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

MATH100 December 2013

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Question 10 (b)
Full-Solution 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 $\displaystyle f(x)=\ln(1-x^{2})$ prove that $\displaystyle -{\frac {5}{32}}\leq \ln \left({\frac {8}{9}}\right)\leq -{\frac {1}{9}}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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 2

Recall that the Lagrange Remainder formula gives that the remainder (error) in the k-th degree Taylor polynomial of f at a is: $R_{k}(x)={\frac {f^{(k+1)}(c)}{(k+1)!}}(x-a)^{k+1}$ where $\displaystyle c$ is some (undetermined) number in the interval $\displaystyle [a,x]$. Since we are using a first-degree Maclaurin polynomial, $a=0$ and $k=1$: $R_{1}(x)={\frac {f''(c)}{2!}}x^{2}$ for some $\displaystyle c$ in the interval $\displaystyle [0,x]$.

Hint 3
To apply the Lagrange Remainder formula, we must find an $x$ such that $\displaystyle f(x)=\ln(1-x^{2})=\ln \left({\frac {8}{9}}\right)$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: ${\begin{aligned}{\frac {f''({\frac {1}{3}})}{18}}&\leq \ln \left({\frac {8}{9}}\right)\leq {\frac {f''(0)}{18}}\\{\frac {1}{18}}\cdot {\frac {-2(({\frac {1}{3}})^{2}+1)}{(1-({\frac {1}{3}})^{2})^{2}}}&\leq \ln \left({\frac {8}{9}}\right)\leq {\frac {1}{18}}\cdot {\frac {-2(0^{2}+1)}{(1-0^{2})^{2}}}\\{\frac {1}{18}}\cdot {\frac {-180}{64}}&\leq \ln \left({\frac {8}{9}}\right)\leq {\frac {1}{18}}\cdot -2\\{\color {blue}-{\frac {5}{32}}}\,&{\color {blue}\leq \ln \left({\frac {8}{9}}\right)\leq -{\frac {1}{9}}}\end{aligned}}$

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