Question

Use a 3rd degree Taylor polynomial centered at x = 1 for f(x) = ln(x) to approximate ln(1.2) to 5 decimal places.

Answer 1

Compute the first 3 derivatives of f(x):

$f(x)=\ln x\implies f'(x)=\dfrac1x\implies f''(x)=-\dfrac1{x^2}\implies f'''(x)=\dfrac2{x^3}$

The 3rd degree Taylor polynomial about $x=1$ is then

$T_3(x)=f(1)+\dfrac{f'(1)}{1!}(x-1)+\dfrac{f''(1)}{2!}(x-1)^2+\dfrac{f'''(1)}{3!}(x-1)^3$

$T_3(x)=(x-1)-(x-1)^2+2(x-1)^3$

$T_3(x)=-4+9x-7x^2+2x^3$

Now use $T_3$ to approximate ln(1.2):

$f(x)\approx T_3(x)\implies \ln1.2\approx T_3(1.2)=\boxed{0.17600}$

(Compare to the actual value, closer to 0.18232)