Let $f:\mathbb{R} \to \mathbb{R}$ be a real analytic function. Assume that $f$ has simple trivial zeros at each nonpositive integer. Then, all the $k$th derivatives $f^{(k)}$ of $f$ have necessarily infinitely many real zeros. Let us consider the functions: $f^{(k)}(12\prod_{j=1}^{k}t_{j})$ for $k=1,..,r$ and $(t_1,t_2,...,t_{r})\in (0,1)^{r}$. We know that these functions have also infinitely many real zeros.
My question is: How I can choose $(t_1,t_2,...,t_{k})∈(0,1)^{k}$ such that
$f^{(k)}(12\prod_{j=1}^{k}t_{j})=0$
and
$f^{(k+1)}(12 \prod_{j=1}^{k} s_{j})\neq 0$
that is, the real $(1  2\prod_{j=1}^{k}s_{j})$ is a simple root of the function $f^{(k)}$?
Take the 2minute tour
×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.



As your question is stated, nothing guarantees that $f^{(k)}$ has a single simple zero. The fact that you introduce extra paremeters cannot change that fact! So, in general, the answer is: there is no way. 

