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1 引言与结果著名的契贝谢夫多项式[1]为 Un(x)=(1)/(2x2-1)[(x+x2-1)n+1-(x-x2-1)n+1]. 因此,(Un(x))/(n!)=(1)/(2x2-1)*(1)/(n!)[(x+x2-1)n+1-(x-x2-1)n+1],∞n=0Un(x)/n!*tn=(1)/(2x2-1)∞n=0(1)/(n!)[(x+x2-1)n+1-(x-x2-1)n+1]tn=(1)/(2x2-1)[(x+x2-1)exp((x+x2-1)t)-(x-x2-1)exp((x-x2-1)t)]