# A question about prime gaps

I have been reading the Wikipedia article about prime gaps (http://en.wikipedia.org/wiki/Prime_gap) and I came across the following:

Hoheisel was the first to show that there exists a constant $$\theta<1$$ such that $$\pi(x+x^\theta)-\pi(x)\sim\frac{x^\theta}{\log x}$$ as $$x$$ tends to infinity.

I have a few questions:

1. What's the meaning of the $$\sim$$ symbol?

2. Does the symbol $$\sim$$ mean that $$\pi(x+x^\theta)-\pi(x)$$ is asymptotic to $$x^\theta/\log x$$? If this is the case, does this mean that $$\pi(x+x^\theta)-\pi(x)$$ is always greater than $$x^\theta/\log x$$ at least for sufficiently large $$x$$? Is $$\pi(x+x^\theta)-\pi(x)$$ the same as $$\pi[x,x+x^\theta]$$, that is to say, the amount of primes in the interval $$[x,x+x^\theta]$$?

I'm not sure what the above expression means, and I'm not sure what it means when they say that a certain function is "asymptotic" to another function or when they say that a certain asymptotic formula holds.

1. Also, in the above expression, is $$\log$$ the natural logarithm or the logarithm in base $$10$$?
• $\sim$ means asymptotic equality, so $$\lim_{x\to\infty} (\pi(x+x^\theta)-\pi(x))\frac{\log x}{x^\theta} = 1.$$ And $\log$ is the natural logarithm. Mar 13 '14 at 23:35
• Thank you for your comment. But does this mean that $\pi(x+x^\theta)-\pi(x)$ is always greater than $x^\theta/\log x$? And is $\pi(x+x^\theta)-\pi(x)$ the same as $\pi[x,x+x^\theta]$? Mar 14 '14 at 1:28
• No, it need not always be greater. It cannot be too much smaller in the long run, however. If $\pi[x,x+x^\theta]$ means the number of primes between $x$ and $x+x^\theta$, then it means the same as $\pi(x+x^\theta)-\pi(x)$. Mar 14 '14 at 1:49
• I think I'm kind of confused and maybe I'm mixing up things. For example, if consider the function $f(x)=1/x$, then its asymptotes are the $x$-axis and the $y$-axis. If consider the positive values for $x$, then as $x$ increases $f(x)=1/x$ gets smaller and smaller but never reaches $0$ (the graph of the function never touches the $x$-axis). This is why I thought that if a function $f(x)$ was asymptotic to another function $g(x)$, then $f(x)$ was always greater than $g(x)$ (or the other way round). What can I do to correctly understand this specific topic? I'm confused. Mar 14 '14 at 2:19
• $f \sim g$ means $$\lim_{x\to\infty} \frac{f(x)}{g(x)} = 1.$$ That is something different from (but not entirely unrelated to) the asymptotes of functions like the axes for $1/x$. Mar 14 '14 at 2:23

$f(x)\sim g(x)$ just means that $\lim_{x\to\infty}f(x)/g(x)=1.$ Informally, $g(x)$ is a good approximation for $f(x)$ (and vice versa) when $x$ is large.
$\pi(x+x^\theta)-\pi(x)\sim x^\theta/\log x$ mean that $\pi(x+x^\theta)-\pi(x)$ is greater than $0.99x^\theta/\log x$ for sufficiently large $x$, and this would remain true (for a different value of "large enough") if you replace 0.99 with any number less than 1. ($\log$ is the natural logarithm, yes.)
$\pi(x+x^\theta)-\pi(x)$ is the number of primes in the interval $(x,x+x^\theta]$, yes.