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work on the transcendence/irrationality of (^n e) - Printable Version

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work on the transcendence/irrationality of (^n e) - JmsNxn - 05/17/2013

Hey everyone, I'm wondering if anyone knows of any work done on proving the irrationality of \( ^n e \) or perhaps transcendence. It seems like a fruitful question in my eyes, and I think that no doubt these constants are probably transcendental. A proof of this would be quite spectacular though, and I imagine it would require some ingenious argument. Irrationality maybe a bit more modest, and I expect easier to prove.

I was wondering because I was trying to prove something similar but slightly stronger, that if \( a_i \in \mathbb{Z} \) \( \sum_i a_i (^i e) = 0\,\,\Leftrightarrow\,\, a_i = 0 \) which I think is perfectly reasonable. I was trying to approach this using ring theory, saying that by contradiction we have some relation and it is the smallest such one, so:

\( \sum_i a_i (^i e) = 0 \) then we can create an isomorphism to the ring \( \mathbb{Z} [X] / p(X) \) where \( p(X) = \sum_i a_i X^i \) by inventing a pointwise multiplication that is compatible with scalar multiplication of integers and has the rule \( (^n e) \times (^m e) = (^{n+m} e) \). However I haven't gotten very far in finding a contradiction.

I also tried using calculus and talking instead about \( f(s) = \sum_i a_i (^i e)^s \) and learning about where the zeroes of such functions are distributed.I think this is probably the more fruitful method. I think it is very unlikely that the zeroes of a function like \( f(s) \) are algebraic. I was going to try and go by induction, since when the biggest term is \( e^s \) the zeroes are transcendental then assume when the biggest term is \( (^{n-1} e)^s \) the zeroes are transcendental and go from there.

Any tips or hints or knowledge would be greatly appreciated. I think proving \( (^n e) \) is transcendental or irrational is a big step in investigating tetration.