Journal of Inequalities in Pure and
Applied Mathematics
APPROXIMATION OF π(x) BY Ψ(x)
MEHDI HASSANI
Institute for Advanced
Studies in Basic Sciences
P.O. Box 45195-1159
Zanjan, Iran.
volume 7, issue 1, article 7,
2006.
Received 07 March, 2005;
accepted 25 August, 2005.
Communicated by: J. Sándor
EMail: mmhassany@srttu.edu
Abstract
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Abstract
In this paper we find some lower and upper bounds of the form Hnn−c for the
P
function π(n), in which Hn = nk=1 k1 . Then, we consider H(x) = Ψ(x + 1) + γ
as generalization of Hn , such that Ψ(x) = dxd log Γ(x) and γ is Euler constant;
this extension has been introduced for the first time by J. Sándor and it helps
x
for the function
us to find some lower and upper bounds of the form Ψ(x)−c
π(x) and using these bounds, we show that Ψ(pn ) ∼ log n, when n → ∞ is
equivalent with the Prime Number Theorem.
Approximation of π(x) by Ψ(x)
Mehdi Hassani
2000 Mathematics Subject Classification: 11A41, 26D15, 33B15.
Key words: Primes, Harmonic series, Gamma function, Digamma function.
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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].
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Dedicated to Professor J. Rooin on the occasion of his 50th birthday
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Contents
1
2
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
x
Bounds of the Form log x−1−
.......................
c
log x
Close
3
5
x
3
Bounds of the Form Hnn−c and Ψ(x)−c
................... 9
4
An Equivalent for the Prime Number Theorem . . . . . . . . . . . . 12
References
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1.
Introduction
As usual, let P be the set of all primes and π(x) = #P∩[2, x]. If Hn =
then easily we have:
(1.1)
γ + log n < Hn < 1 + log n
Pn
1
k=1 k ,
(n > 1),
in which γ is the Euler constant. So, Hn = log n + O(1) and considering the
prime number theorem [2], we obtain:
n
n
π(n) =
+o
.
Hn + O(1)
log n
n
Thus, comparing Hn +O(1)
with π(n) seems to be a nice problem. In 1959,
L. Locker-Ernst [4] affirms that Hnn− 3 , is very close to π(n) and in 1999, L.
2
Panaitopol [6], proved that for n ≥ 1429 it is actually a lower bound for π(n).
In this paper we improve Panaitopol’s result by proving Hnn−a < π(n) for
every n ≥ 3299, in which a ≈ 1.546356705. Also, we find same upper bound
for π(n). Then we consider generalization of Hn as a real value function, which
d
has been studied by
Sándor [7] in 1988; for x > 0 let Ψ(x) = dx
log Γ(x),
R ∞J. −t
x−1
in which Γ(x) = 0 e t dt, is the well-known gamma function [1]. Since
Γ(x + 1) = xΓ(x) and Γ(1) = −γ, we have Hn = Ψ(n + 1) + γ, and this
relation led him to define:

H : (0, ∞) −→ R,
(1.2)
H(x) = Ψ(x + 1) + γ,
Approximation of π(x) by Ψ(x)
Mehdi Hassani
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as a natural generalization of Hn , and more naturally, it motivated us to find
some bounds for π(x) concerning Ψ(x). In our proofs, we use the obvious
relation:
1
Ψ(x + 1) = Ψ(x) + .
x
(1.3)
Also, we need some bounds of the form
x
,
log x−1− logc x
which we yield them by
using the following known sharp bounds [3], for π(x):
1
x
1.8
(1.4)
1+
+
≤ π(x)
(x ≥ 32299),
log x
log x log2 x
and
(1.5)
Mehdi Hassani
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x
π(x) ≤
log x
1
2.51
1+
+
log x log2 x
(x ≥ 355991).
Finally, using the above mentioned bounds concerning π(x), we show that
Ψ(pn ) ∼ log n, when n → ∞ is equivalent with the Prime Number Theorem.
To do this, we need the following bounds [3], for pn :
(1.6)
Approximation of π(x) by Ψ(x)
log n + log2 n − 1 +
log2 n − 2.25
log n
pn
log2 n − 1.8
≤
≤ log n + log2 n − 1 +
,
n
log n
in which the left hand side holds for n ≥ 2 and the right hand side holds for
n ≥ 27076. Also, by log2 n we mean log log n and base of all logarithms is e.
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2.
Bounds of the Form
x
log x−1− logc x
x
Lower Bounds. We are going to find suitable values of a, in which log x−1−
a
log x
≤
π(x). Considering (1.4) and letting y = log x, we should study the inequality
1
1
1
9
≤
1+ + 2 ,
y − 1 − ay
y
y 5y
which is equivalent with
Approximation of π(x) by Ψ(x)
y4
9
≤ y2 + y + ,
2
y −y−a
5
Mehdi Hassani
Title Page
and supposing y 2 − y − a > 0, it will be equivalent with
4
9
9a
2
−a y − a+
y−
≥ 0,
5
5
5
and this forces 45 − a > 0, or a < 45 . Let a = 45 − for some > 0. Therefore
we should study
1
1
9
1
≤
1+ + 2 ,
4
−
y
y 5y
y−1− 5
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y
which is equivalent with:
(2.1)
25y 2 + (25 − 65)y + (45 − 36)
≥ 0.
