中值定理 证明∃n: f(n)/g(n)=f"(n)/g"(n)

hbghlyj

假设f(x)和g(x)在[a ,b]上存在二阶导数,且f(a)=f(b)=g(a)=g(b)=0，g"(x)≠0 证明：至少存在一点n∈(a,b) 使f(n)/g(n)=f"(n)/g"(n)

构造F(x) = f(x)g'(x) - f'(x)g(x)
则F(x)在(a,b)可导，F(a) = F(b)=0
F'(x) = f'(x)g'(x) + f(x)g''(x) - [ f''(x)g(x) + f'(x)g'(x)]
= f(x)g''(x) - f''(x)g(x)

由罗尔定理，存在n∈(a,b)
使得 F'(n) =0
即f(n)g''(n) - f''(n)g(n) =0
即 f(n) / g(n) = f''(n) / g''(n)

原命题得证。

来自@tian27546西西

hbghlyj

Assuming that $f(x)$ and $g(x)$ have second derivatives on $[a,b]$, and $f(a)=f(b)=g(a)=g(b)=0$, and $g''(x)\neq 0$. We want to prove that there exists at least one point $n\in (a,b)$ such that $f(n)/g(n)=f''(n)/g''(n)$.

We construct $F(x) = f(x)g'(x) - f'(x)g(x)$, then $F(x)$ is differentiable on $(a,b)$ and $F(a) = F(b) = 0$. Taking the derivative of $F(x)$ gives:
\begin{align*}
F'(x) &= f'(x)g'(x) + f(x)g''(x) - [f''(x)g(x) + f'(x)g'(x)]\\
&= f(x)g''(x) - f''(x)g(x)
\end{align*}
By the mean value theorem, there exists $n\in (a,b)$ such that $F'(n) = 0$. Therefore,
$$f(n)g''(n) - f''(n)g(n) = 0$$
which implies that $f(n)/g(n) = f''(n)/g''(n)$.
QED