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Math Help - Mobius Inversion problem

  1. #1
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    Mobius Inversion problem

    Suppose that X(n) is defined whenever n is a divisor of 1000, and suppose that the equation SIGMA d|n X(d) = n^2 is satisfied for all n|1000. Use Mobius inversion to find X(1000).

    I know that SIGMA n|1000 X(1000) = n^2

    so X(1000) = n^2 * SIGMA MU( 1000/n) *


    I find all the divisors of 1000, which will give me my n, then sub it into *

    1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000

    so is X(1000) = 1^2 * MU (1000/1) + 2^2 * MU (1000/2) .....

    I know my working out is pretty general.. I'm just after a lead

    Thanks
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  2. #2
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    Quote Originally Posted by ikurwae89 View Post
    Suppose that X(n) is defined whenever n is a divisor of 1000, and suppose that the equation SIGMA d|n X(d) = n^2 is satisfied for all n|1000. Use Mobius inversion to find X(1000).

    I know that SIGMA n|1000 X(1000) = n^2

    so X(1000) = n^2 * SIGMA MU( 1000/n) *


    I find all the divisors of 1000, which will give me my n, then sub it into *

    1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000

    so is X(1000) = 1^2 * MU (1000/1) + 2^2 * MU (1000/2) .....

    I know my working out is pretty general.. I'm just after a lead

    Thanks

    It looks fine...did you notice that up to 50 the value of mu is zero? After that only 4 values get non-zero values and they

    cancel each other...

    Tonio
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