5y 3 5y 2 − 5y + (5 − 4)
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The equation 25y 2 + (25 − 65)y + (45 − 36) = 0 has discriminant 25∆1 with
∆1 = 169+14−1552 , which
is non-negative for −1 ≤ ≤ 169
and the greater
155
√
13−5+ ∆1
2
. Also, the equation 5y − 5y + (5 − 4) = 0 has
root of it, is y1 =
10
discriminant ∆2 = 105√− 100, which is non-negative for ≤ 21
and the greater
20
∆2
1
, 21 } =
root of it, is y2 = 2 + 10 . Thus, (2.1) holds for every 0 < ≤ min{ 169
155 20
21
, with y ≥ max21 {y1 , y2 } = y1 . Therefore, we have proved the following
20
0<≤ 20
theorem.
Theorem 2.1. For every 0 < ≤
21
,
20
the inequality:
Mehdi Hassani
x
log x − 1 −
holds for all:
4
−
5
≤ π(x),
log x
Title Page
√
13−5+ 169+14−1552
10
x ≥ max 32299, e
.
Corollary 2.2. For every x ≥ 3299, we have:
x
log x − 1 +
Approximation of π(x) by Ψ(x)
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1
4 log x
≤ π(x).
Proof. Taking = 21
in above theorem, we yield the result for x ≥ 32299.
20
For 3299 ≤ x ≤ 32298, we check it by a computer; to do this, consider the
following program in MapleV software’s worksheet:
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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 from x = 32298 and verify it, until x = 3299. This completes the proof.
Upper Bounds. Similar to lower bounds, we should search suitable values of
x
b, in which π(x) ≤ log x−1−
b . Considering (1.5) and letting y = log x, we
Approximation of π(x) by Ψ(x)
Mehdi Hassani
log x
should study
1
y
1
251
1+ +
y 100y 2
≤
1
.
y − 1 − yb
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2
Assuming y − y − b > 0, it will be equivalent with
151
251
251b
2
−b y − b+
y−
≤ 0,
100
100
100
which forces b ≥
study
151
.
100
Let b =
151
100
+ for some ≥ 0. Therefore we should
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1
y
1
251
1+ +
y 100y 2
≤
1
y−1−
151
+
100
,
y
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which is equivalent with:
(2.2)
10000y 2 + (10000 + 40200)y + (25100 + 37901)
≥ 0.
100y 3 100y 2 − 100y − (100 + 151)
J. Ineq. Pure and Appl. Math. 7(1) Art. 7, 2006
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The quadratic equation in the numerator of (2.2), has discriminant 40000∆1
with ∆1 = 40401−17801−226002 , which is
non-negative for − 40401
≤≤1
22600
√
−201−50+ ∆1
. Also, the quadratic equation
and the greater root of it, is y1 =
100
in denominator of it, has discriminant 1600∆2 with ∆2 = 44 + 25,
√ which is
44
non-negative for − 25
≤ and the greater root of it, is y2 = 12 + 5∆2 . Thus,
(2.2) holds for every 0 ≤ ≤ min{1, +∞} = 1, with y ≥ max {y1 , y2 } = y2 .
0≤≤1
Finally, we note that for 0 ≤ ≤ 1, the function y2 () is strictly increasing and
so,
√
√
1
44
1
69
6 < e 2 + 5 = ey2 (0) ≤ ey2 () ≤ ey2 (1) = e 2 + 5 < 9.
Approximation of π(x) by Ψ(x)
Mehdi Hassani
Therefore, we obtain the following theorem.
Theorem 2.3. For every 0 ≤ ≤ 1, we have:
π(x) ≤
x
log x − 1 −
151
+
100
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(x ≥ 355991).
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log x
Corollary 2.4. For every x ≥ 7, we have:
π(x) ≤
x
log x − 1 −
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151
100 log x
.
Proof. Taking = 0 in above theorem, yields the result for x ≥ 355991. For
7 ≤ x ≤ 35991 it has been checked by computer [5].
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3.
Bounds of the Form
n
Hn −c
and
x
Ψ(x)−c
Theorem 3.1.
(i) For every n ≥ 3299, we have:
n
< π(n),
Hn − a
in which a = γ + 1 −
1
4 log 3299
≈ 1.5463567.
Approximation of π(x) by Ψ(x)
(ii) For every n ≥ 9, we have:
π(n) <
in which b = 2 +
151
100 log 7
Mehdi Hassani
n
,
Hn − b
Title Page
≈ 2.77598649.
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Proof. For n ≥ 3299, we have
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1
,
γ + log n ≥ a + log n − 1 +
4 log n
and considering this with the left hand side of (1.1), we obtain Hnn−a <
n
1
log n−1+ 4 log
n
and this inequality with Corollary 2.2, yields the first part of theorem.
