# Why Taylor Series works so well for some functions and not for others

 Math Emeritus Sci Advisor Thanks PF Gold P: 39,285 No, even if a function's Taylor series, $\sum (f^{(n)}(a)/a!)(x- a)^n$, converges for all x, it does not necessarily converge to the function itself (except at a). For example, the function $f(x)= e^{-1/x^2}$ if $x\ne 0$, f(0)= 0 is infinitely differentiable at all x and all of its derivatives at 0 are equal to 0. So its Taylor series, about a= 0, is identically 0 while f(x) is 0 only at x= 0. Whether or not a function's Taylor series actually converges to the function itself is a very complicated question. Technically such functions are called "analytic" functions. (Sometimes "real analytic" to distinguish them from the same concept in functions of complex variables where the definition is the same but there are much more complicated consequences.)