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Prime counting function

From Wikipedia, the free encyclopedia

In mathematics, the prime counting function is the function counting the number of prime numbers less than or equal to some real number x. It is denoted by π(x) (this does not refer to the number π).

The 60 first values of π(n)
The 60 first values of π(n)

Contents

[edit] History

Of great interest in number theory is the growth rate of the prime counting function. It was conjectured in the end of the 18th century by Gauss and by Legendre to be approximately

x/\operatorname{ln}(x),\,

in the sense that

\lim_{x\rightarrow +\infty}\frac{\pi(x)}{x/\operatorname{ln}(x)}=1.\,

This statement is the prime number theorem. An equivalent statement is

\lim_{x\rightarrow +\infty}\pi(x) / \operatorname{li}(x)=1,\,

where li is the logarithmic integral function. This was first proved around 1896 by Hadamard and by de la Vallée Poussin (independently), using properties of the Riemann zeta function introduced by Riemann in 1859.

More precise estimates of π(x) are now known; for example

\pi(x) = \operatorname{li}(x) + O \left( x \exp \left( -\frac{\sqrt{\ln(x)}}{15} \right) \right),

where the O is big O notation. Proofs of the prime number theorem not using the zeta function or complex analysis were found around 1948 by Atle Selberg and by Paul Erdős (for the most part independently).

Another conjecture about the growth rate for prime series involving the prime number theorem is

G(n,x)= \sum_{p}^{x} p^{n} \sim \pi(x^{n+1}).

[edit] Table of π(x), x / ln x, and Li(x)

Here is a table that shows how the three functions π(x), x / ln x and Li(x) compare:

x π(x) π(x) − x / ln x Li(x) − π(x) x / π(x)
10 4 −0.3 2.2 2.500
102 25 3.3 5.1 4.000
103 168 23 10 5.952
104 1,229 143 17 8.137
105 9,592 906 38 10.425
106 78,498 6,116 130 12.740
107 664,579 44,158 339 15.047
108 5,761,455 332,774 754 17.357
109 50,847,534 2,592,592 1,701 19.667
1010 455,052,511 20,758,029 3,104 21.975
1011 4,118,054,813 169,923,159 11,588 24.283
1012 37,607,912,018 1,416,705,193 38,263 26.590
1013 346,065,536,839 11,992,858,452 108,971 28.896
1014 3,204,941,750,802 102,838,308,636 314,890 31.202
1015 29,844,570,422,669 891,604,962,452 1,052,619 33.507
1016 279,238,341,033,925 7,804,289,844,393 3,214,632 35.812
1017 2,623,557,157,654,233 68,883,734,693,281 7,956,589 38.116
1018 24,739,954,287,740,860 612,483,070,893,536 21,949,555 40.420
1019 234,057,667,276,344,607 5,481,624,169,369,960 99,877,775 42.725
1020 2,220,819,602,560,918,840 49,347,193,044,659,701 222,744,644 45.028
1021 21,127,269,486,018,731,928 446,579,871,578,168,707 597,394,254 47.332
1022 201,467,286,689,315,906,290 4,060,704,006,019,620,994 1,932,355,208 49.636
1023 1,925,320,391,606,818,006,727 37,083,513,766,592,669,113 7,236,148,412 51.939

The first column is sequence A006880 in OEIS; the second column is sequence A057835; and the third column is sequence A057752.

A distributed computing project computed π(x) for x ≈ 1023, but the project was stopped after a discrepancy of 1 was discovered in the global check. [1]

[edit] Algorithms for evaluating π(x)

A simple way to find π(x), if x is not too large, is to use the sieve of Eratosthenes to produce the primes less than or equal to x and then to count them.

A more elaborate way of finding π(x) is due to Legendre: given x, if p1p2, …, pk are distinct prime numbers, then the number of integers less than or equal to x which are divisible by no pi is

\lfloor x\rfloor - \sum_{i}\left\lfloor\frac{x}{p_i}\right\rfloor + \sum_{i<j}\left\lfloor\frac{x}{p_ip_j}\right\rfloor - \sum_{i<j<k}\left\lfloor\frac{x}{p_ip_jp_k}\right\rfloor + \cdots,

(where \lfloor\cdot\rfloor denotes the floor function). This number is therefore equal to

\pi(x)-\pi\left(\sqrt{x}\right)+1\,

when the numbers p_1, p_2,\dots,p_k are the prime numbers less than or equal to the square root of x.

