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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?

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  • 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.
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  • $\begingroup$ 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! $\endgroup$ – Pspl Aug 13 at 7:41

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