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If a function $f$ is $n$-times differentiable on $\mathbb R$ and $f^{(n)}=0$, prove $f$ is a polynomial of degree $\leq n-1$.

A hint would suffice.

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Hint: Induction –  Jason DeVito Nov 3 '11 at 3:49
@JasonDeVito: Thank you, see the comment on Patrick's post. –  MathMathCookie Nov 3 '11 at 3:59
Just for curiosity, did you cover Taylor series yet? My guess would be not, but one never knows :) –  N. S. Nov 3 '11 at 4:04
@user9176 no I haven't –  MathMathCookie Nov 3 '11 at 4:08
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1 Answer

up vote 1 down vote accepted

Have you tried integrating? Start with the case $n=1$. Which functions have $0$ derivative? With the case $n=2$, which functions have $0$ second derivative? I think looking at it this way will make your life more easier.

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Ah but I have not "learned" integration yet :) –  MathMathCookie Nov 3 '11 at 3:50
Well go by induction on $n$ : have you "learned" induction? –  Patrick Da Silva Nov 3 '11 at 3:51
Thank you. I am trying this; I realized that I do not understand what this: $f^{(n)}=0$ means however. I know I am OP, but this was the way the problem was stipulated on paper... So for $f^{(1)}=0$ (aka base case), what exactly is this saying? –  MathMathCookie Nov 3 '11 at 3:58
$f^{(n)}=0$ means that the nth derivative is the constant zero function. –  N. S. Nov 3 '11 at 4:02
Perhaps understanding the question is the first step to answering it :) maybe you are more familiar with the notation $f'(x) = 0$ and $f''(x) = 0$. Writing $f^{(1)} = 0$ is another way of saying that $f'(x) = 0$ for all $x$ in the domain of $f$. (In this case $\mathbb R$.) –  Patrick Da Silva Nov 3 '11 at 4:08
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