# Approximations using Taylor Polynomials

I have been sitting with this exercise for quite some time now, and I cannot understand how to solve it. The exercise is:

Use the second order Taylor Polynomial of $f(x)= x^{1/3}$ about x = 8 to approximate $9^{1/3}$. Also, estimate the error. Lastly, write the smallest interval you can be sure contains the exact value $9^{1/3}$.

I have already calculated the Taylor Polynomial $P_2(x) = 2 + \frac{x-8}{12} + \frac{(x-8)^2}{288}$ and also the estimated value $P_2(9) = 599/288$. However, what really trips me over is:

• How do I estimate the error?

My thought process is that I need to give the interval of possible errors by maximizing and minimizing the absolute value of the Lagrange Remainder, and then use these values as bounds for the interval. But this did not work.

• How do I calculate the smallest interval that is sure to contain $9^{1/3}$?

Here, I thought I could use the minimum value from the error bounds as a left bound for where the exact value could be, and do the analog for the right bound, to yield $estimate + minError < exactAnswer < estimate + maxError$ However this was incorrect as well.

I would appreciate a thorough answer that explains at least one of these points.

Thank you :)

• Unless I missed something the polynomial $P_2$ should start with $P_2(x)=2+...$ and not $P_2(x)=1+...$. – Adren Feb 12 '17 at 19:54
• Sorry, typo. I will fix that asap! Thank you for spotting that. – Simon Sirak Feb 12 '17 at 19:55

You can let $f(x) = (8+x)^{1/3}$ and use Taylor's theorem with the mean-value form of the remainder to estimate that $f(x) - P_2(x) = R_2(x) = \frac{f^{(3)}(\xi)}{3!} x^{3}$ with some $\xi$ between 0 and 1. Since $\frac{f^{(3)}(\xi)}{3!} = \frac{5}{81 (\xi + 8)^{8/3}}$ and $x^3 = 1$ one can estimate the error to be less than the maximum absolute value of $\frac{5}{81 (\xi + 8)^{8/3}}$ for $\xi$ between 0 and 1, which is $\frac{5}{81 \cdot 8^{8/3}} \simeq 2.4 \cdot 10^{-4}$ . Indeed, $9^{1/3} - 599/288 \simeq 2.2 \cdot 10^{-4}$ so the estimate is pretty good.
The same estimate can be used to calculate the smallest interval. For all values of $0< \xi < 1$ and $x=1$, $R_2(x) > 0$. So with $f(x) = P_2(x) + R_2(x)$ we have $P_2(x) < f(x) < P_2(x) + \max(R_2(x))$, that is $599/288 < 9^{1/3} < 599/288 + \frac{5}{81 \cdot 8^{8/3}}$ or in approximate numbers, $2.0799 < 9^{1/3} < 2.08010$.
This can be improved to $P_2(x) + \min(R_2(x)) < f(x) < P_2(x) + \max(R_2(x))$, that is $599/288 + \frac{5}{81 \cdot 9^{8/3}} < 9^{1/3} < 599/288 + \frac{5}{81 \cdot 8^{8/3}}$ or in approximate numbers, $2.080037 < 9^{1/3} = 2.08008 < 2.08010$.