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Sum over primes asymtotics

  1. Jul 22, 2010 #1
    is the following asymptotic approximation valid whenever dealing sums over primes ?? [tex] \sum_{p\le x}f(p) \sim \int_{2}^{x}\frac{f(x)dx}{log(x)} [/tex]
  2. jcsd
  3. Jul 22, 2010 #2


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    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)
  4. Jul 22, 2010 #3
    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
  5. Jul 22, 2010 #4


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    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
  6. Jul 22, 2010 #5


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    I assume the integral was intended to be [tex]\int_2^x\frac{f(t)dt}{\log t}[/tex].

    Here's a uniformly continuous function supported on x > 0 where the two differ asymptotically:

    [tex]f(x) = 2^{-x}[/tex] for x a positive integer
    [tex]f(x) = 1[/tex] for x a positive half-integer
    [tex]f(x) = (\lfloor x\rfloor-x+1/2)f(\lfloor x\rfloor)+x-\lfloor x\rfloor[/tex] for [tex]\lfloor x\rfloor<x<\lfloor x\rfloor+1/2[/tex]
    [tex]f(x) = x-\lfloor x\rfloor-1/2+(\lfloor x\rfloor+1-x)f(\lfloor x\rfloor+1)[/tex] for [tex]\lfloor x\rfloor+1/2<x<\lfloor x\rfloor+1[/tex]

    [tex]\sum_{p\le x}f(p)\sim1\ldots[/tex] (actually, 0.4146825... + o(1), where the number is Sloane's A051006)
    [tex]\int_2^x\frac{f(t)dt}{\log t}\sim\frac{x}{2\log x}[/tex]

    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: Jul 22, 2010
  7. Jul 22, 2010 #6
    well, how about trying it for WELL BEHAVED function ? like polynomials , trigonometric and all this stuff ..
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