Averages of euler

Example : $sumlimits_dphi(d) = phi(1) + phi(2) + phi(4) = 1 + 1 + 2 = 4$

I was told this has a simple proof. Problem is, I can not think of a way to lớn show this in a very simple & straightforward way.




Bạn đang xem: Averages of euler

*

*

Let $G=langle x angle$ be the cyclic group of order $n$ generated by $x$. Then,eginalignn &= |G| \&= sum_1leq dleq n |\textelements of order d| \&= sum_n|\textelements of order d|endalignby Lagrange"s Theorem. But $G$ has a quality cyclic subgroup of order $d$ for each $d|n$, namely $langle x^n/d angle$. Moreover, each such subgroup has $varphi(d)$ generators, so$$n = sum_dvarphi(d)$$


*

Hint: You can show that $sumlimits_nphi(d) = sumlimits_dphi(fracnd) $ but $phi(fracnd)$ is all numbers with GCD of $d$ with $n$. Therefore the last sum counts all numbers till $n$ & is equal lớn $n$.


*

The simplest proof I know is this:

Consider all proper fractions of the size $a/n$. There are $n$ of those. When you consider their reduced forms you get fractions of the khung $b/d$ with $d|n$ and $(b,d)=1$. By definition, there are $phi(d)$ of those. The result follows.


*

Here"s one approach:

Consider the polynomial $x^n-1$ as a polynomial over the complex numbers $au-79.netbbC$. You can easily see, almost tautologically, that

$$x^n-1=prod_dmid nPhi_d(x)qquad(ast)$$

where $Phi_d(x)$ is the polynomial whose roots are the primitive $d^ extth$ roots of unity (i.e. solutions to $x^d-1$ that bởi vì not satisfy $x^c-1$ for $0




*












*



*







Xem thêm: Người Sinh Năm 74 Mệnh Gì, Tuổi Con Gì, Hợp Hướng Nào Và Hợp Màu Gì Nhất?



27

0

3

3

-3


12

8

2

5

0

1




Staông xã Exchange Network

au-79.netematics Stachồng Exchange works best with JavaScript enabled
*

Accept all cookies Customize settings



Xem thêm: Cách Khắc Chế Morgana Sp - Cách Khắc Chế Tướng Morgana Mùa 11



Chuyên mục: Blockchain