Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Can anybody suggest a good (e.g. "non-technical") introduction to estimating bounds for logarithmic sums of the form

$$\sum_{i=1}^{r}{{\alpha_i}{\log(q_i)}}$$

where the $$\alpha_i$$ are positive integers (not necessarily distinct) and the $$q_i$$ are odd primes.

The reason why I ask this question is because I am trying to estimate (both lower and upper) bounds for the logarithm of a number-theoretic function, specifically $$\sigma_{1}(N)$$

I was able to show (in 2008) that

$$\sigma_{1}({q_i}^{\alpha_i}) \le \frac{2}{3}\frac{N}{{q_i}^{\alpha_i}}$$

for all $i = 1, 2, ..., r = \omega(N)$,

where $N$ is an OPN (i.e. Odd Perfect Number) and ${q_i}^{\alpha_i} || N$ for all $i$. You will be able to get an upper bound for the logarithmic sum given above, by first taking logarithms of both sides of the inequality, then summing over all $i$.

Unfortunately, for the "numbers" $N$ that I am considering, the current literature (on OPNs) do not point to an approach "strong enough" to prove nonexistence of such "numbers" $N$. (This is because the upper bound alluded to in the previous paragraph is still (of course) dependent on $r = \omega(N)$).

(And that is the reason why --) I'd be particularly interested in an analytic-number-theoretic approach. =) (Thanks to Gerry Myerson for encouraging this "clarification".)

share|improve this question
2  
Why does good mean "non-technical" here? Especially in the context of analytic number theory... –  Yemon Choi Dec 3 '10 at 21:05
1  
@Yemon, are you confusing e.g. with i.e.? @Amie, since any finite sum can be written in that form, I think you're going to have to rethink the question. Voting to close. –  Gerry Myerson Dec 3 '10 at 22:04
2  
I guess it is implied that all variables take integer (or algebraic) values here. Of course, the answer is that if we knew how to get good bounds for them in an easy way, the number theory books would be quite different from what they are now. Even the exact Liouville exponent of $\log 2$ (in your language, a "sharp" lower bound for the sum $a+b\log 2$) is unknown as far as I remember. After all, Alan Baker didn't get his Fields medal for nothing. –  fedja Dec 3 '10 at 23:09
    
Thank you, Fedja for articulating what I was trying to hint at (and Baker's name was the first thing that came to mind, but I am unfamiliar with the details of work in this area). –  Yemon Choi Dec 4 '10 at 1:00
1  
@Arnie, thanks, but now it looks like you are asking for bounds on the logarithm of a natural number (since any number can be written as a product of positive integer powers of primes), which means you're just asking for bounds on a natural number (since log is monotone), so the answer is, a sharp lower bound is zero, and there is no upper bound. Question still needs work if it is to have any kind of useful answer. –  Gerry Myerson Dec 4 '10 at 11:56
show 7 more comments

1 Answer 1

up vote 0 down vote accepted

Rather than multiplying, we sum $\forall i \in {1, 2, \ldots \omega(N)}$ to get:

$$\sum_{j = 1}^{\omega(N)}\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N} \le \frac{2\omega(N)}{3}$$

Following Nielsen, we know that (for lack of an "effective" upper bound for $\omega(N)$):

$$\inf\left({\frac{2\omega(N)}{3}}\right) \ge 6$$

Now multiplying, $\forall i \in {1, 2, \ldots \omega(N)}$ we get:

$$\prod_{j = 1}^{\omega(N)}\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N} = \frac{{\displaystyle\prod_{j = 1}^{\omega(N)}{{q_j}^{\beta_j}}\displaystyle\prod_{j = 1}^{\omega(N)}}{{\sigma({q_j}^{\beta_j})}}}{\displaystyle\prod_{j = 1}^{\omega(N)}{N}} = \displaystyle\frac{N{\displaystyle\prod_{j = 1}^{\omega(N)}{\sigma({q_j}^{\beta_j}})}}{\displaystyle\prod_{j = 1}^{\omega(N)}{N}} = \displaystyle\frac{\displaystyle\prod_{j = 1}^{\omega(N)}{\sigma({q_j}^{\beta_j}})}{\displaystyle\prod_{j = 1}^{\omega(N) - 1}{N}} = \displaystyle\frac{\sigma(N)}{N^{\omega(N) - 1}}$$

But:

$$\displaystyle\frac{\sigma(N)}{N^{\omega(N) - 1}} = \displaystyle\frac{2N}{N^{\omega(N) - 1}} = \displaystyle\frac{2}{N^{\omega(N) - 2}} = 2{N^{2 - \omega(N)}}$$

Since $7 \le \omega(N) - 2 = r$, it follows that $N^7 \le N^r$, which gives:

$$N^{-r} \le N^{-7}$$

Consequently:

$$\prod_{j = 1}^{\omega(N)}\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N} \le 2N^{-7} = \displaystyle\frac{2}{N^7}$$

Taking logarithms of both sides of the last inequality:

$$\displaystyle\sum_{j = 1}^{\omega(N)}\log(\displaystyle\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N}) \le \log(2) - 7\log(N)$$

This is as far as I could go, using only elementary notions that are familiar to me. I will let you guys know if I discover anything else in the coming days.

Post-Edit:

I missed it!

$$\displaystyle\sum_{j = 1}^{\omega(N)}\log(\displaystyle\frac{{{q_j}^{\beta_j}}{\sigma({q_j}^{\beta_j})}}{N}) = \displaystyle\sum_{j = 1}^{\omega(N)}{\left(\log({q_j}^{\beta_j}) + \log(\sigma({q_j}^{\beta_j})) - \log(N)\right)}$$ $$= \displaystyle\sum_{j = 1}^{\omega(N)}{\log({q_j}^{\beta_j})} + \displaystyle\sum_{j = 1}^{\omega(N)}{\log(\sigma({q_j}^{\beta_j}))} - \displaystyle\sum_{j = 1}^{\omega(N)}{\log(N)}$$ $$= \log(\displaystyle\prod_{j = 1}^{\omega(N)}{{q_j}^{\beta_j}}) + \log(\displaystyle\prod_{j = 1}^{\omega(N)}{\sigma({q_j}^{\beta_j})}) - \log(\displaystyle\prod_{j = 1}^{\omega(N)}{N})$$ $$= \log(N) + \log(\sigma(N)) - \log(N^{\omega(N)}) \le \log(2) - 7\log(N)$$

which implies that:

$$8\log(N) + \log(\sigma(N)) - \omega(N)\log(N) \le \log(2)$$

Finally, we have:

$$\log(\sigma_{1}(N)) \le \log(2) + \left(\omega(N) - 8\right)\log(N)$$

Equality holds if and only if $\omega(N) = 9$.

share|improve this answer
2  
This question, and this answer, seem dangerously close to using MO as a kind of public blog or attempted polymath.... –  Yemon Choi Mar 5 '11 at 7:22
    
@Yemon Choi, please "define" polymath. And can you explain why you think both my question (asked in Dec 3, and tagged with "reference- request") and answer (which, if I may mention, I painstakingly typed for two continuous hours) "seem dangerously close to using MO as a kind of public blog"? That being said, no offense meant from my side -- just clarifying your point of view, as I am unable to consider it as either constructive or the opposite. Do take note that I answered my own question. =) My original request for a "non-technical" introduction to analytic number theory stands... anyone? –  Jose Arnaldo Dris Mar 5 '11 at 8:11
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.