Faster-than-light communication using Taylor's theorem? I was thinking about Taylor's theorem and how if a function $f(x)$ is analytic at a point $a$ and one can measure all the derivatives at $a$, $f^{(n)}(a)$ then one knows the complete behaviour of the function within the radius of convergence (and maybe, by analytic continuation, everywhere except at singularities).
A blind man standing in a landscape whose height was described by an analytic function could, in principle, by locally measuring derivatives of the height function, know the complete shape of the landscape within a certain radius.
But suppose $x$ is space and $f$ is some physical observable and you are able to locally measure its derivatives. Then under the physical hypothesis that $f$ varies analytically, you should immediately know how $f$ behaves at other points in space in a way that seems only constrained by how many (and how quickly) you can measure these local derivatives and not the speed of light. In this way you would seem to "know" about events before light from those events had reached you. What is wrong with this argument?
 A: Solutions to a wave equation, like those governing most of the waves that transmit information from point to point in the physical universe, need not be analytic, even for "physically reasonable" parameters.
As an example, consider the classical wave equation for a scalar wave $\psi(x, t)$ propagating in one dimension with speed $c$ - could be the speed of light, could be some speed less:
$$\frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} = \frac{\partial^2 \psi}{\partial x^2}$$
Suppose the initial scalar field $\psi(x, 0) := \psi_i(x)$ is started out as a bump function - smooth eveywhere, non-analytic, with compact support, i.e. zero outside of some interval, say $[-1, 1]$. We will assume the first time derivative is zero everywhere as well for simplicity. Now consider some interval suitably far outside this, e.g. $x \in [2, \infty)$. It is easy to see that since $\psi_i(x)$ is 0 there and moreover constant, then $\frac{\partial^2 \psi}{\partial x^2}$ is also zero, and so then also $\frac{\partial^2 \psi}{\partial t^2} = 0$. If the first time derivative is initialized to also be zero as we did, then it will thus remain zero at these points at least for a short time into the future, and likewise at suitably near times the field itself will still be at value zero at these points - but it will be nonzero in a growing region around the initial "hump".  This means we have a solution that is not spatially analytic since it switches between two different types of analytic behavior: being constant on a nontrivial interval yet not constant everywhere. Thus Taylor's theorem does not apply, and the propagation speed $c$ is preserved. Moreover, since the wave equation is linear, if you superpose this on any other "background solution", even an analytic one, you'll get a new nonanalytic solution, which then generically models the creation of a localized disturbance within the field.
A: When you say an event this implies something is happening. Or you could say something is changing. This change will still take time to propagate. 
In other words, you would only be using Taylor's theorem for the function at some point in time, but you would need to wait for the function you are measuring to change at your location to know this event occurred. This change will still take a finite amount of time to happen after the event.
A: The important observation to make is that determination of an infinite order derivative involves the whole function domain. If for the first derivative an interval dx is needed then for the nth one you need ndx. Since in practice dx is finite thus tends to infinity. So to know all derivatives is equivalent to knowing all function values.
