Transcendentals as the Roots of Infinite Polynomials

Question

I have always been taught that the difference between an algebraic and a transcendental number is that the former is the root to a polynomial of ${\bf finite}$ degree with integer coefficients. I did a quick search of the interwebs and I tried to find out if it is possible to express a transcendental as the root of a polynomial with ${\bf infinite}$ degree and I didn't find anything immediately, so I figured I'd ask here. So I guess my question would be:

Is it possible to find a polynomial $f(x)$ of infinite degree with integral coefficients such that $f(\alpha)=0$ where $\alpha$ is transcendental?

What caused my inquiry:

If we consider $\alpha = \pi$, we get a fairly good approximation with $$f(x)=x^3-9x^2+20x-5$$ which has real roots $x\approx 0.285521$ and $x\approx 3.14328$, and this is only a third degree polynomial!

Answer 1

As Patrick points out, you cannot hope in general for such a power series with integral coefficients, for reasons of convergence. But you can do it with rational coefficients - construct the sequence $a_n$ recursively so that $|\sum_{i=0}^n a_i \alpha^i|<10^{-n}$, say.

Answer 2

This one has rational coefficients, not integral, but might still be of interest : $$ \sin(\pi) = 0 = \sum_{k \ge 0} \frac{(-1)^k \pi^{2k+1}}{(2k+1)!}. $$ Notice that you cannot find such a power series for a real number $x$ with $|x| > 1$ with integral coefficients, simply because convergent power series $\sum_{n \ge 0} a_n$ satisfy $a_n \to 0$ when $n \to \infty$ (which means that if the $a_n$ are integers, $a_n \to 0$ implies that your series was in fact a polynomial). So if you want $\alpha$ to satisfy $\sum_{n \ge 0} a_n \alpha^n = 0$ with $a_n$ being integers, you need $|\alpha| < 1$.

Hope that helps,