Is there an easy function without an easy Taylor series expansion?

After class one day, a student approached me with an interesting question:

Is there an easy function without an easy Taylor expansion?

This question really struck me for several reasons.

Most functions do not have an easy Taylor (or Maclaurin) expansion. After all, the formula for a Taylor expansion involves the $n$ th derivative of the original function, and higher-order derivatives usually get progressively messier with each successive differentiation.
Most of the series expansions that are taught in Calculus II arise from functions that somehow violate the above rule, like $f(x) = \sin x$ , $f(x) = \cos x$ , $f(x) = e^x$ , and $f(x) = 1/(1-x)$ .
Therefore, this student was under the misconception that most easy functions have easy Taylor expansions, while in reality most functions do not.

It took me a moment to answer his question, but I answered with $f(x) = tan x$ . Successively using the Quotient Rule makes the derivatives of $tan x$ messier and messier, but $tan x$ definitely qualifies as an easy function that most students have seen since high school. It turns out that the Taylor expansion of $f(x) = \sin x$ can be written as an infinite series using the Bernoulli numbers, but that’s a concept that most calculus students haven’t seen yet.