Fredholm lacunary series ($f_l(z)={\sum }_{n\ge 0}{z}^{l^n},l\ge 2$)

This lacunary series is analytic for $|z|<1$ and satisfies $f_l(z^l)=f_l(z)-z$. At roots of unity of $l$-power order, its radial tail diverges, giving a dense set of singularities on the unit circle and excluding algebraicity as a function. The Mahler method proves transcendence of its values at every algebraic $z$ with $0<|z|<1$. At $z=1/2$, reduced truncations have denominators $2^{l^m}$ and error less than twice their $-l$th power; for $l\ge 3$, the Roth theorem gives a shorter proof.

For $0<p<q$, truncating the Fredholm series at $n$ gives a rational with denominator at most $Q_n=q^{l^n}$ and positive error at most $2Q_n^{-l(1-\log p/\log q)}$ for large $n$. The displayed strict inequality first rules out a rational value by the rational separation bound, and then rules out an algebraic irrational value by Roth theorem. Reduced denominators are at most $Q_n$ and tend to infinity because the truncations are distinct. In particular $l\ge 3$ and $p<q^{\delta }$, $\delta <1/3$, imply transcendence.

A nonzero integer polynomial of bidegree at most $N$ can cancel the first $(N+1{)}^2-1$ Taylor coefficients of the Fredholm series. Functional transcendence makes the resulting analytic function nonzero. If $\alpha$ and $f_l(\alpha )$ were both algebraic, the relation $f_l({\alpha }^{l^r})=f_l(\alpha )-{\sum }_{j<r}{\alpha }^{l^j}$ would keep the auxiliary values in one fixed number field. Their Absolute logarithmic Weil height is at most $N{l}^{r}h(\alpha )(1+1/(l-1))+O_N(r+1)$, whereas their chosen absolute value is $O_N(|\alpha {|}^{T_N{l}^r})$. The Liouville height inequality forces $T_N=O(N)$, contradicting its quadratic lower bound for sufficiently large $N$.