The zeroes of any non-zero analytic function $\mathbb{R} \rightarrow \mathbb{R}$ are isolated. By a generalized zero of a smooth function $f:\mathbb{R} \rightarrow \mathbb{R}$, let us mean a zero of $f^{(n)}$ for some natural number $n \in \{0,1,2\ldots\}$.
Are the generalized zeroes of a non-polynomial analytic function $\mathbb{R} \rightarrow \mathbb{R}$ necessarily isolated?
My guess is 'no', but if the answer is yes, I'm also interested in the complex case where $\mathbb{R}$ is replaced by $\mathbb{C}$.
Let $$f(z) = \frac{e^{z}-e^{-z}}{2}- \sum_{k=0}^\infty \frac{1}{((2k)!)^2} z^{2k}$$ It is not hard to see there is a sequence $z_k\sim \frac{1}{(2k)!} \to 0$ such that $f^{(2k)}(z_k) = 0$