- **Math**
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- - **Chebyshev's Estimates**
(*https://www.mersenneforum.org/showthread.php?t=11380*)

Chebyshev's EstimatesI'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 [tex]\pi(x)[/tex].
For reference, it's Theorem 3 on page 11. The desired upper bound is [tex] \pi(x) \leq \{ \log 4 + \frac{8 \log \log n}{\log n} \} \frac{n}{\log n}[/tex] Using the bound [tex] \prod_{p \leq n} p \leq 4^n [/tex] it's easy to show that for [tex] 1 < t \leq n [/tex] we have [tex] \pi(n) \leq \frac{n \log 4}{\log t} + \pi(t) [/tex] Tenenbaum then gives the bound [tex] \pi(n) \leq \frac{n \log 4}{\log t} + t [/tex] So far, so good. At this point in the proof, Tenenbaum says "The stated result follows by choosing [tex] t = n / (\log n)^2[/tex]" 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. |

[QUOTE=brownkenny;159770]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 [tex]\pi(x)[/tex].
For reference, it's Theorem 3 on page 11. The desired upper bound is [tex] \pi(x) \leq \{ \log 4 + \frac{8 \log \log n}{\log n} \} \frac{n}{\log n}[/tex] Using the bound [tex] \prod_{p \leq n} p \leq 4^n [/tex] it's easy to show that for [tex] 1 < t \leq n [/tex] we have [tex] \pi(n) \leq \frac{n \log 4}{\log t} + \pi(t) [/tex] Tenenbaum then gives the bound [tex] \pi(n) \leq \frac{n \log 4}{\log t} + t [/tex] So far, so good. At this point in the proof, Tenenbaum says "The stated result follows by choosing [tex] t = n / (\log n)^2[/tex]" 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.[/QUOTE] After the substitution factor out n/log(n) and then use partial fractions....? This shouldn't be too bad. |

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 [tex] \frac{8 \log \log n}{\log n} [/tex] comes from. Thanks again, Dr. Silverman. |

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