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jamiebookeater
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I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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The function I'm looking for I'll call \(f(n)\). It obeys this relationship to the zeta function: \[\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}\] Here's the relationship: \[[\zeta(s)]^i = \sum_{n=1}^\infty \frac{f(n)}{n^s}\]
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This is one sort of generalization of the Mobius function \(\mu(n)\) which obeys this relationship for anyone who's interested or curious why I'm interested in this. :P \[[\zeta(s)]^{-1} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s}\]

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