Approximate evaluation of this series (exponential sum)

  • Context: Graduate 
  • Thread starter Thread starter Kevin_spencer2
  • Start date Start date
  • Tags Tags
    Approximate Series Sum
Click For Summary
SUMMARY

The discussion centers on the approximate evaluation of the series \(\sum_{p PREREQUISITES

  • Understanding of prime number theory and the prime counting function \(\pi(N)\)
  • Familiarity with complex analysis, particularly exponential functions and integrals
  • Knowledge of integration techniques, specifically integration by parts
  • Basic concepts of Riemann-Stieltjes Integrals
NEXT STEPS
  • Explore advanced techniques in partial summation for series involving primes
  • Study the properties and applications of Riemann-Stieltjes Integrals
  • Investigate the implications of oscillatory behavior in exponential sums
  • Learn about lattice points on the unit circle and their relation to prime distributions
USEFUL FOR

Mathematicians, number theorists, and researchers interested in prime number distributions and the evaluation of oscillatory series.

Kevin_spencer2
Messages
29
Reaction score
0
Let be the series

\sum_{p<N}e^{2\pi p ix}=f(x) where the sum is intended to be

over all primes less or equal than a given N.

My question is if there are approximate methods to evaluate this series for N big , since for a big prime the exponential sum is very oscillating would it be an 'intelligent' form to evaluate it for big N?, of course we know the trivial bound f(x)<\pi(N) however i think this is rather useless.
 
Physics news on Phys.org
Maybe this is not in the ballpark of what you're looking for, but I believe you can approximate this using partial/Abel summation. We can approximate \pi(N)=\sum_{p<N} 1 and use this to approximate f(x):

In particular, f(x) can be written as the Riemann-Stieltjes Integral
f(x)=\int_{1}^{N} e^{2\pi i t x} d\pi(t)
which then can be evaluated using integration by parts to get
f(x)=\pi(t) e^{2\pi i t x} |^{N}_{1} -2\pi i x \int_{1}^{N} e^{2\pi i t x} \pi(t) dt
Now you can use some approximations of \pi(t) to approximate the integral, and maybe that would give a decent answer. I don't know, I haven't worked it out.
 
Since e^{2\pi \i} is equal to 1, and one 1 the power of anything is equal to one, function is the addition of 1 \pi(N) times. This basically means f(x) is a constant function, but dependent on N. Not sure about my answer though...
 
x doesn't have to be an integer.
 
Gib Z said:
Since e^{2\pi \i} is equal to 1, and one 1 the power of anything is equal to one, function is the addition of 1 \pi(N) times. This basically means f(x) is a constant function, but dependent on N. Not sure about my answer though...

haha, I should have noticed that :redface:. Perhaps the original poster meant e^{2\pi i/p}, which would make the question slightly more interesting.

edit: or even better, what Hurkyl said.
 
Last edited:
Gib Z said:
Since e^{2\pi \i} is equal to 1, and one 1 the power of anything is equal to one, function is the addition of 1 \pi(N) times. This basically means f(x) is a constant function, but dependent on N. Not sure about my answer though...

Just so you're clear on what was meant above: exp{2pi i x} is 1 if and only if x is an integer. It should not be thought of as exp(2 pi i) to the power x. Raising things to powers creates issues anyway with branches.
 
Isn't each term looking for points mod p on the unit circle (you can think of a p lattice on the unit circle, and x maps to some point in the one of the the domains). You are in adding a bunch of number mod different primes in essence, which being all coprime might make it easier.

Anyway, it seemed like going down that path might produce something useful. You could even "unroll" the unit circle into a full axis and put a lattice there if it were easier (not sure it is).

Just some random ideas.
 
Just realized that it was 2 \pi i p and not {2 \pi i \over p}. Not sure anything I said still applies.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K