# Conditional equivalence of expression to cardinality of primes on square intervals

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.

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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 $${\log x\left(1-{1\over\log x}\right)\over \log\left(x-{x\over\log x}\right)}$$