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 $$ f(x)= g(x)b(1)+g(x/2)b(2)+g(x/3)b(3) $$
mobius-inversion
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up vote
0
down vote
favorite
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
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
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
mobius-inversion
asked Sep 2 at 10:54
Jose Garcia
4,02511235
4,02511235
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