# Is it possible to find a real number $a$ such that $f′(a)=0,f′′(a)=0,…,f^{(r)}(a)=0,f^{(r+1)}(a)≠0$

Let $f\colon\mathbb C\to\mathbb C$ be an entire non constant function. We consider its values on the real line. The function $f$ has infinitely many real zeros and there is infinitely many real solutions to the equation $f(s)=w$ for any real number $w$.

Assume that there exist a vector $(u₁,u₂,u₃,...,u_{r},u_{r+1})∈ℝ^{r+1}$ such that $$f′(u₁)=0,f′′(u₂)=0,...,f^{(r)}(u_{r})=0,f^{(r+1)}(u_{r+1})≠0$$

My question is : Is it possible to find a real number $a$ such that $$f′(a)=0,f′′(a)=0,...,f^{(r)}(a)=0,f^{(r+1)}(a)≠0$$

Can we add some conditions to get this result if the answer is NO.

• Of course not. Consider $f(x)=e^x\sin x$. – Hagen von Eitzen Oct 8 '14 at 11:26
• @HagenvonEitzen: Can we add some conditions to get this result if the answer is NO – DER Oct 8 '14 at 11:29
• This may not be true, but I suspect this can happen only if $f$ has a zero of order $r$ at $a$. I'm writing this without having thought about it very much, so I may be completely wrong. I'll consider it after posting this comment. – MPW Oct 8 '14 at 11:31