Any closed form for series like $$F(x)=\sum_{p=2}^{\infty}x^p,\quad p\text{ is prime}$$ or $$F(x)=\sum_{i=0}^{\infty}x^{i!}\quad ?$$

More generally, we can obtain a power series from decimal expansion of a number $r$ (with $0< r<1$) by replacing $$\left(\frac{1}{10}\right)^i$$ with $$x^i$$ like $$\frac{1}{3}=3\left(\frac{1}{10}\right)^1+3\left(\frac{1}{10}\right)^2+\dotsb 3\left(\frac{1}{10}\right)^i+\dotsb.$$ In this example we obtain: $$f(x)=\sum_{i=1}^{\infty}3x^i.$$

  • When $f(x)$ is convergent, what restriction do we have to put on $r$ (if $r$ is c.e. number) to make $f(x)$ have a closed form?
  • When is $f(x)$ algebraic, or transcendental?
  • 2
    $\begingroup$ Your first display makes no sense. Maybe you meant $$\sum_{p=2}^{\infty}x^p,p{\rm\ is\ prime}$$ $\endgroup$ May 24, 2013 at 1:51
  • 15
    $\begingroup$ This might be better expression: $$\sum_{p\textrm{ prime}} x^p$$ $\endgroup$ May 24, 2013 at 2:02
  • 3
    $\begingroup$ XL, you came back here to edit your question, but did nothing to engage with the comments. Why? $\endgroup$ May 24, 2013 at 5:34
  • $\begingroup$ @Gerry,@i707i707,thank both of you very much,your expressions are right,they are what I intend to express $\endgroup$ May 24, 2013 at 5:48
  • 3
    $\begingroup$ Here is an answered related question mathoverflow.net/q/80572/22277. Here we find out F. Carlson has shown in 1921 that power series with integer coefficients and radius of convergence 1 are either rational or have the unit disk as a natural boundary. $\endgroup$ Aug 27, 2022 at 14:02

3 Answers 3


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.

  • $\begingroup$ @i707i707,thank you.But it is only a partial answer to this question,especially,when r is algebraic number $\endgroup$ May 24, 2013 at 5:06
  • $\begingroup$ XL, if the answer is helpful, even if it's not a complete answer, I'd encourage you to vote it up. $\endgroup$ May 24, 2013 at 5:36
  • $\begingroup$ @Gerry,thank you for your reminding $\endgroup$ May 24, 2013 at 5:49
  • 1
    $\begingroup$ Maybe you wanted to link to this post: Uniformly at Random (rather than to list of all posts tagged power-series)? (Of course, since there are only two such posts, anybody can easily find that link.) $\endgroup$ Aug 26, 2022 at 14:36
  • 1
    $\begingroup$ @MartinSleziak, thanks for that link; I tried to guess and got it wrong, because I didn't notice there were two posts. Tiny correction: Uniformly at Random is the name of the blog, whereas Power series with integer coefficients $\endgroup$
    – LSpice
    Aug 26, 2022 at 20:19

Also, the classical Fabry gap theorem tells you that the unit circumference is the natural boundary. Meanwhile, all "elementary" functions can be analytically continued along almost every path on the plane, so give up all hopes for a closed formula of any sort...

  • $\begingroup$ @fedja,thank you,only can one post be chosen as answer. $\endgroup$ May 26, 2013 at 14:16

Functions with a natural boundary tend to satisfy functional equations, and they are sometimes the unique solutions to those functional equations. While a system of functional equations is far from being a closed form expression, systems of functional equations are probably as close as we can get to closed form expressions for these kinds of otherwise pathological functions.

The following observations apply to the function $\sum_{k=0}^{\infty}z^{k!}$ and related functions.

Observation: Suppose that $f$ is a function holomorphic in a neighborhood of $0$ with $f(z)=\sum_{k=0}^{\infty}a_k z^{k}$. Let $m/n$ be a reduced rational number. The following are equivalent:

  1. $f(z\cdot e^{2\pi im/n})-f(z)$ is a polynomial.

  2. For all but finitely many $k$, $a_k=0$ or $k$ is a multiple of $n$ .

  3. There is some $k$ where $f$ satisfies the functional differential equation $$\frac{d^k}{dz^k}\big(f(z\cdot e^{2\pi im/n})-f(z)\big)=0.$$

Observation: Suppose that $f(z)$ is a function that is analytic in a neighborhood of $z=0$. Let $(r_n)_{n=0}^\infty$ be a strictly increasing sequence of positive integers where $r_n$ is always a factor of $r_{n+1}$. Then $f(z)$ can be put in the form $\sum_{k=0}^\infty a_kz^{r_k}$ if and only if $f^{(r)}(0)=0$ for $r<r_0$ and for all $n\geq 0$, $f(z)$ satisfies the functional differential equation $$\frac{d^{r_{n+1}}}{dz^{r_{n+1}}}\big(f(z)-f(z\cdot\exp(2\pi i/r_{n+1}))\big)=0.$$ Suppose now that $f$ satisfies these conditions. Then $$\frac{d^{r_n}}{dz^{r_n}}\big(f(z)-f(z\cdot\exp(2\pi i/r_{n+1}))\big)=a_n\cdot (r_n)!\cdot\big(1-\exp(\frac{2\pi i r_n}{r_{n+1}})\big)$$ for each $n\geq 0$, and the function $f(z)$ is completely determined by these functional differential equations.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.