n-th derivative condition to become polynomial for real function I've seen a similar result in complex analysis, when an entire complex function satisfies $f(z)f^{(n)}(z)=0$ for all $z \in \mathbb{C}$ implies that $f(z)$ is polynomial. 
What if when $f: \mathbb{R} \rightarrow \mathbb{R}$ is a $n$ times differentiable function such that $f(x)f^{(n)}(x)=0$ for all $x\in\mathbb{R}$, does it follow that $f$ is polynomial?
thanks.
what i have tried so far: 
I tried induction: for $n=1$ we have $f(x)f'(x)=0$ which means $f(x)^2=c$ hence $f'(x)=0$ for all $x$. Supposing the statement is true for $n−1$, I want to prove that $f^{(n)}(x)f(x)=0$ will implies $f^{(n)}(x)f'(x)=0$ which implies $f^{(n)}(x)=0$ (by induction hypothesis) .. my idea is to define function $g$ such that i could get $f^{(n)}(x)f'(x)=0$.. i'm stuck here.. I have another try, by analysing $\bigcup A_j$, where $A_j$ is the open interval on which $f(x)$ is not zero (it can be proved it is indeed interval). 
I also have tried taylor theorem. 
 A: Since $f$ is continuous the set $S:=\{x\in{\mathbb R}\ |\ f(x)\ne 0\}$ is open, so it is a countable union of open intervals. Let $I:=\ ]0,h[\ $ be such an interval. As $f^{(n)}(x)\equiv 0$ on $I\ $ there is a polynomial $p$ of degree $\leq n-1$ such that $f(x)=p(x)$ on $I$. Assume $p$ is given by $p(x)=\sum_{k=r}^{n-1} a_k x^k$ with $a_r\ne 0$.
Now $f$ has a Taylor expansion of order $r$ at the origin, which means that there is a polynomial $j(x)=\sum_{k=1}^r c_k x^k$ such that $f(x)=j(x) + o(x^r)$ $\ (x\to 0)$. Since for all $x\in I$ we have $f(x)=p(x)$ it follows that $j(x)=a_r x^r$, and this implies
$$f(x)=(a_r + o(1)) x^r \qquad (x\to 0).$$
It follows that the zero $0$ of $f$ is in fact isolated, whence it is the right endpoint of an interval $I':=\ ]-h,0[\ $ of $S$. Using the Taylor expansion of order $n-1$ at $0$ we can verify that $f(x)=p(x)$  on $I'$ as well. 
This argument carries through for all component intervals of $S$; therefore $f$ has only isolated zeros on ${\mathbb R}$. Any finite interval $\ ]-M,M[\ $ can only contain a finite number of such zeros, and in between we have a single polynomial $p$ that represents $f$. It follows that in fact $f(x)=p(x)$ on all of ${\mathbb R}$.
A: I have a solution for this problem on my blog: http://mathproblems123.wordpress.com/2009/09/08/polynomial/
