In this paper we find some lower and upper bounds of the form Hn n−c for the func- tionπ(n), in whichHn =Pn k=1 1 k

http://jipam.vu.edu.au/

Volume 7, Issue 1, Article 7, 2006

APPROXIMATION OF π(x) BYΨ(x)

MEHDI HASSANI INSTITUTE FORADVANCED

STUDIES INBASICSCIENCES

P.O. BOX45195-1159 ZANJAN, IRAN. mmhassany@srttu.edu

Received 07 March, 2005; accepted 25 August, 2005 Communicated by J. Sándor

Dedicated to Professor J. Rooin on the occasion of his 50th birthday

ABSTRACT. In this paper we find some lower and upper bounds of the form _Hⁿ

n−c for the func- tionπ(n), in whichH_n =Pn

k=1 1

k. Then, we considerH(x) = Ψ(x+ 1) +γas generalization ofHn, such thatΨ(x) = _dx^d log Γ(x)andγ is Euler constant; this extension has been intro- duced for the first time by J. Sándor and it helps us to find some lower and upper bounds of the form _Ψ(x)−c^x for the functionπ(x)and using these bounds, we show thatΨ(pn)∼logn, when n→ ∞is equivalent with the Prime Number Theorem.

Key words and phrases: Primes, Harmonic series, Gamma function, Digamma function.

2000 Mathematics Subject Classification. 11A41, 26D15, 33B15.

1. INTRODUCTION

As usual, letPbe the set of all primes andπ(x) = #P∩[2, x]. IfH_n=Pn k=1

1

k, then easily we have:

(1.1) γ+ logn < Hn<1 + logn (n >1),

in which γ is the Euler constant. So, H_n = logn+O(1) and considering the prime number theorem [2], we obtain:

π(n) = n

H_n+O(1) +o n

logn

.

ISSN (electronic): 1443-5756 c

2006 Victoria University. All rights reserved.

I deem it my duty to thank P. Dusart, L. Panaitopol, M.R. Razvan and J. Sándor for sending or bringing me the references, respectively [3], [5, 6], [2] and [7].

065-05

Thus, comparing _H ⁿ

n+O(1) withπ(n)seems to be a nice problem. In 1959, L. Locker-Ernst [4]

affirms that ⁿ

Hn−³₂, is very close toπ(n)and in 1999, L. Panaitopol [6], proved that forn≥1429 it is actually a lower bound forπ(n).

In this paper we improve Panaitopol’s result by proving _Hⁿ

n−a < π(n)for everyn ≥ 3299, in which a ≈ 1.546356705. Also, we find same upper bound for π(n). Then we consider generalization ofH_nas a real value function, which has been studied by J. Sándor [7] in 1988;

forx > 0letΨ(x) = _dx^d log Γ(x), in whichΓ(x) = R∞

0 e^−tt^x−1dt, is the well-known gamma function [1]. SinceΓ(x+ 1) =xΓ(x)andΓ(1) =−γ, we haveHn = Ψ(n+ 1) +γ, and this relation led him to define:

(1.2)







H: (0,∞)−→R, H(x) = Ψ(x+ 1) +γ,

as a natural generalization of Hn, and more naturally, it motivated us to find some bounds for π(x)concerningΨ(x). In our proofs, we use the obvious relation:

(1.3) Ψ(x+ 1) = Ψ(x) + 1

x. Also, we need some bounds of the form_logx−1−^x c

logx

, which we yield them by using the following known sharp bounds [3], forπ(x):

(1.4) x

logx

1 + 1

logx + 1.8 log²x

≤π(x) (x≥32299), and

(1.5) π(x)≤ x

logx

1 + 1

logx + 2.51 log²x

(x≥355991).

Finally, using the above mentioned bounds concerning π(x), we show that Ψ(p_n) ∼ logn, whenn→ ∞is equivalent with the Prime Number Theorem. To do this, we need the following bounds [3], forp_n:

(1.6) logn+ log₂n−1 + log₂n−2.25 logn ≤ p_n

n ≤logn+ log₂n−1 + log₂n−1.8 logn , in which the left hand side holds forn ≥2and the right hand side holds forn ≥27076. Also, bylog₂nwe meanlog lognand base of all logarithms ise.

2. BOUNDS OF THE FORM _logx−1−^x c logx

Lower Bounds. We are going to find suitable values ofa, in which _logx−1−^x a logx

≤ π(x). Con- sidering (1.4) and lettingy= logx, we should study the inequality

1

y−1− ^a_y ≤ 1 y

1 + 1

y + 9 5y²

, which is equivalent with

y⁴

y²−y−a ≤y²+y+ 9 5,

and supposingy²−y−a >0, it will be equivalent with 4

5 −a

y²−

a+9 5

y− 9a 5 ≥0,

and this forces ⁴₅ −a >0, ora < ⁴₅. Leta= ⁴₅ −for some >0. Therefore we should study 1

y−1− ⁴⁵⁻_y

≤ 1 y

1 + 1

y + 9 5y²

,

which is equivalent with:

(2.1) 25y²+ (25−65)y+ (45−36)

5y³ 5y² −5y+ (5−4) ≥0.

