Is $z \mapsto \sum_{k=0}^\infty \frac{1}{(k!)!} z^{k!}$ an elementary function? I designed it to be analytic but analogous to Liouville's constant, but don't know how to search for this function.
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2Not that it matters much, but $k!!$ denotes the double factorial and not the iterated factorial $(k!)!$? Anyway, it is a lacunary series, and elementary functions don't have lacunary series. [Waits for somebody to point out an elementary function I haven't heard of yet.] – Daniel Fischer Jun 24 '15 at 09:30
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@DanielFischer: I knew of the double factorial but thought it would be obvious what I meant. Anyway why do elementary functions not have lacunary series? – user21820 Jun 24 '15 at 09:50
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I don't know why, it is just the case that elementary functions (trigonometric functions, hyperbolic functions, exponentials, logarithms, powers [including roots], polynomials, rational functions of these, compositions of these) happen to not have lacunary series [to the best of my knowledge]. – Daniel Fischer Jun 24 '15 at 09:55
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@DanielFischer: Wait I just checked up on lacunary series. They aren't analytic, but isn't my function analytic? – user21820 Jun 24 '15 at 10:03
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I used "lacunary series" in the loose sense that the gaps between non-zero terms grow sufficiently fast. If the coefficients converge to $0$ fast enough, that can of course still give entire functions. If the coefficients converge too slowly to $0$ or not at all, then you have a function with natural boundary on its circle of convergence. – Daniel Fischer Jun 24 '15 at 10:06
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1Well, $0$ is surely an elementary function, and it has a lacunary series... sorry, heh. – David C. Ullrich Jun 24 '15 at 17:10