证明设由条件,知(3)因为互不相同,所以方程组(3)的系数行列式则方程组(3)有唯一解,即唯一的次数小于的多项式使得,例3设多项式,,则不可能有非零且重数大于的根。证明反设是的重数大于的根,则=0,进而即(4)把(4)看成关于为未知量的齐次线形方程组则(4)的系数行列式=所以方程组(4)只有零解,从而,所以必有这与矛盾,故没有非零且重数大于的根。附件:(外文资料原文)NewproofoftheVandermondedeterminantandsomeapplications(A):anewmethodofproof:mathematicalinductionWeonthenforthattheinductivemethod.(1)When,Whentheresultisright.(2)TheVandermondedeterminantconclusionassumptionsfortheclass,nowlookatthelevelof.in,Subtractingtherowstimes,thefirstrowsbysubtractingtimes,thatis,abottom-upsequentiallysubtractedfromeachrowonrowtimeshare有ThelatterdeterminantisaVanDearMinddeterminant,accordingtotheinductionassumption,itisequaltoallpossibledifference;ContainsdifferenceallappearinfrontoftheconsequentconclusionvandearMenddeterminantofthelevelntheestablishmentofmathematicalinduction,pleted
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