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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.

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

which 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.

See the blog post: Hadamard’s gap theorem at Uniformly at Random.

It 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.

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.

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.

See the blog post: Hadamard’s gap theorem at Uniformly at Random.

It 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.