Science:Math Exam Resources/Courses/MATH100/December 2015/Question 11 (b)

MATH100 December 2015

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Question 11 (b)
If $f^{(3)}(x)={\frac {x\sin(x)+x^{2}\cos(x)}{10-x^{2}}}$, show that the absolute value of the error when we approximate $f(1)$ using the second Maclaurin polynomial is less than $0.04={\tfrac {1}{25}}$.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall the Taylor's inequality: If $|f^{(n+1)}(x)|\leq M$ for $|x-a|\leq d$, then the remainder $R_{n}(x)$ of the Taylor series of $f$ at $a$ satisfies the inequality $|R_{n}(x)|\leq {\frac {M}{(n+1)!}}|x-a|^{n+1}$ for $|x-a|\leq d$.

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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. Note that Maclaurin series is the special case of Taylor series at $a=0$. Since $f^{(3)}(x)={\frac {x\sin(x)+x^{2}\cos(x)}{10-x^{2}}}$ is given and we know that $|\sin x|\leq 1$ and $|\cos x|\leq 1$ for any $x$ in the real line, we have for $|x-0|\leq 1$ $|f^{(3)}(x)|\leq \left|{\frac {x\sin(x)+x^{2}\cos(x)}{10-x^{2}}}\right|\leq {\frac {|x||\sin(x)|+|x^{2}||\cos(x)|}{10-|x|^{2}}}\leq {\frac {1+1}{10-1}}={\frac {2}{9}}.$ Then, using the Taylor's inequality in the Hint, we have $|f(x)-T_{2}(x)|=|R_{2}(x)|\leq {\frac {1}{3!}}\cdot {\frac {2}{9}}\cdot |x|^{3}$ for $|x|\leq 1$. Therefore, we obtain $\color {blue}|f(1)-T_{2}(1)|\leq {\frac {1}{3!}}\cdot {\frac {2}{9}}\cdot |1|^{3}\leq {\frac {1}{27}}\leq {\frac {1}{25}}.$