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  1. #1
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    average order

    How do you find the value of  \sum\limits_{n \leq x} \mu(n)\Bigl[\frac{x}{n}\Bigr]^{2}
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Chandru1 View Post
    How do you find the value of  \sum\limits_{n \leq x} \mu(n)\Bigl[\frac{x}{n}\Bigr]^{2}
    What do you mean by "find the value"? Do you mean a closed form expression without a  \sum ?
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  3. #3
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    Average order

    well

    Prove that \sum\limits_{n \leq x} \mu(n) \left[\frac{x}{n}\right]= \frac{1}{\zeta(2)} x^{2} + O(x \log{x})
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Chandru1 View Post
    well

    Prove that \sum\limits_{n \leq x} \mu(n) \left[\frac{x}{n}\right]= \frac{1}{\zeta(2)} x^{2} + O(x \log{x})
    Wait, is it  \left[\frac{x}{n}\right] or  \left[\frac{x}{n}\right]^2 ?
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    I'm having trouble getting the desired error term, but I'm close to solving the problem. Maybe you could polish my method to get what you want...

     \sum\limits_{n \leq x} \mu(n)\Bigl[\frac{x}{n}\Bigr]^{2} = \sum\limits_{n \leq x} \mu(n)\left(\frac{x}{n}-\left\{\frac{x}{n}\right\}\right)^{2}

     = \sum\limits_{n \leq x} \mu(n)\frac{x^2}{n^2}-2\sum\limits_{n \leq x} \mu(n)\frac{x}{n}\left\{\frac{x}{n}\right\}+\sum\l  imits_{n \leq x} \mu(n)\left\{\frac{x}{n}\right\}^{2}

     = x^2\sum\limits_{n \leq x}\frac{\mu(n)}{n^2}-2x\cdot O\left(\sum\limits_{n \leq x}\frac{\mu(n)}{n}\right)+O\left(\sum\limits_{n \leq x} \mu(n)\right)

    First sum:  x^2\sum\limits_{n \leq x}\frac{\mu(n)}{n^2} = x^2\sum\limits_{n=1}^{\infty} \frac{\mu(n)}{n^2} - x^2\sum\limits_{n > x}\frac{\mu(n)}{n^2} = \frac{x^2}{\zeta(2)} + O\left(x^2\sum_{n>x}\frac{1}{n^2}\right) \longleftarrow Here's where you need to improve the error term.

    Second sum:  2x\cdot O\left(\sum\limits_{n \leq x}\frac{\mu(n)}{n}\right) = o(x) since  \sum\limits_{n \leq x}\frac{\mu(n)}{n} = o(1) (which is equivalent to the prime number theorem!).

    Third sum:  O\left(\sum\limits_{n \leq x} \mu(n)\right) = O(x) since  |\mu(n)|\leq 1 .


    Enumerating we get  \sum\limits_{n \leq x} \mu(n)\Bigl[\frac{x}{n}\Bigr]^{2} = \frac{x^2}{\zeta(2)} + O(f(x)) + o(x) + O(x) = \frac{x^2}{\zeta(2)} + O(f(x)) . Where  f(x) is the error term that you need to improve upon.
    Last edited by chiph588@; March 17th 2010 at 11:08 AM.
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  6. #6
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    Mistake

    <br /> <br />
    " alt="= \sum\limits_{n \leq x} \mu(n)\frac{x^2}{n^2}+2\sum\limits_{n
    \leq x} \mu(n)\frac{x}{n}\left\{\frac{x}{n}\right\}+\sum\l imits_{n \leq
    x} \mu(n)\left\{\frac{x}{n}\right\}^{2}
    " />


    before this a minus sign..
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  7. #7
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    Done

    Hi--


    i think that i have found the solution, Here i upload it in a file. Just check it.
    Attached Files Attached Files
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Chandru1 View Post
    Hi--


    i think that i have found the solution, Here i upload it in a file. Just check it.
    That looks good, could you show me where you learned  \sum_{n\leq x}\frac{1}{n^s} = \frac{x^{1-s}}{1-s}+\zeta(s)+O(x^{-s}) ?

    The formula looks familiar, but I can't find it anywhere for some reason.
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  9. #9
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    Proof of the formula

    Dear

    Chiph, sorry for the delay in writing. I hope you are aware of the Eulers summation formula which says:

    If f has a continuous derivative on the interval [y,x] then

    \sum\limits_{y<n \leq x}f(n) = \int\limits_{y}^{x} f(t) \ dt + \int\limits_{y}^{x} (t-[t]) f'(t) \ dt + f(x)([x]-x) -f(y)([y]-y)

    Please apply f(n)= \frac{1}{n^s} in the Eulers Summation formula to get the desired result.


    All of these can be found in Tom Apostol's Analytic Number Theory book.
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  10. #10
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Chandru1 View Post
    Dear

    Chiph, sorry for the delay in writing. I hope you are aware of the Eulers summation formula which says:

    If f has a continuous derivative on the interval [y,x] then

    \sum\limits_{y<n \leq x}f(n) = \int\limits_{y}^{x} f(t) \ dt + \int\limits_{y}^{x} (t-[t]) f'(t) \ dt + f(x)([x]-x) -f(y)([y]-y)

    Please apply f(n)= \frac{1}{n^s} in the Eulers Summation formula to get the desired result.


    All of these can be found in Tom Apostol's Analytic Number Theory book.
    Of course! it just slipped my mind!
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  11. #11
    Super Member PaulRS's Avatar
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    Just to point out a combinatorial interpretation of the sum  \sum\limits_{k \leq n} \mu(k)\left\lfloor\frac{n}{k}\right\rfloor^{2}

    It's equal to the cardinal of \left\{(x,y)\in\left(\left[n\right]\right)^2\text{such that gcd}(x,y)=1\right\}<br />
(1)

    The explanation is simple:

    Fix n, let A_p be \left\{(x,y)\in\left(\left[n\right]\right)^2\text{such that } p|x \wedge p|y\right\} (for each prime p\leq n ).

    Now (1) is the set of pairs of integers in [n]=\left\{1,2,...,n\right\}, such that they - the pairs- do not belong to the union \bigcup_{p\leq n}A_p.
    Note that A_{i_1}\cup...\cup A_{i_m}=\left\{(x,y)\in\left(\left[n\right]\right)^2\text{such that }(i_1...i_m)|x\wedge (i_1...i_m)|y\right\} -since we are working with primes- which cardinal is \left\lfloor\frac{n}{i_1\cdot ...\cdot i_m}\right\rfloor^2

    The rest is inclusion-exclusion. Evidently you can substitute that 2 for any m\in\left\{2,3,...\right\}
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