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- Id: 736778
- Posted: 2011-12-14 08:55:28
by danbooru - Size: 1296x1812
- Source: img63.pixiv.net/img/laevateinn495/17211423_big_p3.jpg
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That's right. But to have a generalized proof of that is difficult. The next segment will simplify it with the <a href="http://en.wikipedia.org/wiki/Moebius_function">Moebius Function</a>.
Theorem 10.2 seems like a really neat theorem, but in reality, I don't know if there's any other function in number theory that has that property other than phi(n). To show that, we want to prove that if SUM(F(d), d|n)=n, then F(n) = phi(n). How do we prove that? Given that F(n) is any function in number theory, G(n)=SUM(F(n), d|n) is also a function in number theory. Let's use a concrete example to think about how to obtain F(n) from G(n). Let n=35, then we have d=1, 5, 7, 35... Please, Miss Flan?
[Theorem 10.2] SUM(phi(d), d|n)=n [Proof] Suppose that d is a divisor of n. Further, suppose that we have a natural number x other than 1 or n such that (x, n) = d. Consequently we have x=dx', and n=dn' and (x', n') = 1, and the number of x' is the value of Euler's function for n/d, that is, phi(n/d). Similarly evaluate all n's proper divisors other than 1, and we have SUM(phi(n/d), d|n) = n. Again, n/d is n's divisor, like d. Therefore, we have SUM(phi(d), d|n) = n. I have actually never learned anything about Euler's Function, cyclotonic polynomials, and generators for cyclic groups, so please feel free to correct me.
<a href="http://en.wikipedia.org/wiki/Euler%27s_totient_function">Euler's function</a>
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