Question

Use the Taylor series to find the values of ln 1.4 accurate to 10^-3 . Use the integral remainder.

Answer 1

$$\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+...$$

$$=\displaystyle\sum_{n=1}^{\infin}\dfrac{(-1)^{n+1}x^n}{n}, -1<x\leq 1$$

$$\ln(1.4)=\ln(1+0.4)$$

$$\dfrac{0.4^n}{n}\leq0.001$$

$$\dfrac{0.4^5}{5}=0.002048>0.001$$

$$\dfrac{0.4^6}{6}\approx0.000683<0.001$$

$$\ln(1.4)\approx0.4-\dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}-\dfrac{0.4^6}{6}$$

$$\approx0.336$$