MathOverflow is a question and answer site for professional mathematicians. 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

I would like to know if I could do something like:

$\epsilon (0)\text{:=}0$
$\epsilon (n)\text{:=}\frac{4}{2 n-(-1)^n+1}$

and use it instead of a constant.

As $n\rightarrow \infty$, $\epsilon \rightarrow 0.$

Edit for clarification:

In Hardy and Wright, sixth edition, page 494, (22.19.2), it states, "...there is always a prime $p$ satisfying"

$x < p < (1 + \epsilon) x$

Using this function:

$x(n)\text{:=}\frac{1}{8} \left(2 n (n+2)-(-1)^n+1\right)$

I want to have:

$x(n) < p \leq (1+\epsilon(n)) x(n)$

Edit after the closure: The answer I was looking for is "YES." I have located several proofs that have $\epsilon$ dependent on $x,$ meaning it can vary under my control. Now, I can't use it that way because neither $x$ nor $\epsilon$ are dependent on each other, but both are dependent on $n.$ But, I hope to adapt.

I did not sign off on quid's answer because it was not exactly what I needed. However, I do like the answer very much.


share|cite|improve this question

closed as not a real question by GH from MO, Angelo, Vladimir Dotsenko, Martin Brandenburg, Qiaochu Yuan Jul 19 '12 at 15:08

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Use it for what? The question does not make any sense to me. – Angelo Jul 19 '12 at 10:35
up vote 4 down vote accepted

If this is true for all small $n$ (which I did not check) it is likely true for all $n$. However, to prove this seems presently infeasible.

Leaving the precise details of the constants asside the $x(n)$ is quadratic in $n$ and the $\varepsilon (n)$ being roughly $1/n$, this question amounts to (dropping constants) asking whether there is some prime between $an^2$ and $an^2 + b n$ for certain $a,b$ or to put it differently $x$ and $x + c x^{1/2}$.

Much stronger things are widely believed to be true asymptotically, but the problem for $x$ and $x + c x^{1/2}$ is open (even under GRH).

Various information can be found on the Wikipedia page on Prime gaps

So, asymptoically, this should be true and follows from standard conjectures; but a (unconditional) proof seems out of reach.

share|cite|improve this answer

I don't have H & W handy, but I assume what they say is something like, for every positive $\epsilon$, there is an $N$ depending on $\epsilon$, such that if $x\gt N$, then there is a prime $p$ with $x\lt p\lt(1+\epsilon)x$. So, for the particular $\epsilon$ you have chosen, there is some function $x(n)$ you can use, but it can't be any old function, and you'd certainly have some work to do to show that the $x(n)$ you have chosen in valid.

share|cite|improve this answer
Would be quite nice if OP (or anyone) managed to do this work! :) – user9072 Jul 19 '12 at 12:53

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