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mobius function |
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| Nov14-08, 09:02 AM | #1 |
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mobius function
By using mobius inversion's formula how can I prove that
P(x) = sum_k 1/kp(x^{1/k}) where p(x) = sum_k u(k)/kP(x^{1/k}), here u is the mobius function. here both sums are infinite. |
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