The equation25y²+(25−65)y+(45−36) = 0has discriminant25∆₁with∆₁ = 169+14−

155², which is non-negative for−1 ≤ ≤ ¹⁶⁹₁₅₅ and the greater root of it, isy₁ = ¹³⁻⁵⁺

√∆1

10 .

Also, the equation 5y² − 5y + (5− 4) = 0 has discriminant ∆₂ = 105−100, which is non-negative for≤ ²¹₂₀ and the greater root of it, isy₂ = ¹₂ +

√∆2

10 . Thus, (2.1) holds for every 0 < ≤ min{¹⁶⁹₁₅₅,²¹₂₀} = ²¹₂₀, with y ≥ max

0<≤²¹₂₀

{y₁, y₂} = y₁. Therefore, we have proved the following theorem.

Theorem 2.1. For every0< ≤ ²¹₂₀, the inequality:

x

logx−1− _log⁴⁵⁻_x

≤π(x),

holds for all:

x≥max

32299, e¹³⁻⁵⁺

√

169+14−1552 10

.

Corollary 2.2. For everyx≥3299, we have:

x

logx−1 + _{4 log}¹ _x ≤π(x).

Proof. Taking = ²¹₂₀ in above theorem, we yield the result forx ≥ 32299. For 3299 ≤ x ≤ 32298, we check it by a computer; to do this, consider the following program in MapleV soft- ware’s worksheet:

restart:

with(numtheory):

for x from 32298 by -1 while

evalf(pi(x)-x/(log(x)-1+1/(4*log(x))))>0 do x end do;

Running this program, it starts checking the result fromx= 32298and verify it, untilx= 3299.

This completes the proof.

Upper Bounds. Similar to lower bounds, we should search suitable values ofb, in whichπ(x)≤

x logx−1− ^b

logx

. Considering (1.5) and lettingy= logx, we should study 1

y

1 + 1

y + 251 100y²

≤ 1 y−1− ^b_y. Assumingy² −y−b >0, it will be equivalent with

151 100 −b

y²−

b+251 100

y− 251b 100 ≤0,

which forcesb≥ ¹⁵¹₁₀₀. Letb= ¹⁵¹₁₀₀ +for some≥0. Therefore we should study 1

y

1 + 1

y + 251 100y²

≤ 1

y−1− ¹⁵¹¹⁰⁰_y⁺ ,

which is equivalent with:

(2.2) 10000y²+ (10000+ 40200)y+ (25100+ 37901) 100y³ 100y²−100y−(100+ 151) ≥0.

The quadratic equation in the numerator of (2.2), has discriminant40000∆₁with∆₁ = 40401−

17801−22600², which is non-negative for−⁴⁰⁴⁰¹₂₂₆₀₀ ≤ ≤1and the greater root of it, isy₁ =

−201−50+√

∆1

100 . Also, the quadratic equation in denominator of it, has discriminant1600∆₂with

∆₂ = 44 + 25, which is non-negative for−⁴⁴₂₅ ≤and the greater root of it, isy₂ = ¹₂ +

√∆2

5 . Thus, (2.2) holds for every0 ≤ ≤ min{1,+∞} = 1, withy ≥ max

0≤≤1{y₁, y₂} = y₂. Finally, we note that for0≤≤1, the functiony₂()is strictly increasing and so,

6< e¹²⁺

√ 44

5 =e^y2(0) ≤e^y2() ≤e^y2(1) =e¹²⁺

√ 69 5 <9.

Therefore, we obtain the following theorem.

Theorem 2.3. For every0≤ ≤1, we have:

π(x)≤ x

logx−1− ¹⁵¹¹⁰⁰_log⁺_x

(x≥355991).

Corollary 2.4. For everyx≥7, we have:

π(x)≤ x

logx−1− _{100 log}¹⁵¹ _x.

Proof. Taking= 0in above theorem, yields the result forx≥355991. For7≤x ≤35991it

has been checked by computer [5].

3. BOUNDS OF THE FORM _Hⁿ

n−c AND _Ψ(x)−c^x

Theorem 3.1.

(i) For everyn≥3299, we have:

n

H_n−a < π(n), in whicha=γ+ 1− _{4 log 3299}¹ ≈1.5463567.

