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If we have a function $f:[0,1]\longrightarrow\mathbb{R}$ that is eight times differentiable, and know that

$f'(0)=f''(0)=f'''(0)=f^{(4)}(0)=f^{(5)}(0)=0$ and $f^{(6)}>0$,

then how can we show that there exists a $\delta\in(0,1)$ such that $f(x)\ge f(0) + \frac{f^{(6)}(0)}{721}x^6$ for all $x\in[0,\delta)$?

I thought a sensible approach would be to begin by using Taylor's theorem since that last $x^6$ term looks like a Taylor remainder with the 720 being $6!+1$. However, this hasn't gotten me anywhere.

I also tried to show that for $\epsilon>0$ there exists a $\delta\in(0,1)$ such that for all $x\in(0,\delta)$ we have that $|\frac{f(x)-p_6(x)}{x^6}|\lt\epsilon$ where $p_6(x)$ is the 6th Taylor polynomial, since we should then be able to choose a particular $\epsilon$ that gives us our result, but I also couldn't manage to prove this - does anyone have any hints or advice?

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  • $\begingroup$ The taylor-Lagrange thm is a good idea, but use it with order $7$: $$f(x)=f(0)+\frac{f^{(6)}(0)}{6!}x^6+\frac{f^{(7)}(c)}{7!}x^7$$, with $c$ between $0$ and $x$, and use that as $f^{(7)}$ is continuous, $|f^{(7)}|$ is bounded, say by $M$ on $[0,1]$. You get $f(x)\geq f(0)+\frac{f^{(6)}(0)}{6!}x^6-M\frac{x^7}{7!}$ for all $x\in [0, 1]$ and now it is easy to finish. $\endgroup$
    – Kelenner
    Oct 24, 2017 at 14:17
  • $\begingroup$ I think I follow what you're saying so far, but isn't this completely independent of epsilon or delta? $\endgroup$ Oct 24, 2017 at 14:37
  • $\begingroup$ To finish, you have to find $\delta>0$ such that for $0<x<\delta$ $f(0)+\frac{f^{(6)}(0)}{6!}x^6-M\frac{x^7}{7!}\geq f(0)+\frac{f^{(6)}(0)}{6!+1}x^6$, or $\frac{f^{(6)}(0)}{6!}-M\frac{x}{7!}\geq \frac{f^{(6)}(0)}{6!+1}$ $\endgroup$
    – Kelenner
    Oct 24, 2017 at 14:43
  • $\begingroup$ Sorry I don't think I quite follow... also how come the $\frac{f^{(7)}(c)}{7!}$ became -M? $\endgroup$ Oct 24, 2017 at 14:58
  • $\begingroup$ I have used that for $x\in [0,1]$ we have $|\frac{f^{(7)}(c)}{7!}x^7]\leq \frac{M}{7!}x^7$, so $f(0)+\frac{f^{(6)}(0)}{6!}x^6+\frac{f^{(7)}(c)}{7!}x^7\geq f(0)+\frac{f^{(6)}(0)}{6!}x^6-\frac{M}{7!}x^7$ $\endgroup$
    – Kelenner
    Oct 24, 2017 at 15:04

1 Answer 1

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The Taylor-Lagrange thm is a good idea, but use it with order $7$: $$f(x)=f(0)+\frac{f^{(6)}(0)}{6!}x^6+\frac{f^{(7)}(c)}{7!}x^7$$, with $c$ between $0$ and $x$. Now we use that (as $f$ est $8$ times differentiable) the function $f^{(7)}$ is continuous, hence $|f^{(7)}|$ is bounded, say by $M>0$ on $[0,1]$.

We have $|\frac{f^{(7)}(c)}{7!}x^7|\leq \frac{M}{7!} x^7$ for $x\in [0,1]$ hence we get that $f(x)\geq f(0)+\frac{f^{(6)}(0)}{6!}x^6-M\frac{x^7}{7!}$

To have our result, it suffices now to find a $\delta>0$, such that

$$\frac{f^{(6)}(0)}{6!}-\frac{f^{(6)}(0)}{1+6!}\geq M \frac{x}{7!}$$ for $0\leq x<\delta$, that is easy, we can take $\delta=\frac{7!}{M}(\frac{f^{(6)}(0)}{6!}-\frac{f^{(6)}(0)}{1+6!})$.

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  • $\begingroup$ Thinking about what happens at f(0), is this a local extrema? I feel like it has to be but can't think of how to explain why $\endgroup$ Oct 24, 2017 at 17:32

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