The function $\ln(1-x^2)$ is approximated about $x=0$ by an nth degree Taylor's Polynomial. Find n such that
$|Error|<0.1$ on $0\leq x\leq0.5$.
My Try: So I know to evaluate the remainder term of the Taylor polynomial, I can use the following expression
$$R_n (x)=\frac{x^{n+1}}{(n+1)!}\cdot f^{n+1}(\xi)$$
My doubt is, how do I find the $(n+1)th$ derivative of $\ln(1-x^2) $? I don't see any pattern in its smaller derivatives. How do I evaluate the same?
On
you know the expansion of $$\ln(1-x) = - x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4 }- \cdots$$
replace $x$ with $x^2$ $$\ln(1-x^2) = - x^2 - \frac{x^4}{2} - \frac{x^6}{3} - \frac{x^8}{4 }- \cdots$$
nth term is $$a_n=-\frac{x^{2n}}{n}\hspace{10pt}n\ge 1$$
you want $$\left|-\frac{x^{2n+2}}{n+1}\right| \le 0.1 \hspace{10pt}, 0\le x\le 0.5$$
Hint: I would write $$\ln(1-x^2)=\ln(1-x)+\ln(1+x)$$