ln 与 exp

liu_cheng_ao 2018-07-10 21:07:11 2018-07-10 22:26:23

对于交换环 R ,且 \mathbb{Z}^* 元素在 R 中的同态可逆,那么对于 x\in R 我们定义 \ln \exp

\begin{matrix} e^x & = & \sum_{i=0}^\infty \frac{x^i}{i!} \\ \ln x & = & \sum_{i=1}^\infty \frac{(-1)^{i-1}(x-1)^i}{i} \\ \end{matrix}

那么:

\begin{matrix} e^xe^y & = & (\sum_{i=0}^\infty \frac{x^i}{i!})(\sum_{i=0}^\infty \frac{y^i}{i!}) \\ & = & \sum_{i=0,j=0}^\infty \frac{x^iy^j}{i!j!}\\ & = & \sum_{i=0,j=0}^\infty \frac{x^iy^j\binom{i+j}{i}}{(i+j)!}\\ & = & \sum_{s=0}^\infty \frac{x^iy^{s-i}\binom{s}{i}}{s!}\\ & = & \sum_{s=0}^\infty \frac{(x+y)^s}{s!}\\ & = & e^{x+y} \\ \\ \ln e^x & = & \sum_{i=1}^\infty \frac{(-1)^{i-1}(\sum_{j=1}^\infty \frac{x^j}{j!})^i}{i} \\ & = & \sum_{i=1}^\infty (-1)^{i-1}\frac{\sum_{s=\{a_1,a_2,\cdots,a_{\lvert s\rvert}\},\lvert s\rvert = i}\frac{x^{\sum_j s_j}}{\prod_j s_j!}}{i} \\ & = & \sum_{i=1}^\infty (-1)^{i-1}\frac{\sum_{s=\{a_1,a_2,\cdots,a_{\lvert s\rvert}\},\lvert s\rvert = i}\frac{x^{\sum_j s_j}\binom{\sum_j s_j}{s_1,s_2,\cdots,s_{\lvert s\rvert}}}{(\sum_j s_j)!}}{i} \\ & = & \sum_{c=1}^\infty \frac{x^c}{c!} \sum_{s=\{a_1,a_2,\cdots,a_{\lvert s\rvert}\},\sum_j s_j=c} (-1)^{\lvert s\rvert -1}\binom{\sum_j s_j}{s_1,s_2,\cdots,s_{\lvert s\rvert}}\frac{1}{\lvert s\rvert} \\ & = & \sum_{c=1}^\infty \frac{x^c}{c!} [c=1] \\ & = & x \\ \\ e^{\ln x} & = & \dots \\ & = & x \\ \end{matrix}

共 1 条回复

Secret_tL

请填之前没填完的坑orz!