usig mobius inversion to solve this problem?

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let be the functional equation



$$ g(x) = f(x/2)log(2) +f(x/3)log(3)+ f(x/4)log(4).... $$



where log is the logarithm in basis 'e'



how could i use mobius inversion formula to obtain the function$ f(x)$ from the



equation above ?



i believe that if we call $$ frac1zeta ' (s) = sum_n=1^infty fracb(n)n^s $$



so $$ f(x)= g(x)b(1)+g(x/2)b(2)+g(x/3)b(3) $$










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    let be the functional equation



    $$ g(x) = f(x/2)log(2) +f(x/3)log(3)+ f(x/4)log(4).... $$



    where log is the logarithm in basis 'e'



    how could i use mobius inversion formula to obtain the function$ f(x)$ from the



    equation above ?



    i believe that if we call $$ frac1zeta ' (s) = sum_n=1^infty fracb(n)n^s $$



    so $$ f(x)= g(x)b(1)+g(x/2)b(2)+g(x/3)b(3) $$










    share|cite|improve this question























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      down vote

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      let be the functional equation



      $$ g(x) = f(x/2)log(2) +f(x/3)log(3)+ f(x/4)log(4).... $$



      where log is the logarithm in basis 'e'



      how could i use mobius inversion formula to obtain the function$ f(x)$ from the



      equation above ?



      i believe that if we call $$ frac1zeta ' (s) = sum_n=1^infty fracb(n)n^s $$



      so $$ f(x)= g(x)b(1)+g(x/2)b(2)+g(x/3)b(3) $$










      share|cite|improve this question













      let be the functional equation



      $$ g(x) = f(x/2)log(2) +f(x/3)log(3)+ f(x/4)log(4).... $$



      where log is the logarithm in basis 'e'



      how could i use mobius inversion formula to obtain the function$ f(x)$ from the



      equation above ?



      i believe that if we call $$ frac1zeta ' (s) = sum_n=1^infty fracb(n)n^s $$



      so $$ f(x)= g(x)b(1)+g(x/2)b(2)+g(x/3)b(3) $$







      mobius-inversion






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      asked Sep 2 at 10:54









      Jose Garcia

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