For n ≥ 9, we have
n
.
Hn −b
Considering this, with Corollary 2.4, completes the proof.
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151
> 1 + log n
b + log n − 1 −
100 log n
and considering this with the right hand side of (1.1), we obtain log n−1−n
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151
100 log n
<
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Theorem 3.2.
(i) For every x ≥ 3299, we have:
x
< π(x),
Ψ(x) − A
in which A = 1 −
Ψ(3299)
3298
−
3299
13192 log 3299
≈ 0.9666752780.
(ii) For every x ≥ 9, we have:
Approximation of π(x) by Ψ(x)
in which B = 2 +
151
100 log 7
x
π(x) <
,
Ψ(x) − B
Mehdi Hassani
− γ ≈ 2.198770832.
Title Page
Proof. Let Hx be the step function defined by Hx = Hn for n ≤ x < n + 1.
Considering (1.2), we have H(x − 1) < Hx ≤ H(x).
For x ≥ 3299, by considering part (i) of the previous theorem, we have:
π(x) >
x
x
x
≥
=
.
Hx − a
H(x) − a
Ψ(x + 1) + γ − a
Ψ(3299) +
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x−1
x−1
x
π(x) >
≥
≥
,
1
1
Ψ(x) − A
Ψ(x) + x + γ − a
Ψ(x) + 3299 + γ − a
3299
3298
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Thus, by considering (1.3), we obtain:
in which A = Ψ(3299) −
3299
.
13192 log 3299
Contents
1
3299
+γ−a
= 1−
Ψ(3299)
3298
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For x ≥ 9, by considering second part of previous theorem, we obtain:
π(x) <
x+1
x
x
x
<
=
=
,
Hx+1 − b
H(x − 1) − b
Ψ(x) + γ − b
Ψ(x) − B
in which B = b − γ = 2 +
151
100 log 7
− γ, and this completes the proof.
Approximation of π(x) by Ψ(x)
Mehdi Hassani
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4.
An Equivalent for the Prime Number Theorem
Theorem 3.2, seems to be nice; because using it, for every x ≥ 3299 we obtain:
x
x
+ A < Ψ(x) <
+ B.
π(x)
π(x)
(4.1)
Moreover, considering this inequality with (1.4) and (1.5), we yield the following bounds for x ≥ 355991:
log x
1 + log1 x +
2.51
log2 x
+ A < Ψ(x) <
log x
1 + log1 x +
1.8
log2 x
+ B.
Also, by putting x = pn , nth prime in (4.1), for n ≥ 463 we yield that:
(4.2)
Mehdi Hassani
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pn
pn
+ A < Ψ(pn ) <
+ B.
n
n
Considering this inequality with (1.6), for every n ≥ 27076 we obtain:
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log2 n − 2.25
log n + log2 n + A − 1 +
log n
< Ψ(pn ) < log n + log2 n + B − 1 +
Approximation of π(x) by Ψ(x)
Close
log2 n − 1.8
.
log n
This inequality is a very strong form of an equivalent of the Prime Number
Theorem (PNT), which asserts π(x) ∼ logx x and is equivalent with pn ∼ n log n
(see [1]). In this section, we have another equivalent as follows:
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Theorem 4.1. Ψ(pn ) ∼ log n, when n → ∞ is equivalent with the Prime
Number Theorem.
Proof. First suppose PNT. Thus, we have pn = n log n + o(n log n). Also, (4.2)
yields that Ψ(pn ) = pnn + O(1). Therefore, we have:
Ψ(pn ) =
n log n + o(n log n)
+ O(1) = log n + o(log n).
n
Conversely, suppose Ψ(pn ) = log n + o(log n). By solving (4.2) according to
pn , we obtain:
nΨ(pn ) − Bn < pn < nΨ(pn ) − An.
Therefore, we have:
pn = nΨ(pn ) + O(n) = n log n + o(log n) + O(n) = n log n + o(n log n),
which, this is PNT.
Approximation of π(x) by Ψ(x)
Mehdi Hassani
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References
[1] M. ABRAMOWITZ AND I.A. STEGUN, Handbook of Mathematical Functions: with Formulas, Graphs, and Mathematical Tables, Dover Publications, 1972.
[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.
Approximation of π(x) by Ψ(x)
Mehdi Hassani
[4] L. LOCKER-ERNST, Bemerkungen über die verteilung der primzahlen, Elemente der Mathematik XIV, 1 (1959), 1–5, Basel.
[5] L. PANAITOPOL, A special case of the Hardy-Littlewood conjecture,
Math. Reports, 4(54)(3) (2002), 265–258.
[6] L. PANAITOPOL, Several approximation of π(x), Math. Inequal. & Applics., 2(3) (1999), 317–324.
[7] J. SÁNDOR, Remark on a function which generalizes the harmonic series,
C. R. Acad. Bulgare Sci., 41(5) (1988), 19–21.
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