The Cauchy's theorem is a fundamental result that extends the Lagrange's theorem by introducing a relationship between two functions.
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Cauchy's Theorem
Let \(f, g : [a,b] \to \mathbb{R}\) be continuous functions on \([a,b]\) and differentiable on \((a,b)\), with \(g' \neq 0\) on \((a,b)\). Then there exists \(\xi \in (a,b)\) such that:
\[\frac{f'(\xi)}{g'(\xi)} = \frac{f(b) - f(a)}{g(b) - g(a)}\]
Notice that \(g(b) - g(a) \neq 0\) because of the hypothesis \(g' \neq 0\).
Proof. Let us consider the auxiliary function:
\[h(x) = f(x)[g(b) - g(a)] - g(x)[f(b) - f(a)]\]
This function satisfies:
- \(h\) is continuous on \([a,b]\) (as it is a combination of continuous functions)
- \(h\) is differentiable on \((a,b)\) (since both \(f\) and \(g\) are differentiable)
Now, let us evaluate \(h\) at the endpoints:
For \(x = a\):
\begin{align} h(a) &= f(a)[g(b) - g(a)] - g(a)[f(b) - f(a)] \\ &= f(a)g(b) - f(a)g(a) - g(a)f(b) + g(a)f(a) \\ &= f(a)g(b) - g(a)f(b)\end{align}
For \(x = b\):
\begin{align} h(b) &= f(b)[g(b) - g(a)] - g(b)[f(b) - f(a)] \\ &= f(b)g(b) - f(b)g(a) - g(b)f(b) + g(b)f(a) \\ &= f(a)g(b) - g(a)f(b)\end{align}
Therefore, \(h(a) = h(b)\). By Rolle's Theorem, there exists \(\xi \in (a,b)\) such that \(h'(\xi) = 0\).
Let us compute \(h'(x)\):
\[h'(x) = f'(x)[g(b) - g(a)] - g'(x)[f(b) - f(a)]\]
For \(x = \xi\), we have that \(h'(\xi) = 0\) implies:
\[f'(\xi)[g(b) - g(a)] - g'(\xi)[f(b) - f(a)] = 0\]
\[f'(\xi)[g(b) - g(a)] = g'(\xi)[f(b) - f(a)]\]
Since \(g' \neq 0\) on \((a,b)\), we can divide both sides by \(g'(\xi)\):
\[\frac{f'(\xi)}{g'(\xi)} = \frac{f(b) - f(a)}{g(b) - g(a)}\]
This completes the proof of Cauchy's theorem.