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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
up vote 4 down vote accepted

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.

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+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
This is a nice example that makes me want to modify the question to include integration. I think even if you allow integrals of elementary function, there are lots of entire functions that can not be represented that way. – Lukas Geyer Oct 30 '12 at 15:31
@LukasGeyer: I would think so too. – Hans Lundmark Oct 30 '12 at 16:34
@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
@J.M., how do you show that your series is not elementary? (I believe it is true, but I don't think it is obvious.) – Lukas Geyer Nov 25 '12 at 14:42

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