The Unapologetic Mathematician

Mathematics for the interested outsider

Taylor’s Theorem

Like I said yesterday, because of extraneous terms the higher differentials don’t transform well, and so they’re not going to be useful for many of our purposes. However, there’s one thing it’s really good for: generalizing Taylor’s theorem. Specifically, the version of Taylor’s theorem that resembles the mean value theorem. And we’ll even use an approach like we did for the extension of that result to higher dimensions.

So let’s say that f has continuous partial derivatives up to order m in an open region X\subseteq\mathbb{R}^n. If a and b are two points in X so that the whole line segment (a,b) is contained in X, then there exists some point \xi along that segment so that

\displaystyle f(b)-f(a)=\sum\limits_{k=1}^{m-1}\frac{1}{k!}d^kf(a;b-a)+\frac{1}{m!}d^mf(\xi;b-a)

Just like with the mean value theorem, we’ll define a new function h(t)=f(a+t(b-a)) for t in the closed interval [0,1]. This is a composite of the function f with the function g(t)=a+t(b-a). This function is clearly differentiable, with constant derivative g'(t)=b-a. And so the chain rule tells us that


where we’ll fudge the distinction between the derivative and the differential of h because it’s a real-valued function of a single real variable.

But now we can do the same thing to take the second derivative.

\displaystyle h''(t)=\frac{d}{dt}\left[D_if\right](t)(b^i-a^i)=\left[D_{j,i}f\right](t)(b^j-a^j)(b^i-a^i)=d^2f(g(t);b-a)

and so on, with each derivative of h being given by a similar formula

\displaystyle h^{(k)}(t)=d^kf(g(t);b-a)

up until the index k=m. Beyond that we don’t know that the higher differentials exist.

Now we take all these derivatives and stick them into the usual one-variable Taylor theorem, which tells us that

\displaystyle h(1)=h(0)+\sum\limits_{k=1}^{m-1}\frac{1}{k!}h^{(k)}(0)(1-0)^k+\frac{1}{m!}h^{(m)}(\tau)(1-0)^m

for some \tau in the interval (0,1). With our formulæ for h and its derivatives, this becomes

\displaystyle f(b)-f(a)=\sum\limits_{k=1}^{m-1}\frac{1}{k!}d^kf(a;b-a)+\frac{1}{m!}d^mf(\xi;b-a)

where \xi=a+\tau(b-a) is a point on the line segment (a,b), and the theorem is proved.

October 20, 2009 Posted by | Analysis, Calculus | 2 Comments