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We know that the two times differentiable function $g : \mathbb{R} \rightarrow \mathbb{R}$ is such that $g(0) = 999$, $g'(0) = 1000$ and $|g''(x)| \le 10000$ for every $x \in \mathbb{R}$. Let $K := g(\frac{1}{1000})$. How to prove that $K$'s second digit after the decimal point equals to $9$ or $0$? What are its other digits?

egreg
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malex
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    I suppose $g^{''}$ is constantly continous but I don't think I would help. I just don't know how to begin :( – malex Mar 25 '15 at 10:04

1 Answers1

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From the fact that

$g(10^{-3})=g(0)+g'(0)10^{-3}+\frac{g''(\theta)}{2}(10^{-3})^2$

for some $\theta\in[0,10^{-3}]$ and the related values for $g,g'$ and the bound of $g''$ we have

$|g(10^{-3})-1000|\leq 0.005$

This means that

$999.095\leq g(10^{-3})\leq 1000.005$

and the proof is completed.

RTJ
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  • Could you be more specific? I mean how did you get first equation? – malex Mar 25 '15 at 12:49
  • It's Taylor's theorem: A $k+1$-times differentiable function can be written as $f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f(k)(a)}{k!}(x-a)^k++\frac{f(k)(\xi)}{(k+1)!}(x-a)^{k+1}$ for some $\xi\in[a,x]$. – RTJ Mar 25 '15 at 13:03