# Minimum value of $n$ for Lagrange reminder on Taylor polynomial of $\frac{1}{x}$

I'm trying to solve the following question:

"Find the minimum value of $$n$$ for which is guaranteed $$T_1^n\left(\frac{1}{x}\right)$$ approximates $$\frac{1}{x}$$ with an error less than $$10^{-3}$$ on the interval $${x}\in{[\,0.9\,,\;1.1\,]}$$ using Lagrange reminder."

Here's what I've done so far:

First I found a closed form for the derivatives of $$f(x)=\frac{1}{x}$$:

$$f^{(n)}(x)=(-1)^n\frac{n!}{x^{n+1}}\,,\;\forall{n}\in{\mathbb{N}_0}$$

Then I found the Lagrange reminder $$R_1^n\left(\frac{1}{x}\right)$$:

$$R_1^n\left(\frac{1}{x}\right)=\left|\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-1)^{n+1}\right|\,,\;{x}\in{[\,0.9\,,\;1.1\,]}\,,\;{\xi}\in{V_{|x-1|}^{(1)}}$$

I know I have to maximize $$|f^{(n+1)}(\xi)|$$ by some $$M\gt0$$ but, as

$$f^{(n+1)}(\xi)=(-1)^{n+1}\frac{(n+1)!}{\xi^{n+2}},$$

$${x}\in{[\,0.9\,,\;1.1\,]}$$, and $${\xi}\in{V_{|x-1|}^{(1)}}$$, I simply can't find $$M$$ (because $$\xi$$ is, at least $$0.9$$, the absolute value of $$f^{(n+1)}(\xi)$$ could be as great as we want).

Am I missing something?

• I think $$R_1^n(1/x)$$ should be denoted $$R_1^n(x)$$.
• You need to bound the remainder: $$| R_1^n(x) | = \left| \frac{(-1)^{n+1}}{\xi^{n+2}} (x - 1)^{n+1} \right| \leq \left| \frac{1}{\xi} \left( \frac{0.1}{\xi} \right)^{n+1} \right|$$ The maximum is indeed reached for $$\xi = 0.9$$. $$n = 4$$ gives you the accuracy you want.
• Got it! I was considering $1.1^{n+1}$ as the numerator. That's why I wasn't able to reach the maximum. Thanks! By the way, I'm from Portugal and here we write $R_c^n[f(x)]$, eh eh! – Pspl Aug 13 at 7:41