mersenneforum.org > Math Chebyshev's Estimates
 Register FAQ Search Today's Posts Mark Forums Read

 2009-01-22, 00:06 #1 brownkenny     Jan 2009 2×3 Posts Chebyshev's Estimates I've been working through Tenenbaum's book "Introduction to Analytic and Probabilistic Number Theory" and I'm stuck on the proof of an upper bound for $\pi(x)$. For reference, it's Theorem 3 on page 11. The desired upper bound is $\pi(x) \leq \{ \log 4 + \frac{8 \log \log n}{\log n} \} \frac{n}{\log n}$ Using the bound $\prod_{p \leq n} p \leq 4^n$ it's easy to show that for $1 < t \leq n$ we have $\pi(n) \leq \frac{n \log 4}{\log t} + \pi(t)$ Tenenbaum then gives the bound $\pi(n) \leq \frac{n \log 4}{\log t} + t$ So far, so good. At this point in the proof, Tenenbaum says "The stated result follows by choosing $t = n / (\log n)^2$" and leaves the details to the reader. As much as I've looked at it, I still can't figure out how he arrives at the desired result. Any tips/suggestions? Thanks in advance.
2009-01-22, 12:48   #2
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

22·1,877 Posts

Quote:
 Originally Posted by brownkenny I've been working through Tenenbaum's book "Introduction to Analytic and Probabilistic Number Theory" and I'm stuck on the proof of an upper bound for $\pi(x)$. For reference, it's Theorem 3 on page 11. The desired upper bound is $\pi(x) \leq \{ \log 4 + \frac{8 \log \log n}{\log n} \} \frac{n}{\log n}$ Using the bound $\prod_{p \leq n} p \leq 4^n$ it's easy to show that for $1 < t \leq n$ we have $\pi(n) \leq \frac{n \log 4}{\log t} + \pi(t)$ Tenenbaum then gives the bound $\pi(n) \leq \frac{n \log 4}{\log t} + t$ So far, so good. At this point in the proof, Tenenbaum says "The stated result follows by choosing $t = n / (\log n)^2$" and leaves the details to the reader. As much as I've looked at it, I still can't figure out how he arrives at the desired result. Any tips/suggestions? Thanks in advance.
After the substitution factor out n/log(n) and then use partial fractions....?

 2009-01-22, 17:21 #3 brownkenny     Jan 2009 2×3 Posts Thanks for the suggestion, Dr. Silverman. When I made the substitution I wound up with a log(log(n)) term in the denominator that I'm not sure how to get rid of. I tried estimating some more, but I still can't figure out where the factor of 8 in the term $ \frac{8 \log \log n}{\log n}$ comes from. Thanks again, Dr. Silverman.

 Similar Threads Thread Thread Starter Forum Replies Last Post Dr Sardonicus Number Theory Discussion Group 6 2022-01-15 12:13 danaj Computer Science & Computational Number Theory 9 2018-03-31 14:59 henryzz GMP-ECM 8 2009-12-31 17:51 10metreh Factoring 48 2009-04-08 01:54 henryzz Msieve 27 2009-01-21 18:37

All times are UTC. The time now is 22:32.

Sat Feb 4 22:32:03 UTC 2023 up 170 days, 20 hrs, 1 user, load averages: 0.92, 0.91, 0.94