# Every transcendental number satisfies a power series

I was reading this expository paper by Yves André in which he states a nice result: every transcendental number is the root of a power series over $$\mathbb Q$$. He accredits this theorem to Hurwitz in paragraph 2.3, but doesn't give a reference for it. I haven't been able to find the relevant paper (or another exposition of te proof) myself, so I'm hoping someone here knows where to look!

• @David C. Ullrich has already answered, so I'll do the usual (for me) and give some references. I posted some references in this 18 December 2006 sci.math post archived at Math Forum. The 1890/1891 Hurwitz paper was probably not online at the time (maybe it was, but I suspect I would have given a link if it had been), but you can now find it here. – Dave L. Renfro Dec 26 '18 at 21:37

Edit: There's a straightforward proof below for the result exactly as mentioned in the OP. However, the paper linked to cites a stronger result, specifying that the power series should define an entire function of exponential growth. I don't see how the argument below gives that; see Comments at the bottom.

Say $$\alpha\in\Bbb C$$, $$\alpha\ne0$$.

First, if $$\alpha$$ is real it's trivial that it's a zero of some power series with rational coefficients: If $$r_0,\dots,r_n\in\Bbb Q$$ have been chosen, there exists $$r_{n+1}\in\Bbb Q$$ with $$|r_0+\dots+r_{n+1}\alpha^{n+1}|<1/n;$$hence $$r_0+r_1\alpha+\dots=0.$$

Now say $$\alpha=\rho e^{it}$$, $$\rho>0$$, $$t\in\Bbb R$$. If $$t/\pi$$ is rational then there exists a positive integer $$N$$ so that $$\beta=\alpha^N\in\Bbb R$$; so $$\beta$$ is a root of some rational power series, hence so is $$\alpha$$.

Finally, suppose $$t/\pi$$ is irrational. Then $$\{e^{ikt}:k=1,2\dots\}$$ is dense in the unit circle. Hence for every $$n$$ the set $$\{r\alpha^k:r\in\Bbb Q, k=n+1,n+2,\dots\}$$ is dense in $$\Bbb C$$ (to approximate $$z$$ by $$r\alpha^k$$, first choose $$k$$ so as to get the argument approximately right, then choose $$r$$ to fix up the modulus). So as above we can recursively construct a sequence $$r_j$$ of rationals and a strictly increasing sequence $$n_j$$ of positive integers so that $$\sum r_j\alpha^{n_j}=0.$$

If $$\alpha$$ is real this is no problem: Say wlog $$\alpha>1$$ to keep the inequalities clean and replace the main inequality above by $$|r_0+\dots+r_{n+1}\alpha^{n+1}|<1/(n+2)!;$$it follows that $$r_n=O(1/n!)$$, hence the power series is an entire function of exponential growth.
And so we're done if $$\alpha^N$$ is real. But the case $$t/\pi$$ irrational is not so simple, as far as I can see. We can make $$\left|\sum_{j=0}^k r_j\alpha^{n_j}\right|$$ as small as we want as a function of $$k$$, but the doesn't help; saying for example $$\left|\sum_{j=0}^k r_j\alpha^{n_j}\right|\le1/k!!!$$ says nothing about the radius of convergence. The problem is that in order to make $$\left|\sum_{j=0}^k r_j\alpha^{n_j}\right|$$ small, by the trivial argument above, we may be forced to take $$n_k$$ large, so we don't get anything analogous to$$\left|\sum_{j=0}^k r_j\alpha^{n_j}\right|\le1/(n_k)!,$$which is what we need.
Perhaps one can fix the argument above, showing that you don't need to take $$n_k$$ too large to make $$\left|\sum_{j=0}^k r_j\alpha^{n_j}\right|$$ small.
Or something I just thought of. The proof that the sum of two algebraic numbers is algebraic is fairly simple if you look at it right, but it's not a priori obvious. Maybe some extension of that argument shows that if $$\alpha=x+iy$$ then the existence of suitable power series for $$x$$ and for $$y$$, proved above, implies the same for $$\alpha$$?
• Very nice! I was hoping there was some way to distill some kind of "representative" power series for a transcendental number out of the proof, similar to how the minimal polynomial of an algebraic number $a$ is a "representative" for all polynomials vanishing at $a$, but this seems a rather hopeless task. Thank you anyway for the insight! – user Dec 20 '18 at 22:35