# Proofs of Taylor-series expansions

For differentiable functions, one would think that the values of successive derivatives at a single point are so “very local” properties of a function, that they cannot possibly determine its values at other points lying “very far away”. As is well-known, for many useful functions this is not so, as revealed by the Taylor-series expansion theorem, of functions exponential, trigonometric, logarithmic etc.

My problem is that in all proofs I know of, this overwhelmingly wonderful & miraculous phenomenon is seemingly extinguished by a rather dry technical manoeuver: by repeated application of the mean-value theorem, for all x the remainder is bounded by something tending to zero as n tends to infinity, so the expansion is valid.

I wonder, where has the “miracle-point” gone? Can someone give an intuitive explanation that captures the essence of these “by-a-single-point-uniquely -determinable” functions?

• The passage from local to global is done by using some global condition. For example, for $\sin$ you know that its derivative is bounded and then the $n!$ overpowers the $(\theta-0)^n$ for all $\theta\in[-R,R]$. For $e^x$ on the interval $[-R,R]$ you know that the $n!$ outweighs the factor $e^{\theta}(\theta-0)^n$ for all $\theta\in[-R,R]$. It is always some condition that holds for all $\theta$. The sequence of derivatives at a single point alone is not enough. – YAlexandrov Mar 14 '18 at 13:32
• The classes of functions in which the derivatives at a point determines the function (within the class) are called quasi-analytic classes of functions. This generalizes the class of analytic functions, which among other properties have the property that you were talking about (being determined by their derivatives at a point). $e^x,\sin(x)$ are analytic functions and the class of analytic functions is quasi-analytic. – YAlexandrov Mar 14 '18 at 13:39