Question

The nth Taylor polynomial Tn(x) for f(x)=\ln(1-x) based at b=0?Find the smallest value of n such that Taylor's inequality guarantees that |\ln(x)-\ln(1-x)|<0.01 for all x in the interval l=[-\frac{1}{2},\frac{1}{2}]

Polynomials
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asked 2021-05-07

The nth Taylor polynomial Tn(x) for \(f(x)=\ln(1-x)\) based at b=0?
Find the smallest value of n such that Taylor's inequality guarantees that \(|\ln(x)-\ln(1-x)|<0.01\) for all x in the interval \(l=[-\frac{1}{2},\frac{1}{2}]\)

Answers (1)

2021-05-08

Solution: Given: \(f(x)=\ln(1-x)\)
About b=0: the maclaurins series is:
\(f(x)=\ln(1-x)=(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-...\infty)=(-1)\sum_{n=1}^\infty\frac{x^n}{n}\)
Over [\(-\frac{1}{2},\frac{1}{2}\)] by the Taylor's inequality:
\(|R_n|\leq|\frac{f^{(n+1)}e}{(n+1)!}\cdot|x-b|^{n+1}\)
Now: \(|x-b|=\frac{1}{2}\) and \(f'(x)=\frac{1}{1-x}\)
So: \(f^{n+1}(x)\) over \([\frac{-1}{2},\frac{1}{2}]\) is maximised
\(\Rightarrow(n+1)>+\infty\Rightarrow n>99\)

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