Prime-counting function

**NonCommAlg** · October 24th 2009, 05:17 PM

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 $x$ ?

**chisigma** · October 24th 2009, 10:34 PM

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

$\chi$ $\sigma$

**NonCommAlg** · October 25th 2009, 12:10 AM

Originally Posted by chisigma

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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