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I was lectured on this however I did not understand what should I do exactly. How can I find this interval?

Find a neighbourhood ($-\delta,\delta$) of $0$ for which the $3rd$ order Taylor polynomial $P_{3,0}$ of $f(x)=e^x$ is within $1/200$ of $f(x)$.

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  • $\begingroup$ You wrote "$3$rd order Taylor polynomial $P_{5,0}$." Did you want third or fifth order here? $\endgroup$ – Mark Viola Mar 10 '16 at 5:31
  • $\begingroup$ @Dr.MV A typo, I meant third order... $\endgroup$ – NeoXx Mar 10 '16 at 5:44
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From the extended mean value theorem, there exists a number $\xi \in (0,x)$ ($\xi \in (x,0)$)for $x>0$ ($x<0$) such that

$$e^x=1+x+\frac12x^2+\frac16 x^3+\frac1{24}e^{\xi}x^4$$

Then, the error $E(x)$ between the exponential function and the third order approximation is

$$E(x)=\frac{1}{24}e^{\xi}x^4$$

Note that we want to find a number $\delta$ such that $x\in (-\delta,\delta)$ implies $E(x)<1/200$ or

$$E(x)=\frac{1}{24}e^{\xi}x^4<1/200 \tag 1$$

Taking $x<(3/25)^{1/4}$ we have from $(1)$

$$\begin{align} E(x)&=\frac{1}{24}e^{\xi}x^4\\\\ &<\frac{1}{24}e^{(3/25)^{1/4}}x^4\\\\ &<1/200\\\\ &\implies x<e^{-(1/4)(3/25)^{1/4}}(3/25)^{1/4}\\\\ \end{align}$$

We may choose a smaller interval and the error will still be bounded. So, since $e^{-(1/4)(3/25)^{1/4}}\ge 1-(1/4)(3/25)^{1/4}$, if $x<(3/25)^{1/4}\,(1-(1/4)(3/25)^{1/4})$, then $E(x)<1/200$.

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