广义牛顿二项式定理

$${r \choose n}= \begin{cases} 0， & n<0\\ 1， & n=0\\ \frac{r(r-1)\cdots (r-n+1)}{n!}， & n>0 \end{cases}$$

$(x+y)^{\alpha}=\sum_{n=0}^\infty{\alpha \choose n}x^{n}y^{\alpha-n}$

$(x+y)^{\alpha}=y^{\alpha}(1+z)^{\alpha}，\mid z\mid <1$

$f^{(n)}(z)=\alpha (\alpha -1)\cdots (\alpha -n+1)(1+z)^{\alpha -n}$

$(1+z)^{\alpha} =1+\frac{\alpha}{1!}z+\frac{\alpha (\alpha -1)}{2!}z^{2}+\cdots +\frac{\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}z^{n}+r_n(0;z)$

$(1+z)^{\alpha}=1+{\alpha \choose 1}z+{\alpha \choose 2}z^{2}+\cdots +{\alpha \choose n}z^{n}+r_n(0;z)$

$r_n(0;z)=\frac{\alpha (\alpha -1)\cdots (\alpha -n)}{n!}(1+ξ)^{\alpha -n}(z-ξ)^{n}z$

$r_n(0;z)=\alpha (1-\frac{\alpha}{1})\cdots (1-\frac{\alpha}{n})(1+ξ)^{\alpha}(\frac{z-ξ}{1+ξ})^{n}z$

$\mid \frac{z-ξ}{1+ξ}\mid =\frac{\mid z\mid -\mid ξ\mid}{\mid 1+ξ\mid}\leq \frac{\mid z\mid -\mid ξ\mid}{1-\mid ξ\mid}=1-\frac{1-\mid z\mid}{1-\mid ξ\mid}\leq 1-(1-\mid z\mid)=\mid z\mid$

posted @ 2019-08-25 01:59  Asika391  阅读(4652)  评论(0编辑  收藏  举报