Question
Using Taylor's Inequality, find how many terms of the Maclaurin series for $\ln(1+x)$ you need to use to estimate $\ln 1.4$ to within $0.001$?
I have found that $|f^{n+1}(x)|=\frac{n!}{(1+x)^{n+1}}$
I need to find the upper bound on $|f^{n+1}(x)|$ or in other words the upper bound on the second, third, fourth… derivatives (because $n$ starts from $1$ and goes to infinity). This upper bound is called $M$. So I am trying to find $M$ such that $|f^{n+1}(x)|\leq M$. Most importantly, we need to define an interval of $x$ so that $M$ has meaning. This interval is $-d\leq x\leq d$ where $d$ is a positive number. This is where I get confused. What is $d$ in this problem?
Best Answer
Your Taylor series is centered at $1$. You want to evaluate $\ln(1 + 0.4)$, so you want your estimate to be good out to a radius of $0.4$ from the center, so your estimate will be good for all $x$ with $1 - 0.4 \leq x \leq 1 + 0.4$.
I also note that each of your derivatives, $f^{(n)}$, is decreasing in $x$ on this interval, so each of them is bounded by their value at the left end of the interval, that is, at $x = 1-0.4 = 0.6$. (You know that the graph of $1/x^{n+1}$ decreases on $(0, \infty)$. Shifting it left and scaling it vertically by a positive number doesn't change that.)