The nth Taylor polynomial Tn(x) for&nbsp;f(x)=ln⁡(1−x)&nbsp;is the base at zero?Find the value of n such that Taylor's inequality guarantees the lowest value&nbsp;|ln⁡(x)−ln⁡(1−x)|&lt;0.01&nbsp;for all x in the interval&nbsp;l=[−12,12]

Solution: Given: f(x)=ln⁡(1−x)About b=0: the maclaurins series is:f(x)=ln⁡(1−x)=(−x−x22−x33−x44−...∞)=(−1)∑n=1∞xnnOver [−12,12] by the Taylor's inequality:|Rn|≤|f(n+1)e(n+1)!⋅|x−b|n+1Now: |x−b|=12 and f′(x)=11−xSo: fn+1(x) over [−12,12] is maximised⇒(n+1)&gt;+∞⇒n&gt;99

Question: "The nth Taylor polynomial Tn(x) forf(x)=ln⁡(1−x)is the base..."

High School

ankarskogC

Answered question

2021-05-07

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

Answer & Explanation

FieniChoonin

Skilled2021-05-08Added 102 answers

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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