# Is $\pi$ approximately algebraic?

As we know, $\pi$ is transcendental, meaning that there is no rational numbers $a_0,\ldots,a_n\in\mathbb{Q}$ such that $$a_0+a_1\pi+\cdots+a_n\pi^n=0.$$ But I was wondering if we can get this as a limiting process: Is there a sequence of polynomials $\{p_n(x)\}_{n=1}^\infty$ with rational coefficients such that the first positive root of $p_n(x)$ tends to $\pi$ as $n\to\infty$?

• Sure, $p_1(x)=x-3$, $p_2(x)=x-3.1$, $p_3(x)=x-3.14$, and so on. – vadim123 Mar 4 '15 at 16:22
• I think this has almost nothing to do with "algebraic number", because it involves a limit. So perhaps the title is not so good. – Dietrich Burde Mar 4 '15 at 16:26
• Vadim's comment above is exactly to the point. $\pi$ is not merely "approximately algebraic" in this sense; it is approximately rational. You can fix $n=1$ and find any number of sequences of first-degree polynomials with the desired property. To make the question interesting you need to include some measure of how well-approximated $\pi$ is by the roots of the polynomials. The irrationality measure is a measure of this type. – MJD Mar 4 '15 at 16:27

## 1 Answer

Yes, $$\sin\pi=0$$ so the first positive root of $$p_n(x)=x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\cdots+\frac{(-1)^n}{(2n+1)!}x^{2n+1}$$ tends to $\pi$ as $n\to\infty$.

• There is more involved here. Let $\{f_n:\Bbb R\to\Bbb R\}$ be a sequence of functions, and suppose that every function of the sequence has a least positive root, say $a_n$. Is it true that if $f_n\to f$ and $f$ has also a first positive root $a$, then $a_n$ converges and its limit is $a$? I don't find it trivial at all. – ajotatxe Mar 4 '15 at 16:28