|
|
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 |
|
