作业帮f(x)=sin[π/3(x+1)]-根号3cos[π/3(x+1)],则f(1)+f(2)+f(3)+……+f(2008)等于?
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![f(x)=sin[π/3(x+1)]-根号3cos[π/3(x+1)],则f(1)+f(2)+f(3)+……+f(2008)等于?](/uploads/image/z/2407771-19-1.jpg?t=f%28x%29%3Dsin%5B%CF%80%2F3%28x%2B1%29%5D-%E6%A0%B9%E5%8F%B73cos%5B%CF%80%2F3%28x%2B1%29%5D%2C%E5%88%99f%281%29%2Bf%282%29%2Bf%283%29%2B%E2%80%A6%E2%80%A6%2Bf%282008%29%E7%AD%89%E4%BA%8E%3F)
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f(x)=sin[π/3(x+1)]-根号3cos[π/3(x+1)],则f(1)+f(2)+f(3)+……+f(2008)等于?
f(x)=sin[π/3(x+1)]-根号3cos[π/3(x+1)],则f(1)+f(2)+f(3)+……+f(2008)等于?
f(x)=sin[π/3(x+1)]-根号3cos[π/3(x+1)],则f(1)+f(2)+f(3)+……+f(2008)等于?
∵f(x)=sin[π(x+1)/3]-√3cos[π(x+1)/3]
=sin(πx/3+π/3)-√3cos(πx/3+π/3)
=2sin(πx/3+π/3-π/3)
=2sin(πx/3)
∴f(x)周期T=2π/(π/3)=6
又f(1)+f(2)+f(3)+f(4)+f(5)+f(6)=0
且2008=334×6+4
故f(1)+f(2)+f(3)+……+f(2008)=f(1)+f(2)+f(3)+f(4)=2√3
若f(x)=sinπ/2x,则f(1)+f(2)+f(3)+...+f(2012)+f(2013)=
傅里叶级数作图f（x）=2sin[x] - sin[2x] + 2/3sin[3x] - 1/2sin[4x]我用mathematica输入程序Plot[{2sin[x],-2sin[x],2sin[x] - sin[2x],-2sin[x] + sin[2x],2sin[x] - sin[2x] + 2/3sin[3x],-2sin[x] + sin[2x] - 2/3sin[3x],2sin[x] - sin[2x] + 2/3si
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