Timeline for What is the simplest, most elementary proof that a particular number is transcendental?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 15, 2010 at 14:20 | comment | added | Daniel Briggs | Or, increasing from 2(n-1)!, the first factorial sum to be seen is n!+1!. (And decreasing from 2(n-1)!, it's (n-1)!+(n-2)!, which is very far away, and this shows that the factorial sums can't conspire to make "magic" 0s.) Similarly, with mL^3, all the products involving at least one 1 from the n! place on make less than 3m 10^-n! L^2 (the 3 is from choosing the 1 to be in the n! place in the first, second, third L, and the "less than" from the microscopic overcounting); moving left from here, the first thing we see is m at the 3(n-1)! place, which is sooner than in a multiple of L^2. | |
| Dec 15, 2010 at 13:53 | comment | added | Daniel Briggs | Given mL^2 for an integer m, let's go back from the n! spot towards the 2(n-1)! spot. Near the n! spot there can be contributions of the form 2m 10^-n!10^-k! for small k, but notice that the effect of the 10^-k! is to move right, rather than left, and the smallest values k! can take on are 1, 2, 6, 24, ... . So .2m+.02m+.000002m+... will be seen here, but that makes at most one more positive digit left than .2m does, and it's the same near the n! spot for any n. Once you get past it, there's nothing going back to 2(n-1)!, and then there's something (if m has final 0s, once we get past them). | |
| Dec 15, 2010 at 1:08 | comment | added | Justin Lanier | Daniel, thanks so much for your thoughtful answer. A few clarifying questions: 1) How do you know that a 4 doesn't pop up somewhere in L^2, or more generally that the sum of some factorials doesn't equal the sum of some others? 2) Do you mean if we look between the 2(n-1)! and the 2n! place for large enough n, we'll see the integer multiplier bare at the end of that stretch? If so, how do we know that there aren't any 2's cropping up along the way that would mess things up? 3) What with the 2's, I feel lost on how you're calculating the length of the swaths of zeros. Could you expand on this? | |
| Dec 14, 2010 at 6:55 | comment | added | Daniel Briggs | @David Feldman: Agreed! It was, for me, a very thought-provoking question. | |
| Dec 14, 2010 at 6:45 | history | edited | Daniel Briggs | CC BY-SA 2.5 |
added 63 characters in body; added 3 characters in body
|
| Dec 14, 2010 at 6:40 | history | answered | Daniel Briggs | CC BY-SA 2.5 |