Polynomial Function and $n$-times differentiable

Question

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.

Just for curiosity, did you cover Taylor series yet? My guess would be not, but one never knows :)

Patrick Da Silva · Accepted Answer · 2011-11-03 03:49:03Z

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.

answered Nov 3 '11 at 3:49

Patrick Da Silva
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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

1

$f^{(n)}=0$ means that the nth derivative is the constant zero function. – N. S. Nov 3 '11 at 4:02

1

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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Polynomial Function and $n$-times differentiable

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