设f(x)为n+1阶可导函数,求证:f(x)为n次多项式的充要条件是f(n+1)(x)≡0,f)(n)(x)≠0.

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问题 设f(x)为n+1阶可导函数,求证:f(x)为n次多项式的充要条件是f(n+1)(x)≡0,f)(n)(x)≠0.

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答案由带拉格朗日余项的n阶泰勒公式得 f(x)=f(0)+f’(0)x+…+[*]f(n)(0)xn+[*]xn+1. 若fn+1(x)≡0,f(n)(x)≠0,由上式[*] f(x)=f(0)+f’(0)x+…+[*]f(n)(0)xn是n次多项式. 反之,若f(x)=anxn+an-1xn-1+…+a1x+a0(an≠0)是n次多项式,显然 f(n)(x)=ann!≠0,f(n+1)(x)≡0.

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