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Divisor function

  1. Dec 24, 2008 #1
    Let t(n) be the divisor function, i.e., the function gives the positive divisors in n including 1 and n. Then I want to show that


    I have tried different ideas but nothing is working can somebody please give some hints.
  2. jcsd
  3. Dec 24, 2008 #2


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    I'm sorry, I don't understand what it is you want to prove. Is that last formula just [itex]2nn^{1/2}= 2n^{3/2}[/itex] and what does it have to do with t(n)?
  4. Dec 25, 2008 #3
    soory .

    t(n) = divisor function, i.e., number of positive divisors of n including n and 1.

    Then I want to show that t(n) < 2*(n)^{1/2}. I mean 2 times squareroot of n.
  5. Dec 29, 2008 #4
    If the prime factorization of n is p1^k1 . p2^k2 . p3^k3 . ..., then t(n) = (k1+1)(k2+2)(k3+3)... Using this, and the fact that the logarithm function is monotonic, maybe (just maybe) taking logs on both sides of the inequality might help. Just an idea.
  6. Jan 16, 2009 #5
    You started in interesting direction, but [tex]f(n) = k![/tex] where [tex]k=\sum(k_i)[/tex]

    Let's start iterative theorem prove:
    then [tex]t(n)=1[/tex] and [tex]1<2\sqrt{n}[/tex] for any [tex]n>1[/tex].

    Prove for k+1
    Now we have to prove that [tex]k!(k+1) < 2\sqrt{n}[/tex]
    Let's suppose that it's not true. Divide both sides on k+1, than we'll have [tex]k! > \frac{2}{k+1}\sqrt{n}[/tex] (1). But we know that [tex]k!<2\sqrt{n}[/tex] and [tex]\frac{2}{k+1}\sqrt{n}<2\sqrt{n}[/tex].

    This means that (1) is impossible and [tex]k!(k+1) < 2\sqrt{n}[/tex] is true.
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