Science:Math Exam Resources/Courses/MATH100/December 2013/Question 10 (b)/Solution 1

To apply the Lagrange Remainder Formula, we first need to find an $x$ such that $\displaystyle f(x)=\ln(1-x^{2})=\ln \left({\frac {8}{9}}\right)$, that is, $\displaystyle 1-x^{2}={\frac {8}{9}}\Leftrightarrow x=\pm {\frac {1}{3}}$.

Since $\displaystyle \ln(1-x^{2})$ is an even function, we choose the positive root $\displaystyle x={\frac {1}{3}}$ without loss of generality.

By Taylor's Remainder theorem (as applied to Maclaurin polynomials),

$T_{1}(x)+{\text{min}}\{R_{1}(x)\}\leq f(x)\leq T_{1}(x)+{\text{max}}\{R_{1}(x)\}$

where $\displaystyle {\text{min}}\{R_{1}(x)\}$ and $\displaystyle {\text{max}}\{R_{1}(x)\}$ denote the minimum and maximum values of $\displaystyle R_{1}(x)$ on the interval $\displaystyle [0,x]$, respectively.

Since we know that $\displaystyle T_{1}(x)=0$ from part (a), we simply consider:

${\text{min}}\{R_{1}(x)\}\leq \ln \left({\frac {8}{9}}\right)\leq {\text{max}}\{R_{1}(x)\}$

We now calculate the upper and lower bounds of the error. To do so, we substitute $\displaystyle x={\frac {1}{3}}$ into the Lagrange Remainder Formula:

$R_{1}\left({\frac {1}{3}}\right)={\frac {f''(c)}{2!}}\left({\frac {1}{3}}\right)^{2}={\frac {f''(c)}{18}}$

for some $c$ in in the interval $[0,{\frac {1}{3}}]$. To find the error bounds, we maximize this function as a function of $c$. But observe that

$f''(x)={\frac {-2(x^{2}+1)}{(1-x^{2})^{2}}}$

is monotonically decreasing on the interval $[0,{\frac {1}{3}}]$ (the denominator is increasing while the numerator is decreasing (note the minus sign), so overall $\displaystyle f''(x)$ is decreasing). Hence the minimum and maximum values of the error are attained at $c={\frac {1}{3}}$ and $\displaystyle c=0$, respectively.

Plugging this in we obtain the required lower and upper bound for the error:

${\begin{aligned}{\frac {f''({\frac {1}{3}})}{18}}&\leq \ln \left({\frac {8}{9}}\right)\leq {\frac {f''(0)}{18}}\\{\frac {1}{18}}\cdot {\frac {-2(({\frac {1}{3}})^{2}+1)}{(1-({\frac {1}{3}})^{2})^{2}}}&\leq \ln \left({\frac {8}{9}}\right)\leq {\frac {1}{18}}\cdot {\frac {-2(0^{2}+1)}{(1-0^{2})^{2}}}\\{\frac {1}{18}}\cdot {\frac {-180}{64}}&\leq \ln \left({\frac {8}{9}}\right)\leq {\frac {1}{18}}\cdot -2\\{\color {blue}-{\frac {5}{32}}}\,&{\color {blue}\leq \ln \left({\frac {8}{9}}\right)\leq -{\frac {1}{9}}}\end{aligned}}$