Can you express every algebraic number in terms of $\pi$ I was doing my physics homework, when I noticed that the answer I just calculated was very close to $\sqrt{3}$. The number was $\frac{441\pi}{800} \approx 1.73180… .$
To put it into context,
$\sqrt{3} - \frac{441\pi}{800} \approx 2.478 \times10^{-4}$
, percentage error $\approx0.0143$
This got me thinking if there is way to express algebraic numbers, if not all irrationals, as
$$n = \sum_{r=1}^{\infty} a_r\pi^r$$
where $a_i \in\mathbb{Q},n\in\mathbb{R}$ and $n$ is algebraic?
 A: $n = \sum_{r=1}^{\infty} a_r\pi^r$
The problem is $\pi^r \rightarrow \infty$ as $r \rightarrow \infty$. So you have 2 options. You can either define $a_r$ to be $0$ for $r >$ some natural number $n$. If you do that, then you can never achieve equality because that would be the equivalent of finding a polynomial which spits an algebraic number given $pi$, which is impossible, since the algebraic numbers are closed under polynomial arithmetic.
The second option is you can define $a_r$ to approach $0$ as $r \rightarrow \infty$. In that case, what your doing is taking equating the limit of a sequence to be an algebraic number. Given you can choose $a_r$ arbitrarily, then it's trivially possible, for example, just choose any rational number that returns $z \cdot 10^{-z}$ after being multiplied by $\pi^r$, where z is the $z^{th}$ digit to the right of the algebraic number's decimal point (doesn't matter what the remaining decimals are just ignore them), and adjust for the next rational number $a_{r+1}$. You don't need the $\pi^r$, infact you can achieve equality for any real number, not just algebraic numbers. Infinite sequences are neat that way.
A: For every real number $x$, define by induction a sequence $(a_n)$, such that
$$a_0 = 0$$
$$a_{r+1} \in \mathbb{Q} \quad \text{ such that } \quad \frac{1}{\pi^{r+1}} \left(x - \sum_{k=0}^r a_k \pi^k - \frac{1}{r+1}\right) \leq a_{r+1} \leq \frac{1}{\pi^{r+1}} \left(x - \sum_{k=0}^r a_k \pi^k\right)$$
Such a sequence can always be constructed by density of $\mathbb{Q}$ is $\mathbb{R}$. And you have
$$x- \frac{1}{r+1}  \leq \sum_{k=0}^{r+1} a_k \pi^k \leq x$$
so the series $\sum a_k \pi^k$ converges to $x$.
