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

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 $$l=[-\frac{1}{2},\frac{1}{2}]$$

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
$$\Rightarrow(n+1)>+\infty\Rightarrow n>99$$