Let $f:ℝ→ℝ$ be a real analytic function. Assume that $f$ has simple trivial zeros at each nonpositive integer. Then, all the $k^{th}$ derivative $f^{(k)}$ of $f$ have necessarily infinitely many real zeros. Let us consider the functions: $f^{(k)}(1-2∏_{j=1}^{k}t_{j})$ for $k=1,..,r$ and $(t₁,t₂,...,t_{r})∈(0,1)^{r}$. We know that these functions have also infinitely many real zeros.
My question is: How I can choose $(t₁,t₂,...,t_{k})∈(0,1)^{k}$? such that $f^{(k)}(1-2∏_{j=1}^{k}t_{j})=0$ and $f^{(k+1)}(1-2∏_{j=1}^{k}s_{j})≠0$

that is, the real $(1-2∏_{j=1}^{k}s_{j})$ is a simple root of the function $f^{(k)}$

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 $f^{(k)}$?

Let $f:ℝ→ℝ$ be a real analytic function. Assume that $f$ has simple trivial zeros at each nonpositive integer. Then, all the $k^{th}$ derivative $f^{(k)}$ of $f$ have necessarily infinitely many real zeros. Let us consider the functions: $f^{(k)}\left(1-2\prod_{j=1}^{k}t_{j}\right)$ for $k=1,..,r$ and $(t₁,t₂,...,t_{r})∈(0,1)^{r}$. We know that these functions have also infinitely many real zeros.
My question is: How I can choose $(t₁,t₂,...,t_{r})∈(0,1)^{r}$? such that $f^{(k)}\left(1-2\prod_{j=1}^{k}t_{j}\right)=0$ and $f^{(k+1)}\left(1-2\prod_{j=1}^{k}t_{j}\right)≠0$

that is, the real $(1-2\prod_{j=1}^{k}t_{j})$ is a simple root of the function $f^{(k)}$

Let $f:ℝ→ℝ$ be a real analytic function. Assume that $f$ has simple trivial zeros at each nonpositive integer. Then, all the $k^{th}$ derivative $f^{(k)}$ of $f$ have necessarily infinitely many real zeros. Let us consider the functions: $f^{(k)}(1-2∏_{j=1}^{k}t_{j})$ for $k=1,..,r$ and $(t₁,t₂,...,t_{r})∈(0,1)^{r}$. We know that these functions have also infinitely many real zeros.
My question is: How I can choose $(t₁,t₂,...,t_{k})∈(0,1)^{k}$? such that $f^{(k)}(1-2∏_{j=1}^{k}t_{j})=0$ and $f^{(k+1)}(1-2∏_{j=1}^{k}s_{j})≠0$

that is, the real $(1-2∏_{j=1}^{k}s_{j})$ is a simple root of the function $f^{(k)}$

Let $f:ℝ→ℝ$ be a real analytic function. Assume that $f$ has simple trivial zeros at each nonpositive integer. Then, all the $k^{th}$ derivative $f^{(k)}$ of $f$ have necessarily infinitely many real zeros. Let us consider the functions: $f^{(k)}(1-2∏_{j=1}^{k}t_{j})$ for $k=1,..,r$ and $(t₁,t₂,...,t_{r})∈(0,1)^{r}$. We know that these functions have also infinitely many real zeros.
My question is: How I can choose $(t₁,t₂,...,t_{r})∈(0,1)^{r}$? such that $f^{(k)}(1-2∏_{j=1}^{k}t_{j})=0$ and $f^{(k+1)}(1-2∏_{j=1}^{k}s_{j})≠0$

that is, the real $(1-2∏_{j=1}^{k}s_{j})$ is a simple root of the function $f^{(k)}$

Let $f:ℝ→ℝ$ be a real analytic function. Assume that $f$ has simple trivial zeros at each nonpositive integer. Then, all the $k^{th}$ derivative $f^{(k)}$ of $f$ have necessarily infinitely many real zeros. Let us consider the functions: $f^{(k)}\left(1-2\prod_{j=1}^{k}t_{j}\right)$ for $k=1,..,r$ and $(t₁,t₂,...,t_{r})∈(0,1)^{r}$. We know that these functions have also infinitely many real zeros.
My question is: How I can choose $(t₁,t₂,...,t_{r})∈(0,1)^{r}$? such that $f^{(k)}\left(1-2\prod_{j=1}^{k}t_{j}\right)=0$ and $f^{(k+1)}\left(1-2\prod_{j=1}^{k}t_{j}\right)≠0$

that is, the real $(1-2\prod_{j=1}^{k}t_{j})$ is a simple root of the function $f^{(k)}$