【学习笔记】数论、数学—补充证明 (1)

【学习笔记】数论、数学—补充证明 (1)

\(\text{结论:}\)

\[\sum_{d|n}\varphi(d)=n\\ (\varphi*1=\operatorname{id}) \]

\(\text{证明:}\ \text{设}\ f=\varphi*1\)

\(\forall p \in \{Prime\},\varphi(p)=p-1,\ \text{则有:}\)

\(\begin{align} f(p^k)&=\sum_{d|p^k}\varphi(d)\\ &=\sum_{i=0}^{k}\varphi(p^i)\\ &=\varphi(1)+\sum_{i=1}^{k}\varphi(p^i)\\ &=1+\sum_{i=1}^{k}\varphi(p)*p^{i-1}\\ &=1+\varphi(p)\sum_{i=1}^{k}p^{i-1}\\ &=1+\varphi(p)\left({\frac{p^0(1-p^k)}{1-p}}\right)\\ &=1+(p-1)\left({\frac{p^k-1}{p-1}}\right)\\ &=p^k\\ \end{align}\)

\(\text{当}\ \gcd(x,y)=1\ \text{时},\ f(xy)=\sum_{d|xy}\varphi(d)=\left(\sum_{d|x}\varphi(d)\right) * \left(\sum_{d|y}\varphi(d)\right)=f(x)*f(y)\)

\(\text{所以}\ f(n)=\prod_{i=1}^{k}f({p_i}^{c_i})=\prod_{i=1}^{k}{p_i}^{c_i}=n\)

\(\text{即:}\ \varphi*1=\operatorname{id}\)

\(\text{证毕。}\)

posted @ 2019-12-05 11:28  辰星凌  阅读(143)  评论(0编辑  收藏