Science:Math Exam Resources/Courses/MATH100/December 2016/Question 13 (b)

MATH100 December 2016

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Question 13 (b)
Now let $T_{n}(x)$ be the $n$th degree Taylor polynomial centred at $x=1$ for the function ${\begin{aligned}f(x)&=\log(x).\end{aligned}}$ (Remember that $\log x=\log _{e}x=\ln x$.) For which value(s) of $n$ will $T_{n}(1.1)$ give an underestimate of $\log(1.1)$? You must justify your answer.

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 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
Use Taylor's theorem and the mean-value form of the remainder.

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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. By differentiating $\log x$ and noticing the pattern, we see that for all $n\geq 0$, $f^{(n+1)}(x)=(-1)^{n}n!x^{-(n+1)}$. From the mean-value form of the remainder from Taylor's theorem, we know that there exists some $c$ between 1 and 1.1 such that ${\begin{aligned}f(1.1)-T_{n}(1.1)&={\frac {f^{(n+1)}(c)}{(n+1)!}}(1.1-1)^{n+1}\\&={\frac {(-1)^{n}}{(n+1)c^{n+1}}}\left(0.1\right)^{n+1}\\&={\frac {-1}{(n+1)c^{n+1}}}(-0.1)^{n+1}\end{aligned}}$ If $T_{n}(1.1)$ is an underestimate, then $T_{n}(1.1)<f(1.1)$, so $f(1.1)-T_{n}(1.1)>0$; since $c>0$, the condition $f(1.1)-T_{n}(1.1)>0$ is equivalent with ${\begin{aligned}0<{\frac {-1}{(n+1)c^{n+1}}}(-0.1)^{n+1}\Longleftrightarrow 0>{\frac {1}{(n+1)c^{n+1}}}(-0.1)^{n+1}\Longleftrightarrow 0>(-0.1)^{n+1}.\end{aligned}}$ This occurs precisely when $n+1$ is odd; that is, when $n$ is even. So, when $n$ is even, $T_{n}(1.1)$ is an underestimate for $f(1.1)$; in other words, when $\color {blue}n\in \{2m:m\in \mathbb {N} \}=\{0,2,4,6,8,\dots \}.$