Let $\displaystyle \pi(x)$ be the prime-counting function and $\displaystyle f(x)$ be a non-constantrationalfunction. Can we have $\displaystyle \pi(x)=f(x)$ for infinitely many values of $\displaystyle x$?

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- Oct 24th 2009, 05:17 PMNonCommAlgPrime-counting function
Let $\displaystyle \pi(x)$ be the prime-counting function and $\displaystyle f(x)$ be a non-constant

*rational*function. Can we have $\displaystyle \pi(x)=f(x)$ for infinitely many values of $\displaystyle x$? - Oct 24th 2009, 10:34 PMchisigma
A 'rational function' $\displaystyle \rho (x)$ can be written as...

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

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

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

... or equivalently...

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

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

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

... or equivalently...

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Oct 25th 2009, 12:10 AMNonCommAlg