Heine’s expansion, whose integral form is reproved in [10]:

Retaining the tuple $k_i=i$ and using $\prod _{d=1}^{N-1}(1-q^d{)}^{2(N-d)}\ge P^{2N}$ gives the lower bound below. For the upper bound put $k_i=i+{\lambda }_i$, where the ${\lambda }_i$ are nonnegative and nondecreasing. Factoring the smaller power from each Vandermonde difference gives the exact exponent

Every weight $2N-1-2i$ is at least $1$. Since $0<q<1$, the remaining power of $q$ is at most $q^{\sum _i{\lambda }_i}$. Also $c_{i+\lambda }/c_i\le (\lambda +1{)}^3$. Discard the remaining Vandermonde factors, each bounded by $1$, and enlarge the nonnegative sum by dropping the ordering restriction on the ${\lambda }_i$. It now factors into $N$ copies of $\sum _{\lambda \ge 0}(\lambda +1{)}^3{q}^{\lambda }$, which gives

Both constants are positive and finite, so $V_N^{\ast }(q)>0$ and $\log (V_N^{\ast }(q)/(C_N{q}^{B_N}))=O_q(N)$, including $N=0$ under the empty determinant convention. Here $N\to \mathrm{\infty }$ with $q$ fixed; the constants are not uniform as $q\uparrow 1$. The formal-order calculation instead fixes $N$ and expands at $q=0$. No joint uniform limit is asserted. The measure depends on $q$, not on the moment index or rank: one measure represents the whole sequence at that base. This is an analytic representation, not a denominator factor for the 2004 polynomial forms. Section 3.2 replaces the band $O_q(N)$ by an exact asymptotic with an explicit positive constant, using the same weights.