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Intuitive Number theory:

  1. Mar 22, 2006 #1
    "Intuitive" Number theory:

    Now i would like to play a game called "conjecture"..we have that for asymptotic behaviour:

    [tex] \pi(x)=li(x) [/tex] where here "li" means the Logarithmic integral..

    my conjecture is that for the sum:

    [tex] \sum_{p}^{x}p^{n}=li(x^{n+1} [/tex]

    i have checked it for n=-1,0,1 and it seems to work, the "justification" is that for example for sum over integers:

    [tex] \int_{0}^{n}x^{k}=1+2^{k}+3{k}+........... k\rightarrow{\infty} [/tex]

    for the primes case there is an extra weight function [tex]\pi(x)-\pi(x-1) [/tex] so our sum would be equal to the integral:

    [tex]\int_{0}^{n}(\pi(x)-\pi(x-1))x^{k} [/tex]

    but using PNT [tex]\pi(x)-\pi(x-1)=1/ln(x) [/tex] so all this becomes:

    [tex]\int_{0}^{n}x^{k}/ln(x)\rightarrow{Li(n^{k+1} [/tex]

    where the last expression comes from using tables to compute the integral, of course for any analyitc function on R we have:

    [tex] \sum_{p}f(p)=\sum_{n=0}^{\infty}a_{n}li(x^{n+1} [/tex]

    for x------->oo
     
    Last edited: Mar 22, 2006
  2. jcsd
  3. Mar 22, 2006 #2

    matt grime

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    But, eljose, it is quite clearly a false conjecture, you cannot possibly mean for that sum to *equal* that integral. Those two quantities are not the same. The correct notation for asymptotic is [itex]\sim[/tex], or at least one of the correct notations is that.
     
  4. Mar 22, 2006 #3
    Then sorry...i,m a physicist so for us the concept or asymptotic or equal are not always clear perhaps i should have stablished that my "conjecture" is in the form:

    [tex]\pi(x)\sim{li(x)} [/tex] PNT

    [tex]\sum_{p}p^{n}\sim{li(x^{n+1})} [/tex]

    where we have used the asymptotic property:

    [tex] \int_{o}^{x}t^{n}\sim{1+2^{n}+3^{n}+4^{n}+...... [/tex]

    hope now is much clearer :) :) of course the idea of "Asymptotic" is that

    [tex] f(x)\sim{g(x)}\rightarrow{f(x)/g(x)\sim{1} [/tex]

    By the way Matt..you and other mathematicians have told me that the asymptotic notation does not imply that if f is asymptotic to g then:

    f(x)-g(x)=0 then what does it is good the use of asymptotyc for?..according to you is useless to have an asymptotic expansion for n! if for big n the factorial can not be calculated
     
    Last edited: Mar 22, 2006
  5. Mar 22, 2006 #4

    shmoe

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    This can't work for n=-1, the sum of the reciprocals of the primes diverges, yet Li(x^0) is a constant.

    Why not try to make your 'justification' rigorous, i.e. use partial summation and the prime number theorem including keeping track of error terms.
     
  6. Mar 22, 2006 #5

    matt grime

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    You make that sound like an accusation, that we're lying when we say that f~g does not mean f-g tends to zero. A physicist will tell you exactly the same thing, as will anyone who's actually looked at the definition of asymptotic.

    And when have I, or anyone, said that asymptotic expansions of anything are useless? They have uses, none of them does what you thought they did because you thought that asymptotic meant convergent.

    Asymptotics is about relative error, not absolute error, that's all. Plenty of applicable mathematics uses asymptotics, though I don't anything about what they do with it.
     
  7. Apr 10, 2006 #6

    CRGreathouse

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    If what you meant was that [tex]\lim_{x\rightarrow\infty}\left(f(x)-g(x)\right)=0[/tex] then just say that. If you don't mean that, then why would you give a conjecture with asymptotic notation if you don't believe it's of any use?
     
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