In a series of articles published between 1870 and 1885, Ernst Meissel described (and used) a practical combinatorial way of evaluating π(x). Let p1p2, …, pn be the first n primes and denote by Φ(m,n) the number of natural numbers not greater than m which are divisible by no pi. Then

\Phi(m,n)=\Phi(m,n-1)-\Phi\left(\left[\frac{m}{p_n}\right],n-1\right),\,

Given a natural number m, if n=\pi\left(\sqrt[3]{m}\right) and if \mu=\pi\left(\sqrt{m}\right)-n, then

\pi(m)=\Phi(m,n)+n(\mu+1)+\frac{\mu^2-\mu}{2}-1-\sum_{k=1}^\mu\pi\left(\frac{m}{p_{n+k}}\right).\,

Using this approach, Meissel computed π(x), for x equal to 5×105, 106, 107, and 108.

In 1959, Derrick Henry Lehmer extended and simplified Meissel's method. Define, for real m and for natural numbers n, and k, Pk(m,n) as the number of numbers not greater than m with exactly k prime factors, all greater than pn. Furthermore, set P0(m,n) = 1. Then

\Phi(m,n)=\sum_{k=0}^{+\infty}P_k(m,n),\,

where the sum actually has only finitely many nonzero terms. Let y denote an integer such that \sqrt[3]{m}\le y\le\sqrt{m}, and set n = π(y). Then P1(m,n) = π(m) − n and Pk(m,n) = 0 when k ≥ 3. Therefore

π(m) = Φ(m,n) + n − 1 − P2(m,n).

The computation of P2(m,n) can be obtained this way:

P_2(m,n)=\sum_{y<p\le\sqrt{m}}\left(\pi\left(\frac mp\right)-\pi(p)+1\right).\,

On the other hand, the computation of Φ(m,n) can be done using the following rules:

  1. \Phi(m,0)=\lfloor m\rfloor;\,
  2. \Phi(m,b)=\Phi(m,b-1)-\Phi\left(\frac m{p_b},b-1\right).\,

Using his method and an IBM 701, Lehmer was able to compute \pi\left(10^{10}\right).

Chinese mathematician Hwang Cheng, in a conference about prime number functions at the University of Bordeaux used the following identities:

e^{(a-1)\Theta}f(x)=f(ax), \,
J(x)=\sum_{n=1}^{\infty}\frac{\pi(x^{1/n})}{n},

and setting x = et, Laplace-transforming both sides and applying a geometric sum on enΘ got the expression

\frac{1}{2{\pi}i}\int_{c-i\infty}^{c+i\infty}g(s)t^{s}\,ds = \pi(t),
(1-e^{\Theta})\frac{\ln \zeta(s)}{s}=g(s),
\Theta(s)=s\frac{d}{ds}.

[edit] Other prime counting functions

Other prime counting functions are also used because they are more convenient to work with. One is Riemann's prime counting function, denoted Π(x) or J(x). This has jumps of 1/n for prime powers pn, with it taking a value half-way between the two sides at discontinuities. That added detail is because then it may be defined by an inverse Mellin transform. Formally, we may define J by

J(x) = \frac12 \bigg(\sum_{p^n < x} \frac1n\ + \sum_{p^n \le x} \frac1n\bigg)

where p is a prime.

We may also write

J(x) = \sum_{n=1}^\infty \frac1n \pi(x^{1/n}) and
\ln( \zeta (s))=s\int_{0}^{\infty} J(x) x^{-s+1}\,dx

Also using Möbius inversion formula then:

\pi (x)= \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} J(x^{1/n})

and knowing the relationship between log of the Riemann function and the von Mangoldt function Λ, and the Perron formula we have

J(x)= \sum_{2}^{x} \frac{ \Lambda (n)}{\ln(n)}

except where we have discontinuities at prime powers.

The Chebyshev function weights primes or prime powers pn by ln p:

\theta(x)=\sum_{p\le x}\ln p,
\psi(x) = \frac12 \bigg(\sum_{p^n < x} \ln p\ + \sum_{p^n \le x} \ln p\bigg).

Apart from the discontinuities at prime powers, we have

\psi(x)=\sum_{n=1}^\infty\theta(x^{1/n})=\sum_{n\le x}\Lambda(n),

where Λ(n) is the von Mangoldt function.

[edit] Formulas for prime counting functions

These come in two kinds, arithmetic formulas and analytic formulas. The latter are what allow us to prove the prime number theorem. They stem from the work of Riemann and von Mangoldt, and are generally known as explicit formulas.

