Archive for May, 2012

Removing Schanuel

May 14, 2012

Let u_{a,b} and v_{a,b} be coprime integers such that v_{a,b} is positive and \dfrac{u_{a,b}}{v_{a,b}} = \displaystyle \sum_{i=a}^b \dfrac{r_i}{i s_i}, where \{r_i\}_{i \in \mathbb{N}} and latex \{s_i\}_{i \in \mathbb{N}} are assumed to be given periodic sequences of integers. That is, for all i we have r_{i+t} = r_i for some t \in \mathbb{N} and s_{i+t'} = s_i for some t' \in \mathbb{N}. In my paper I prove the following statement:

If Schanuel’s Conjecture holds, then for every a \in \mathbb{N}, there exists an integer b > a, such that v_{a,b} < v_{a,b-1}. Furthermore, if \max_i |s_i| = 1, then the smallest such b grows at most linearly in a.

I now believe I can prove the above, without assuming Schanuel’s Conjecture! Although I haven’t fully convinced myself that my ideas work, I’m getting more confident by the minute. I will now present an (admittedly very brief) overview of my ideas. I will expand on this as time goes on. Also, for definitions and lemma’s I refer to, check my preprint.

I use Schanuel’s Conjecture for two things:

1) To get an arbitrarily large prime p dividing X_b. (Theorem 2 in my preprint)
2) To make sure \gcd(X_{a,b}, ls_b) < p.

It is, if I’m not mistaken, not hard to do the second part without Schanuel. We can prove that l < p (as I do on page 14, beginning with ‘Now that we have shown that X_n has arbitrarily large prime divisors’), and since \gcd(X_{a,b}, ls_b) < p is immediate when ls_b < p, we may assume l > ps_b^{-1}. Since p is arbitrarily large, l is arbitrarily large too. In particular, it must be divisible by a large prime power. On the other hand, we can use Lemma 4, 5, or 6 to make sure that X_{a,b} is not divisible by a large power of that prime, and we are done.

A proof of the first statement is somewhat more involved, but the idea is to take the following assumption:

There exists a finite set of primes, \{p_1, p_2, .., p_m\} such that, for all n, |X_n| can be written as |X_n| = p_1^{e_{1,n}} \cdots p_m^{e_{m,n}}.

And derive a contradiction  by using nested intervals I_1 \supset I_2 \supset .. \supset I_m such that for all k \in \{1,2,..,m\}, there exists a j \in \{1,2,..,m\}, such that p_k^{e_{k,n}} < n for all n \in I_j. In particular, p_k^{e_{k,n}} < n for all k and all n \in I_m, which is, for large enough n, a contradiction, since |X_n| > n^m, by Lemma 3.

To do this, realize that the last sentence implies that, for all n large enough, there exists an index k = k(n), such that p_k^{e_{k,n}} > n. But if c is an arbitrary constant, n is large enough and n' is the smallest integer larger than n such that p_{k(n)}^c divides n', then p_{k(n)}^{e_{k(n),n'}} < n'. Now, if we take c to be suitably large, then we have a large interval where p_{k(n)}^{e_{k(n),n''}} < n'' for all n'' inside this interval. Then, inside this interval, we repeat this trick with a different index k', and a slightly smaller constant c, such that inside this interval, we find a slightly smaller interval, such that for all n''' inside this smaller interval, we have p_{k'}^{e_{k',n'''}} < n'''. Repeat this process m times, and we have that p_k^{e_{k,n''''}} < n'''' for all k and all n'''' in the last, smallest, interval (which, to be clear, is thus contained in all the others).

Hopefully the above made some sense. If not, just have a little patience and look forward to more elaboration, which will come in due time. In the mean time, try to embrace the glorious notation I laid out in this post.

EDIT (9 december 2012): Have a look here.

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