数学问题——解方程解方程（x+1
(x+1)^n=(x-1)^n
--->[(x+1)/(x-1)]^n=1
所以(x+1)/(x-1)=Ek,
其中Ek是1的n次方根
Ek=cos(2kpi/n)+isin(2kpi/n) k=0,1,2,……,n-1。
所以xEk-Ek=x+1
x(Ek-1)=Ek+1
x=(Ek+1)/(Ek-1)
=[1+cos(2kpi/n)+isin(2kpi/n)]/[-1+cos(2kpi/n)+isin(2kpi/n)]
={2[cos(kpi/n)]^2+2isin(kpi/n)cos(kpi/n)}
/{-2[sin(kpi/n)]^2+2isin(kpi/n)cos(kpi/n...全部
(x+1)^n=(x-1)^n
--->[(x+1)/(x-1)]^n=1
所以(x+1)/(x-1)=Ek,
其中Ek是1的n次方根
Ek=cos(2kpi/n)+isin(2kpi/n) k=0,1,2,……,n-1。
所以xEk-Ek=x+1
x(Ek-1)=Ek+1
x=(Ek+1)/(Ek-1)
=[1+cos(2kpi/n)+isin(2kpi/n)]/[-1+cos(2kpi/n)+isin(2kpi/n)]
={2[cos(kpi/n)]^2+2isin(kpi/n)cos(kpi/n)}
/{-2[sin(kpi/n)]^2+2isin(kpi/n)cos(kpi/n)}
=cos(kpi/n)[cos(kpi/n)+isin(kpi/n)]
/{isin(kpi/n)[cos(kpi/n)+isin(kpi/n)]}
=cos(kpi/n)/[isin(kpi/n)]
=-icot(kpi/n)。
k=0,1,2,3,……,n-1。收起
