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Let $f: [0,1] \to \mathbb{R}$ be a function and suppose that $f$ is differentiable on $[0,1]$. Suppose that $f'(0)=f''(0)=\ldots=f^{(5)}(0)=0$ and $f^{(6)}(0)>0$. Show that there exists a $\epsilon >0$ such that $ f(x) \ge f(0)+\displaystyle\frac{f^{(6)}(0)}{721}x^6$ for all $x \in [0,\epsilon)$

Here is what I have got so far, using remainder theorem of Taylor polynomial we have $f(x) - \left(f(0)+\displaystyle\frac{f^{(6)}(0)}{721}x^6\right)=\displaystyle\frac{f^{(7)}(0)}{7!}x^7$ for $ 0 \le c \le x$. The problem is that I need to show $f^{(7)}(c) \ge 0$. This is where I got stuck. Any hint would be greatly appreciated.

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No, you need to show that, close to zero, $x^6$ wins when compared with $x^7$ –  user20266 May 20 '12 at 14:01
    
@Thomas So, what do I need to prove then. Cheers –  Viet Hoang Quoc May 20 '12 at 14:05
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2 Answers 2

It is to tedious to write that as a comment.

Note that $6!= 720$. So you have a strict inequality $$\frac{1}{721} x^6 < \frac{1}{6!} x^6$$ if $x\neq 0$. So from Tayler you get a little amount more than what you need to show at order $6$. You need to use that little amount to dominate the remainder in a neighbourhood of $0$.

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To write out Thomas' answer: By Taylor's theorem one has $$f(x)=f(0)+{f^{(6)}(0)\over 720} x^6+o(x^6)=f(0)+{f^{(6)}(0)\over 721} x^6+f^{(6)}(0) x^6\left({1\over 519120}+o(1)\right)\qquad (x\to 0)\ .$$ Here the last term is $>0$ for sufficiently small $|x|>0$.

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