2
$\begingroup$

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 $n\to\infty$?

$\endgroup$
3
  • 8
    $\begingroup$ Sure, $p_1(x)=x-3$, $p_2(x)=x-3.1$, $p_3(x)=x-3.14$, and so on. $\endgroup$ – vadim123 Mar 4 '15 at 16:22
  • $\begingroup$ I think this has almost nothing to do with "algebraic number", because it involves a limit. So perhaps the title is not so good. $\endgroup$ – Dietrich Burde Mar 4 '15 at 16:26
  • 4
    $\begingroup$ 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. $\endgroup$ – MJD Mar 4 '15 at 16:27
8
$\begingroup$

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 $n\to\infty$.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ – ajotatxe Mar 4 '15 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.