SCHLÖMILCH’S THEOREM

 

SCHLÖMILCH’S THEOREM

 

In Section 4.3, we provide a modern English-language translation of Schlomilch paper [22]. Since the basis of his argument comes from the Mean Value Theorem for Integrals, which he refers to as the “well-known” theorem, we will provide a few preliminaries in Section 4.2 by introducing and arguing the following three results: (1) Rolle’s Theorem, which will be used to prove (2) the Mean Value Theorem, then extend that to (3) the Mean Value Theorem for Integrals.

 

4.1     Oscar Schlomilch

 

¨Oscar Xavier Schlomilch (13 April 1823 - 7 February 1901) was a German mathematician. ¨ He studied mathematics and physics primarily in Jena, Berlin and Vienna. Most of his work was strongly influenced by Johann Peter Gustav Lejeune Dirichlet (1805-1859). In 1844, Schlomilch received his doctorate from Friedrich-Schiller-Universität. From 1851 to 1874, he was a professor at Dresden teaching Higher Mathematics and Analytic Mechanics [13].

 

4.2    Mean Value Theorem

 

Theorem 4.2.1 (Rolle’s Theorem [18]). Let f: [a, b] → R be a continuous function such that f is differentiable on (a, b) where a < b. If f(a) = f(b), then there exists some point x ∈ (a, b) where f ⁰(x) = 0.

Proof. Let f(a) = f(b). Since f is continuous on [a, b], then f attains a maximum at some point t ∈ [a, b] and a minimum at some point s ∈ [a, b].

Suppose first that s, t is both endpoints of [a, b]. Since f(a)= f(b), then the maximum and the minimum are equivalent, which means f is a constant function on [a, b]. In other words, f(x) = 0, then for each x ∈ (a, b), f ⁰(x) = 0, in which case, we are done.

But now consider when (I) s is not an endpoint of [a, b], or when (II) t is not an endpoint of [a, b].

(I)                 If s is not an endpoint of [a, b], then s ∈ (a, b), where f has a local maximum at s, and therefore f ⁰(s) = 0.

(II)               If t is not an endpoint in [a, b], then t ∈ (a, b) and f has a local minimum at t, and therefore f ⁰(t) = 0.

We have proved that in all cases for some point x ∈ (a, b), f ⁰(x) = 0.                    

 

Theorem 4.2.2 (Mean Value Theorem [18]). Suppose f: [a, b] → R be a continuous function such that f is differentiable on (a, b). Then, there is some point t ∈ (a, b) such that

f(b)− f(a) =fʹ(t)(b−a).

Proof. Let

y(x) = (x-a) +f(a)

Notice that this is the slope of the secant of the graph of  f on [a,b]. Now let

h(x) = f(x) - (x-a)-f(a).

and note that

h(a) = h(b) = 0

and h is continuous on [a, b] and differentiable on (a, b). Applying Rolle’s Theorem, there is some t ∈ (a, b) such that hʹ(t) = 0. But since

hʹ(t) = fʹ(t)-  = 0

then

 fʹ(t) =   .