# Expressing $\pi(x) = \frac{P(x)}{Q(x)}$, where $P$ and $Q$ are polynomials

In Tom Apostol's Analytic Number Theory book there is a problem which states:

• That there do not exists polynomials $P(x)$ and $Q(x)$ such that $$\pi(x) = \frac{P(x)}{Q(x)}$$ for all $x \in \mathbb{N}$.

I tried this problem, but couldn't find a solution. Here is what i attempted. Since $\pi(x) \sim x\log{x}$ as $x \to \infty$, i saw what happens to the Right hand side $\frac{P(x)}{Q(x)}$ as $x \to \infty$. I couldn't conclude anything. If there are interesting proof's for this result, i shall be happy to see it.

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Note that in fact $\pi(x) \sim \frac{x}{\log x}$ (in particular it is less than $x$ for sufficiently large $x$, not greater than $x$!) Both of the answers so far seem to have taken the OP literally at his word, so seem to require minor modifications. – Pete L. Clark Jan 12 '11 at 16:58

It seems like you've already gotten the answer. By asymptotic considerations, we must have $\deg P(x) > \deg Q(x)$. Moreover, if we have $\deg P(x) = \deg Q(x) + 1$, then we have $$\frac{\pi(x) Q(x)}{P(x)} \sim \log x \frac{xQ(x)}{P(x)},$$ but the $xQ(x)/P(x)$ is asymptotic to the ratio of the leading coefficients, so this can't be asymptotic to $1$, i.e. $P(x)/Q(x)$ grows too slowly. On the other hand, if $\deg P(x) = \deg Q(x)+2$, then we similarly see that $P(x)/Q(x)$ grows too quickly to be asymptotic to $x \log x$.
HINT $\$ A rational function is asymptotic to $\rm\ x^n,\$ for some $\rm\ n \in \mathbb Z\:.\$ But $\rm\ x \log(x)\$ lies strictly between $\rm\ x\$ and $\rm x^2$ asymptotically, i.e. $\rm\ (x\ log(x))/x = log(x)\to\infty,\ \ x^2/(x log(x)) = x/log(x)\to \infty$
How does what you say here, that $\rm\ (x\ log(x))/x = log(x)\to\infty,\ \ x^2/(x log(x)) = x/log(x)\to \infty$ imply that $x \log x$ is not rational? – Pablo Aug 25 '14 at 17:33
@Henry Asymptotically, $\rm\,x\log(x)\,$ is strictly stronger than $\,x\,$ but strictly weaker than $\,x^2.\,$ No rational function has this property, since it is asymptotic to $\,x^n\,$ for some integer $\,n.\$ – Bill Dubuque Aug 25 '14 at 17:54
would it change anything if $\pi(x) \sim \frac{x}{\log x}$? – Pablo Aug 25 '14 at 18:37
@Henry Similarly, that's strictly between $\,x^0\,$ and $\,x^1.\$ Note I'm using a rougher equivalence where $\,x^n\,$ and $\,c x^n\ (0\ne c\in\Bbb R)\,$ are considered equivalent. – Bill Dubuque Aug 25 '14 at 18:49