Answer to Question #181556 in Calculus for Emmanuel Papa

Question

Obtain a four-term Taylor polynomial approximation valid near x=,0 for each (1+x)^-1/3. Estimate the range of x over which three term polynomials will give two decimal accuracy

Answer 1

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

for a four term Taylor polynomial centered at x=0

$f(x)= (1+x)^{-1/3}\implies f(0)=1$

$f'(x)=-\frac 13 (1+x)^{-4/3}\implies f'(0)=-\frac 12$

$f''(x)= -\frac 49(1+x)^{-7/3} \implies f''(0)= \frac 49$

$f'''(x)= -\frac {28}{27}(1+x)^{-10/3}\implies f'''(0)= -\frac {28}{27}$

now the four term Taylor polynomial approximation to f(x) valid near x=0 is;

$f(x) =(1+x)^{-1/3} = 1-\frac 13x + \frac{\frac 49}{2!}x^2- \frac {\frac {28}{27}}{3!}x^3$

$= 1-\frac 13 x + \frac 29 x^2 - \frac {14}{81}x^3$

for the range of values of x over which these term polynomials will give a two decimal accuracy we may use the Lagrange error bound;

$m\frac {|x-a|^{n+1}}{(n+a)!}<=$  maximum admissible error

where $m= max _{x\isin\Iota}|f^{n+1}(x)|$

$\therefore m \frac {|x-0|^3}{3!}<= 0.01$ or $|x|^3<= \frac {0.01*3!}{m}$

$|x|<= \frac {0.39}{\sqrt [3]m}$

$\implies -0.4<= x<= 0.4$ is the range desired.