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

It's true that I'm not familiar with too many exotic functions, but I don't understand why there exist functions that cannot be described by a Taylor series? What makes it okay to describe any particular functions with such a series? Is there any difference for different number sets? In the case of complex numbers maybe? Could somebody provide an example?

• A Taylor series exists if and only if the function is infinitely differentiable at some a. – Doug M Feb 2 '17 at 3:46
• And even then, the Taylor series doesn't always converge to the function in some neighbor. @DougM – Thomas Andrews Feb 2 '17 at 3:48
• @ThomasAndrews Indeed, I thought about going there, but then decided to keep it to the point. Radius of convergence is then a whole 'nuther thing. – Doug M Feb 2 '17 at 3:50
• – Winther Feb 2 '17 at 3:52
• The Taylor series represent the function when the Taylor remainder tends to zero, otherwise the Taylor series doesnt represent the function. – Masacroso Feb 2 '17 at 4:03

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.

• I've never understood why so many find this surprising; I guess it is because many are taught (incorrectly) that Taylor Series must converge to the function. I was taught from the beginning that a Taylor Series simply provides an approximation in terms of derivatives around a point and might happen to converge. In the neighborhood of $0$, the function $e^{-1/x^2}$ is approximated quite well by $0$ – Brevan Ellefsen Feb 2 '17 at 4:02
• @BrevanEllefsen: That's not really why. It's because you would intuitively expect a smooth function to be predictable, and it seems weird that the derivatives of a function do not contain enough information to predict it at a nearby point. – Mehrdad Feb 2 '17 at 7:59
• Yes, but that function does have a Taylor series, it just doesn't agree with the function anywhere but at 0. – Klangen Feb 2 '17 at 14:21
• @BrevanEllefsen But any infinitely differentiable function in complex numbers does have a converging Taylor series, so the question for reals is interesting. I wouldn't say it is surprising that there are counter-examples in the reals, but it is interesting. – Thomas Andrews Feb 2 '17 at 15:24
• One of my favorite "proofs" that complex numbers "exist" is that the Taylor series at the real number $a$ for the function $\frac{1}{1+x^2}$ has radius of convergence $\sqrt{1+a^2}$. This is entirely a statement about real numbers, but it suggests there is some root of $1+x^2$ that is $\sqrt{1+a^2}$ distance away from each real number $a$. (If $p(x)$ has all real roots, then the radius of convergence at $a$ of $\frac{1}{p(x)}$ is always the distance to the nearest root to $a$.) @BrevanEllefsen – Thomas Andrews Feb 2 '17 at 15:43

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.

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.

The existence of functions that cannot be described by Taylor series is actually completely intuitive; take the indicator function of the rational numbers viewed as a subset of the reals, for example. Try to keep in mind that functions can be really... arbitrary.

Much more subtle is the existence of smooth functions that aren't analytic; Thomas Andrews gives the standard example of such a beast. Fwiw, my understanding of why this is possible is that okay, there's functions that change behaviour suddenly at a point, BUT the change in behaviour at that point is so gradual, so gentle, so smooth, that none of the function's derivatives can see the change happening; therefore, the Taylor series can't, either.

In addition to all the comments here, I would like to add the curious Weierstrass function, which is known for its quality of being nowhere differentiable despite the fact that it is continuous everywhere:

$$W(x) = \sum_{n=0}^\infty a^n\cos(b^n\pi x)$$

Consequently, it does not have a Taylor series.

You can find a visualization of $W$ here.

• This is a nice fractal! Are all fractals no differentiable? – dmtri Aug 23 '18 at 10:55
• @dmtri while I'm unsure of a formal definition of a fractal, differentiable functions must look "linear" if you "zoom in enough". Fractals self similarity on "zooming in" seems to rule this out. – Mark Aug 23 '18 at 16:02