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Suppose $ f(x)$ that is infinitely differentiable in $[a,b]$.

For every $c\in[a,b] $ the series $\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n $ is a polynomial.

Is true that $f(x)$ is a polynomial?

I can show it is true if for every $c\in [a,b]$, there exists a neighborhood $U_c$ of $c$, such that
$$f(x)=\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n\quad\text{for every }x\in U_c,$$ but, this equality is not always true.

What can I do when $f(x)\not=\sum\limits_{n=0}^\infty \cfrac{f^{(n)}(c)}{n!}(x-c)^n$?

  • 9
    Two solutions starting from weaker assumptions are given [in this MO thread](http://mathoverflow.net/questions/34059/if-f-is-infinitely-differentiable-then-f-coincides-with-a-polynomial)2011-12-22
  • 5
    Put $F_n:=\bigcap_{k\geq n}\{x\in [a,b], f^{(k)}(x)=0\}$ and apply Baire's category theorem.2011-12-22
  • 6
    I'm left wondering if the stronger assumptions here permit some more elementary proof.2011-12-22
  • 2
    @t.b. Would you (or @Davide) mind typing up a correct answer (possible just taken from MO), perhaps as community wiki? (Or, I can do it if no one else wants to). There are currently 10 incorrect answers (some deleted), and no correct answers.2012-10-22

2 Answers 2