2

Suppose you a function $f$ which is differentiable,

with the property that

$$ f^{(n)} (0) = (n!)^2 $$

And in general

$$ f^{(n)} (a) = O((n!)^2)$$

For any $a \in \mathbb{R}$.

This function then is nowhere well defined by a taylor series. But I don't immediately accept that this function "doesn't exist". What if we have some degenerate super fast growing function (like an analytically continued nested ackermann or something along those lines).

Can anyone give an example of a function that satisfies the above equations, or generally a function such that

$$ f^{(n)}(a) = 2^{O(\Omega(n \ln (n) ))} $$

Sidharth Ghoshal
  • 16,771
  • cf. http://math.stackexchange.com/questions/63050/every-power-series-is-the-taylor-series-of-some-c-infty-function – ThomasR Jan 28 '17 at 16:06

1 Answers1

3

There is no holomorphic function (on any neighborhood of $0$) with $f^{(n)}(0) = (n!)^2$, but you can find a $C^\infty$ smooth function with arbitrarily prescribed partial derivatives (this is a theorem of Borel). In the complex setup, you can for example ask for a smooth function satisfying $$ \frac{\partial^n f}{\partial z^n}(0) = (n!)^2 \qquad \frac{\partial^{k+l} f}{\partial \bar z^k \partial z^l}(0) = 0 $$ for all $l$ and $k \ge 1$

mrf
  • 43,639
  • 1
    Shouldn't you make a distinction between prescribing partial derivatives at a single point (possible) as opposed to everywhere (impossible). The asker did ask for the derivatives at every $a \in \mathbb{R}$ to have a certain asymptotic growth presumably with constant independent of $a$. – user21820 Jul 20 '16 at 09:38
  • So if the function isn't holomorphic then asking if it has derivative of a certain value is meaningless to begin with – Sidharth Ghoshal Jul 28 '16 at 00:17