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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?

rtybase
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BLUNT
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    $\pi$ is irrational, as such ${n\pi+k\mid n,k\in\mathbb{Z}}$ is dense in $\mathbb{R}$. So, it is possible to approximate not only $\sqrt{3}$. – rtybase Sep 03 '20 at 11:44
  • @rtybase But the question is whether we can achieve equality, not just an arbitary good approximation. – Peter Sep 03 '20 at 12:05
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    @rtybase So, how will this theorem help in proving it for all algebraic numbers though? I know that the case of $\sqrt{3}$ is a trivial one I just encountered, which sparked this entire thought. Also, I undertand that $\pi$ is not so special here, and that if this holds true for $\pi$, then it will hold for any real – BLUNT Sep 03 '20 at 12:06
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    One more https://en.wikipedia.org/wiki/Non-integer_base_of_numeration – rtybase Sep 03 '20 at 12:11
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    @Peter I think the answer is trivially yes, or trivially no, depending on what you mean. –  Sep 03 '20 at 12:12
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    If you want to express $x\in\mathbb R$ (doesn't even need to be algebraic) this way, just choose $a_r$ such that $\vert x-\sum_{r=1}^n a_r\pi^r\vert<1/n$. Then the series converges to $x$. Allowing the coefficients to be chosen from the rationals makes this possible, because the rationals are dense in the reals. – Vercassivelaunos Sep 03 '20 at 12:13

2 Answers2

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$n = \sum_{r=1}^{\infty} a_r\pi^r$

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.

  • Interesting, so you are saying that $$ n = \sum_{r=1}^{k} a_r\pi^r, k \in \mathbb{N}, n \in \mathbb{R}$$ Also, this argument can be further extended to any irrational number, for example $e$ too right? – BLUNT Sep 03 '20 at 14:02
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    What - That is the exact opposite of what I said lol. You can either swap $=$ and $n \in \mathbb{R}$ for $\neq$ and $n \in \mathbb{A}$, or swap $k \in \mathbb{N}$ for $k = \infty$. –  Sep 03 '20 at 16:18
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    Unless you mean that the finite sum is a real number, which I would agree would be absolutely astonishing! –  Sep 03 '20 at 16:23
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    Unless you mean that the finite sum is a real number, which I would agree would be absolutely astonishing! –  Sep 03 '20 at 16:23
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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 $x$.

TheSilverDoe
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