Question

-1/3 (i) Find the third-degree Maclaurin polynomial Tz(x) for f(x)= (1+4x). You can use differentiation or derive the polynomial using binomial series. (ii) Find approximation errors|f(x)– Tz(x) at x =0.1 and x=1.

Answer 1

$f(x)=(1+4x)^{-1/3}$

Using Binomial series

$(1+x)^{n}=1+nx+\frac{n(n-1)}{2!}x^{2}+\frac{n(n-1)(n-2)}{3!}x^{3}+...$

we can apply this to given function and obtain

$(1+4x)^{-1/3}\approx 1-\frac{4}{3}x+\frac{32}{9}x^{2}-\frac{896}{81}x^{3}$

$T_{3}(x)=1-\frac{4}{3}x+\frac{32}{9}x^{2}-\frac{896}{81}x^{3}$

This is the required function that we now need to comapre with exact values.

x = 0.1

$f(0.1)=(1.4)^{-1/3}=0.89390353509$

$T_{3}(0.1)=1-\frac{4}{30}+\frac{32}{900}-\frac{896}{81000}=0.89116049382$

$\left | f(x)-T_{3}(x) \right |_{x=0.1}=0.002743$

x = 1

$f(1)=(5)^{-1/3}=0.58480354764$

$T_{3}(1)=1-\frac{4}{3}+\frac{32}{9}-\frac{896}{81}=-7.83950617284$

$\left | f(x)-T_{3}(x) \right |_{x=1}=8.42431$

This approximation fails, because of the limitation of Maclaurin series that converges only till finite points