work on the transcendence/irrationality of (^n e) JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/17/2013, 06:22 PM (This post was last modified: 05/17/2013, 06:33 PM by JmsNxn.) 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. « Next Oldest | Next Newest »