TITLE: On what interval does a Taylor series approximate (or equal?) its function?
QUESTION [20 upvotes]: Suppose I have a function $f$ that is infinitely differentiable on some interval $I$.
When I construct a Taylor series $P$ for it, using some point $a$ in $I$, does $f(x) = P(x)$ for all $x$ in $I$?
I'm confused as to whether Taylor series approximate (or equal - I'm not about that either) functions on one point, or on an interval.

REPLY [5 votes]: Since you speak about intervals (on the real line), perhaps it should also be mentioned that the "natural habitat" for power series is really the complex plane; computing a power series involves only +, -, *, /, and limits, which are well defined operations on complex numbers. And for so-called "complex analytic" (or "holomorphic") functions, which includes most functions that you encounter in calculus, it is a fact that the Taylor series at any point in the complex plane is convergent (equal to the function) in a circle around that point. The size of this circle is such that it exactly reaches out to the nearest singularity of the function. (Circles in the complex plane are the counterparts to intervals on the real line in this context.)
A simple example is $f(x)=1/(1+x^2)$. If you just look at the graph of this function (for $x$ real) it looks perfectly nice, and there seems to be no reason why the Taylor series at $x=0$ only manages to converge to the function in the interval $(-1,1)$. But if you think of $x$ as a complex variable, it's clear that there are singularities (division by zero) at the points $x=\pm i$, which lie at distance one from the origin, and that explains why the Taylor series converges inside the circle with radius one. (Note that the intersection of this disk with the real axis is just the interval $(-1,1)$.)
The same thing goes for $g(x)=\arctan x$. Its derivative is $f(x)$ above, and where the derivative has singularitites, the function has too. So the Taylor series for $g(x)$ at $x=0$ is also convergent inside the unit circle.