# Which of the following is 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 $l = [\dfrac {- 1}{2}, \dfrac {1}{2} ]$

The aim of this question is to find the $n^{th}$ Taylor polynomial of the given expression. Furthermore, the smallest value of a variable that satisfies Taylor’s inequality of a specific expression with a given interval also needs to be understood.

Moreover, this question is based on the concepts of arithmetics. The $nth$ Taylor polynomial of a function is a partial sum that is formed by the first $n + 1$ terms of the Taylor series, moreover, it is a polynomial of degree $n$.

As we have,

$f (x) = ln (1 – x)$

Moreover, when $b = 0$, the Taylor polynomial and the  Maclaurin’s series become equal. Therefore, we have used Maclaurin’s series as follows.

$f (x) = ln(1 – x)$

The right side of the equation can be extended as,

$ln(1 – x) = (- x – \dfrac{x^2}{2} – \dfrac{x^3}{3} – \dfrac{x^4}{4} – \dfrac{x^5}{5} -, … , \infty)$

$(- x – \dfrac {x^2}{2} – \dfrac{x^3}{3} – \dfrac{x^4}{4} – \dfrac{x^5}{5} -, … , \infty) = (-1) \sum_{n = 1}^{\infty} \dfrac{x^n}{n}$

The Taylor’s inequality over the given interval of $[- \dfrac{1}{2}, \dfrac{1}{2} ]$,

$R_n \ge | \dfrac {f^{n + 1}e}{(n + 1)! } | . |x – b|^{n + 1}$

Therefore,

$|x – b| = \dfrac{1}{2}$

and the first derivative of the given expression can be calculated as,

$f'(x) = \dfrac{1}{1 – x}$

Hence,

$f^{n + 1} (x) \text{ over } [ \dfrac{-1} {2}, \dfrac{1} {2} ] \text { is maximised}$

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

## Numerical Results:

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

$(n) > 99$

## Example:

Find the Taylor series for $f(x) = x^3 – 10x^2 + 6$ about $x = 3$.

Solution:

To find the Taylor series, we need to calculate the derivatives upto $n$.

$f^0 (x) = x^3 – 10x^2 + 6$

$f^1 (x) = 3x^2 – 20x$

$f^2 (x) = 6x -20$

$f^3 (x) = 6$

As the derivative of constant is 0. Therefore, the further derivatives of the expression is zero.

Moreover, as $x = 3$, therefore, $f^0 (3), f^1 (3), f^2 (3), f^3 (3)$, are -57, -33, -3, and 6, respectively.

Hence by Taylor series,

$f(x) = x^3 – 10x^2 + 6 = \sum_{0}^{ \infty} \dfrac{f^n (3)}{n!} (x – 3)^3$

$= -57 – 33(x – 3) – (x – 3)^2 + (x – 3)^3$

\[= 42 – 33x – (x – 3)^2 + (x – 3)^3 \