Cosx+Cos2x+Cos3x+……Cosnx=?如何用复数的 方法解

e^(ix)=cosx+isinx e^(ix)+e^(i2x)+e^(i3x)+……+e*(inx)=(cosx+cos2x+……+cosnx)+i(sinx+sin2x+……+sinnx) =[e^(inx+ix) -e^(ix)]/[e^(ix)-1]; 将最后一个等号右端分成实部和虚部(分母和分子同乘以 (cosx-1)-isinx),与等号左端实部和虚部对应相等...

乘以2sinx, 积化和差就变成了 sin2x-0+sin3x-sinx+sin4x-sin2x+...+ sinnx-si(n-2)x+sin(n+1)x-sin(n-1)x =sin(n+1)x+sinnx-sinx 再除以2sinx,即为答案,[sin(n+1)x+sinnx-sinx]/2sinx

利用 e^(ix)=cosx+isinx; e^(ix)+e^(i2x)+e^(i3x)+……+e*(inx)=(cosx+cos2x+……+cosnx)+i(sinx+sin2x+……+sinnx) =[e^(inx+ix) -e^(ix)]/[e^(ix)-1]; 将最后一个等号右端分成实部和虚部(分母和分子同乘以 (cosx-1)-isinx),与等号左端实部和虚部...

dfh

cosx+cos2x+cos3x+....+cosnx =sin(x/2)*[ cosx+cos2x+cos3x+....+cosnx] / sin(x/2) ( 将sin(x/2) 移入方括号里并化简) = {sin[x(2n+1)/2] - sin(x/2) }/ [2sin(x/2)]

cosx+cos2x+......+cosnx =1/2sin(x/2)*(cosx*2sin(x/2)+cos2x*2sin(x/2)+......+cosnx*2sin(x/2)) =1/2sin(x/2)*(sin(3x/2)-sin(x/2)+sin(5x/2)-sin(3x/2)+......+sin(n+1/2)x-sin(n-1/2)x) =1/2sin(x/2)*(sin(n+1/2)x-sin(x/2)) =1/2sin(x/2)*...

n(n+1)(2n+1)/12

#include #include float sum(float x,int n) //定义一个求和函数sum { int i; double sum=0.0; for(i=1;i

我提供一个求均值的解答,方差是类似地。

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