What kind of functions cannot be described by the Taylor series? Why is this?

Solution 1:

We have the somewhat famous function:

$$f(x)=\begin{cases}e^{-1/x^2}&x\neq 0\\ 0&x=0 \end{cases}$$

is infinitely differentiable at $0$ with $f^{(n)}(0)=0$ for all $n$, so, even though the function is infinitely differentiable, the Taylor series around $0$ does not converge to the value of the function for any $x>0$.

Technically, any function that is infinitely differentiable at $a$ has a Taylor series at $a$. Whether you find that Taylor series useful depends on what you want the series to do.

For example, if given a $g$ infinitely differentiable at $0$, the we know that there exists $C,\epsilon>0$ such that:

$$\left|g(x)-\sum_{k=0}^{n} \frac{g^{(k)}(0)}{k!}x^k\right|<Cx^{n+1}$$

for all $|x|<\epsilon$.

So the finite terms of the Taylor series are in some sense always the "best" polynomial for agreeing with function.

So what happens to our function $f$ above is that $f(x)$ converges to $0$ faster than any function $x^n$.

What we don't always get, for real functions, is a Taylor series that converges to the function in the interval.

In complex numbers, things become intriguing. It turns out, if you define differentiation on complex functions in a relatively simple way, then any function which is differentiable at a point is infinitely differentiable at that point, and the Taylor series converges in some "ball" centered on that point.

Solution 2:

If the limit of the Lagrange Error term does not tend to zero (as $n \to \infty $), then the function will not be equal to its Taylor Series.

You can also read more on this in Appendix $1$ in Introduction to Calculus and Analysis $1$ by Courant and John. Hope it helps.

Solution 3:

I think the intuition you want is the fact that functions that are not complex-differentiable* (also known as holomorphic) are not described by a Taylor series.

And to give another example that is perhaps even more unexpected than the one given by Andrew:

$$f(z) = \begin{cases} e^{-\frac{1}{z}} && \text{if } z > 0 \\ 0 && \text{otherwise}\end{cases}$$

This function is smooth and zero over an infinitely long interval, and yet nonzero, because it is not holomorphic.

*If you're not familiar with complex differentiation, it's like real differentiation, with $h$ complex:

$$f'(z) = \lim_{h \to 0} \frac{f(z + h) - f(z)}{h}$$

For details, see here.