The question is to find $P_3(x)$ for $ f(x) = (x-1) \; \ln x $ about $x_0 = 1$ and find the upper bound on the error for $P_3(0.5)$ used to estimate $f(0.5)$.
I got $$ f^{(4)}(x)= \frac{2}{x^3} + \frac{6}{x^4} $$ and $$ P_3(x) = (x – 1)^2 – \frac{1}{2} (x – 1)^3 $$ and $$ R_3(x) = \left( \frac{2}{(\xi(x))^3} + \frac{6}{(\xi(x))^4} \right) \frac{1}{4!} (x-1)^4. $$
I chose to replace $\xi(x)$ with $0.5$, since that would maximize $\frac{2}{(\xi(x))^3} + \frac{6}{(\xi(x))^4} $. I'm confused on whether I should choose the same value for $x$ as well, because if I use $0.5$ for $x$ also, the error bound would be $0.291\bar6$, which is the answer the book has. But shouldn't I use $0$ for $x$, since that would maximize $(x-1)^4$?
Best Answer
I figured it out when I realized that $\xi(x)$ has to be between $x_0$ and $x$, so it has to be on the interval $[0.5, 1]$. Using $0.5$ for $\xi$ and $x$ will maximize $R_3(x)$, which is $0.291\bar6$. They happened to be the same for this problem.