## Tuesday, August 29, 2006

### Cauchy's Mean Value Theorem

I use Cauchy's Mean Value Theorem in my proof of L'Hopital's Rule (proof to be added later). Augustin-Louis Cauchy used this proof to as part of his effort to make calculus more rigorous. For those interested in learning more about Cauchy's role in reworking the foundations of calclulus, check out The Origins of Cauchy's Rigorous Calculus by Judith V. Grabiner.

Theorem Cauchy's Mean Value Theorem

If f(x),g(x) are continuous functions on the closed interval [a,b] and differentiable on (a,b)

Then, there exists a number c in (a,b) such that:

[f(b) - f(a)]g'(c) = [g(b)-g(a)]f'(c)

Proof:

(1) Let us define a function h(x) such that:

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

(2) h(a) = h(b) = 0 since:

h(a) = [g(b) - g(a)]*[f(a) - f(a)] - [f(b) - f(a)]*[g(a) - g(a)] = [g(b) - g(a)]*0 - [f(b) - f(a)]*0 = 0

h(b) = [g(b) - g(a)]*[f(b) - f(a)] - [f(b) - f(a)]*[g(b) - g(a)] = 0

(3) h(x) is continuous in the closed interval [a,b] since:

(a) Using the Constant Law (see Lemma 1, here), we know that we can treat the values f(a),f(b),g(a),g(b), -1 as continuous functions.

(b) Using the Multiplication Law (see Lemma 3, here), we can treat -g(a), -f(a) as continuous functions.

(c) Using the Addition Law (see Lemma 2, here), we know that the following are continuous functions:

g(b) + [- g(a)] = g(b) - g(a)
f(x) + [- f(a)] = f(x) - f(a)
f(b) + [-f(a)] = f(b) - f(a)
g(x) + [-g(a)] = g(x) - g(a)

(d) Using the Multiplication Law and the Addition Law, we can see that h(x) is a continuous function

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

(4) h(x) is also differentiable on (a,b) since:

NOTE: The detail here is parallel to the detail in step #3.

(a) We know that all constants are differentiable (see Lemma 1, here) so this means that f(a),f(b),g(a),g(b), -1 are all differentiable on (a,b)

(b) We know that the product of all differentiable functions are differentiable (see Lemma 4, here) so this means that -g(a), -f(a) are differentiable on (a,b)

(c) We also know that the addition of differentiable functions are differentiable (see Lemma 3, here) so this means that the following are all differentiable on (a,b):

g(b) + [- g(a)] = g(b) - g(a)
f(x) + [- f(a)] = f(x) - f(a)
f(b) + [-f(a)] = f(b) - f(a)
g(x) + [-g(a)] = g(x) - g(a)

(d) Finally, from the principles that we have already reviewed we know that h(x) is differentiable on (a,b) since:

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

(e) We can now define h'(x)

Let u(x) = [g(b) - g(a)]*[f(x) - f(a)]

Let v(x) = [f(b) - f(a)]*[g(x) - g(a)]

u'(x) = [g(b) - g(a)][f'(x) - 0] + [0 - 0]*[f(x) - f(a)] = [g(b) - g(a)]f'(x) [See Lemma 4, here]

v'(x) = [f(b) - f(a)][g'(x) - 0] + [0 - 0]*[g(x) - g(a)] = [f(b) - f(a)]g'(x) [See Lemma 4, here]

h(x) = u(x) - v(x)

h'(x) = u'(x) - v'(x) = [g(b) - g(a)]f'(x) - [f(b) - f(a)]g'(x) [See Lemma 3, here]

(5) Using Rolle's Theorem, we know that there exists a point c such that:

h'(c) = 0 and c in (a,b)

(6) Now, combining our result for h'(x) in step #4 with step #5, we have:

h'(c) = [g(b) - g(a)]f'(c) - [f(b) - f(a)]g'(c) = 0

(7) Adding [f(b) - f(a)]g'(c) to both sides gives us:

[g(b) - g(a)]f'(c) = [f(b) - f(a)]g'(c)

QED

Corollary: Mean Value Theorem

If f(x) is a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point c in (a,b) such that:

f(b) - f(a) = f'(c)(b-a)

Proof:

(1) Let g(x) = x

(2) Then g'(x) = 1

(3) Using Cauchy's Mean Value Theorem above, we see that there exists a value c such that:

[f(b) - f(a)]g'(c) = [g(b)-g(a)]f'(c)

(4) Since g'(c) = 1, we see that:

f(b) - f(a) = [g(b) - g(a)]f'(c)

(5) Since g(x)=x, we see that:

f(b) - f(a) = (b - a)f'(c)

QED

References