2020年12月7日 星期一

證明: sin(π/n)×sin(2π/n)×sin(3π/n)×...×sin((n-1)π/n) =n/2^(n-1)

正弦連乘積公式

sinπnsin2πnsin(n1)πn=n2n1
證明:
cosθ+isinθ=eiθsinθ=eiθeiθ2isinkπn=eikπneikπn2i=eikπn2i(ei2kπn1)sinπnsin2πnsin(n1)πn=eiπn(1+2++(n1))(2i)n1(ei2πn1)(ei4πn1)(ei2(n1)πn1)=eiπ(n1)2(2i)n1(ei2πn1)(ei4πn1)(ei2(n1)πn1)=(2i)1neiπ(n1)2(ω1)(ω21)(ωn11),ω=ei2πn=21neiπ2(1n)eiπ(n1)2(ω1)(ω21)(ωn11)(i=eiπ2i1n=eiπ2(1n))=21nei(1n)π(ω1)(ω21)(ωn11)=21n(1)(1n)(ω1)(ω21)(ωn11)=(2)1n(ω1)(ω21)(ωn11)=(2)1n(1)1n(1ω)(1ω2)(1ωn1)=21n(1ω)(1ω2)(1ωn1)f(x)=xn1=(x1)(1+x+x2++xn1)f(x)=01,ω,ω2,,ωn1,ω=ei2πnf(x)=(x1)(1+x+x2++xn1)=(x1)(xω)(xω2)(xωn1)1+x+x2++xn1=(xω)(xω2)(xωn1)x=1n=(1ω)(1ω2)(1ωn1)sinπnsin2πnsin(n1)πn=21n(1ω)(1ω2)(1ωn1)=21nn=n2n1,

沒有留言:

張貼留言