Sum over primes asymtotics

  • Thread starter Thread starter zetafunction
  • Start date Start date
  • Tags Tags
    Primes Sum
zetafunction
Messages
371
Reaction score
0
is the following asymptotic approximation valid whenever dealing sums over primes ?? \sum_{p\le x}f(p) \sim \int_{2}^{x}\frac{f(x)dx}{log(x)}
 
Physics news on Phys.org
f(x)=1 if x is a prime number, 0 otherwise

You'll definitely need at least some assumptions like continuity, and probably uniform continuity, otherwise you could just make f(x) approximately zero except for at prime numbers, where it forms sharper and sharper spikes over time

Even then, if f(x) only had finite support, say f(x)=1 for x<6, f(x)=7-x if 6<x<7, 0 if x>7, then you will never get the difference to become small or the ratio of the two to become close to each other (I'm not sure exactly what you mean by ~ in this case but since for x>7 the two stop changing and are not equal to each other, it's obviously not the case that they become asymptotically close for any definition of asymptotic)
 
i mean ASYMPTOTIC that is the quotient between the series and the integral will tend to 1 for big x

for example if you set f(x)=1 you get the Prime Number Theorem
 
And if you set f(x) to be either of the examples that I gave in my post you don't get such a result
 
I assume the integral was intended to be \int_2^x\frac{f(t)dt}{\log t}.Here's a uniformly continuous function supported on x > 0 where the two differ asymptotically:

f(x) = 2^{-x} for x a positive integer
f(x) = 1 for x a positive half-integer
f(x) = (\lfloor x\rfloor-x+1/2)f(\lfloor x\rfloor)+x-\lfloor x\rfloor for \lfloor x\rfloor&lt;x&lt;\lfloor x\rfloor+1/2
f(x) = x-\lfloor x\rfloor-1/2+(\lfloor x\rfloor+1-x)f(\lfloor x\rfloor+1) for \lfloor x\rfloor+1/2&lt;x&lt;\lfloor x\rfloor+1

\sum_{p\le x}f(p)\sim1\ldots (actually, 0.4146825... + o(1), where the number is Sloane's A051006)
while
\int_2^x\frac{f(t)dt}{\log t}\sim\frac{x}{2\log x}What about restricting f to monotonic functions? I mean with all the other stuff -- without uniform continuity I think I can make it fail even for monotonics.
 
Last edited:
well, how about trying it for WELL BEHAVED function ? like polynomials , trigonometric and all this stuff ..
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
Back
Top