Estimating $\ln(0.9)$ using a Taylor polynomial about $x=1$ what is the least degree of the polynomial that assures an error smaller than $0.0010$?

(The $n^{th}$ derivative of $(-1)^{n-1}\cfrac{(n-1!)}{x^n}$ for $n\ge1$

Then Khan Academy says, the Lagrange bound for the error assures that $\left|R_n(1.4)\right|\le\left|\cfrac{(-1)^n\cfrac{n!}{z^{n+1}}}{(n+1)!}(1.4-1)^{n+1}\right|$

Then it says $\cfrac{0.4^{n+1}}{{(n+1)}{z^{n+1}}}\le\cfrac{0.4^{n+1}}{(n+1)}$

In the sentence before this one, why do they get rid of $z^{n+1}$?

  • $\begingroup$ Because $1\leq z$, therefore if you replace it by $1$ the denominator gets smaller, and the fraction larger. Take into account that I don't know what $R_n(1.4)$ and therefore, $z$, because you haven't defined it, but it looks like after you do, that is going to be the explanation. $\endgroup$ – user578878 Jul 28 '18 at 1:32

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