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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.

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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 $n\geqslant 8$.

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  • $\begingroup$ wouldn't the largest error be when $\xi = 1/2$? I thought the error should be $2^{n+2}\frac{1}{4}^{n+1}$ $\endgroup$
    – mXdX
    Jan 13, 2020 at 16:39

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