Removing Schanuel

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.

This entry was posted on May 14, 2012 at 02:40 and is filed under Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

Woett's Blog