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Math Help - Prime-counting function

  1. #1
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    Prime-counting function

    Let \pi(x) be the prime-counting function and f(x) be a non-constant rational function. Can we have \pi(x)=f(x) for infinitely many values of x?
    Last edited by NonCommAlg; October 24th 2009 at 09:21 PM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    A 'rational function' \rho (x) can be written as...

     \rho(x)= a \cdot \frac{N(x)}{D(x)} (1)

    ... where N(x) and D(x) are polynomial in x of degree n and m so that we can write...

    \rho(x) \sim k\cdot x^{n-m} (2)

    ... or equivalently...

    \lim_{x \rightarrow \infty} \frac{\rho (x)}{k \cdot x^{n-m}} =1 (3)

    At the end of nineteenth century it has been demonstrated that is...

    \pi(x) \sim \frac{x}{\ln x} (4)

    ... or equivalently...

    \lim_{x \rightarrow \infty} \frac{\pi (x)}{\frac{x}{\ln x}} =1 (5)

    Combining (3) and (5) we find that is...

    \lim_{x \rightarrow \infty} \frac{\rho(x)}{\pi(x)}= \left\{\begin {array}{cc}\infty,&\mbox{if }n-m>0\\0,&\mbox{if }n-m \le 0\end{array}\right. (6)

    ... so that the situation described by NCA is impossible...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by chisigma View Post
    A 'rational function' \rho (x) can be written as...

     \rho(x)= a \cdot \frac{N(x)}{D(x)} (1) we don't need that a.

    ... where N(x) and D(x) are polynomial in x of degree n and m so that we can write...

    \rho(x) \sim k\cdot x^{n-m} (2)

    ... or equivalently...

    \lim_{x \rightarrow \infty} \frac{\rho (x)}{k \cdot x^{n-m}} =1 (3)

    At the end of nineteenth century it has been demonstrated that is...

    \pi(x) \sim \frac{x}{\ln x} (4)

    ... or equivalently...

    \lim_{x \rightarrow \infty} \frac{\pi (x)}{\frac{x}{\ln x}} =1 (5)

    Combining (3) and (5) we find that is...

    \lim_{x \rightarrow \infty} \frac{\rho(x)}{\pi(x)}= \left\{\begin {array}{cc}\infty,&\mbox{if }n-m>0\\0,&\mbox{if }n-m \le 0\end{array}\right. (6)

    ... so that the situation described by NCA is impossible...

    Kind regards

    \chi \sigma
    nice! before using the prime number theorem we need to explain why if \pi(x)=\rho(x) has infinitely many solutions, then the set of solutions must be unbounded.
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