# The Unapologetic Mathematician

## Consequences of the Mean Value Theorem

So now that we have the mean value theorem what can we do with it? First off, we can tell something that seems intuitively obvious. We know that a constant function has the constant zero function as its derivative. It turns out that these are the only functions with zero derivative.

To see this, let $f$ be a differentiable function on $(a,b)$ so that $f'(x)=0$ for all $x\in(a,b)$. Let $x_1$ and $x_2$ be any points between $a$ and $b$ with $x_1. Then $f$ restricts to a continuous function on the interval $\left[x_1,x_2\right]$ which is differentiable on the interior $(x_1,x_2)$. The differentiable mean value theorem then applies, and it tells us that there is some $c\in(x_1,x_2)$ with $f'(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}$. But by assumption this derivative is zero, and so $f(x_2)=f(x_1)$. Since the points were arbitrary, $f$ takes the same value at each point in $(a,b)$.

What about a function $f$ for which $f'(x)>0$ on some interval $(a,b)$? Looking at the graph it seems that the slope of all the tangent lines should be positive, and so the function should be increasing. Indeed this is the case.

Specifically we have to show that if $x_2>x_1$ for two points in $(a,b)$ then $f(x_2)>f(x_1)$. Again we look at the restriction of $f$ to a continuous function on $\left[x_1,x_2\right]$ which is differentiable on $(x_1,x_2)$. Then the mean value theorem tells us that there is some $c\in(x_1,x_2)\subseteq(a,b)$ with $f'(x)=\frac{f(x_2)-f(x_1)}{x_2-x_1}$. By assumption this quantity is positive, as is $x_2-x_1$, and so $f(x_2)>f(x_1)$. Similarly we can show that if $f'(x)<0$ on an interval $(a,b)$ then the function is decreasing there.

January 24, 2008 - Posted by | Analysis, Calculus

1. One consequence that’s made it to popular media is the following:
For each great circle on the earth’s surface, there is some pair of antipodal points sharing the same temperature.

For this to work, we need to assume that temperatures vary smoothly across the earth’s surface. However, once we do have that, pick some point and consider the function on the half circle from that point, in some direction, to its antipode that sends each point x to the difference between the temperature at x and the temperature at the antipode of x. Unless the temperature of the start and end points are equal, in which case we are done, we will have some smooth function on an interval, which starts and ends with different signs. Hence, at some point in the middle, this function will be zero; which means that that point and its antipode have the same temperature.

Comment by Mikael Vejdemo Johansson | January 27, 2008 | Reply

2. Temperature and pressure, even, but that gets into more than real-valued functions, so I was going to delay.

Comment by John Armstrong | January 27, 2008 | Reply

3. It looks like the MVT Mikael is talking about is the one by Bolzano, saying that a continuous function on a real interval hits all its intermediate values, so differentiability is not essential in his example, continuity is quite enough. The same is true for the temperature + pressure on a sphere, no differentiability is needed.

As for the relation between the MVT of this section (i.e. for the derivative) and its relation to the fact that the functions with positive derivatives increase, this relation is rather misleading in my opinion. The proof John gives in this post uses the “pure existence” MVT to demonsrate the fact that is much more computational in nature and can be proved more directly. I will present several direct proofs together with an analysis of the underlying assumptions later. You can also see the article by Mark Bridger at http://www.math.neu.edu/~bridger/LBC/lbcswp.pdf for a mild-mannered discussion of some related ideas.

Comment by Michael Livshits | January 28, 2008 | Reply

4. Yes, Michael, you’ve already demonstrated time and again that you knowfar more than I do about mathematics and its exposition.

Comment by John Armstrong | January 28, 2008 | Reply

5. Well, John, just let me know if you want no more remarks from me and I will stop making them. I just thought that they might be of interest to the intended audience of your blog. Maybe I was wrong. I apologize for any offences that I have committed. I didn’t intend to compete with you.

Comment by Michael Livshits | January 28, 2008 | Reply

6. Your whole tone is competitive, Michael. You assert that my approach is not only different from yours, but actually incorrect and misleading.

Comment by John Armstrong | January 28, 2008 | Reply

7. [...] One of the consequences of the mean value theorem we worked out was that two differentiable functions and on an interval differ by a constant if [...]

Pingback by Antiderivatives « The Unapologetic Mathematician | January 28, 2008 | Reply

8. I’m not saying that your approach is incorrect, I’m saying that in your approach there are too many explanations of simple things in terms of complicated things. We have a somewhat different perspectives, and they are complimentary rather than competitive. Your approach is geared towards mathematical abstractions and “generality”, while I care more about the computational aspect of the subject and simplicity. I don’t claim that mine is better, it depends on what you want to do with it.

In the introduction to his recent book “Lectures on PDE” V.I. Arnold says: “Instead of the usual in mathematical books principle of the most generality the author tried to follow the principle of the least generality according to which every idea has to be first clearly understood in the simplest situation and only then the developed method can be generalized to the more complicated situations. Although the more general fact usually can be proven simpler than
its numerous special cases, the content of a mathematical theory for a student is no more than a collection of the examples that he understood clearly and completely.”

It looks like a reasonable approach, especially if you want to explain some mathematical ideas to non-specialists. I’m sorry if I’m pushing this idea a bit too vigorously, but I’m not your boss, and you don’t have to follow my suggestions. I also think that people who read your blog may be interested in different views and opinions.

Comment by Michael Livshits | January 28, 2008 | Reply

9. When I said that using MVT to prove that a function with positive derivative is increasing is misleading, I meant that there are more direct proofs that don’t use MVT. In fact it is a common fallacy that MVT is necessary here. It is common because too many people know only one proof and think that it is the only one there is.

Comment by Michael Livshits | January 28, 2008 | Reply

10. I didn’t say it was necessary. And “misleading”, whatever your intent, implies that your method is better, no matter how you try to disclaim it after the fact.

In fact, almost every one of your comments is pushing the glories of a computational, analytic proof over my conceptual (and wrong-headed) one, and my response is always that your approach has its own benefits and penalties, and I’ve chosen this one. It just gets tiresome after a while, so I’m not even going to bother arguing the points anymore. Yes, you know far more than I, che siete in piccoletta barca.

On the other hand, I suppose I’m obligated to continue my own exposition so that you can have a place to expound the proper method.

Comment by John Armstrong | January 28, 2008 | Reply

11. In case anyone is interested, my proof of the mean value theorem can be found at:

http://www.geocities.com/john_gabriel/mvtdef.html

Also check out my calculus book at:

http://www.geocities.com/john_gabriel

Chapter 4 contains my two finite difference formulas that provide the connection between finite differences and derivatives for the first time since Isaac Newton.

Comment by John Gabriel | February 15, 2008 | Reply

12. [...] off, the derivative of is , which is always positive for positive . Thus it’s always strictly increasing. That is, if then . So no two distinct numbers ever have the same natural logarithm, and the [...]

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13. [...] the derivative is itself differentiable, then the differential mean-value theorem tells us that since is nondecreasing. This leads us back to the second derivative test to distinguish [...]

Pingback by Differentiable Convex Functions « The Unapologetic Mathematician | April 16, 2008 | Reply

14. [...] the derivative of is again, and takes only positive values. And so we know that is everywhere increasing. What does this mean? Well, if then , and so . On the other hand if [...]

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