The Unapologetic Mathematician

Mathematics for the interested outsider

The Mean Value Theorem

Here’s a nice technical result we may have call for from time to time: a higher-dimensional version of the differential mean value theorem. Remember that this says that if we’ve got a function f continuous on the closed interval \left[a,b\right]\subseteq\mathbb{R} and differentiable on its interior, there is some point \xi in the middle where the derivative of the function is the same as the average — the mean — rate of change of the function over the interval. In more than one dimension we’re going to modify this a bit to make it clearer what it means.

First of all, instead of talking about the closed interval \left[a,b\right], we’re going to use the closed straight line segment. That is, the collection of all the points between a and b in a straight line, and including the endpoints. We first look at the total displacement b-a from one point to the other. Then we start at a and move some portion of this displacement towards b. That is, the closed line segment \left[a,b\right] consists of all points of the form a+t(b-a) for t in the closed interval \left[0,1\right]. Setting t=0 gives us the point a, and t=1 gives us the point b. Similarly, the open line segment \left(a,b\right) consists of all points of the form a+t(b-a) for t in the open interval \left(0,1\right).

Next, we have to be clear about the average rate of change. As we move from a to b, the value of the function f changes by f(b)-f(a). It takes a displacement of \lVert b-a\rVert to get there, so on average the rate of change is

\displaystyle\frac{f(b)-f(a)}{\lVert b-a\rVert}

Finally, we don’t just have a single value for the instantaneous rate of change, we have a differential df(\xi). But we can use it to find directional derivatives. Specifically, we’ll consider the derivative of f in the direction pointing from a to b. We’ll pick out this direction with the unit vector we get by normalizing the displacement

\displaystyle\frac{b-a}{\lVert b-a\rVert}

So the mean value theorem will tell us that if f is differentiable in some open region S that contains the whole closed line segment \left[a,b\right]. Then there is some point \xi in the open line segment \left(a,b\right) so that the average rate of change of f from a to b is equal to the directional derivative of f at \xi in the direction pointing from a to b:

\displaystyle\frac{f(b)-f(a)}{\lVert b-a\rVert}=df(\xi)\left(\frac{b-a}{\lVert b-a\rVert}\right)

or, more simply

\displaystyle f(b)-f(a)=df(\xi)\left(b-a\right)

We’ll get at this by changing to a function of one variable so we can bring the one-dimensional version to bear. To that end, we define h(t)=f(a+t(b-a)) for t in the closed interval \left[0,1\right]. Then f(b)-f(a)=h(1)-h(0), and we can also show that h is differentiable everywhere inside the interval. Indeed, we can evaluate the difference quotient

\displaystyle\frac{h(s)-h(t)}{s-t}=\frac{f((a+t(b-a))+(s-t)(b-a))-f(a+t(b-a))}{s-t}

Taking the limit as s approaches t, we find

\displaystyle h'(t)=\left[D_{b-a}f\right](a+t(b-a))=df(a+t(b-a))(b-a)

which exists since f is differentiable.

So our old differential mean-value theorem tells us that there is some \tau\in\left(0,1\right) so that

\displaystyle\begin{aligned}f(b)-f(a)&=h(1)-h(0)\\&=\frac{h(1)-h(0)}{1-0}\\&=h'(\tau)\\&=df(a+\tau(b-a))(b-a)\\&=df(\xi)(b-a)\end{aligned}

where \xi=a+\tau(b-a) is a point in the open line segment (a,b).

October 13, 2009 Posted by | Analysis, Calculus | 4 Comments

   

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