# Are there non-trivial examples of an 'infinitieth' derivative? What is the criteria for 'convergence'?

Let me preface this by saying that 'convergence' refers to $$\lim_{n\to\infty}\frac{d^nf(x)}{dx^n}$$ being well defined. This stands in contrast to functions which, despite having derivatives of all orders, do not approach any particular value as each successive derivative is taken (e.g. $$\cos{x}$$ or $$1/x$$).

There are two trivial cases for the 'infinitieth' derivative - namely $$ke^{x+c}$$, for which $$\frac{d^\infty f(x)}{dx^\infty}=ke^{x+c}$$, and $$\frac{d^\infty f(x)}{dx^\infty}=0$$, which is the case for any function with some constant $$n^\text{th}$$ derivative (e.g. power/polynomial functions).

Intuitively, I can think of a few reasons why there wouldn't be any other examples, but then there might be some remarkable special function which defies all intuition.

Are there any non-trivial examples of functions (real or complex) where the 'infinitieth' derivative exists?

• Presumably by "power functions" you mean "polynomials"? – Rob Arthan Feb 20 at 1:59
• Polynomial functions, yes. – R. Burton Feb 21 at 0:07

Assume that in some neighbourhood of $$x=0$$, the function $$f$$ is given by a convergent power series $$f(x) = \sum_{n \in \Bbb N} a_n x^n$$. Then the existence of $$\lim_{n\to \infty} f^{(n)}(0)$$ is equivalent to the existence of

$$(*) \qquad \displaystyle\lim_{n\to \infty} n! \cdot a_n$$.

Conversely, given a sequence $$(a_n)_n$$ such that $$(*)$$ exists, the function $$f(x) := \sum_{n \in \Bbb N} a_n x^n$$, by comparison with the exponential series, actually converges on all of $$\Bbb R$$, and the limit of its derivatives also exists everywhere.

The set of functions thus defined by sequences satisfying $$(*)$$ is closed under addition and scalar multiplication, and includes all examples so far (for polynomials, $$a_n = 0$$ for $$n \gg 0$$; for Robert Israel’s basic case $$f(x) = Ae^{cx}$$ with $$c \in (-1, 1]$$, we have $$a_n = \frac{A c^n}{n!}$$); but also many more. For example:

(1) $$f(x) = \sum_{n \in \Bbb N} \frac{1}{(n!)^2} x^n$$, or $$f(x) = \sum_{n \in \Bbb N} \frac{1}{(n!)!} x^n$$

(2) take any subset $$S \subsetneq \Bbb N$$ and set $$f(x) = \sum_{n \in S} \frac{1}{n!} x^n$$

or any combination of these ideas (Robert Israel's general example is included here too).

For the analogous question over $$\Bbb C$$, these should be all solutions. If you do not impose stronger conditions, on $$\Bbb R$$ there are more, e.g. $$f(x) = e^{-1/x^2}$$ (with the discontinuity at $$0$$ removed) at least has $$\lim_{n\to \infty} f^{(n)}(0) = 0$$.

Also $$f(x) = P(x) e^{cx}$$ where $$|c|<1$$ and $$P$$ is a polynomial has $$\lim_{n \to \infty} f^{(n)}(x) = 0$$. And don't forget that linear combinations of solutions are solutions.