Question

高数………………

设函数y=f(x)在x=0的某邻域内具有n阶导数，且f(0)=f'(0)=……=f^(n-1)(0)=0，试用柯西中值定理证明：f(x)/x^n=f^(n)(ax)/n! (0<a<1)

要过程 Thanks

Answer 1

令g(x) = x^n,则 g^(k)(x) = A(n,k) x^(n-k),其中A(n,k) 为排列数，即A(n, k) = n! / (n-k)!.
则g(0)=g'(0)=……=g^(n-1)(0)=0, g^(n) = n!.
f(x)/x^n = f(x)/g(x) = [f(x) - f(0)] / [g(x) - g(0)]
因为 g(b1 * x) <> 0,b1在(0,1)之间，因此柯西中值定理可得
f(x)/g(x) = [f(x) - f(0)] / [g(x) - g(0)] = f'(a1 * x) / g'(a1 * x)，a1在（0,1）之间
反复用n次柯西定理可得
[f(x) - f(0)] / [g(x) - g(0)] = f'(a1 * x) / g'(a1 * x) =...=f^(n)(an * x) / g^(n)(an * x),an 在(0, 1)之间
因为g^(n) (x) = n!
所以
f(x)/g(x) = [f(x) - f(0)] / [g(x) - g(0)] = f^(n)(an * x) / g^(n)(an * x) = f^(n) (an * x) / n!
原题得证

Answer 2

能用泰勒中值吗？我会用泰勒做