# Science:Math Exam Resources/Courses/MATH100/December 2016/Question 13 (b)/Solution 1

By differentiating ${\displaystyle \log x}$ and noticing the pattern, we see that for all ${\displaystyle n\geq 0}$, ${\displaystyle f^{(n+1)}(x)=(-1)^{n}n!x^{-(n+1)}}$.

From the mean-value form of the remainder from Taylor's theorem, we know that there exists some ${\displaystyle c}$ between 1 and 1.1 such that

{\displaystyle {\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 ${\displaystyle T_{n}(1.1)}$ is an underestimate, then ${\displaystyle T_{n}(1.1), so ${\displaystyle f(1.1)-T_{n}(1.1)>0}$; since ${\displaystyle c>0}$, the condition ${\displaystyle f(1.1)-T_{n}(1.1)>0}$ is equivalent with

{\displaystyle {\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 ${\displaystyle n+1}$ is odd; that is, when ${\displaystyle n}$ is even.

So, when ${\displaystyle n}$ is even, ${\displaystyle T_{n}(1.1)}$ is an underestimate for ${\displaystyle f(1.1)}$; in other words, when ${\displaystyle \color {blue}n\in \{2m:m\in \mathbb {N} \}=\{0,2,4,6,8,\dots \}.}$