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[watermark]X^9+X^8+X^7+X^6+X^5+X^4+X^3+X^2+X+1,由于它的表达式有点复杂,所以有必要进行"化简".
注意到它是首项为1,公比为X,的等比数列的前十项和.
根据等比数列的前n项和的求和公式,得:
X^9+X^8+X^7+X^6+X^5+X^4+X^3+X^2+X+1=(X^10-1)/(X-1)
先考虑对X^10-1进行因式分解
令X^10-1=0,既:X^10=1=1(cos0+isin0)
根据棣莫佛定理,得:x[k]=cos(2kπ/10)+isin(2kπ/10),(k=0,1,2,3,4,5,6,7,8,9)
x[k]是1的10个原根,它把单位圆十等份.
x0=1
x1=cos36°+icos36°
x2=cos72°+icos72°
x3=cos108°+icos108°
x4=cos144°+icos144°
x5=cos180°+icos180°=-1
x6=cos216°+icos216°=-(cos36°+icos36°)
x7=cos252°+icos252°=-(cos72°+icos72°)
x8=cos288°+icos288°=-(cos108°+icos108°)
x9=cos324°+icos324°=-(cos144°+icos144°)
x10=cos360°+icos360°
所以X^10-1=(X-1)(X+1)[X-(cos36°+icos36°)][X-(cos72°+icos72°)][X-(cos108°+icos108°)][X-(cos144°+icos144°)][X+(cos36°+icos36°)][X+(cos72°+icos72°)][X+(cos108°+icos108°)][X+(cos144°+icos144°)]
所以X^9-1=(X+1)
*[X-(cos36°+icos36°)][X-(cos72°+icos72°)][X-(cos108°+icos108°)][X-(cos144°+icos144°)]
*[X+(cos36°+icos36°)][X+(cos72°+icos72°)][X+(cos108°+icos108°)][X+(cos144°+icos144°)]
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