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Math Help - A question about proof of prime number theorem

  1. #1
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    A question about proof of prime number theorem

    The following is the basic steps in one proof of the prime number theorem (many steps are left out)

    \pi(x)\sim \frac{x}{\ln(x)}\quad\text{iff}\quad \psi(x)\sim x

    Now,

    \psi_0(x)=-\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\zeta'(s)}{\zeta(s)}\fra  c{x^s}{s}ds

    where:
    \psi_0(x)=\left\{\begin{array}{ccc} \psi(x) & \text{for}& x\ne p^m \\<br />
                                     \psi(x)-1/2\ln(p) & \text{for}& x=p^m<br />
                                     \end{array}\right.<br />

    That is, \psi_0(x) differs from \psi(x) only when x is a prime power, the difference being 1/2\ln(p). Now, via residue integration:

    \psi_0(x)=x-\sum_{\rho}\frac{x^{\rho}}{\rho}-\ln(2\pi)-1/2\ln\left(1-1/x^2\right)

    where the sum is over all the non-trivial zeros of the zeta function. Dividing through by x and letting x tend to infinity:

    \frac{\psi(x)}{x}\to 1-\lim_{x\to\infty}\frac{1}{x}\sum_{\rho}\frac{x^{\r  ho}}{\rho}

    I think it can be shown that: \sum_{\rho}\frac{x^{\rho}}{\rho}=\textbf{O}(\sqrt{  x}) and therefore:

    \frac{\psi(x)}{x}\sim 1 and thus \pi(x)\sim\frac{x}{\ln(x)}

    I'm not sure about the order of the sum. Can someone confirm this or explain further how this sum is bounded?
    Last edited by shawsend; September 11th 2008 at 10:20 AM. Reason: corrected formula, explained psi_0 and psi
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  2. #2
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    Hello,

    I am not an expert, this is what I found in books. (Mainly "The theory of the Riemann zeta-function" by S.J.Patterson).
    The explicit formula for psi_0(x) is due to von Mangoldt.

    If we let S(x, T)=\Sigma_\rho \frac{X^{\rho}}{\rho} where \rho runs the zeros of the zeta function with |Im(\rho)|<T, then |x^(rho)|<=x, 1/rho=O(1/T), there are O(log T) such zeros. Thus, S(x, T)=O((x log T)/T).

    I don't know where you got \lim_{T\to\infty}S(x, T)=O(\sqrt{x}).

    Bye.
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  3. #3
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    Ok. I was wrong (I thought it might be \sqrt{x}). Thanks a bunch.

    I'll try to find that book. I got a question about the number of zeros: I thought the number of roots between 0 and T is approximately:

    \frac{T}{2\pi}\ln\left(\frac{T}{2\pi}\right)-\frac{T}{2\pi}. Can someone explain to me how that's \textbf{O}(\ln(T))?
    Last edited by shawsend; September 12th 2008 at 11:54 AM. Reason: added question about number of zeros
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  4. #4
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    Hello,

    Quote Originally Posted by shawsend View Post
    I got a question about the number of zeros: I thought the number of roots between 0 and T is approximately:

    \frac{T}{2\pi}\ln\left(\frac{T}{2\pi}\right)-\frac{T}{2\pi}. Can someone explain to me how that's \textbf{O}(\ln(T))?
    Sorry, I was wrong. Forget my first post. I hope someone wiser might help.

    Bye.
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  5. #5
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    Hey guys, Wikipedia under Chebyshev function gives:

    \sum_{\rho}\frac{x^{\rho}}{\rho}=\textbf{O}(\sqrt{  x}\ln^2 x)

    when this is substituted into the expression for \frac{\psi(x)}{x}, I get:

    \lim_{x\to\infty}\frac{\textbf{O}(\sqrt{x}\ln^2 x)}{x}\to 0

    which is what one expects.

    Would be interesting to show how this order is determined. I'll try.
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  6. #6
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    Hello,

    Quote Originally Posted by shawsend View Post
    Hey guys, Wikipedia under Chebyshev function gives:
    \sum_{\rho}\frac{x^{\rho}}{\rho}=\textbf{O}(\sqrt{  x}\ln^2 x)
    The Wikipedia says that you can prove this estimate "if the Riemann Hypothesis is TRUE." (In fact, the estimate is equivalent to RH.) Prove it, and you get a prize!

    Bye.
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  7. #7
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    Hey Wisterville. I think the two are separate and Wikipedia is alluding to the fact the sum would not be of this order if other zeros outside the critical line were included.

    I believe the sum can be considered completely independently of the Riemann Hypothesis like this: What is the order of the sum \sum_{\rho}\frac{x^{\rho}}{\rho} assuming \rho=1/2+it and the density of the set \{\rho_n\} in the range (0,T) is of order \frac{T}{2\pi}\ln\frac{T}{2\pi}-\frac{T}{2\pi}.

    Note: The sum is taken symmetrically over the zeros:

    \sum_{\rho}\frac{x^{\rho}}{\rho}=\lim_{T\to\infty}  \sum_{|t|\leq T}\frac{x^{\rho}}{\rho};\quad \rho=1/2+it
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