Can $\epsilon$ be a generating function? - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-25T08:48:08Z http://mathoverflow.net/feeds/question/102632 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102632/can-epsilon-be-a-generating-function Can $\epsilon$ be a generating function? Fred Kline 2012-07-19T09:49:51Z 2012-07-20T07:20:40Z <p>I would like to know if I could do something like:</p> <p>$\epsilon (0)\text{:=}0$<br> $\epsilon (n)\text{:=}\frac{4}{2 n-(-1)^n+1}$</p> <p>and use it instead of a constant.</p> <p>As $n\rightarrow \infty$, $\epsilon \rightarrow 0.$</p> <p>Edit for clarification:</p> <p>In Hardy and Wright, sixth edition, page 494, (22.19.2), it states, "...there is always a prime $p$ satisfying" </p> <p>$x &lt; p &lt; (1 + \epsilon) x$</p> <p>Using this function: </p> <p>$x(0)\text{:=}0$<br> $x(n)\text{:=}\frac{1}{8} \left(2 n (n+2)-(-1)^n+1\right)$ </p> <p>I want to have: </p> <p>$x(n) &lt; p \leq (1+\epsilon(n)) x(n)$</p> <p>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.</p> <p>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.</p> <p>Thanks.</p> http://mathoverflow.net/questions/102632/can-epsilon-be-a-generating-function/102655#102655 Answer by Gerry Myerson for Can $\epsilon$ be a generating function? Gerry Myerson 2012-07-19T12:21:22Z 2012-07-19T12:21:22Z <p>I don't have H &amp; 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. </p> http://mathoverflow.net/questions/102632/can-epsilon-be-a-generating-function/102659#102659 Answer by quid for Can $\epsilon$ be a generating function? quid 2012-07-19T12:48:04Z 2012-07-19T12:48:04Z <p>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. </p> <p>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 $x + c x^{1/2}$. </p> <p>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). </p> <p>Various information can be found on the Wikipedia page on <a href="http://en.wikipedia.org/wiki/Prime_gap" rel="nofollow">Prime gaps</a> </p> <p>So, asymptoically, this should be true and follows from standard conjectures; but a (unconditional) proof seems out of reach. </p>