Cauchy's Theorem: Statement and Proof

Cauchy's theorem is a fundamental result that extends Lagrange's theorem by introducing a relationship between two functions.

Contents

• Cauchy's Theorem
• Proof

Let \(f, g : [a,b] \to \mathbb{R}\) be functions continuous on \([a,b]\) and differentiable on \((a,b)\), with \(g' \neq 0\) on \((a,b)\). Then there exists \(\xi \in (a,b)\) such that:

\[\frac{f'(\xi)}{g'(\xi)} = \frac{f(b) - f(a)}{g(b) - g(a)}\]

Note that \(g(b) - g(a) \neq 0\) by virtue of the hypothesis \(g' \neq 0\).

\[h(x) = f(x)[g(b) - g(a)] - g(x)[f(b) - f(a)]\]

This function satisfies:

• \(h\) is continuous on \([a,b]\) (as a combination of continuous functions)
• \(h\) is differentiable on \((a,b)\) (since both \(f\) and \(g\) are differentiable)

We 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 \(h'(\xi) = 0\).

We calculate \(h'(x)\):

\[h'(x) = f'(x)[g(b) - g(a)] - g'(x)[f(b) - f(a)]\]

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.