(ii) For everyn≥9, we have:

π(n)< n H_n−b, in whichb= 2 + _{100 log 7}¹⁵¹ ≈2.77598649.

Proof. Forn≥3299, we have

γ+ logn≥a+ logn−1 + 1 4 logn, and considering this with the left hand side of (1.1), we obtain _Hⁿ

n−a < _logn−1+ⁿ 1 4 logn

and this inequality with Corollary 2.2, yields the first part of theorem.

Forn ≥9, we have

b+ logn−1− 151

100 logn >1 + logn and considering this with the right hand side of (1.1), we obtain ⁿ

logn−1−_{100 log}¹⁵¹ _n < _Hⁿ

n−b. Con-

sidering this, with Corollary 2.4, completes the proof.

Theorem 3.2.

(i) For everyx≥3299, we have:

x

Ψ(x)−A < π(x),

in whichA= 1− ^Ψ(3299)₃₂₉₈ − 13192 log 3299³²⁹⁹ ≈0.9666752780.

(ii) For everyx≥9, we have:

π(x)< x Ψ(x)−B, in whichB = 2 + _{100 log 7}¹⁵¹ −γ ≈2.198770832.

Proof. LetHxbe the step function defined byHx =Hnforn≤x < n+ 1. Considering (1.2), we haveH(x−1)< H_x≤H(x).

Forx≥3299, by considering part (i) of the previous theorem, we have:

π(x)> x

H_x−a ≥ x

H(x)−a = x

Ψ(x+ 1) +γ−a. Thus, by considering (1.3), we obtain:

π(x)> x−1

Ψ(x) + ¹_x +γ−a ≥ x−1

Ψ(x) + ₃₂₉₉¹ +γ−a ≥ x Ψ(x)−A, in whichA= Ψ(3299)− ³²⁹⁹₃₂₉₈ Ψ(3299) + ₃₂₉₉¹ +γ−a

= 1−^Ψ(3299)₃₂₉₈ − 13192 log 3299³²⁹⁹ . Forx≥9, by considering second part of previous theorem, we obtain:

π(x)< x+ 1

H_x+1−b < x

H(x−1)−b = x

Ψ(x) +γ−b = x Ψ(x)−B,

in whichB =b−γ = 2 + _{100 log 7}¹⁵¹ −γ, and this completes the proof.

4. AN EQUIVALENT FOR THEPRIMENUMBERTHEOREM

Theorem 3.2, seems to be nice; because using it, for everyx≥3299we obtain:

(4.1) x

π(x)+A <Ψ(x)< x

π(x)+B.

Moreover, considering this inequality with (1.4) and (1.5), we yield the following bounds for x≥355991:

logx 1 + _log¹_x + _log^2.512x

+A <Ψ(x)< logx 1 + _log¹_x +_log^1.82x

+B.

Also, by puttingx=p_n,n^thprime in (4.1), forn ≥463we yield that:

(4.2) p_n

n +A <Ψ(pn)< p_n n +B.

Considering this inequality with (1.6), for everyn ≥27076we obtain:

logn+ log₂n+A−1 + log₂n−2.25 logn

<Ψ(p_n)<logn+ log₂n+B−1 + log₂n−1.8 logn . This inequality is a very strong form of an equivalent of the Prime Number Theorem (PNT), which asserts π(x) ∼ _log^x_x and is equivalent with pn ∼ nlogn (see [1]). In this section, we have another equivalent as follows:

Theorem 4.1. Ψ(p_n)∼logn, whenn → ∞is equivalent with the Prime Number Theorem.

Proof. First suppose PNT. Thus, we have p_n = nlogn+o(nlogn). Also, (4.2) yields that Ψ(p_n) = ^p_nⁿ +O(1). Therefore, we have:

Ψ(p_n) = nlogn+o(nlogn)

n +O(1) = logn+o(logn).

Conversely, supposeΨ(p_n) = logn+o(logn). By solving (4.2) according top_n, we obtain:

nΨ(p_n)−Bn < p_n < nΨ(p_n)−An.

Therefore, we have:

p_n=nΨ(p_n) +O(n) =n logn+o(logn)

+O(n) =nlogn+o(nlogn),

which, this is PNT.

REFERENCES

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[2] H. DAVENPORT, Multiplicative Number Theory (Second Edition), Springer-Verlag, 1980.

[3] P. DUSART, Inégalités explicites pour ψ(X), θ(X), π(X) et les nombres premiers, C. R. Math.

Acad. Sci. Soc. R. Can., 21(2) (1999), 53–59.

[4] L. LOCKER-ERNST, Bemerkungen über die verteilung der primzahlen, Elemente der Mathematik XIV, 1 (1959), 1–5, Basel.

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