Timeline for Why could Mertens not prove the prime number theorem?
Current License: CC BY-SA 3.0
6 events
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| May 6, 2012 at 7:04 | comment | added | Nilotpal Kanti Sinha | @ David, Thank you for the explanation. Now the gap is clear to me. I have accepted your answer as the answer to this question. | |
| May 4, 2012 at 17:07 | comment | added | David E Speyer | So, $|R'(x)|$ is of the same order as $d (\log \log x)/dx = 1/(x \log x)$, even though $R(x) << \log \log x$. When you compute $N(t) = \int t d(\log \log t)/dt \cdot dt + \int t R'(t) dt$, the two terms are of the same order of magnitude, and you have no control over $N(t)$. | |
| May 4, 2012 at 16:57 | comment | added | David E Speyer | The problem is that $R$ small does not force $R'$ small. In the scenario of my answer, $R(x) = O(1/\log x)$. Between $9 \times 10^k$ and $10 \times 10^k$, the function $R$ changes by $\log \log (10 \times 10^k) - \log \log (9 \times 10^k) \approx c/\log 10^{k+1}$, where $c$ is a constant I don't care to work out. So $R' \approx c/(10^k \log 10^{k+1})$ or $R'(x) \approx c/(x \log x)$. For $10^k < x < 9 \times 10^k$, you get $R'(x) \approx d/(x \log x)$, for a different constant $d$. (continued) | |
| May 4, 2012 at 16:50 | comment | added | David E Speyer | First a minor note: $R$ and $N$ aren't actually differentiable functions; they have jump discontinuities at integers. But you can fix that by replacing $R'(x)$ by $(R(x+h)-R(x))/h$ where $h$ is much smaller than $x$ but much larger than $1$. I'll refer to $R'$ below, but what I really mean is "the size of $(R(x+h)-R(x))/h$" for $h$ in the right range. | |
| May 4, 2012 at 11:39 | history | edited | Nilotpal Kanti Sinha | CC BY-SA 3.0 |
added 2 characters in body; added 2 characters in body
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| May 4, 2012 at 6:27 | history | answered | Nilotpal Kanti Sinha | CC BY-SA 3.0 |