# Why doesn't the identity theorem for holomorphic functions work for real-differentiable functions?

I've been fascinated by the idea of analytic continuation and I came across the identity theorem for holomorphic functions. (http://en.wikipedia.org/wiki/Identity_theorem)

On wikipedia it states: "Thus a holomorphic function is completely determined by its values on a (possibly quite small) neighborhood in D. This is not true for real-differentiable functions"

Is there a simple way or an accessible example to show why this is so? I'm an enthusiast with a undergraduate math background, but I have not studied complex analysis.

• Differentiability in $\mathbb C$ is a much, much, much stronger condition than differentiability in $\mathbb R$. – MPW Mar 19 '14 at 2:24
• The basic reason is that every complex-differentiable function has a power series and the power series about a point defines a unique power series about every other point within its radius of convergence. – Kevin Carlson Mar 19 '14 at 2:46
• I just thought that I'd point out that the current Purdue Problem of the Week (Number 11) is related to this idea. You should give it a shot: math.purdue.edu/pow – Joel Apr 2 '14 at 19:28

Consider the function $$f(x) = \mathrm{e}^{-1/x^2}$$ on the reals. Although not defined at zero, it clearly has a limit at zero and that limit is zero. Now look at its derivatives: $$f'(x) = \frac{2}{x^3} f(x)$$ $$f''(x) = \frac{4-6x^2}{x^6} f(x)$$ $$f'''(x) = \frac{8-36x^2+24x^4}{x^9} f(x)$$ and so on, with the exponential overwhelming the rational function as we take limits at zero. We find, in fact, that this limiting process gives us zeroes for every derivative at zero.