7
$\begingroup$

How do we show that For $x,y \ge 0$ real numbers, there exists a constant C suchthat: $\pi(x+y)-\pi(y) \le \frac{1}{3}x+C$ Where $\pi(.)$ denotes thes prime counting function, is true?

the hint is to sieve n with $y< n \le x+y $:

$\pi (x+y) \le 1+ \sum _{n \le x+y} 1+1-1 - \sum_{2|n}1 - \sum_{3|n}1 + \sum_{6|n}1 + \sum_{n\le x+y} 1 = $ $1+ \sum _{n \le x+y} 1+1-1 - \sum_{2|n}1 - \sum_{3|n}1 + \sum_{6|n}1 + [x+y]$

because: $1\le n = dm \le x+y \Leftrightarrow \frac{1}{d}\le m \le \frac{x+y}{y} $

so: $\sum_{n\le x+y , d|n}1 = [\frac{x+y}{d}]$

then that gives: $\pi (x+y) < 1+ [x+y] - [\frac{x+y}{2}] - [\frac{x+y}{3}] + [\frac{x+y}{6}]$

so that will give: $\pi (x+y) < \frac{x+y}{3} + 3$ but also we get : $\pi(y) < \frac{y}{3} + 3$ so for any constant $C\ge 0$ it will surely hold that:

$\pi(x+y) - \pi(y) < \frac{x}{3} \le \frac{x}{3} + C$

Is this correct?

  • 1
    Thank you. The prime number theorem wasnt proven yet, so the only thing that comes in question is the sieve of eratosthenes (as TMM suggests, sieving n numbers in the interval of y but it looks like I did it wrong (I believe).2012-03-15

2 Answers 2

3

Hint: Which numbers modulo $6$ can be prime? (Certainly not those divisible by $2$ or $3\ldots$)

So on an interval of width $x$ from $y$ to $y + x$, how many primes ($\pi(y+x) - \pi(y)$) do we expect at most on this interval?

  • 0
    I put my comment to your answer in the edit of my question, thanks.2012-03-15
2

You are asking us to show that your inequality holds for any $x, y \ge 0$ and for any constant $C$, which is obviously false. Perhaps this is what you mean:

Show that there exists a constant $C$ such that for any real numbers $x, y \ge 0$, $\pi(x+y)-\pi(y) \le \frac{1}{3}x+C$.

  • 0
    @VVV: Still not right, I'm afraid: now your statement is vacuously *true*. I suppose that's an improvement!2012-03-16