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 brownkenny 2009-01-22 00:06

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.

 R.D. Silverman 2009-01-22 12:48

[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 $$\pi(x)$$.

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.[/QUOTE]

After the substitution factor out n/log(n) and then use partial fractions....?

 brownkenny 2009-01-22 17:21

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.

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