When does a function equal its Taylor series?

Taylor’s theorem says

f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n

When does the thing on the left equal the thing on the right?

A few things could go wrong:

  • Maybe not all the terms on the right side exist, i.e. the function f might not be infinitely differentiable.
  • Maybe f is infinitely differentiable but the series diverges.
  • Maybe f is infinitely differentiable but the series converges to something other than f.

The canonical example of the last problem is the function g(x) defined to be exp(-1/x2) for positive x and 0 otherwise. This function is so flat when it approaches the origin that all derivatives are zero there. So all the coefficients in Taylor’s formula are zero. But the function g is positive for positive x.

Everything above can be found in a standard calculus text, but the following results are hardly known.

Being infinitely differentiable is necessary but not sufficient for a function to be real analytic. What additional requirements would be sufficient?

We start by examining what went wrong with the function g(x) above. It has a Taylor series at every point, but the radii of convergence go to zero as we get close to the origin. At the origin it is not analytic because the radius of convergence collapses to 0.

Alfred Pringsheim claimed that it is enough to require that the radii of convergence be bounded below on an interval. If we let ρ(x) be the radius of convergence for a series centered at x, then a function f is analytic in an open interval J if ρ(x) has some lower bound δ > 0 in J. Pringsheim’s theorem was correct, but his proof was flawed. The proof was accepted for 40 years until Ralph Boas discovered the flaw.

Here is a generalization of Pringsheim’s theorem. (It’s not clear to me from my source [1] who first proved this theorem. Perhaps Boas, but in the same context Boas mentions the work of others on the problem.)

Let f be infinitely differentiable in an open interval J. Suppose the radius of convergence ρ(x) for a Taylor series centered at x is positive for each x in the interval J. Suppose also that for every point p in J we have

\liminf_{x\to p} \frac{\rho(x)}{|x-p|} > 1.

Then f is analytic on the interval J.

Note that if ρ(x) is bounded below by δ > 0 then the limit above is infinite and so Pringsheim’s theorem follows as a special case.

The function g(x) above does not satisfy the hypothesis of the theorem on any open interval around 0 because if we set p = 0, the limit above is 1, not greater than 1.

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[1] R. P. Boas. When Is a C Function Analytic? Mathematical Intelligencer, Vol 11, No. 4, pp. 34–37.


One thought on “When does a function equal its Taylor series?

  1. Very interesting result. Is there any simple example of an analytic function whose convergence radius for a given p goes to zero so slowly that the limit is larger than one?

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