## $\mathbb{C} \succ \mathbb{R}$

- Mon, 28 Dec, 2009

Suppose I have two functions $f$ and $g$, on the real numbers (or some compact interval if it makes you feel warmer inside). Is it reasonable to expect that if their derivatives match at some point, i.e. $f(x)=g(x)$, $f^\prime(x)=g^\prime(x)$, and so on, then they are equal? Furthermore, suppose this is true for all values $y \le x$?

As a counter-example, consider the function: $$f(x) = \begin{cases}0 & x \le 0 \\ \exp(-1/x) & x \gt 0 \end{cases}$$ This function is continuous at $x=0$, and its derivatives there are all zero. In other words, at $x=0$ it "exactly looks like" the constant function $g(x)=0$.

In a way, it seems perverse that we can't "sense" the impeding rise as we move through the origin. Another way to say it is that the derivatives of $f$ are well defined, and so can be used to form a Taylor series; however, the function does not equal its Taylor expansion, even though the latter exists.

This all makes a bit more sense when discussed in the context of complex analysis. The function $f(z)$, regarded as a function over $\mathbb{C}$ has an essential singularity at $z=0$. This is an example of the fact that although smooth functions over the reals seems nicer, from an elementary point of view, smooth functions over the complex numbers enjoy more "globally" nice properties, e.g. over $\mathbb{C}$ the existence of a Taylor expansion is the same as being smooth and defined almost everywhere (see Wikipedia's entry on this theorem).