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Math Help - Liouville's Function

  1. #1
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    Liouville's Function

    Let

    f(n)=\underset{d|n}{\sum}d^{2}\lambda(d)

    where \lambda(n) is Liouville's function.

    How do I evaluate the sum

    \underset{n=1}{\overset{\infty}{\sum}}f(n)\: n^{-4}

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Cairo View Post
    Let

    f(n)=\underset{d|n}{\sum}d^{2}\lambda(d)

    where \lambda(n) is Liouville's function.

    How do I evaluate the sum

    \underset{n=1}{\overset{\infty}{\sum}}f(n)\: n^{-4}

    Thanks in advance.
    Here's a broad hint: First, do you know the identity

    \frac{\zeta(2s)}{\zeta(s)} = \underset{n=1}{\overset{\infty}{\sum}}\frac{\lambd  a(n)}{n^s}

    ? It may be found in the Wikipedia article on Liouville's function, or as Theorem 300 in Hardy and Wright.

    Next, write your sum as

    \underset{n=1}{\overset{\infty}{\sum}}\underset{d|  n}{\sum}\frac{d^{2}\lambda(d)}{n^4}

    Write out the first nine or so terms and collect together terms with \lambda(1), \lambda(2), \lambda(3) and so on. Simplify, using the identity

    \zeta(4) = \underset{n=1}{\overset{\infty}{\sum}}\frac{1}{n^4  }

    where \zeta is the Riemann zeta function. You should end up with an expression that involves the identity that I mentioned at the beginning of this post (with s = 2). Finally, use the well-known identities

    \zeta(2) = \frac{\pi^2}{6}

    and

    \zeta(4) = \frac{\pi^4}{90}

    to arrive at the final answer.
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  3. #3
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    offtopic but can you tell me the book by hardy and right which you are suggesting here ?
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  4. #4
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    Quote Originally Posted by kira View Post
    offtopic but can you tell me the book by hardy and right which you are suggesting here ?
    An Introduction to the Theory of Numbers by Hardy and Wright.
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  5. #5
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    Thanks for your hint, Petek.

    I'm not sure I follow

    You should end up with an expression that involves the identity that I mentioned at the beginning of this post (with s = 2).
    When I write out the terms of the sum, I can see that this is equal to \zeta(4), but can't see how to evaluate it from here.

    Am I right in thinking that

    \frac{\zeta(2s)}{\zeta(s)}=\underset{n=1}{\overset  {\infty}{\sum}}\frac{\lambda(n)}{n^{s}}=\zeta(4)

    so that when s=2

    \underset{n=1}{\overset{\infty}{\sum}}\frac{\lambd  a(n)}{n^{4}}=\zeta(2)\zeta(4)=\frac{\pi^{6}}{540}

    Regards
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  6. #6
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    Here's what I had in mind:

    \underset{n=1}{\overset{\infty}{\sum}}\frac{f(n)}{  n^4}
    = f(1) + \frac{f(2)}{2^4} + \frac{f(3)}{3^4} + \frac{f(4)}{4^4} + \cdots
    = 1 + \frac{1 + 2^2\lambda(2)}{2^4} + \frac{1 + 3^2\lambda(3)}{3^4} + \frac{1 + 2^2\lambda(2) + 4^2\lambda(4)}{4^4} + \cdots
    = (1 + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots)
    + (\frac{2^2}{2^4} + \frac{2^2}{4^4} + \cdots)\lambda(2)
    + (\frac{3^2}{3^4} + \frac{3^2}{6^4} + \cdots)\lambda(3)
    + (\frac{4^2}{4^4} + \frac{4^2}{8^4} + \cdots)\lambda(4)
    + ...

    Can you take it from there? If not, what did you try?
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  7. #7
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    I wrote the same and concluded that the sum is equal to zeta(4). I then used the identity you gave to reach the answer I provided above.
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  8. #8
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    OK, now I see where we're not connecting. In one of your above posts you wrote

    \frac{\zeta(2s)}{\zeta(s)}=\underset{n=1}{\overset  {\infty}{\sum}}\frac{\lambda(n)}{n^{s}}=\zeta(4)
    However, for s = 2, we get \frac{\zeta(4)}{\zeta(2)}, not \zeta(4). Making that change should give you the same answer that I got.
    Last edited by Petek; March 7th 2011 at 12:30 PM. Reason: Typo
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  9. #9
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    Quote Originally Posted by Petek View Post
    OK, now I see where we're not connecting. In one of your above posts you wrote

    However, for s = 2, we get \frac{\zeta(4)}{\zeta(2)}, not \zeta4). Making that change should give you the same answer that I got.
    I'm not sure I'm following you here.
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  10. #10
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    First, remember that we're trying to evaluate

    \underset{n=1}{\overset{\infty}{\sum}}f(n)\: n^{-4}

    Refer to my equations in post #6 and let's continue from there:

    \underset{n=1}{\overset{\infty}{\sum}}\frac{f(n)}{  n^4}

    = \zeta(4) \frac{\lambda(1)}{1^2}

    + \frac{2^2}{2^4}(1 + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots)\lambda(2)

    + \frac{3^2}{3^4}(1 + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots)\lambda(3)

    + \frac{4^2}{4^4}(1 + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots)\lambda(4)

    + \cdots

    = \zeta(4)\frac{\lambda(1)}{1^2} + \zeta(4)\frac{\lambda(2)}{2^2} + \zeta(4) \frac{\lambda(3)}{3^2} + \cdots

    = \zeta(4) \underset{n=1}{\overset{\infty}{\sum}}\frac{\lambd  a(n)}{n^{2}}

    (Since \frac{2^2}{2^4} = \frac{1}{2^2},\ \frac{3^2}{3^4} = \frac{1}{3^2}, and so on.)

    =\frac{\zeta^2(4)}{\zeta(2)}
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  11. #11
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    pi^6 / 1350 ?

    Thanks
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  12. #12
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    Quote Originally Posted by Cairo View Post
    pi^6 / 1350 ?

    Thanks
    That's what I got. You're welcome.
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