Errors and Taylor Polynomials

Question

For $g(x)=x^{1/3}$, $a=1$, degree $3$ I found the Taylor polynomial:

$$p_3(x) = 1 + (x-1)/3 - ((x-1)^2)/9 + (5(x-1)^3)/81$$

How do I use the error formula for the Taylor polynomial of degree 3 to obtain an upper bound on:

$$|g(x) - p_3(x)|, \quad 1 \le x \le 1.5$$ which is as small as possible?

So far I found the required derivative: $g''''(x) = -80x^{-11/3}/81$

Answer 1

we will make a change of variable $$ x = 1 + u, u = 1 - x.$$ then $$(1 + u)^{1/3}= 1 + \frac 13u -\frac 13 \frac 23 \frac{1}{2!}u^2 + \frac 13 \frac 23 \frac 53 \frac{1}{3!}u^3 - \frac 13 \frac 23 \frac 53 \frac 83\frac{1}{5!}u^5+\cdots $$ this is an alternating series so the error is bounded above by the first neglected term $$ |error| \le \frac{2\cdot5 \cdot8}{3^4 \cdot5!} \frac1{2^5} $$

Answer 2

Since you have found the next derivative, determine whether the function $g^{4}(x)=\frac{-80x^{\frac{-11}{3}}}{81}$ is increasing or decreasing. Check the value at the endpoints and take the largest value.

This then becomes $ |E| \leq \frac{8}{2*81*120}$ where E is the error based off this website, you can see how I got my answer.

http://www.millersville.edu/~bikenaga/calculus/remainder-term/remainder-term.html

Answer 3

Here is the reminder term in the Taylor series

$$ R_n = \frac{ f^{(n+1)} (\eta) }{ (n+1)! } (x-1)^{n+1},\quad 1< \eta < x $$

$$ \implies |R_n| = \frac{80}{81\, 4!} \left|\eta^{-11/3}\right||x-1|^4 \leq \frac{80}{81\, 4!}(0.5)^4 = \frac{5}{1944}. $$