umurjUniversity of Michigan Undergraduate Research Journal2640-8988133410-Estimates of the Bounds of π(x) and π((x + 1)2) − π(x2)10.3998/umurj.1334Estimates of the Bounds of π(x) and π((x + 1)2) − π(x2)Connor Paul WilsonEstimates of the Bounds of π(x) and π((x + 1)2) − π(x2)WilsonConnor Pauldpoae@umich.eduUniversity of MichiganContact: Connor Paul Wilson <dpoae@umich.edu>3108202115CC BY-NC-ND 4.0
We show the following bounds on the prime counting function π(x) using principles from analytic number theory, giving an estimate
log2≤liminfx→∞π(x)x/logx≤limsupx→∞π(x)x/logx≤2log2
We also conjecture about the bounding of π((x + 1)2) − π(x2), as is relevant to Legendre’s conjecture about the number of primes in the aforementioned interval such that:
⌊12((x+1)2log(x+1)−x2logx)−(logx)2log(logx)⌋≤π((x+1)2)−π(x2)≤⌊12((x+1)2log(x+1)−x2logx)−log2xloglogx⌋
Introduction
Recall the definition:
π(x):=∑p≤xpprime1,
and let us define the following:
Θ(x):=∑p≤xpprimelogp,Ψ(x):=∑1≤n≤xΛ(n)=∑pm≤xm≥1pprimelogp,Λ(n)={logpifn=pkforsomeprimepandintegerk≥10otherwise.
The following are simple statements from real analysis that are required for rigorousness’ sake: let {xn} be a sequence of real numbers and L be a real number with the following two properties: ∀ϵ > 0, ∃N such that xn < L + ϵ, ∀n ≥ N. ∀ϵ > 0 ^ N ≥ 1, ∃n ≥ N with xn > L – ϵ. We thus define L as:
limsupn→∞xn=L
Thus on the contrary we must have:
liminfn→∞xn=−limsupn→∞−xn
Necessary Preliminary Results
Theorem 2.1.For all α ∈ (0, 1), and all x ≥ x0:
Θ(x)log(x)≤Ψ(x)log(x)≤π(x)≤Θ(x)αlog(x)+xα
Proof. Clearly Θ(x) ≤ Ψ(x), such that
limsupn→∞Ψ(x)x≥limsupn→∞Θ(x)x
Also, if p is a prime and pm ≤ x < pm+1, then log p occurs in the sum for Ψ(x) exactly m times. [1]
Ψ(x)=∑pm≤xpprimem≥1logp=∑p≤xpprime[logxlogp]logp≤∑p≤xpprimelogx=π(x)logxlimsupx→∞Ψ(x)x≤limsupx→∞π(x)x/logx
Now fix α ∈ (0, 1). Given x > 1,
Θ(x)=∑p≤xpprimelogp≥∑xa<p≤xpprimelogp.
It is clear that all p from the second sum satisfy: log p > α log x.
Using Theorem 2.1, we get:
liminfx→∞π(x)x/logx=liminfx→∞Ψ(x)x≥liminfx→∞S(x)x>limx→∞log(x/(x+2))−2xlog(x+2)+log2=log2
To complete the proof, we will need some auxiliary results taken from Murty’s Analytic Number Theory [1] in the form of three lemmas:
Lemma 3.2.
ordp(m!)=∑r≥1[mpr],∀m∈ℤ+,primep
Proof. Fix an exponent r. The positive integers no larger than m that are multiples of pr are
pr,2pr,…,[mpr]pr
and those that are multiples of pr+1 are
pr+1,2pr+1,…,[mpr+1]pr+1
Thus there are precisely [m/p]r−[m/p]r+1 positive integers n ≤ m with ordp(n) = r.
∴
1>(2n+1)Pn2,
giving the upper bound. For the lower bound:
1−1(2i−1)2<1
∀i ≥ 1, such that
1>∏i=2n(1−1(2i−1)2)=∏i=2n(2i−1)2−1(2i−1)2=∏i=2n(2i−2)(2i)(2i−1)2=14nPn2
yielding our lemma. □
Lemma 3.4.∀n∈ℤ+,
Θ(n)<2nlog2
Proof. By Lemma 3.3,
log((2nn)12)=log((2nn))−log2<2nlog2−12log(2n+1)−log2=(2n−1)log2−12log(2n+1)
since
(2nn)12=(2n)!(n!)2n2n=(2n−1)!n!(n−1)!=(2n−1n−1).
by Lemma 3.2:
log((2nn)12)=log(2n−1n−1)=∑pprimeordp((2n−1)!)logp−∑pprimeordp((n−1)!)logp−∑pprimeordp(n!)logp=∑pprimelogp∑r≥1[(2n−1)/pr]−[n/pr]≥∑pprimen<p≤2n−1logp=Θ(2n−1)−Θ(n)
Where
Θ(2n−1)−Θ(n)<(2n−1)log2−12log(2n+1)
We now proceed by induction. Proceeding from the trivialities, suppose m > 2 and the lemma is true for n < m,n,m∈ℕ. If m is odd, then m = 2n – 1 for some integer n ≥ 2 since m > 2. Thus by induction,
Θ(m)=Θ(2n−1)<Θ(n)+(2n−1)log2−12log(2n+1)<2nlog2+(2n−1)log2−12log(2n)=(4n−1)log2−12log(2n)≤(4n−2)log2(sincen≥2)=2mlog2
If m is even, then m = 2n for some integer n with m > n ≥ 2 and m is composite. Clearly Θ(m) = Θ(m − 1) and we know:
Θ(m)=Θ(m−1)<2(m−1)log2<2mlog2
Lemma 3.4 gives
2log2≥limsupx→∞Θ(x)x.□
The desired lower bound follows from Theorem 2.1 □
On Primes in the Gaps between Squares
The following is relatively aleatory compared to the previous workings, but it is worth mentioning considering the importance of the statement.
By Hassani [2], we have
⌊12((x+1)2log(x+1)−x2logx)−log2xloglogx⌋≤π((x+1)2)−π(x2)12(x2logn−32log3)−∑j=3x−1log2xloglogk<π(x2)−π(32)
And thus:
∑j=3x−1⌊12((j+1)2log(j+1)−j2logj)−log2jloglogj⌋<∑j=3n−1π((j+1)2)−π(j2)