## Saturday, July 19, 2008

### The Middle Value Theorem

I admit that this is strange, to write about integrals and then to start talking about the Middle value theorem without finishing the previous topic. However, there is a reason for this. Firstly, I intended to write about both, and secondly, to write about integrals means to use latex. Unfortunately, because of the fluctuations in the weather in Jerusalem my head hurts to much for the thought to write in Latex to enter it. So today I will write about the Middle value theorem.

The Middle value theorem:
Let f(x) be a continues function from [a,b] to R. Suppose that f(b)>f(a). Than for any f(b)>y>f(a) there are exists b>x>a such that f(x)=y.

The proof I am going to give is the historical proof, there are others way to get to this result. However, Bolzano who originally proved this theorem used this method. To prove the theorem we firstly need to proof another statement:
Suppose that M is a property which is correct for some, but not all, numbers. Also, there is a number u with the property that M is correct for all numbers strictly less than u. Than there exists a number U with the property that M is correct for all numbers less than U, but there are no numbers large than U with the property that M is correct for all numbers lesser than them.

This is how Bolzano formulated this - in modern notation:

A={u| for all x less than u M is correct}, U=supA

Now to the proof. Balzano proofed this by simple construction. From the problem he new that A is not empty, and that there is a number that is larger than all of the numbers in A (because otherwise M is true for all numbers) we will mark it b. Lets look on the series:

$a_{n}=u+\frac{b}{2^n}$

If U>u we will pass it on some step. If this doesn't happen than u=U. If however we passed U on the N step, we can now look on the series:

$b_{n}=u+\frac{b}{2^N}+\frac{b}{2^n}$

Again, we repeat the previous step. If we will have to repeat it infinite number of times, we will have a series of numbers (which are all less than U) that is bounded (by U) and increases monotonically - and therefore it has a limit, which is clearly U. This part I am leaving without a proof - it is easy to show using elementary theorems. Bolzano proofed a statement which can be read as "Cauchy series converges" in order to proof this part, but this proof is omitted because it is identical to the proof given in first year Calculus - which means it requires epsilon and thus latex.
It is also interesting to note that as it was given by Bolzano the above proof is incorrect, because without the axiom of completeness he couldn't say that the limit of the Cauchy series is a real number.

Now it is time to use this to prove the Middle value theorem. The proof is very simple - let M be the property: y>f(x). Because this is not true for x=b, we get from the statement that we just proved that there exists U such that for all x less than U, M is true, and for all x greater than U M is false. Therefore f(U)=y.

This proof, unlike the previous one, if perfectly valid. It can be used on an exam today. A question to the reader: Where in the proof the property that f is continues is used?