Question

Use the Taylor series to find the values of ln 0.4 accurate to 10^-3 . Use the integral remainder.

Answer 1

We use f(x)=ln(1+x);

Taylor formula:

$f(x)=f(a)+f^{(1)}(a)\cdot \frac{(x-a)^1}{1!}+...+f^{(n)}(a)\cdot \frac{(x-a)^n}{n!}+\frac{1}{n!}\int_a^x(x-t)^n\cdot f^{(n+1)}(t)dt$

We take a=1,dx=x-a=-0.6

f(a)=ln(1)=0

$f^{(1)}(x)=\frac{1}{x},f^{(1)}(1)=1;\\ f^{(2)}(x)=-\frac{1}{x^2},f^{(2)}(1)=-1;\\ f^{(3)}(x)=\frac{2!}{x^3},f^{(3)}(1)=2!;\\ ,,,,,,,,,\\ f^{(n)}(x)=(-1)^{n-1}\frac{(n-1)!}{x^n}, \frac{f^{(n)}(1)(0.4-1)^n}{n!}=-\frac{0.6^n}{n!};\\$

$f^{(n+1)}(t)=(-1)^n\frac{n!}{t^{n+1}}$

$f(x)=-0.6-\frac{0.6^2}{2}-...-\frac{0.6^n}{n}+(-1)^{n}\frac{n!}{n!}\int_1^{x}\frac{(0.4-t)^n}{t^{n+1}}dt=\\ -0.6-\frac{0.6^2}{2}-...-\frac{0.6^n}{n}+(-1)^{n}\int_1^{0.4}\left( \frac{0.4}{t}-1\right)^n\frac{1}{t}dt$

$|(1-\frac{0.4}{t}|^n\cdot {1}{|t|}\le(1-\frac{0.4}{1})^n\cdot \frac{1}{0.4}=0.6^n\frac{1}{0.4},\space t\in(0.4,1)$

$|(-1)^{n}\int_1^{0.4}\left( \frac{0.4}{t}-1\right)^n\frac{1}{t}dt|\le 0.6^n\cdot |0.4-1|\cdot \frac{1}{0.4}=\frac{0.6^{n+1}}{0.4}<0.001$

$0.6^{14}/0.4>0.001,\space 0.6^{15}/0.4<0.001\\ n=14;\\ ln(0.4)=-0.6-\frac{0.6^2}{2}-\frac{0.6^3}{3}-\frac{0.6^4}{4}-\frac{0.6^5}{5}-\frac{0.6^6}{8}-\frac{0.6^7}{7}-\frac{0.6^8}{8}- \\ -\frac{0.6^9}{9}-\frac{0.6^{10}}{10}-\frac{0.6^{11}}{11}-\frac{0.6^{12}}{12}-\frac{0.6^{13}}{13}-\frac{0.6^{14}}{14}-=-0.916$