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It's known that $\pi$ is not algebraic i.e. there is no finite "polynomial" $P(x) = \sum_{n=0}^{k} a_n x^{c_n}$ s.t. $P(\pi)=0$ where $a_n, c_n$ are rational.

We might instead ask, what about $\sum_{n=0}^{k} a_n x^{c_n}$ where $a_n, c_n$ are algebraic? (Which we can can call level-2 algebraic (if it is)).

To my knowledge this isn't known (and I expect the answer is no). It's worth pointing out that some rather transcendental looking numbers such as Gelfond's Constant $e^{\pi}$ are level-$2$ algebraic (via $P(x) = x^{i} + 1$ ).

Sidharth Ghoshal
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    The problem with non-integer exponents is that $x\mapsto x^\alpha$ is not a well-defined function. When $\alpha$ is rational, it has finite values, but for $\alpha$ irrational or not real, the map has infinite values. That just makes defining it hard. – Thomas Andrews May 27 '25 at 18:37
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    You can make this more precise by asking if $e^y=\pi,$ is there some algebraic $a_i,c_i$ a root of $$f(yz)=\sum_{i}a_ie^{c_iy}.$$ Of course, there are only countably many "2nd level algebraic" numbers, so "most" numbers are not. – Thomas Andrews May 27 '25 at 18:46
  • @ThomasAndrews. If there is an answer to your problem above it is somehow within the Gelfond Schneider Theorem. Where we get to write $2^\sqrt{2}$ as transcendental because is each element in the set $x^{1/\sqrt{2}}=2$ is transcendental. It is true that as we scale the problem up it seems we talking about intersections of sets using the language of real numbers. TLDR. For sure that is true. Seems surmountable. – Mason May 27 '25 at 20:06

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Let the numbers you've described be the set $A_2$. And let $a, b \in A_1$ the set of algebraic numbers. This is the notation used here. This framework was also discussed on math overflow here.

Your question is indeed open, but we can make some progress with Schanuel's conjecture. After an even stronger unproven assumption called the Schanuel Subset Conjecture written about here we may say (Theorem 3) that $\pi \ne a^b$. Another paper seems to make the same claim with no more than Schanuel's Conjecture.

It would be nice to learn that this allows us to demonstrate $\pi\not \in A_2$ which as you've noted is surely true but not neccessarily demonstrable.

What we know is that $e^a \notin A_2$. This follows from Lindemann Weierstrass Theorem. And as you noted Gelfond's constant is not a member.

Mason
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    Curousity: Is there an organic number demonstrably not in $ \cup A_k$? Its countable and could be recursively enumerable) we can construct one through diagonalization but maybe numbers with some irrationality measure criteria can be shown not in $A_k$ – Mason May 27 '25 at 19:43
  • how do you know that e does not belong to A2? Can you explain a little bit? – Md Asfaque Jun 15 '25 at 13:00
  • @MdAsfaque I know it by the Lindemann Weierstrass theorem. What machinery powers Lindemann Weierstrass Theorem? I probably can't do much better than the wiki linked above... But $e$ is very special in that $\frac{d}{dx}{e^x}=e^x$. And this allows for the theorem to go through... Feel free to ask a more specific question and I'll try my best. – Mason Jun 15 '25 at 15:12