## Approximation of pi(x)

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• Hello! It s known that Li(x)=integral(dt/ln(t),t,0,x) is a good approximation for pi(x), the number of primes less than or equal to x. [Sure, if x 1, we take
Message 1 of 2 , Apr 7, 2001
Hello!

It's known that

Li(x)=integral(dt/ln(t),t,0,x)

is a good approximation for pi(x), the number of primes less
than or equal to x.
[Sure, if x>1, we take the mean value of integral, i.e.
Li(x)=lim(integral(dt/ln(t),t,0,1-eps)+integral(dt/ln(t),t,1+eps,x),eps->0)].

But Riemann suggested yet better approximation for pi(x):

R(x)=sum(mu(n)*Li(x^(1/n)/n),n,1,+infinity),

where mu(n) is Moebius' function.

R(x) works really good. We can transform it into better
form, using some series:

Li(exp(x))=Euler+ln(x)+sum(x^k/k!/k,k,1,+infinity), where
Euler=0.57721...;
sum(mu(k)/k^n,k,1,+infinity)=1/zeta(n), where zeta(n) is
Riemann's zeta-function;
sum(mu(k)*ln(k)/k,k,1,+infinity)=-1.

The result is:

R(x)=1+sum(ln(x)^n/n!/n/zeta(n+1),n,1,+infinity).

Note that 1 might be expressed as ln(x)^0/0!/0/zeta(1) if
believed that 0*zeta(1)=1 (so n would run from 0 to
infinity). Note also that lim(R(eps),eps->0)=0.

If we look at the difference R(x)-pi(x), we will notice that
it oscillates around some function with rising amplitude. To
find this function (which seems to be very close to zero
point), let consider

integral(dt*(R(t)-pi(t)/x,t,0,x),

which gives us the "mean" difference between R(t) and pi(t)
for 0<t<x. It also oscillates around some function yet
closer to zero point. We can continue this process (similar
to finding generalized sum of divergent series), and we'll
see that it's (heuristically) indeed zero point.

Note that if we do such process for Li(x)-pi(x), we'll
obtain something similar to sqrt(x)/ln(x), but not zero
point.

Now about amplitude of oscillations of R(x)-pi(x). At the
beginning amplitude seems to be less than C*sqrt(x)/ln(x)
for some small positive C. But we know that Li(x) isn't
always greater than pi(x), hence, C is not always less than
1, it becomes higher, i.e. C is function.

Using statistics and assuming that probability of N being
prime is 1/ln(N), Cramer showed that heuristically

|Li(x)-pi(x)|<sqrt(2*x*lnln(x)/ln(x)).

If we replace 1/ln(N)=Li'(N) by

R'(N)=sum(mu(k)/N^(1-1/k)/k,k,1,+infinity)/ln(N),

we'll obtain something similar to

|R(x)-pi(x)|=O(sqrt(x/ln(x))),

i.e. C=O(sqrt(ln(x))).

Maybe, this is a good approximation when x is enough big.
However, for small x (less that 10^16) we can see that C can
be approximated much better (by a constant); it even goes
down a bit when x goes up.
On the other hand, |R(x)-pi(x)| seems to be closely bound by
some definite function A(x), our C*sqrt(x)/ln(x) being only
approximation of this function from above. It would be
useful and interesting to find A(x) more exactly, however,
even RH (Riemann Hypothesis) gives us only rough
ceiling-limit O(sqrt(x)*ln(x)). Perhaps, some heuristic
methods (as Cramer's one) will give some better results?

Andrey
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• Hello! ... must be: integral(dt*(R(t)-pi(t))/x,t,0,x). Note that I ve missed a parentheses. Sorry, Andrey ... Профессиональная
Message 2 of 2 , Apr 8, 2001
Hello!

There was a typo in my posting about pi(x):

>integral(dt*(R(t)-pi(t)/x,t,0,x),

must be:

integral(dt*(R(t)-pi(t))/x,t,0,x).

Note that I've missed a parentheses.

Sorry,

Andrey
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