# The accuracy of approximating $f(x) = x^{2/5}$ for $0.9 \le x \le 1.1$ using the cubic Taylor polynomial

For the equation $f(x) = x^{2/5}$, $a=1$, $n=3$, $0.9 \le x \le 1.1$

I was able to approximate f by the following Taylor polynomial: $$F_3(x) = 1 + \frac2 5 (x-1) - \frac3{25}(x-1)^2 + \frac8{125}(x-1)^3$$

When using Taylor's Inequality to estimate the accuracy of the approximation $f(x) \approx T_n(x)$ I got:

$$R_3(x) \le \frac{\frac{624}{625}0.9^{\frac{-18}{3}}}{24}(0.1)^4 \approx 0.00000783$$

This answer is incorrect. Can someone please tell me what I've done wrong in my calculation?

• You used the fourth derivative of $f$ there, right? What formula do you have for it? – user147263 Oct 5 '14 at 5:22
• What are all those $5$'s doing? Or is your function something other than $x^{2/3}$? – André Nicolas Oct 5 '14 at 5:23
• @AndréNicolas My bad, I've edited the question. I meant to write $x^{2/5}$. – Alex Oct 5 '14 at 20:02
• @CareBear my fourth derivative was $\frac {-624}{625} x^{\frac {-18}{5}}$ – Alex Oct 5 '14 at 20:03

You made a mistake in the development since, from binomial theorem,$$x^a=\Big(1+(x-1)\Big)^a=1+a (x-1)+\frac{1}{2} a(a-1) (x-1)^2+\frac{1}{6} a(a-1) (a-2) (x-1)^3+\frac{1}{24} a(a-1) (a-2) (a-3) (x-1)^4+O\left((x-1)^5\right)$$ If $a=\frac{2}{3}$, you then have $$x^{\frac{2}{3}}=1+\frac{2 (x-1)}{3}-\frac{1}{9} (x-1)^2+\frac{4}{81} (x-1)^3-\frac{7}{243} (x-1)^4+O\left((x-1)^5\right)$$
By the way, you could have checked that, for $x=1.1$, the expansion you used give a result equal to $1.02915$ for an exact value of $1.06560$ which is what you get using the correct expansion limited to the first terms.