# error term in the Taylor expansion of 1/x

What is the error in the Taylor polynomial of degree 5 for $$f(x)=1/x$$ using $$x_0=3/4$$ for $$x\in [1/2, 1]$$?

Since the $$(n+1)th$$ derivative of $$f(x)=1/x$$ is $$\frac{(-1)^{n+1}(n+1)!}{x^{n+2}}$$, I think the remainder term would be

$$R_{6}(x)=\frac{1}{\xi^{7}}(x-3/4)^{6}$$. (is this correct?)

Now how many terms must be taken to get an error of less than $$10^{-2}? 10^{-4}$$? This one I am not sure how to do.

You are correct in that $$R_6(x) = \frac1{\xi^7}(x-3/4)^6$$, where $$\xi$$ is a real number between $$3/4$$ and $$x$$. Now, \begin{align} \sup_{x,\xi\in[3/4,1]}|R_6(x)| &= \sup_{x,\xi\in[1/2,3/4]}\frac1{\xi^7}|x-3/4|^6\\ &= \frac1{(3/4)^7}(1-3/4)^6\\ &= \left(\frac 43\right)^7\left(\frac14\right)^6\\ &= \frac 4{3^7}\\ &= \frac 4{2187}\approx 0.001828989. \end{align} In general, $$\sup_{x,\xi\in[3/4,1]}|R_{n+1}(x)| = \frac 4{3^{n+2}}.$$ So to get the remainder less than $$10^{-4}$$ we have $$\frac 4{3^{n+2}} < 10^{-4}\iff n > \frac1{\log 3}(\log 40000 - \log 9)\approx 7.645473.$$ Hence we need $$n\geqslant 8$$.
• wouldn't the largest error be when $\xi = 1/2$? I thought the error should be $2^{n+2}\frac{1}{4}^{n+1}$