MATH100 December 2013
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q4 (c) • Q5 • Q6 • Q7 • Q8 (a) • Q8 (b) • Q8 (c) • Q8 (d) • Q9 • Q10 (a) • Q10 (b) • Q11 •
Question 10 (b)

FullSolution Problem. Justify your answer and show your work. Full simplification of numerical answers is required unless explicitly stated otherwise.
Using the Lagrange Remainder Formula for $\displaystyle f(x)=\ln(1x^{2})$ prove that $\displaystyle {\frac {5}{32}}\leq \ln \left({\frac {8}{9}}\right)\leq {\frac {1}{9}}$.

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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1

Use the Maclaurin polynomial obtained from part (a) and the Lagrange remainder formula to calculate the error.

Hint 3

To apply the Lagrange Remainder formula, we must find an $x$ such that $\displaystyle f(x)=\ln(1x^{2})=\ln \left({\frac {8}{9}}\right)$.

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Solution

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To apply the Lagrange Remainder Formula, we first need to find an $x$ such that $\displaystyle f(x)=\ln(1x^{2})=\ln \left({\frac {8}{9}}\right)$, that is, $\displaystyle 1x^{2}={\frac {8}{9}}\Leftrightarrow x=\pm {\frac {1}{3}}$.
Since $\displaystyle \ln(1x^{2})$ 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
 $f''(x)={\frac {2(x^{2}+1)}{(1x^{2})^{2}}}$
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:
${\begin{aligned}{\frac {f''({\frac {1}{3}})}{18}}&\leq \ln \left({\frac {8}{9}}\right)\leq {\frac {f''(0)}{18}}\\{\frac {1}{18}}\cdot {\frac {2(({\frac {1}{3}})^{2}+1)}{(1({\frac {1}{3}})^{2})^{2}}}&\leq \ln \left({\frac {8}{9}}\right)\leq {\frac {1}{18}}\cdot {\frac {2(0^{2}+1)}{(10^{2})^{2}}}\\{\frac {1}{18}}\cdot {\frac {180}{64}}&\leq \ln \left({\frac {8}{9}}\right)\leq {\frac {1}{18}}\cdot 2\\{\color {blue}{\frac {5}{32}}}\,&{\color {blue}\leq \ln \left({\frac {8}{9}}\right)\leq {\frac {1}{9}}}\end{aligned}}$

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