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 $f(x)$.

  • $\begingroup$ 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

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


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 $E(x)<1/200$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.