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: $f^{(k)}(1-2\prod\_{j=1}^{k}t\_{j})$ 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 
$f^{(k)}(1-2\prod\_{j=1}^{k}t\_{j})=0$ 
and 
$f^{(k+1)}(1-2 \prod\_{j=1}^{k} s\_{j})\neq 0$
that is, the real $(1 - 2\prod\_{j=1}^{k}s\_{j})$ is a simple root of the function $f^{(k)}$?



How I can choose $(t_1,t_2,...,t_{r}) \in (0,1)^{r}$ such that $f^{(k)}\left(1-2\prod_{j=1}^{k}t_{j}\right)=0$?
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.


