*infinite*number of derivatives, you know how

*each*derivative changes, so you should be able to predict the

*full global behaviour of the function*, assuming it is infinitely differentiable (smooth) throughout.

Everything is nice and dandy in this picture. But then you come across two disastrous, life-changing facts that make you cry for those good old days:

**Taylor series have**If I can predict the behaviour of a function up until a certain point, why can't I predict it a bit afterwards? It makes sense if the function becomes rough at that point, like if it jumps to infinity, but even functions like $1/(1+x^2)$ have this problem. Sure, we've heard the explanation involving complex numbers, but why should we care about the complex singularities (here's a question: do we care about quaternion singularities?)? Specifically, a Taylor series may have a zero radius of convergence. Points around which a Taylor series has a zero radius of convergence are called*radii of convergence*--**Pringsheim points**.**Weird crap --**Like $e^{-1/x}$. Here, the Taylor series*does*converge, but it converges to the wrong thing -- in this case, to zero. Points at which the Taylor series doesn't equal a function on any neighbourhood, despite converging, are called**Cauchy points**.

In this article, we'll address the

**weird crap --**$e^{-1/x}$ (or "$e^{-1/x}$ for $x>0$, 0 for $x= 0$" if you want to be annoyingly formal about it) will be the example we'll use throughout, so if you haven't already seen this, go plot it on Desmos and get a feel for how it looks near the origin.*Terminology:*We'll refer to**smooth non-analytic functions**as**defective functions**.
The thing to realise about $e^{-1/x}$ is that the Taylor series -- $0 + 0x + 0x^2 + ...$ --

*isn't wrong*. The truncated Taylor series of degree $n$ is the*best polynomial approximation*for the function near zero, and none of the logic here fails for $e^{-1/x}$. There is honestly no other polynomial that better approximates the shape of the function as $x\to 0$.
If you think about it this way, it isn't too surprising that such a function exists -- what we have is a function that

**goes to zero**as $x\to 0$**faster than any polynomial**does. I.e. a function $g(x)$ such that
$$\forall n, \lim\limits_{x\to0}\frac{g(x)}{x^n}=0$$

This is not fundamentally any weirder than a function that escapes to infinity faster than all polynomials. In fact, such functions are quite directly connected. Given a function $f(x)$ satisfying:

$$\forall n, \lim\limits_{x\to\infty} \frac{x^n}{f(x)} = 0$$

We can make the substitution $x\leftrightarrow 1/x$ to get

$$\forall n, \lim\limits_{x\to0} \frac{1}{x^n f(1/x)} = 0$$

So $\frac1{f(1/x)}$ is a valid $g(x)$. Indeed, we can generate plenty of the standard smooth non-analytic functions this way: $f(x)=e^x$ gives $g(x)=e^{-1/x}$, $f(x)=x^x$ gives $g(x)=x^{1/x}$, $f(x)=x!$ gives $g(x)=\frac1{(1/x)!}$ etc.

To better study what exactly is going on here, consider Taylor expanding $e^{-1/x}$ around some point other than 0, or equivalently, expanding $e^{-1/(x+\varepsilon)}$ around 0. One can see that:

$$\begin{array}{*{20}{c}}{f(0) = {e^{ - 1/\varepsilon }}}\\{f'(0) = \frac{1}{{{\varepsilon ^2}}}{e^{ - 1/\varepsilon }}}\\{f''(0) = \frac{{ - 2\varepsilon + 1}}{{{\varepsilon ^4}}}{e^{ - 1/\varepsilon }}}\\{f'''(0) = \frac{{6{\varepsilon ^2} - 6\varepsilon + 1}}{{{\varepsilon ^6}}}{e^{ - 1/\varepsilon }}}\\ \vdots \end{array}$$

Or ignoring higher-order terms for our purposes,

$$f^{(N)}(0)\approx(1/\varepsilon)^{2N}e^{-1/\varepsilon}$$

Each derivative $\frac{e^{-1/\varepsilon}}{\varepsilon^{2N}}\to0$ as $\varepsilon\to0$, but they each approach zero

*slower*than the previous derivative, and somehow that is enough to give the sequence of derivatives the "kick" that they need in the domino effect that follows -- from somewhere at $N=\infty$ (putting it non-rigorously) -- to make the function grow as $x$ leaves zero, even though all the derivatives were zero at $x=0$.*But*we can still make it work -- by letting $N$, the upper limit of the summation approach $\infty$

*first*, before $\varepsilon\to 0$. In other words, instead of directly computing the derivatives $f^{(n)}(0)$, we consider the terms

$$\begin{array}{*{20}{c}}{f_\varepsilon^{(0)} = f(0)}\\{{{f}_\varepsilon^{(1)} }(0) = \frac{{f(\varepsilon ) - f(0)}}{\varepsilon }}\\{{{f}_\varepsilon^{(2)} }(0) = \frac{{f(2\varepsilon ) - 2f(\varepsilon ) + f(0)}}{{{\varepsilon ^2}}}}\\{{{f}_\varepsilon^{(3)} }(0) = \frac{{f(3\varepsilon ) - 3f(2\varepsilon ) + 3f(\varepsilon ) - f(0)}}{{{\varepsilon ^3}}}}\\ \vdots \end{array}$$

And write the generalised

**Hille-Taylor series**as:
$$f(x) = \mathop {\lim }\limits_{\varepsilon \to 0} \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}f_\varepsilon ^{(n)}(0)} $$

Then $N\to\infty$ before $\varepsilon\to0$ so you "reach" $N\to\infty$ first (or rather, you get large $n$th derivatives for increasing $n$) before $\varepsilon$ gets to 0.

Another way of thinking about it is that the "local determines global" stuff makes sense to predict the value of the function at $N\varepsilon$, countable $N$, but it's a stretch to talk about uncountably many $\varepsilon$s away, which is what a finite neighbourhood is. But with these difference operators in the Hille-Taylor series, one ensures that each neighbourhood is a finite multiple of $h$ away at any point, so the differences determine $f$.

**Very simple (but fun to plot on Desmos) exercise:**use $e^{-1/x}$ or another defective function to construct a "

**bump function**", i.e. a smooth function that is 0 outside $(0, 1)$, but takes non-zero values everywhere in that range.

Similarly, construct a "

**transition function**", i.e. a smooth function that is 0 for $x\le0$, 1 for $x\ge1$. (hint: think of a transition as going from a state with "none of the fraction" to "all of the fraction")

If you're done, play around with this (but no peeking):

**desmos.com/calculator/ccf2goi9bj**

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