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I Prime Number Theorem: the meaning of the limit

  1. May 27, 2016 #1
    Hi All.

    I have a doubt concerning the limit:

    $$ \lim_{n \to \infty} \frac{\pi (n)}{Li(n)} = 1 $$.

    This mathematical statement does not imply that both functions converge to the same value. The main reason is that both tend to infinity as n tend to infinity. I would like to ask you if it is correct to infer that when n tends to infinity, ##\pi (n) ## and ## Li(n) ## grow at the same rate, possibly being separed by a constant.

    If I am correct, is this constant known?

    Best regards,

    DaTario
     
  2. jcsd
  3. May 27, 2016 #2

    andrewkirk

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    How do you want to interpret the growth rate of ##\pi(n)##? It is a step function, which is discontinuous and hence constant except at a countable collection of points, at which it is not differentiable. But most interpretations of growth rate involve differentiation.
     
  4. May 27, 2016 #3

    mfb

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    The limit statement is even weaker. As an example: $$\lim_{n \to \infty} \frac{\ln (n) + \ln (\ln (n))}{\ln(n)} = 1$$
    The difference between numerator and denominator diverges, but the ratio still goes to 1.

    For this particular case, it is known that ##\pi(n) > Li(n)## and ##\pi(n) < Li(n)## switch infinitely often, the first time before 10371. Looks like there is no upper bound on the absolute deviation.
    More about it at mathworld
     
  5. May 27, 2016 #4
    Thank you, mfb.

    Answering andrewkirk, the function ##\pi (n)## is forced, by its definition, to increase by one unity at a time. As is goes to infinity, the relative weight of this discreteness goes to zero, allowing one to consider it as a continous function. Its logarithmic nature also allows us to trust that no great departure from smoothness may happen for high values of ##n##. So the derivative of ##\pi(n)## may be taken as the average of:
    $$\frac{\pi(n+1) - \pi(n)}{(n+1) - n} $$

    for, say, a million of values of ##n## located at the vicinity of the point whose derivative we wish to calculate.
    Note that our region of interest is ##n## large.

    (someone, please correct me if I am wrong...)

    Best wishes,

    DaTario
     
  6. May 27, 2016 #5

    mfb

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    There is no need to smoothen the function arbitrarily.
     
  7. May 27, 2016 #6

    andrewkirk

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    Let's formalise that. Your definition of growth rate is then

    $$G(n)=10^{-6}\times \sum_{k=1}^{1000000}\frac{\pi(n-k+1)-\pi(n-k)}{(n-k+1)-(n-k)}=
    10^{-6}\times\Big(\pi(n)-\pi(n-1000000)\Big)$$

    That will not attain any limit as ##n\to\infty## because it will be 0 for many large ##n## but ##10^{-6}## for some and, rarely, ##m\times 10^{-6}## for values of ##m## greater than 1.

    The growth rate of ##Li(n)## is ##H(n)\equiv\frac d{dn}\int_2^n\frac1{\log t}dt= \frac1{\log n}##, which smoothly tends towards a limit of zero as ##n\to\infty##. So neither ##G(n)-H(n)## nor ##\frac{G(n)}{H(n)}## has a limit as ##n\to\infty##.
     
  8. May 27, 2016 #7
    Perhaps if we try to make a linear regression with a million of points of ##(n,\pi(n))##, with n ranging from ##n_0 - 5 \times 10^5 ## to ##n_0 + 5 \times 10^5 ##.
    I would also accept a polynomial interpolation or a convolution with some gaussian.:smile:

    My point is that I was taking the limit of the theorem as an inference that both function would be identical as n tends to infinity. I now see that what seems to happen is very different from this.
     
  9. May 28, 2016 #8

    mfb

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    Right. There are always points where the functions differ by more than 10, or more than 100, or probably even more than a million. Removing the steps of 1 doesn't help to get convergence.
     
  10. May 28, 2016 #9
    This consequence of the Riemann hypothesis gives some hint about this difference:

    NumberedEquation20.gif
     
  11. May 28, 2016 #10

    mfb

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    It is an upper limit on the difference, but that limit goes to infinity. It is sufficient to see that the ratio converges to one, but it doesn't tell us if it has a fixed upper limit.
     
  12. May 28, 2016 #11
    Yes.

    Thank you.
     
  13. May 30, 2016 #12

    chiro

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    Basically they try and use it to get the idea of how the prime counting function behaves.

    The better the approximation - the better you can predict what the primes are.

    The prime counting function though - is often elusive because of how "random" it seems.
     
  14. May 31, 2016 #13
    Thank you, chiro.
     
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