Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is an exercise to show that

$$\frac{\pi((x+1)^2) - \pi(x^2)}{\pi(x- \pi (x)) } \sim 1 $$

assuming the unproven hypothesis: $\displaystyle \pi (x^2, x^2+x^{2( \theta)}) \sim \frac{x^{2( \theta)}}{\ln x^2} $ with $ \theta = \frac{1}{2}$.

With that assumption, if $\displaystyle \pi(x^2 , x^2+ x) \sim \frac{x}{\ln x^2}$ then $\displaystyle \pi (x^2, x^2+ 2x) \sim \frac{2x}{\ln x^2} = \frac{x}{\ln x}$

As an aside, with the hypothesis, the number of primes on a square interval beginning at $x^2$ of length 2x would be asymptotically equal to that on the interval $(0,x)$, a notion supported by Mathematica for the numbers in its reach.

The hypothesis $\theta = 1/2$ is stronger than current unconditional results.

So we want to show that $\displaystyle \pi(x - \pi(x)) \sim \frac{x}{\ln x}$. I see no reason not to use the PNT here... $\pi(x) \sim \frac{x}{\ln x}$. Applying this twice,

$$\frac{x}{\ln x } \sim \frac{ (x - \frac{x}{\ln x})} { \ln ( x - \frac{x}{\ln x})}$$

The exercise is maybe trivial at this point because the numerator of the r.h.s. of this expression might be interpreted as the number of composites on an interval as x gets large, which we know will be close to x for large x. Then the entire fraction looks like the number of primes $\leq$ x.

$\displaystyle \frac{ \ln x (x - \frac{x}{\ln x})} {x \ln ( x - \frac{x}{\ln x})} \sim 1$. But

$$ \begin{eqnarray*} \frac{ \ln x (x - \frac{x}{\ln x})} {x \ln ( x - \frac{x}{\ln x})} &=& \frac{x \ln x - \frac{x \ln x}{\ln x}} {x \ln ( x - \frac{x}{\ln x})} &=& \frac{x \ln x - x} {x \ln ( x - \frac{x}{\ln x})}\\ = \frac{x (\ln x - 1)} {x \ln ( x - \frac{x}{\ln x})} &=& \frac{\ln x - 1} { \ln ( x - \frac{x}{\ln x})} &=& \frac{\ln x - 1} { \ln (x(1 - \frac{1}{\ln x}))}\\ &=& \frac{\ln x - 1} { \ln x + \ln (1 - \frac{1}{\ln x})} &\sim& 1 \end{eqnarray*} $$

So the question is, after editing-- is the calculation correct as far as it goes?

Edit: The motivation for this was that I thought the iterated $\pi$ expression was a better approximation of card(primes on square interval) than $\frac{x}{\ln x}$. But both ratios $\displaystyle \frac{\pi((x+1)^2) - \pi(x^2)}{\pi(x- \pi (x)) } $ and $\displaystyle \frac{\pi((x+1)^2) - \pi(x^2)}{(\frac{\ln x }{x}) } $ are oscillatory and the result depends on the precise value of x chosen.

share|cite|improve this question
up vote 1 down vote accepted

It looks fine, although a little more complex than necessary. From $${\log x\left(x-{x\over\log x}\right)\over x\log\left(x-{x\over\log x}\right)}$$ you can immediately cancel a factor of $x$ top and bottom to get to $${\log x\left(1-{1\over\log x}\right)\over \log\left(x-{x\over\log x}\right)}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.