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From: Leroy Quet on 11 Jul 2010 15:52
Leroy Quet wrote:
\...

\ sum{j=1 to n} sum{k|GCD(j,n)} a(k) * cos(2 pi j/n) =
\
\ a(n),
\ for all positive integers n.

This easily leads to a less-well-known inversion formula:
({b(k)} is any sequence.)

a(n) =

sum{k|n} mu(n/k) b(k)

= (tuh duh!...)

sum{k=1 to n} cos(2 pi k/n) b(GCD(k,n)).

Here, sum{k|n} a(k) = b(n) for all positive integers n, of course.

The inversion formula also is:

a(n) =
sum{k=1 to n} exp(i 2 pi k/n) b(GCD(k,n)),

obviously.

Thanks,
Leroy Quet

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