Cauchy’s Theorem

Introduction

Cauchy’s Theorem establishes a relationship between the changes of two functions over a given interval. Specifically, if $f ( x )$ and $g ( x )$ are continuous on a closed interval $[ a , b ]$ and differentiable in its interior, with $g^{‘} ( x ) \neq 0$, then there exists at least one point $c$ in $( a , b )$ where the ratio of their derivatives matches the ratio of their overall change across the interval:

Statement

The Cauchy’s theorem states the following. Let $f ( x )$ and $g ( x )$ be two functions such that:

• $f ( x )$ and $g ( x )$ are continuous on the interval $[ a , b ]$.
• $f ( x )$ and $g ( x )$ are differentiable at every point in the interior of the interval.
• $g^{‘} ( x ) \neq 0$ for every $x$ in the interior of $[ a , b ]$.

Then, there exists at least one point $c$ in the interior of the interval $[ a , b ]$ such that:

that is, the ratio of the increments of the functions $f ( x )$ and $g ( x )$ over the interval $[ a , b ]$ is equal to the ratio of their respective derivatives evaluated at a point $c$ in the interior of the interval.

Setting $g ( x ) = x$ reduces Cauchy’s theorem directly to Lagrange’s theorem. With $g^{‘} ( x ) = 1$ and $g ( b ) - g ( a ) = b - a$, the conclusion takes the form: \(\frac{f^{'} ( c )}{1} = \frac{f ( b ) - f ( a )}{b - a}\) Lagrange’s theorem is therefore a special case of Cauchy’s theorem, obtained when one of the two functions is the identity.

Cauchy’s theorem provides the theoretical basis for the proof of L’Hôpital’s Rule.

Proof of Cauchy’s theorem

To prove the theorem, define a new function $\varphi ( x )$ where $\lambda$ is a constant to be determined.:

We want to choose $\lambda$ such that $\varphi ( a ) = \varphi ( b )$. From this condition, we obtain:

Now the function $\varphi ( x )$ is continuous on $[ a , b ]$ and differentiable on $( a , b )$, with $\varphi ( a ) = \varphi ( b )$. We can apply the Rolle’s Theorem, which guarantees that there exists at least one point $c \in ( a , b )$ such that $\varphi^{‘} ( c ) = 0$. By calculating the derivative of $\varphi ( x )$ and setting $\varphi^{‘} ( c ) = 0$, we obtain:

\(\varphi^{'} ( x ) = f^{'} ( x ) - \lambda g^{'} ( x )\) \(f^{'} ( c ) = \lambda g^{'} ( c )\)

Substituting the value of $\lambda$, we get:

By dividing both sides by $g^{‘} ( c )$, we obtain:

Thus, we have proven the conclusion of the theorem.

Example 1

Let’s verify, for example, that the theorem is applicable to the functions $f ( x )$ and $g ( x )$ in the interval [1,3]:

\(f ( x ) = 2 x^{2} - 4 x + 2\) \(g ( x ) = x^{2}\)

The two functions are polynomials, therefore they are continuous and differentiable for every $x \in \mathbb{R}$. Moreover, $g^{‘} ( x ) = 2 x \neq 0$. This satisfies the hypotheses of the theorem.

Let’s now verify the existence of a point $c \in ( 1 , 3 )$ such that:

We obtain:

Now let’s calculate:

From $1$, the equality becomes:

Since $c = 2 \in [ 1 , 3 ]$, the theorem is verified.

A note on the hypothesis $g^{‘} ( x ) \neq 0$

Consider the particular case in which $g ( b ) = g ( a )$. In this situation, the denominator of the following ratio is zero, and the expression is undefined: \(\frac{f ( b ) - f ( a )}{g ( b ) - g ( a )}\) For this reason, the constant cannot be introduced: \(\lambda = \frac{f ( b ) - f ( a )}{g ( b ) - g ( a )}\) As a result, the proof utilising the following auxiliary function is no longer valid: \(\varphi ( x ) = f ( x ) - \lambda g ( x )\) Under the hypotheses of Cauchy’s theorem, this situation does not arise. The condition $g^{‘} ( x ) \neq 0$ within the interior of the interval ensures that g is strictly monotonic, which in turn guarantees that $g ( b ) \neq g ( a )$. Therefore, the case $g ( b ) = g ( a )$ is precluded by the theorem’s assumptions.

Selected references

• University of Florida J. Keesling. The Cauchy Mean Value Theorem
• University of Maryland. The Cauchy Mean Value Theorem and Consequences
• UC Berkeley A. Vizeff. Mean Value Theorem and L’Hôpital’s Rule