# The Unapologetic Mathematician

## The Differential Mean Value Theorem

Let’s say we’ve got a function $f$ that’s continuous on the closed interval $\left[a,b\right]$ and differentiable on $(a,b)$. We don’t even assume the function is defined outside the interval, so we can’t really set up the limit for differentiability at the endpoints, but they don’t matter much in the end.

Anyhow, if we look at the graph of $f$ we could just draw a straight line from the point $(a,f(a))$ to the point $(b,f(b))$. The graph itself wanders away from this line and back, but the line tells us that on average we’re moving from $f(a)$ to $f(b)$ at a certain rate — the slope of the line. Since this is an average behavior, sometimes we must be going faster and sometimes slower. The differential mean value theorem says that there’s at least one point where we’re going exactly that fast. Geometrically, this means that the tangent line will be parallel to the secant we drew between the endpoints. In formulas we say there is a point $c\in(a,b)$ with $f'(c)=\frac{f(b)-f(a)}{b-a}$.

First let’s nail down a special case, called “Rolle’s theorem”. If $f(a)=0=f(b)$, we’re asserting that there is some point $c\in(a,b)$ with $f'(c)=0$. Since $\left[a,b\right]$ is compact and $f$ is continuous, the extreme value theorem tells us that $f$ must take a maximum and a minimum. If these are both zero, then we’re looking at the constant function $f(x)=0$, and any point in the middle satisfies $f'(c)=0$. On the other hand, if either the maximum or minimum is nonzero, then we have a local extremum at a point $c\in(a,b)$ where $f$ is differentiable (since it’s differentiable all through the open interval). Now Fermat’s theorem tells us that $f'(c)=0$ since $c$ is a local extremum! Thus Rolle’s theorem is proved.

Now for the general case. Start with the function $f$ and build from it the function $g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a)$. On the graph, this corresponds to applying an “affine transformation” (which sends straight lines in the plane to other straight lines in the plane) to pull both $f(a)$ and $f(b)$ down to zero. In fact, it’s a straightforward calculation to see that $g(a)=0=g(b)$. Thus Rolle’s theorem applies and we find a point $c$ with $g'(c)=0$. But applying our laws of differentiation, we see that $g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}$. And so $f'(c)=\frac{f(b)-f(a)}{b-a}$, as desired.

January 22, 2008 - Posted by | Analysis, Calculus

1. The “trick” behind proving the Mean Value theorem is basically to write $g(x) = f(x) - h(x)$, where $y = h(x)$ is the equation of the straight line connecting the points $(a, f(a))$ and $(b, f(b))$. And this is what you did above. I know it might not sound very interesting, but I have always wondered if we can obtain another useful result (just like the mean value theorem, for instance) if we let $y = h(x)$ be the equation of a parabola passing through the points $(a, f(a))$ and $(b, f(b))$. And, do we obtain more “useful” results if $y = h(x)$ is some function other than a straight line or a parabola?
(That was just a random thought.)

Comment by Vishal | January 22, 2008 | Reply

2. Indeed you do, Vishal, but you ask that the function be twice-differentiable and you match another parameter: the derivative of $f$ at $a$. Then those three parameters specify a unique parabola, which has constant second derivative. And there’s some point in between $a$ and $b$ at which the function actually has that second derivative.

There’s a whole sequence of these “generalized mean value theorems”, which can be used as Taylor series remainder approximations. Your exercise is to work out the analogues of Rolle’s theorem you guessed were around, and to see if you can prove them in general. I’ll come back to this when I’m doing Taylor series.

Comment by John Armstrong | January 22, 2008 | Reply

3. I had a hunch earlier that we could indeed derive more of those “generalized mean value theorems” and somehow relate those to the Taylor series approximations. Your comment, it seems, has confirmed that hunch. I will be looking forward to your future post(s) on Taylor series.

Comment by Vishal | January 22, 2008 | Reply

4. This theorem is also known as the Lagrange mean value theorem.

Comment by Michael Livshits | January 23, 2008 | Reply

5. It also has an interesting corollary that says that the derivative of any function that is differentiable on an interval has the mean value property, i.e. it hits all its intermediate values between any 2 of the values that it assumes, even if that derivative is not continuous.

In partcular, a simple step function H(x) that is 0 for x1 is not a derivative of any function F differentiable on (-1,1), so H(x) has no primitive on (-1,1).

Comment by Michael Livshits | January 23, 2008 | Reply

6. In comment #5 H(x)=0 for x0, also known as the Heaviside function.

Comment by Michael Livshits | January 23, 2008 | Reply

7. There must be a bug in the system, that messed up my posts, again (now using LaTex): $H(x)=0$ for $x \le 0$
and $0$ for $x>0$.

Comment by Michael Livshits | January 23, 2008 | Reply

8. There must be a bug in the system, that messed up my posts, again (now using LaTex): $H(x)=0$ for $x \le 0$
and $=0$ for $x>0$.

Comment by Michael Livshits | January 23, 2008 | Reply

9. Michael, you’re having trouble with the < symbol. That’s not a bug in the system.

The bug in the system is that I can’t seem to log in to WordPress at all. Whether this has to do with a comment flood or not, I have no idea. Just in case, could you try in the future to think carefully about what you want to say and then say it all in one comment?

Comment by John Armstrong | January 23, 2008 | Reply

10. Sorry, I got troubles with > that seems to have some special meaning on this site and also with LaTex preprocessor here that messed up my formula. I wish we could edit our entries.
I doubt that your trouble logging in is related to the number of comments on your page.

Comment by Michael Livshits | January 23, 2008 | Reply

11. It’s not this site. > has a special meaning in HTML, which is sort of the basis of the web. WordPress’ processor can’t tell the difference between using it to mean “greater than” and “close a tag”.

But still, you made three separate comments before the litany of trying to fix the problem with the > sign. Slow down, take a deep breath, and decide what you want to say before saying it. This isn’t Twitter.

Comment by John Armstrong | January 23, 2008 | Reply

12. Actually I wanted to say that H(x) is the Heaviside function that jumps from 0 to 1 at x=0.

Comment by Michael Livshits | January 23, 2008 | Reply

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