# Transcendentals as the Roots of Infinite Polynomials

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!

• What do you mean by polynomial of infinite order? Commented Jun 6, 2014 at 0:12
• @MarcinŁoś I meant degree. Oops! Commented Jun 6, 2014 at 0:22
• Well, that's not a problem. My point is, polynomials by definition have finite degree, hence my question. Do you mean power series? Commented Jun 6, 2014 at 0:25
• @AJStas : Any irreducible polynomial of degree more than $1$ will give you such $\alpha$'s, you don't need power series for that. If you just want to find solutions of power series which are irrational, they are just screaming for you out there! A transcendental element would have to be an element of $\mathbb R \backslash \overline{\mathbb Q}$, i.e. a non-algebraic real number. Commented Jun 6, 2014 at 0:34
• Polynomials of infinite degree are called entire functions
– mike
Commented Jun 6, 2014 at 0:37

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.
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 $|\alpha| < 1$.