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Estimates of the Bounds of ?(x) and ?((x + 1)2) ? ?(x2)

Author: Connor Paul Wilson (University of Michigan)

  • Estimates of the Bounds of ?(x) and ?((x + 1)2) ? ?(x2)

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    Estimates of the Bounds of ?(x) and ?((x + 1)2) ? ?(x2)

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Wilson, C., (2021) “Estimates of the Bounds of ?(x) and ?((x + 1)2) ? ?(x2)”, University of Michigan Undergraduate Research Journal 15. doi: https://doi.org/10.3998/umurj.1334

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We show the following bounds on the prime counting function π(x) using principles from analytic number theory, giving an estimate

log2liminfxπ(x)x/logxlimsupxπ(x)x/logx2log2

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

1. Introduction

Recall the definition:

π(x):=pxpprime1,

and let us define the following:

Θ(x):=pxpprimelogp,

Ψ(x):=1nxΛ(n)=pmxm1pprimelogp,

Λ(n)={logpifn=pkforsomeprimepandintegerk10otherwise.

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, ∀nN. ∀ϵ > 0 ^ N ≥ 1, ∃nN with xn > Lϵ. We thus define L as:

limsupnxn=L

Thus on the contrary we must have:

liminfnxn=limsupnxn

2. 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

limsupnΨ(x)xlimsupnΘ(x)x

Also, if p is a prime and pmx < pm+1, then log p occurs in the sum for Ψ(x) exactly m times. [1]

Ψ(x)=pmxpprimem1logp=pxpprime[logxlogp]logppxpprimelogx=π(x)logx

limsupxΨ(x)xlimsupxπ(x)x/logx

Now fix α ∈ (0, 1). Given x > 1,

Θ(x)=pxpprimelogpxa<pxpprimelogp.

It is clear that all p from the second sum satisfy: log p > α log x.

Θ(x)>αlogxxα<px1=αlogx(π(x)π(xα))>αlogx(π(x)xα)

϶

Θ(x)x>απ(x)x/logxαlogxx1α

α ∈ (0, 1) we have:

limxαlogxx1α=0.

Combining these we get:

limsupxΘ(x)xlimsupxαπ(x)x/logx

Once again, since our statement is true ∀α ∈ (0, 1),

limsupxΘ(x)xlimsupxπ(x)x/logx

Similarly:

liminfxΨ(x)xliminfxΘ(x)x

liminfxπ(x)x/logxliminfxΨ(x)x

Once again, we apply the same method:

liminfxΘ(x)xliminfxπ(x)x/logx,

and have thus proven Theorem 2.1. □

3. Main Result

Theorem 3.1. We have:

log2liminfxπ(x)x/logxlimsupxπ(x)x/logx2log2.

Proof. First the lower bound. Take:

S(x):=1nxlogn21nx/2logn.

Then,

1dnd|nΛ(d)=pr|dd|npprimelogp=i=1lr=1eilogpiwheren=p1e1plel=i=1leilogpi=logn

S(x)=1nxd|nΛ(d)21nx/2d|nΛ(d)

Clearly {d, 2d, …, qd} is the set of n satisfying 1 ≤ n ≤ x and d | n (we can see this easily by writing x = r + qd with 0 ≤ r < d).

S(x)=1dxΛ(d)[xd]21dx/2Λ(d)[x2d]=1dx/2Λ(d)([xd]2[x2d])+(x/2)<dxΛ(d)[xd]1dx/2Λ(d)+(x/2)<dxΛ(d)=Ψ(x).

So,

Ψ(x)xS(x)x=1x1nxlogn2x1nx/2logn.

log t is increasing,

1x+1logtdt1nxlogn,

1[x]logtdt1nxlogn.

Actually, assumingx+,

S(x)x1x1xlogtdt2x1(x/2)+1logtdt=1x(xlogxx+1)2x(x+22log(x+22)x+22+1)=logx+1xx+2xlog(x+2)+x+2xlog2>log(xx+2)2xlog(x+2)+log2

Using Theorem 2.1, we get:

liminfxπ(x)x/logx=liminfxΨ(x)xliminfxS(x)x>limxlog(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!)=r1[mpr],m+,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.

ordp(m!)=n=1mordp(n)=r11nmordp(n)=rr=r1r([m/pr][m/pr+1])=r1r[m/pr]r1r[m/pr+1]=r1r[m/pr]r1(r1)[m/pr]=r1[mpr]

Lemma 3.3.n+,

22n2n<(2n2)<22n2n+1

Proof.

Pn:=in(2i1)(2i)=(2n)!22n(n!)2=(2n2)122n

Since:

(2i1)(2i+1)(2i)2<1

for all i ≥ 1.

1>(2n+1)Pn2,

giving the upper bound. For the lower bound:

11(2i1)2<1

i ≥ 1, such that

1>i=2n(11(2i1)2)=i=2n(2i1)21(2i1)2=i=2n(2i2)(2i)(2i1)2=14nPn2

yielding our lemma. □

Lemma 3.4.n+,

Θ(n)<2nlog2

Proof. By Lemma 3.3,

log((2nn)12)=log((2nn))log2<2nlog212log(2n+1)log2=(2n1)log212log(2n+1)

since

(2nn)12=(2n)!(n!)2n2n=(2n1)!n!(n1)!=(2n1n1).

by Lemma 3.2:

log((2nn)12)=log(2n1n1)=pprimeordp((2n1)!)logppprimeordp((n1)!)logppprimeordp(n!)logp=pprimelogpr1[(2n1)/pr][n/pr]pprimen<p2n1logp=Θ(2n1)Θ(n)

Where

Θ(2n1)Θ(n)<(2n1)log212log(2n+1)

We now proceed by induction. Proceeding from the trivialities, suppose m > 2 and the lemma is true for n < m,n,m. If m is odd, then m = 2n – 1 for some integer n ≥ 2 since m > 2. Thus by induction,

Θ(m)=Θ(2n1)<Θ(n)+(2n1)log212log(2n+1)<2nlog2+(2n1)log212log(2n)=(4n1)log212log(2n)(4n2)log2(sincen2)=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(m1)<2(m1)log2<2mlog2

Lemma 3.4 gives

2log2limsupxΘ(x)x.

The desired lower bound follows from Theorem 2.1

4. 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)log2xloglogxπ((x+1)2)π(x2)12(x2logn32log3)j=3x1log2xloglogk<π(x2)π(32)

And thus:

j=3x112((j+1)2log(j+1)j2logj)log2jloglogj<j=3n1π((j+1)2)π(j2)

12((x+1)2log(x+1)x2logx)(logx)2log(logx)π((x+1)2)π(x2).

And by the prime number theorem, which gives us the asymptotic estimate for some

F(x):=π((x+1)2)π(x2)~12((x+1)2log(x+1)x2logx)

We propose:

π((x+1)2)π(x2)12((x+1)2log(x+1)x2logx)+log2xloglogx

by the same method.

References

Ram Murty, R., Problems in Analytic Number Theory, Second Edition — Graduate Texts in Mathematics, Springer, 2001.

Hassani, M., Counting primes in the interval (n2, (n + 1)2), AMS, 1997.