We have the following expression for ψ:

\psi(x) = x - \sum_\rho \frac{x^\rho}{\rho} - \ln 2\pi - \frac12 \ln(1-x^{-2}).

Here ρ are the zeros of the Riemann zeta function in the critical strip, where the real part of ρ is between zero and one. The formula is valid for values of x greater than one, which is the region of interest, and the sum over the roots is conditionally convergent, and should be taken in order of increasing absolute value of the imaginary part.

For J we have a more complicated formula

J(x) = \operatorname{li}(x) - \sum_{\rho}\operatorname{li}(x^{\rho}) - \ln 2 + \int_x^\infty \frac{dt}{t(t^2-1) \ln t}.

Again, the formula is valid for x > 1, and ρ are the nontrivial zeros of the zeta function ordered according to their absolute value. The first term li(x) is the usual logarithmic integral; however, it is not easily describable what li means in the other terms. The best way to think about it is to consider the expression \operatorname{li}(x^\rho) as an abbreviation for \operatorname{Ei}(\rho\ln x), where Ei is the analytic continuation of the exponential integral function from positive reals to the complex plane with branch cut along the negative reals.

The following simple approximation can be useful: round(exAlnx + B) where A = 1.007 , B = − .05245.[citation needed]

An explicit integral expression involving sum over non-trivial zeros and Mertens function is:

\pi (x) = \int_{1}^{\infty} du G(u) u^{-2}( \operatorname{li}(v) - \sum_{\rho}\operatorname{li}(v^{\rho}) - \ln 2 + \int_v^\infty \frac{dt}{t(t^2-1) \ln t})


x1 / u = v(x)


u M'(u)=G(u)= \sum_{\rho}\frac{u^{\rho}}{\zeta ' (\rho)}+\sum_{n=1}^{\infty} \frac{(-1)^{n}}{n(2n-1)! \zeta (2n+1)}( \frac{2 \pi }{u})^{2n} Where M(u) is the Mertens function[citation needed][dubious ]

[edit] Inequalities

Here are some useful inequalities for π(x).

\pi(x) > \frac {x} {\log x} for x ≥ 17.
\pi(x) < 1.25506 \frac {x} {\log x} for x > 1.
\frac {x} {\log x + 2} < \pi(x) <  \frac {x} {\log x - 4} for x ≥ 55.

Here are some inequalities for the nth prime, pn.

n\ \ln n + n\ln\ln n - n  < p_n <  n \ln n + n \ln \ln n for n ≥ 6.

The left inequality holds for n ≥ 1 and the right inequality holds for n ≥ 6.

An approximation for the nth prime number is

p_n = n \ln n +  n \ln \ln n - n + \frac {n \ln \ln n - 2n} {\ln n} +  O\left( \frac {n (\ln \ln n)^2} {(\ln n)^2}\right).

[edit] The Riemann hypothesis

The Riemann hypothesis is equivalent to a much tighter bound on the error in the estimate for π(x), and hence to a more regular distribution of prime numbers,

\pi(x) = \operatorname{li}(x) + O(\sqrt{x} \log{x}).

[edit] Relation to prime sums

if we had a sum of a function over all primes :

f(x)
p

and we wish to accelerate its convergence we can write it as:

\sum_{n=1}^{\infty}(-1)^{n}(\pi(n)-\pi(n-1)+1)f(n)=2f(2)-\sum_{p}f(x)+\sum_{n=1}^{\infty}(-1)^{n}f(n)

for the series on the left we could apply Euler transform for alternating series, providing that f(n)>f(n+1) and that the 2 series converges, it also relates an alternating series to its prime sum counterpart, the main task of using this is that we can give a good approximation using only a few values of the prime number counting function.

[edit] References

  • Bach, Eric; Shallit, Jeffrey (1996). Algorithmic Number Theory. MIT Press, volume 1 page 234 section 8.8. ISBN 0-262-02405-5. 
  • Dickson, Leonard Eugene (2005). History of the Theory of Numbers I: Divisibility and Primality. Dover Publications. ISBN 0-486-44232-2. 
  • Ireland, Kenneth; Rosen, Michael (1998). A Classical Introduction to Modern Number Theory, Second edition, Springer. ISBN 0-387-97329-X. 
  • Hwang H. Cheng Prime Magic conference given at the University of Bordeaux (France) at year 2001 Démarches de la Géométrie et des Nombres de l'Université du Bordeaux
  • Titchmarsh, E. C. The Theory of Functions, 2nd ed. Oxford, England: Oxford University Press, 1960.
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