九、(7分)设函数f(x)在(-∞，+o∞)上有任意阶导数，且对任意实数x及n= 0,1,2…满足fm(x川≤nlx斗.求证：f(x)=0. 证明由|fm)(x川≤nlx可知对任意自然数n有fm(O)=0.对于x∈(-1,1) 根据Taylor展开存在9∈(0,1)使得 f()=f(0)+f'(0)x+…+ -0))o)() (n-1)！ n! n! 因此 |f(xl≤Iox·lz≤xm+1 在此式中令n→o得f(x)=0(x∈(-1,1).由连续性可知 f(x)=0,x∈[-1,1] (4分) 假设f(x)=0,x∈【-k,.令g(x)=f(x+),h(x)=f(x-).则g(x)和h(x)在0 点的任意阶导数为0.根据Taylor展开存在01,2∈(0,1)使得 g=gm6四g=fo8r+月 n! n！ h(x)=了 o=fo6t-月gn n! n! 因此 lg(x川≤l9x+·lzm, h(x)川≤l2x+kA·lzm. 令n→o得g(x)=0,h(x)=0,x∈(-1,1).于是f(x)=0,x∈【-k-1,k+1小.根 据归纳原理可知f(x)=0,x∈R. (7分) 第6页（共6页）Ê!(7 ©) ¼ê f(x) 3 (−∞, +∞) þk?¿ê, …é?¿¢ê x 9 n = 0, 1, 2 · · · ÷v |f (n) (x)| 6 n!|x|. ¦y: f(x) = 0. y² d |f (n) (x)| 6 n!|x| Œé?¿g,ê n k f (n) (0) = 0. éu x ∈ (−1, 1) Šâ Taylor Ðm3 θ ∈ (0, 1) ¦ f(x) = f(0) + f 0 (0)x + · · · + f (n−1)(0) (n − 1)! x n−1 + f (n) (θx) n! x n = f (n) (θx) n! x n . (1 ©) Ïd |f(x)| 6 |θx| · |x| n 6 |x| n+1 . 3dª¥- n → ∞  f(x) = 0 (x ∈ (−1, 1)). dëY5Œ f(x) = 0, x ∈ [−1, 1]. (4 ©) b f(x) = 0, x ∈ [−k, k]. - g(x) = f(x + k), h(x) = f(x − k). K g(x) Ú h(x) 3 0 :?¿ê 0. Šâ Taylor Ðm3 θ1, θ2 ∈ (0, 1) ¦ g(x) = g (n) (θ1x) n! x n = f (n) (θ1x + k) n! x n . h(x) = h (n) (θ2x) n! x n = f (n) (θ2x − k) n! x n . Ïd |g(x)| 6 |θ1x + k| · |x| n , |h(x)| 6 |θ2x + k| · |x| n . - n → ∞  g(x) = 0, h(x) = 0, x ∈ (−1, 1). u´ f(x) = 0, x ∈ [−k − 1, k + 1]. Š â8BnŒ f(x) = 0, x ∈ R. (7 ©) 1 6 £
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