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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 $f^{(k)}$?

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Please try to make sure the typesetting outputs as expected,. In its present form your question is unfortunately unreadable. Try backquoting your expressions with `, as explained in the margin "How to write math". –  Loïc Teyssier Apr 15 '13 at 11:30
    
@Loïc Teyssier: Ok I edit the question. But I notice that this is the first version. Something changes my text! –  Shpigle Apr 15 '13 at 13:41
    
I guess there is a typo: the index in the product should be $r$ instead of $k$, right? –  Loïc Teyssier Apr 15 '13 at 15:16
    
Yes, thank you. Fixed. –  Shpigle Apr 15 '13 at 15:19
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I tried to fix the formatting; I hope everything is as intended. I think the problem is that you mix the use of tex/mathjax and the usage of special characters; for the product and for subscripts and so on. –  quid Apr 15 '13 at 15:29
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1 Answer

up vote 2 down vote accepted

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.

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