Real Analysis – Why Doesn’t a Taylor Series Always Converge?

Question

The Taylor expansion itself can be derived from mean value theorems which themselves are valid over the entire domain of the function. Then why doesn't the Taylor series converge over the entire domain? I understand the part about the convergence of infinite series and the various tests. But I seem to be missing something very fundamental here..

Answer 1

It is rather unfortunate that in calc II we teach Taylor series at the same time as we teach Taylor polynomials, all the while not doing a very good job of stressing the distinction between an infinite series and a finite sum. In the process we seem to teach students that Taylor series are a much more powerful tool than they are, and that Taylor polynomials are a much less powerful tool than they are.

The main idea is really the finite Taylor polynomial. The Taylor series is just a limit of these polynomials, as the degree tends to infinity. The Taylor polynomial is an approximation to the function based on its value and a certain number of derivatives at a certain point. The remainder formulae tell us about the error in this approximation. In particular they tell us that higher degree polynomials provide better local approximations to a function.

But the issue is that what "local" means really depends on the degree $n$. Looking at the Lagrange remainder, the error in an approximation of degree $n$ is

$$\frac{f^{(n+1)}(\xi_n) (x-x_0)^{n+1}}{(n+1)!}$$

where $\xi_n$ is between $x_0$ and $x$. So the ratio of the errors between step $n$ and step $n-1$ is*

$$\frac{f^{(n+1)}(\xi_n) (x-x_0)}{f^{(n)}(\xi_{n-1}) (n+1)}$$

where similarly $\xi_{n-1}$ is between $x$ and $x_0$. So the error is smaller where this quantity is less than $1$. From this form we can see that we can choose $x$ close enough to $x_0$ to guarantee that this quantity is less than $1$, because for fixed $n$ we have something bounded times $x-x_0$. But if

$$\frac{f^{(n+1})(\xi_n)}{f^{(n)}(\xi_{n-1})} > \frac{n+1}{x-x_0}$$

then the approximation of degree $n$ will be worse than the approximation of degree $n-1$ at $x$. If this keeps happening over and over, then there is no hope of the Taylor series converging to the original function anywhere except the point of expansion. In other words, if the derivatives near $x_0$ (not necessarily just at $x_0$) grow way too fast with $n$, then Taylor expansion has no hope of being successful, even when the derivatives needed exist and are continuous.

*Here I am technically assuming that $f^{(n)}(\xi_{n-1}) \neq 0$. This assumption can fail even when $f$ is not a polynomial; consider $f=\sin,x_0=\pi/2,n=1$. But this is a "degenerate" situation in some sense.