Algebraic functions have a discrete set of singularities. [Lacunary functions](http://en.wikipedia.org/wiki/Lacunary_function), e.g. $f(z)=\sum\_{n=0}^\infty z^{2^n}$, have a continuum of singularities at every point of the boundary of their disk of convergence. The [Ostrowski–Hadamard gap theorem](http://en.wikipedia.org/wiki/Ostrowski-Hadamard_gap_theorem) gives a sufficient condition for the exponents with non-vanishing coefficients for a power series which defines a lacunary function.


Can one also construct a lacunary function using only a finite number of algebraic operations, i.e. arithmetics and solving plane algebraic equations over the complex numbers, and a finite number of analytic operations, particularly applications of the exponential function and the logarithm?



Finite construction of lacunary functions using algebraic and certain analytic operations
1. You have to define what a "lacunary function" means. There are various lacunary conditions in the literature.
2. If you mean sufficiently large lacunas so that your series have a whole circle of
singularities, like in the example you give, such functions cannot be constructed as
a finite composition of algebraic, exponential, and logarithmic ones.


One reason for this is the "Iversen property". Which is the following. Let $(f,z\_0)$
be a germ of an analytic function. If for every curve $\gamma(t)$ 
starting at $z\_0$, and 
every $\epsilon >0$, there exists a curve $\gamma'(t)$ starting at $z\_0$ such that $|\gamma(t)-\gamma'(t)|<\epsilon$ for all $t$, and $(f,z\_0)$ has an analytic continuation
along $\gamma'$, we say that $f$ has the Iversen property.


Composition of functions with this property also has it. Algebraic, entire and log functions have it. But a function with a whole line of singularities, evidently does not have it.


