Skip to main content
5 of 6
Rollback to Revision 3
LSpice
  • 12.9k
  • 4
  • 45
  • 69

See this blog post: https://uniformlyatrandom.wordpress.com/tag/power-series/

contains a proof of the result by Fatou:

A function whose power series expansion has integer coefficients and radius of convergence 1 is either rational(in $\mathbb{Q}(x)$) or transcendental(over $\mathbb{Q}(x)$).

If $r$ is rational, then the decimal expansion will be eventually periodic. So we have rational function. (Indeed this can be done explicitly)

Otherwise, when $r$ is irrational, then the resulting function cannot be rational(plug in $1/10$, then you get irrational number). Thus, we have transcendence of $f$.

In particular, your functions $F$ in the beginning are transcendental. However, getting closed form will be extremely hard for those examples.

Sungjin Kim
  • 3.3k
  • 25
  • 28