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Math Help - Showing existence of 2 constants

  1. #1
    Senior Member Dinkydoe's Avatar
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    Showing existence of 2 constants

    I want to show there exists 2 constants c_1,c_2> 1 such that:

    c_1^n\leq z_n\leq c_2^n where z_n=\mathrm{lcm}(1,\cdots,n)=\prod_{p\leq n}p^{b_p}},

    where b_p is the largest number such that p^{b_p}\leq n

    I accomplished to show the following: (more or less trivial)

    z_{n+1}=z_n\cdot n as n is prime
    z_{n+1}=z_n\cdot p as n+1=p^{b_p} (i.e a power of only one prime)
    z_{n+1}=z_n otherwise (i.e. n+1 has at least 2 prime-divisors)

    I believe these are useful observations, but I'm more or less stuck here. I hope someone can give me a little push in the right direction here...
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  2. #2
    Senior Member roninpro's Avatar
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    Have you considered taking c_1=2 and c_2=n?
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by roninpro View Post
    Have you considered taking c_1=2 and c_2=n?
    Is it true that 2^3 \leq \mbox{lcm}(1,2,3) ? Moreover n is not a constant.
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  4. #4
    Senior Member roninpro's Avatar
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    Thank you. I hadn't checked small cases.

    After rereading the post, I see that c has not been restricted to an integer and that c_1, c_2 are to be fixed.
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  5. #5
    Senior Member Dinkydoe's Avatar
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    I think I solved it
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Dinkydoe View Post
    I think I solved it
    You can't just say that! You have to share!
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  7. #7
    Senior Member Dinkydoe's Avatar
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    Epic Fail! But i got it now :P

    Ill share it soon!
    Last edited by Dinkydoe; November 17th 2010 at 06:09 AM.
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  8. #8
    Senior Member Dinkydoe's Avatar
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    OK, so i'll remind you that \pi(n) is bounded the following way:  \frac{1}{2}\frac{n}{\log n}\leq \pi(n)\leq \frac{3n}{\log n}

    Here \pi(n) denotes the number of primes p, with p\leq n. Recall that \mathrm{lcm}(1,\cdots,n)=\prod_{p\leq n}p^{b_p}

    Thus we have \sqrt{n}^{\pi(n)}\leq \mathrm{lcm}(1,\cdots,n)\leq n^{\pi(n)}. This leads to the following:

    \left(e^{\frac{1}{4}}\right)^n= \sqrt{n}^{\frac{n}{2\log n}}\leq \mathrm{lcm}(1,\cdots,n)\leq n^{\frac{3n}{\log n}}= (e^3)^n

    This proves the claim.

    I don't have to bother myself with the boundaries for \pi(n). The proof for this is in my course-notes.
    Last edited by Dinkydoe; November 20th 2010 at 05:54 AM.
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