狄利克雷特征函数

青青子衿 · 2020-8-17 18:13

这个引理怎么证明呢？
另外所得结果为分数可以同余吗？

Sum[DirichletCharacter[n, 1, i]/i^2, {i, 1, n - 1}] /. {n -> 5}
Sum[DirichletCharacter[n, 1, i]/i^2, {i, 1, n - 1}]/n /. {n -> 5}

tommywong · 2020-10-31 22:07

同餘用逆嘅概念處理分數，$ab\equiv 1\pmod{n}$，則$a^{-1}\equiv b\pmod{n}$
所以如果a,n唔係互質，a就會冇逆，算式就冇意義
$1^{-2}+2^{-2}+3^{-2}+4^{-2}\equiv 1^2+3^2+2^2+4^2\equiv 0\pmod{5}$

tommywong · 2020-11-12 21:57

$ab\equiv 1\pmod{n},~(a,b)=1$
當$a=b$時，a嘅逆就是a
當$a\neq b$時，a嘅逆是b，b嘅逆是a，只係換咗位
於是簡化剩餘系嘅元素取逆後仲係同一個簡化剩餘系
所以$\displaystyle\sum_{1\le i< n\atop (i,n)=1}\frac{1}{i^2}
\equiv\sum_{1\le i< n\atop (i,n)=1}i^2\pmod{n}$

我搵到哩個和等於乜嘢，不過我唔知點解
mathoverflow.net/questions/112113/sum-of-inte … e-to-and-less-than-n
$\displaystyle\sum_{1\le i< n\atop (i,n)=1}i^2
=\frac{1}{3}n^2\varphi(n)+\frac{n}{6}\prod_{p\mid n}(1-p)$
哩個答案就可以解釋晒點解你果三種情況都會整除

tommywong · 2020-11-13 20:20

我搵到方法喇，係Introduction to Analytic Number Theory嘅25頁
faculty.math.illinois.edu/~hildebr/ant/main1.pdf

$\displaystyle\sum_{1\le m\le n\atop (m,n)=1}f(m)
=\sum_{1\le m\le n}f(m)e((m,n))
=\sum_{1\le m\le n}f(m)\sum_{d|(m,n)}\mu(d)$
$\displaystyle
=\sum_{d|n}\mu(d)\sum_{1\le m\le n\atop d|m}f(m)
=\sum_{d|n}\mu(d)\sum_{1\le k\le \frac{n}{d}}f(dk)$

$e(n)=\begin{cases}1 & n=1\\0 & \text{otherwise}\end{cases}$

$\displaystyle\sum_{1\le m\le n\atop (m,n)=1}m^2
=\sum_{d|n}\mu(d)\sum_{1\le k\le \frac{n}{d}}d^2 k^2$
$\displaystyle
=\frac{n}{6}\sum_{d|n}d\mu(d)
\left(\frac{n}{d}+1\right)\left(\frac{2n}{d}+1\right)$
$\displaystyle
=\frac{n}{6}\sum_{d|n}\frac{d^2+3dn+2n^2}{d}\mu(d)$
$\displaystyle
=\frac{n}{6}\prod_{p|n}(1-p)+\frac{n^2}{2}e(n)+\frac{n^2}{3}\varphi(n)$

哩個方法亦都可以求歐拉函數，而且比較簡單

$\displaystyle\varphi(n)=\sum_{1\le m\le n\atop (m,n)=1}1
=\sum_{d|n}\mu(d)\sum_{1\le k\le \frac{n}{d}}1$
$\displaystyle
=\sum_{d|n}\mu(d)\frac{n}{d}
=n\prod_{p|n}\left(1-\frac{1}{p}\right)$

$\displaystyle\sum_{1\le m\le n\atop (m,n)=1}m
=\sum_{d|n}\mu(d)\sum_{1\le k\le \frac{n}{d}}dk$
$\displaystyle=\frac{n}{2}\sum_{d|n}\mu(d)
\left(1+\frac{n}{d}\right)
=\frac{n}{2}\left(e(n)+\varphi(n)\right)$