考研数学微积分题海拾贝

考研数学微积分题海拾贝

设\(f(x)\)在\([0,1]\)内可导,且\(f(0)=0\),\(f(1)=1\). 且\(f(x)\)在\([0,1]\)上严格递增。
证明:存在\(\xi_i \in (0,1) (1 \le i \le n)\),使\[ \frac{1}{f'(\xi_1)} + \frac{1}{f'(\xi_2)} + \cdots +\frac{1}{f'(\xi_n)} = n \]成立。


设有\( \xi_i \in (0,1) \),对f(x)在区间\([0,x_1]\),\([x_1,x_2]\),\([x_2,x_3]\),\( \cdots \),\([x_n,1]\),上分别使用拉格朗日中值定理可得

\[ f'(\xi_1) = \frac{f(x_1)-f(0)}{x_1-0} = \frac{f(x_1)}{x_1} \Rightarrow \exists \xi_1 \in (0,x_1) , \frac{1}{f'(\xi_1)} = \frac{x_1}{f(x_1)}\]

\[ f'(\xi_2) = \frac{f(x_2)-f(x_1)}{x_2-x_1} \Rightarrow \exists \xi_2 \in (x_1,x_2) , \frac{1}{f'(\xi_2)} = \frac{x_2-x_1}{f(x_2) – f(x_1)}\]

\[ \vdots\]

\[ f'(\xi_n) = \frac{f(1)-f(x_{n-1})}{1-x_{n-1}} \Rightarrow \exists \xi_n \in (x_{n-1},1) , \frac{1}{f'(\xi_n)} = \frac{1-x_{n-1}}{1- f(x_{n-1})}\]

要证\[ \frac{1}{f'(\xi_1)} + \frac{1}{f'(\xi_2)} + \cdots +\frac{1}{f'(\xi_n)} = n \]

只需\[ f(x_1) = \frac{1}{n},f(x_2) = \frac{2}{n}, \cdots, f(x_{n-1}) = \frac{n-1}{n} \]

又\(f(0)=0\),\(f(1)=1\),根据连续函数的介值定理,存在\( x_k \in (0,1) \),\( k \in[1,n-1]\),使\(f(x_k) = \frac{k}{n}\)成立,

故得证。


发表评论

电子邮件地址不会被公开。 必填项已用*标注