**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 $n$.

## Expert Answer:

\[ f (x) = ln (1 – x) \]

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:

**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,

## 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 \