# Are all analytic functions on the whole complex plane “generated” by the polynomials and the exponential?

Take $S$ to be the set containing all polynomials and $e^z$ over $\mathbb C$. If we add, subtract, multiply, divide (if the denominator is non-zero everywhere) and compose functions in our set $S$ we get analytic functions.

My questions is: do all analytic functions on the whole complex plane arise this way?

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This is a sweet question! Thanks for sharing. –  uncookedfalcon Oct 29 '12 at 21:29
@Lukas: I have no idea. That is why I said "shouldn't" instead of "can't." –  Qiaochu Yuan Oct 29 '12 at 21:59
Perhaps a Baire category argument would help? –  Nate Eldredge Oct 29 '12 at 22:12
I thought one should be able to argue that all functions obtained this way have integer (or infinite) growth order, so functions like $\cos \sqrt{z}$ with growth order $1/2$ would not be in this class, but I don't quite have a proof yet. –  Lukas Geyer Oct 29 '12 at 23:15
Yes, $\cos \sqrt{z}$ is entire. The apparent singularity at $0$ isn't a problem because $\cos$ is an even function. You can also see it by plugging in $\sqrt{z}$ into the series to get $\cos \sqrt{z} = 1 - \frac{z}{2!} + \frac{z^2}{4!}-\frac{z^3}{6!} \pm \ldots$ –  Lukas Geyer Oct 30 '12 at 3:58

As an alternative to the other (perfectly fine) answer, we can also use the fact that an entire function can grow arbitrarily fast on the real line, and that functions generated by $S$ (including iterated integrals of those) can only grow as fast as finitely iterated exponentials.

In order to make the argument rigorous, let $e_1(x) = e^{x}$, and inductively $e_{k+1}(x) = e_k(e^x)$ be the iterated exponentials. Now define $$f(z)=a_0+\sum_{k=1}^\infty \left(\frac{z}{k}\right)^{n_k}$$ where $a_0 = e_1(1)=e$, and $(n_k)$ is a strictly increasing sequence of natural numbers chosen such that $$\left(\frac{k}{k-1}\right)^{n_{k-1}}\ge e_{k}(k)$$ for all $k \ge 2$. Then $f$ is an entire function (since $|z/k|<1$ for $k > |z|$, the tail of the series is dominated by a geometric series for any fixed $z$), and $f(k) \ge e_k(k)$ for all $k\ge 1$ (since the $k$-th term in the series is $\ge e_k(k)$, and all other terms are positive.) It is easy to see by induction that each $e_n$ is increasing, and that $e_k(x) \ge e_m(x)$ for $k \ge m$ and all $x\ge 0$. This implies that $f(k) \ge e_k(k) \ge e_m(k)$ for all $k \ge m$, so $$\limsup_{x\to\infty} \frac{f(x)}{e_k(x)} \ge 1$$ for all $k$.

On the other hand, all functions $g \in S$ satisfy $$\limsup_{x\to\infty} \frac{|g(x)|}{e_2(x)} = 0,$$ and by induction any function $g$ that arises from functions in $S$ by $k$ operations (addition, multiplication, division, composition, integration) satisfies $$\limsup_{x\to\infty} \frac{|g(x)|}{e_{k+2}(x)} = 0.$$ This shows that $f$ is not in the class of entire functions generated by $S$.

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$\int_0^z e^{w^2} dw$ is an entire function which is not elementary, so no.
+1: This is also an interesting example of an étale surjective (because odd) holomorphic map $\mathbb C\to \mathbb C$ which is not an isomorphism: such examples are not so easy to come by. –  Georges Elencwajg Oct 30 '12 at 9:10
@Lukas, if you don't include infinite sums, then consider $\sum\limits_{k=1}^\infty \frac{x^k}{k^k}$. One could ostensibly come up with infinite products and infinite continued fractions that are analytic and not expressible in terms of known functions... –  Ｊ. Ｍ. Nov 25 '12 at 12:20